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					Math 215 Principles of Mathematics




                      Lecture Notes



                          Math 215

                 Principles of Mathematics




Pierce College                               MAP
Math 215 Principles of Mathematics                                                                                                        i


                                           Table Of Contents


Lecture     1 ............................................................................................................................1
Lecture     2 ........................................................................................................................ 19
Lecture     3 ....................................................................................................................... 36
Lecture     4 ....................................................................................................................... 55
Lecture     5 ....................................................................................................................... 72
Lecture     6 ....................................................................................................................... 94
Lecture     7 ...................................................................................................................... 112
Lecture     8 ......................................................................................................................129
Lecture     9 ......................................................................................................................142
Lecture     10.....................................................................................................................156
Lecture     11 .....................................................................................................................179
Lecture     12.....................................................................................................................195
Lecture     13.....................................................................................................................214




Pierce College                                                                                                                       MAP
Math 215 Principles of Mathematics                                             1



Lecture 1

Sections 1.1 and 1.2 – Problem Solving and Patterns

Problem Solving

Mathematics is used to solve a wide variety of problems in the physical
sciences (e.g. physics and chemistry), engineering (electrical, civil),
economics and finances to everyday problems such as answering questions
like: “How much flour is required if I double the recipe for the cake I’m
baking?”.

All problems involving mathematical solutions can be approached in the same
general manner using a procedure or problem solving process. This process is
commonly first encountered (at least formally) in elementary algebra when
solving ‘word problems’. The four steps to this process are as follows:

    1. Understanding the Problem
       a. Internalize or ‘re-state the problem in your own words’.
       b. What is being sought? i.e. what are you being asked to find or do?
       c. What are the unknowns? Note: some of these may be ‘common
           knowledge and therefore assumed e.g. number of seconds in a
           minute. Identifying the unknowns supports and helps with part (b)
           above.
       d. What information do you obtain from the problem? What
           information is supplied in the problem statement? Take notes!
       e. What information is missing or not needed? Common knowledge
           type information is often assumed and not stated. Also irrelevant
           information may be contained in the problem statement e.g.
           temperature in different cities in a problem involving distance
           between cities.
    2. Devising a Plan (possible strategies)
      a. Look for patterns – useful in problems involving sequences of
          numbers or repeating sequences of objects.
      b. Lessons learned by previous or similar problems.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          2


      c. Examining a simpler case of the problem – useful for determining
          patterns or problems involving many parameters e.g. traveling
          salesman problem.
      d. Making tables, lists, or figures – i.e. taking notes and organizing
          problem supplied data.
      e. Writing an equation. Relating the data mathematically or writing
          down potentially useful formulas e.g. area of a circle = πr2.
      f. Use ‘guess and check’ i.e. guess at the answer and then verify that
          it solves the problem.
      g. Work backward. Sometimes the solution may be known or easily
          guessed and working ‘backwards’ may lead to the solution. This
          approach is sometimes good for proofs.
      h. Identify a sub-goal or sub-goals. Sometimes finding the solution to
          part of the problem combined with solutions to previous problems
          provides a solution. Example: What’s the area in a sheet of paper
          with a circular hole in it? If you know how to compute the area of
          circle then this can be combined with knowledge about the area of a
          rectangle to compute the required area.
      i. Use indirect reasoning – it may be easier to show that something is
          false rather than true.
      j. Direct reasoning – proceed directly from the given information to
          the solution.
    3. Carrying Out the Plan
       a. Implement or try one of the strategies from (2) above. Note: the
           initial strategy may fail causing a re-thinking or revising of the
           plan. Engineers often say they know what doesn’t work rather than
           what does. The process of eliminating possible ‘solutions’ as
           incorrect or unworkable often narrows the field and leads to the
           desired solution.
       b. Work neatly and keep records. This often helps in finding errors
           and mistakes.
    4. Looking Back
       a. Verify that the answer satisfies the original question. If the
           question is how much paint is required to paint a house, the total
           number of square feet of wall area in the house does not answer
           the question.
       b. Does the answer make sense? If the problem is to compute the
           distance between Los Angeles and New York and the answer is 30

Pierce College                                                            MAP
Math 215 Principles of Mathematics                                               3


            miles, something is clearly wrong. Always make sure that the
            solution truly solves the problem.
         c. See if there is another way to solve the problem. If so, try this
            also and make sure the answer is the same. In some cases, it may
            be possible to (easily) estimate the answer. This provides another
            method for verifying the answer.
         d. Lessons learned. Is the solution or method employed useful in
            solving other problems?



Examples

Gauss’s Problem. This problem comes with a bit of lore. As the story goes,
Gauss’s (Carl Gauss 1777-1855) teacher was angry with his class and to
disciple he class, he gave them the task of adding the numbers 1 to 100.
Gauss was able to solve this problem in short order, to the surprise of his
teacher. Using the four-step procedure we look for a solution.

Step 1. Understanding the problem.

In this case, the problem is straightforward: sum the numbers between 1
and 100. That is 1 + 2 + 3 + … + 100 = ?

Step 2. Devising a plan.

Clearly a brute force approach will work here. However summing the
numbers pair-wise is laborious, time consuming, and error prone. However if
we make the following observation:

                       1+   2+ 3+       4+    5+    + 98 + 99 + 100
                    100 + 99 + 98 + 97 + 96 +       +   3+    2+    1
                    101 + 101 + 101 + 101 + 101 +   + 101 + 101 + 101

That is if we write the sum twice, once in increasing order and once in
reverse order, and pair the numbers in each sum, then each pair adds to 101.
We can thus convert the problem to one involving sums to one involving
multiplication. The answer is easily found by dividing the result by two. We
note that this plan involves finding a pattern.

Pierce College                                                                MAP
Math 215 Principles of Mathematics                                            4


Step 3. Executing the Plan.

In this case, the result is straight forward. The answer is just 101 times 100
divided by 2:

                                       101× 100
                            Answer =            = 5050
                                          2



Step 4. Looking Back.

In this case, we have a numerical answer and we need to check that it is
reasonable. Clearly, if correct, the answer satisfies the question. One thing
we can do is estimate the range where our answer should lie. That is, that if
we were to add 1, 100 times the answer would be 100. Likewise if we were to
add 100, 100 times the answer would be 10,000. Our answer must therefore
lie between 100 and 10,000 which it does. This does not mean our answer is
correct, but rather that it is not wrong since in lies in the valid range where
the correct answer should lie. Another thing we could do is solve a simpler
problem, for example adding the numbers between 1 and 10. In this case we
use the technique above as well as brute force, and compare the results. If
we get the same answer, this will give us additional confidence that the
answer is correct.

Finally, if we were to try and solve several simpler such problems like
summing the numbers between 1 and 5, 1 and 10, and 1 and 20 we could find
a generalization or formula for the solution to the general problem of
summing the numbers between 1 and n, where n is any number greater than
one. By using similar reasoning as above we can show that:

                                              n ( n + 1)
                              1+ 2 +   +n =
                                                  2




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                             5


Magic Square Problem. Arrange the numbers 1 to 9 into a square subdivided
into nine smaller squares (i.e. a 3 by 3 arrangement) so that the sum of every
row, column, and diagonal is the same. The resulting arrangement is called a
magic square.

Step 1. Understanding the Problem.

We want to place the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 into the following
square:




So that the sum of each row, column, and the two diagonals is equal to the
same number, say m:

                                            m

                                            m

                                            m

                                            m

                                            m
                              m   m   m


Step 2. Devising a plan.

There are several strategies we can employ here. One is to guess at a
solution and then see if it works. If not, we can re-arrange the numbers
(guess again) and re-check. This strategy is not particularly efficient, but if
we base our next guess on the results of our previous guess, this approach
might not require too many guesses.

Another approach would be to define a sub-goal. In this case, finding m
would be a useful sub-goal since it would aid us in identifying valid triples of
numbers for placing in each row, column, and diagonal. Now, if we knew the
solution, then the sum of each of the three rows would be 3m since each row

Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             6


sums to m. However, the sum of the three rows is just 1 + 2 + 3 + 4 + 5 + 6 +
7 + 8 + 9 since each number appears only once in the magic square and we can
re-arrange the order of the numbers in the summation. We therefore have:

                     1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 3m

Using brute force (or the results from Gauss’s problem above) we have that:



                  1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 = 3m

We therefore have that m = 15. Next we need to determine which numbers
go in which square. We note that the numbers in the four corners will each
appear in 3 sums (a row, a column, and a diagonal). Also the number in the
center of the square will appear in four sums (a row, a column, and both
diagonals). If we make a list of all possible sums of 3 numbers between 1 to
9 whose sum is 15, we can make a table of how many times a number appears
in these sums. Using this table we look for numbers that appear three and
four times. These numbers will be candidates for the corners and center.

Step 3. Executing the Plan.

Fist we make our list:

                                9 + 5 + 1 = 15
                                9 + 4 + 2 = 15
                                8 + 6 + 1 = 15
                                8 + 5 + 2 = 15
                                8 + 4 + 3 = 15
                                7 + 6 + 2 = 15
                                7 + 5 + 3 = 15
                                6 + 5 + 4 = 15

Note: we made this list by starting with 9 as the first number. Since the
sum must contain 3 numbers and equal 15, we rule out 6, 7, and 8 and possible
second numbers since these would result in sums greater than 15. Thus the
largest value for the second number is 5. We then decrease the second
number to 4, and then 3. However with 3, the only possibility for the third

Pierce College                                                            MAP
Math 215 Principles of Mathematics                                            7


number is 3 and since three can appear only once, we must rule this out. We
can also eliminate 2 and 1 for second numbers since this would just be a re-
ordering of the first two results. This exhausts the possibilities with 9. We
then proceed to 8 and follow the same reasoning. The result is the list
above.

We now make a table showing how many times each number appears in the
above list:

Number                                          1   2 3 4     5   6 7    8    9
Number of Sums Containing the Number            2   3 2 3     4   3 2    3    2

From our table we note that only one number appears 4 times: 5. Thus 5
must appear in the center of our magic square. We also note that there are
only four numbers (2, 4, 6, and 8) that appear 3 times. These must be in the
corners.

Now if we place 2 in the upper left corner then 8 must be in the lower right
corner since the sum of the diagonal must equal 15. That is 2 + 5 + 8 = 15. If
we place 6 in the upper right corner we have the following arrangement:


                                2       6
                                    5
                                4       8

The remaining numbers can be filled in starting with the top row, bottom
row, and finally middle row by noting that the sums of each row are 15. We
therefore obtain:


                                2   7   6
                                9   5   1
                                4   3   8




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                               8


Step 4. Looking Back.

In this case we can verify our answer directly by summing each row, column,
and diagonal and making sure we get the same number i.e. 15. We also note
that our answer is not unique. That is, it’s not the only correct answer. For
example:

                                 4   9   2
                                 3   5   7
                                 8   1   6


is also a solution. We might need or want to find all the solutions to insure
that the original question is satisfied.



Checkerboard Problem. Given a checkerboard with two squares removed on
opposite corners and a set of dominos such that one domino covers two
squares, is it possible to completely cover the board with dominos so that no
dominos are hanging over or off the board?

Step 1. Understanding the Problem.

We draw a picture of the board with the two removed (in this case red)
squares:

                                                   domino




Can we cover the squares two at a time with dominos such that all the
squares are covered by a domino and no dominos are hanging off the board?




Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                          9


Step 2. Devising a plan.

We use indirect reasoning here. We assume that the board can be covered
without having any dominos hanging off the board and with no square left
uncovered. We will show this leads to a contradiction, and therefore our
assumption is incorrect and hence, the board cannot be so covered by the
dominos.



Step 3. Executing the Plan.

We note that the complete checkerboard contains 64 squares, 32 of them
black and the other 32 white. The board in the problem however, has two
squares removed and therefore has only 62 squares. However, the two
squares removed were both red squares therefore there are 32 black
squares and 30 red squares.

Now a domino placed (completely) on the board must cover one black square
and one red square. However there are 62 squares and since a domino covers
two squares, 31 dominos are required to cover the board. But since there
are only 30 red squares one of the dominos cannot cover a red square. This
contradicts the fact that a domino must cover one red and one black square.
Our initial assumption that board could be covered by the dominos has
therefore led to a contradiction and must therefore be incorrect. Thus the
checkerboard cannot be so covered by the dominos.



Step 4. Looking Back.

In this case we can try placing the dominos on the board to verify that the
board cannot be covered. Note that this is not definitive unless we try all
possible placements. We also note that the reasoning in our solution implies
that if we remove one black square and one red square from the corners, we
should be able to cover the board (or at lest it might be possible). We could
try this also.




Pierce College                                                            MAP
Math 215 Principles of Mathematics                                          10


Patterns

Patterns and their and study and use in mathematics is a broad topic
spanning several branches of mathematics. In this course we will
concentrate on the following topics:

    •    Inductive Reasoning
    •    Arithmetic Sequences
    •    Geometric Sequences
    •    The Fibonacci Sequence
    •    Figurate/Triangle Numbers

Inductive Reasoning

Inductive reasoning is the process of making generalizations based a
pattern. We saw an example of this when stating the formula for Gauss’s
problem. Consider the following sums:

                          1   +   2   =3
                          1   +   2   +3=6
                          1   +   2   + 3 + 4 = 10
                          1   +   2   + 3 + 4 + 5 = 15

And so on. By examining this pattern we can deduce Gauss’s summation
formula for the sum of the numbers 1 to n (where n is some number greater
than 1):

                                                  n ( n + 1)
                                  1+ 2 +   +n =
                                                      2

Note: that although this formula works for the cases we have examined, it
has not been shown to work in all cases. This important point should be
carefully noted. Although the above formula can in fact be shown to be true
in all case i.e. proved to be true, we cannot assume this is the case based
solely on inductive reasoning. Inductive reasoning is useful in generating a
conjecture which then needs to be proved (i.e. shown to be true in all cases).
A conjecture may be proved false by finding a single case, called a
counterexample, where it is wrong. Finding a counterexample is often


Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                           11


difficult, but it should be noted that the lack of a counterexample is not
sufficient to prove the conjecture true.

Finally we note that inductive reasoning should not be confused with proof
by induction. The later is a formal mathematical technique that uses
inductive-like reasoning along with direct mathematical reasoning to prove a
conjecture. Proof by induction is a mathematically valid method for
constructing a formal proof. Inductive reasoning on the other hand is a
technique for generating an (unproven) conjecture.

Example

                                                                n ( n + 1)
Prove Gauss’s summation formula i.e. 1 + 2 +             +n =                where n is a whole
                                                                    2
number greater than or equal to one.

Proof. We will prove Gauss’s formula using proof by induction. Using
                                                      n ( n + 1)
inductive reasoning we have the formula 1 + 2 + + n =            . For the case
                                                           2
where n = 1 we have:

                                      1(1 + 1)       2
                                1=               =     =1
                                         2           2

And therefore the formula is correct for this case. Now for the case where
n = 2 we have:

                                      2 ( 2 + 1)       2⋅3
                             1+ 2 =                =       =3
                                          2             2

Which is also correct. Thus we have shown the formula is true when n is
equal to one or two. We can clearly continue in this manner, however since
there are an infinite number of case, we will never finish. However since we
have established the formula as correct for two cases we will assume it is
correct for all numbers between 1 and say an arbitrary whole number m. We
now show by direct reasoning that this implies the formula is correct for
m + 1 i.e. the next number. Now by our assumption:



Pierce College                                                                              MAP
Math 215 Principles of Mathematics                                                            12


                                                        m ( m + 1)
                                    1+ 2 +        +m=
                                                             2

So that we must have:

                                                        m ( m + 1)
                           1+ 2 +    + m + ( m + 1) =                + ( m + 1)
                                                             2

Therefore:

                                                       m ( m + 1)
                       1+ 2 +       + m + ( m + 1) =              + ( m + 1)
                                                           2
                                                       m ( m + 1) + 2 ( m + 1)
                                                   =
                                                                  2

                                                   =
                                                       ( m + 1)( m + 2 )
                                                                 2

which is identical to the result we would have obtained by substituting m+1
for n in Gauss’s formula. Thus we have proved that Gauss’s formula is true in
general.



We now give an example where inductive reasoning fails. Consider the
formula:

                                         p = n2 + n + 11

where n represents the natural numbers: 1, 2, 3, 4,. . . Making a table of
values using this formula we obtain:

n                 1    2         3           4          5             6            7    8    9
p                13   17        23           31         41           53           67   83   101

We note that all the values of p in the table are prime numbers i.e. a number
that is evenly divisible only by itself and one. Using inductive reasoning we
can form the conjecture that the above formula generates only prime
numbers. This conjecture is false which can be demonstrated by
counterexample. For example, n = 10 give p = 121. But 121 is divisible by 11

Pierce College                                                                               MAP
Math 215 Principles of Mathematics                                              13


i.e. 11 × 11 = 121 and is therefore not prime. Thus the conjecture that this
formula generates only prime numbers is incorrect. This demonstrates the
difference between inductive reasoning and proof by induction and also
illustrates the ‘danger’ of making conjectures without carefully and
thoroughly testing them. Note: even though we tested the formula for 9
cases, the next case i.e. the 10th case would have prevented us from making
this conjecture if we had continued. Of course, it is often the case (as in
this example) that it is impossible to test all the cases. We are therefore
forced to cut-off or stop our testing at some point. However, this means we
must formally prove the conjecture before accepting it as fact i.e. true.

Arithmetic Sequences

We define a sequence as an ordered arrangement of numbers, symbols,
figures, or other objects. For some sequences (but not all) there may exist
are relationships between successive members of the sequence. That is, if
we know the first few members of the sequence and the relationship
between successive members we can generate the entire sequence from this
information. This is especially useful when the sequence is not finite.

Example

Consider the natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,. . . The natural
numbers form a sequence since they are an order arrangement of numbers.
Moreover each successive member can be generated by adding one to the
previous member. That is if Nk is the kth natural number then:



                                   Nk+1 = Nk + 1

Knowing the first member of the sequence i.e. 1 in this case, and this
relationship allows us to generate the entire sequence. The natural numbers
are an example of an arithmetic sequence. An arithmetic sequence is one in
which each successive member is obtained from the previous member by
adding or subtracting a fixed number, called the difference. In the case of
the natural numbers the difference is 1.




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                   14


Arithmetic sequences are also examples of a recursive pattern. That is, the
sequence is generated by applying the same operation over and over again
upon its members. Recursive operations are often used in computer
programming and in spreadsheets e.g. Excel.

Example

Let an = 4n + 3 where n = 1, 2, 3, 4, . . . The notation says that an is the nth
member of the sequence. If we create a table of the first 10 members of
this sequence we get:



n            1   2       3      4       5          6        7       8      9        10
an           7   11     15      19      23        27        31     35     39        43

We note that the difference between successive member of this sequence is
4, suggesting that it is an arithmetic sequence. We can prove this by noting
that:

                 an+1 - an = 4(n+1) + 3 - 4n – 3 = 4n + 4 – 4n = 4

Geometric Sequences

A geometric sequence is one in which each successive member is obtained
from the previous member by multiplying by a fixed number, called the ratio.
An example of a geometric sequence results from considering the ancestry
of a child. The child has two (biological) parents. Each parent in turn also
has two (biological) parents, and so on. The number of pro-genitive ancestors
increases by a factor of 2 for each generation we go back in the child’s
ancestry. The diagram below illustrates this fact:

                                 1 Child – 0th Generation




                                             2 Parents – 1st Generation

                                                  4 Grandparents – 2nd Generation

                                                       8 Great-Grandparents – 3rd Generation



Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                               15



We therefore have the following sequence which gives the number of
ancestors at each successive generation: 1, 2, 4, 8, . . . This sequence can be
expressed as:

                          an = 2n,     n = 1, 2, 3, 4, . . .



                                                     an +1 2n +1
Note that this is a geometric sequence since              = n = 2 . We can express
                                                      an    2
any geometric sequence in terms of its first term, a1, and its defining ratio, r
as follows:



                        an = a1rn-1,    n = 1, 2, 3, 4, . . .



The Fibonacci Sequence

The Fibonacci sequence is named after its discover Leonardo Fibonacci
(1125-1250). This sequence is neither arithmetic or geometric and, because
of long standing associations found between the sequence and proportions in
art and phenomenon in nature, it ‘sits’ in a category by itself. The first few
members of the Fibonacci sequence are:

                  1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . .




The Fibonacci sequence can be generated by noting that its first two
members are 1 and each successive member is the sum of the previous two
members:

                                  Fn = Fn-1 + Fn-2




Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                            16


Applications

Population Growth. Suppose we have a pair of rabbits, one male, one female.
Rabbits mate and produce offspring (one male and one female) at one month
intervals so that at the end of the second month a female can produce
another pair of rabbits. Assuming that our rabbits don’t die and that the
female always produces one male and one female every month from the
second month on, How many pairs will there be in one year? The answer is
that the number of rabbits is given by the Fibonacci sequence.

Occurrences in Nature. The flowers on many plants have petals that are a
Fibonacci number: buttercups - 5 petals; lilies and iris - 3 petals; delphiniums
– 8 petals; corn marigolds - 13 petals; asters - 21 petals; daisies - 34, 55 or
89 petals.

The human Body. The human body has 2 hands containing 5 fingers. Each
finger has 3 parts connected by 2 knuckles. The human torso has 5
appendages: the head, two arms, and two legs. There are 5 senses (taste,
touch, sound, sight, smell).

Proportion. The Fibonacci sequence is related to a number called the Golden
Ratio, which is denoted by the Greek letter phi: φ. The number phi is:

                                   1+ 5
                              ϕ=        ≈ 1.618033989
                                     2

                                             n
                            ϕ n − (1 − ϕ )
It can be shown that Fn =            . In art and architecture the ratios or
                               5
proportions of parts of an object often are equal to the golden ratio.
Examples include:

    •    The ratio of the height to the width in building such as the Acropolis
         in Athens is equal to the golden ratio.
    •    Other buildings with proportions equal to the golden ratio include the
         Pantheon in Rome, and the United Nations Building in New York.
    •    The proportions of various parts of Michelangelo’s David are equal to
         the golden ratio.


Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             17


In general proportions that are equal to the golden ratio are judged to be
aesthetically pleasing.

Figurate/Triangle Numbers

Figurate numbers are sequences that are neither arithmetic or geometric
(like the Fibonacci sequence). Such numbers can be represented as dots
arranged in various geometric shapes. The number 1 is the beginning of most
Figurate sequences. Triangular numbers provide one example of these
sequences as shown in the figure below.




                 1 dot       3 dots        6 dots            10 dots


Numerically we have the following sequence: 1, 3, 6, 10, 15, . . . Note this
sequence is not arithmetic, geometric and is not the Fibonacci sequence.
However if we list the difference between successive members in this
sequence we have:



                         1        3         6       10       15



                             2         3        4        5

 And we see that the difference forms an arithmetic sequence. In the
original sequence, the second term is obtained from the first by adding 2;
the third term is obtained from the second by adding 3; and so on. In
general, because the nth triangular number has n dots in the nth row, it is
equal to the sum of the dots in the previous triangular number plus the n
dots in the nth row. Following this pattern we get:

                                      an = 1 + 2 + 3 + . . . + n



Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                          18


                                   n ( n + 1)
which we recognize must be an =                 from Gauss’s problem.
                                         2

Now consider the following sequence: 5, 6, 14, 29, 51, 80. . . Again this
sequence is not arithmetic or geometric. We try the same approach as
above:



                   5       6       14        29        51        80



                       1       8        15        22        29



Again we note that the difference sequence is arithmetic since each
successive member is obtained from the previous member by adding 7. The
next term in the sequence is obtained by computing the next number in the
difference sequence which is 29 + 7 = 36 and then adding this to 80 to get
116.

Finally we note that when asked to find a pattern for a given sequence we
first check to see if the sequence is arithmetic, geometric or the Fibonacci
sequence. If not, we can try taking successive differences to see if a
pattern emerges in the difference sequence. If not, we can try taking the
difference of the difference sequence to see if there is a pattern in that
sequence, and so on. We note that it is entirely possible that none of these
strategies will work. See your text book for additional examples.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                               19




Lecture 2

Sections 1.3 and 1.4 – Algebraic Thinking and Logic

Although all real-world problems (as well as many exercises in school) are
presented or given in descriptive format i.e. as word problems, the solution
of the problem often requires solving an equation derived from the
particulars of the problem statement. The text terms the process of solving
equations as ‘algebraic thinking’.

Algebraic thinking involves three concepts:

    1. Variables. Variables are placeholders for something else, usually
       numbers. The name comes from the fact that the placeholder can
       assume any value i.e. vary. We often use symbols to represent
       variables; these are typically letters of the alphabet e.g. ‘x’ but can in
       principle be anything. In higher mathematics letters from the Greek
       alphabet are often used, especially to represent angles.
    2. Notation. Notation involves variables, operators (multiplication,
       division, subtraction, and addition), and numbers (natural, whole,
       rational, irrational, real). As the text points out, notation can be
       confusing if not clearly defined and understood.
    3. Rules. These are more commonly known as properties or laws. For
       example the addition property or cancellation property. These rules
       tell us how we can manipulate the symbols and numbers to transform
       an equation into an equivalent equation. By using these rules we can
       find the solution for the equation.

As the text points out, the equal sign ‘=’ is not an operator. Because of
modern computing devises (e.g. calculators) the equal sign is often confused
with the equal-button (labeled with an equal sign) on these devices. On a
calculator pressing the equal-button performs an operation i.e. displays the
results of a calculation. In an equation the equal sign expresses the
relationship between both sides of the equation. It is not an operator but
rather a type of association. Other types of association are given by the
inequality symbols i.e. greater-than ‘>’, less-than ‘<’, greater-than-or-equal ‘≥’

Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                           20


and less-than-or-equal ‘≤’. The equal symbol belongs in the same category as
these symbols since they all express the relationship between the
expressions to the left and right.

The association expressed by the symbols ‘=’, ‘≤’, ‘≥’, “<’, and ‘>’ can be
evaluated and the equation assigned a truth value i.e. ‘true’ or ‘false’
depending on the results of the evaluation. If one (or both sides) of the
equation contains a variable, then the truth value of the equation is
indeterminate until a value is substituted into the equation for the variable.
Consider the following examples:

                         2 = 2 – Truth Value = TRUE
                         1 = 2 – Truth Value = FALSE
                          2 ≤ 2 – Truth Value = TRUE
                         2 < 2 – Truth Value = FALSE

                 1 + x = 2 – Truth Value = INDETERMINATE

In the last example, the truth value is determined when we substitute a
value for the viable x. If one is substituted for x, the equation will be true,
if any other number is used e.g. 3, the equation will be false. For equations
containing variables, we define the solution to the equation as the value(s) of
the variable(s) that, when substituted into the equation, make the truth
value of the equation TRUE. One method of finding the solution to an
equation is by substituting different values for the variable until one is
found that makes the equation true. This is called guess and check and
provides a simple albeit inefficient method for finding a solution.
Fortunately by applying a simple set of rules for manipulating the symbols
and numbers in an equation, the solution can be found more efficiently and
quicker.

Abu Ja'far Muhammad ibn Musa Al-Khwarizmi (790 – 840) is
credited with discovering/inventing algebra. Fibonacci
introduced Al-Khwarizmi’s work (and hence algebra) into
Europe. Since variables are placeholders for numbers, the
symbols used to represent the variables are subject to the
same rules of arithmetic as ordinary numbers. In particular if
we have the following simple equation:

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                              21



                                               2=2

Then adding the same number to both sides of this equation does not change
the truth value of the equation. Thus for example adding three to both
sides of this equation yields an equation which is still true:

                                             2+3=2+3

The text uses the metaphor of the balanced scale to illustrate this concept.
The equal sign can be viewed as balance scale used to weigh or compare the
weight of different objects:



____ = _____                     →



The idea here is that in order for the equation to be true it must ‘balance’,
that is both sides must be equal. On the balance scale, if the weight of the
objects placed on each side is the same, the scale will balance, that is be
level and not tilted to one side or the other. If an object’s weight is
unknown, we can place it on one side of the scale and then add weights of
known value to the other side until both sides are balanced, as illustrated
below:



                     known               1
                                                                  unknown

                                                        ?


                                 2

                 1     1     2       2




This is analogous to using guess and check to solve an equation. We note
however, that in balancing the scale, we are not confined to adding weights
to just one side. That is we can add weights to both sides and then

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                           22


determine the weight of the unknown by subtracting the total value of the
known weights from both sides:




                           1   2                       ?       2




                           ?       =   1   +   2   -       2


Adding and subtracting the weights is equivalent to moving the weights from
one side of the scale to the other. Thus, once the scale is balanced we can
add or subtract the same amount of weight from both sides and the scale
will still remain balanced. In the example pictured above, instead of adding
the known weights on the left and right and then subtracting them to find
the unknown, we can remove the equivalent amount of weight from both
sides of the scale, until only the unknown remains on the left side. The total
weight remaining on the right side is then equal to the weight of the
unknown.

The metaphor of the balanced scale can be used to help us define the rules
or properties of algebra and allows us to manipulate the numbers and
symbols in an equation to find the solution. An equation with a variable
represents a ‘balanced scale’ with the numbers on both sides playing the
roles of the known weights and the variable the role of the unknown weight.
Like the scale, the equation will remain true (i.e. in balance) if we add or
subtract the same number from both sides. Note: adding or removing
multiples of the same weight is the same as multiplying or dividing by the
value of that weight. We therefore have the following property of
equivalence for any equation:

General Property of Equivalence. The truth value of an equation remains
unchanged when the same operation is performed to both sides of the
equation using the same number or variable.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            23


The text expresses the General Property of Equivalence as four separate
properties:

The Addition Property of Equality. For any numbers a, b, and c, if a = b,
then a + c = b + c.

The Multiplication Property of Equality. For any numbers a, b, and c if a =
b, then ac = bc.

The Cancellation Property. For any numbers a, b, and c, if a + c = b + c, then
a = b. If c ≠ 0 and ac = bc, then a = b.

The Substitution Property. If for any numbers a, b, and c, if a = b and b = c,
then a = c. That is, we can substitute c for b.



By applying these properties to an equation with a variable, we can isolate
the variable on one side of the equation and transform the equation into the
following equivalent form:

                                variable = number

and thereby find the solution to the equation. In order to transform an
equation into this form we may need to apply the same property more than
once and/or apply one or more of the properties listed above. Each time one
of these properties is applied to the equation, we transform it to an
equivalent equation. That is, an equation whose solution is the same as the
original equation. The transformation from the original form to the value
form of variable equals number therefore proceeds in a series of steps, with
each step corresponding to an application of one of the four properties
listed above to the equation. Using the balanced scale metaphor, we are
transforming the scale from the balanced state:



                        1   2                  ?    2




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                           24


to the equivalent balanced state:




                                                     ?
                            1




We note that at each step, the scale always remains in balance. The same is
true for our equation. By using the four properties listed above we insure
that the equation is transformed in such a way that it remains in balance at
all times. That is, the solution of each transformed equation is the same as
the original equation.

Once the application of these properties to solving an equation is mastered,
these rules tend to become ‘second-nature’ and are applied routinely and
(unfortunately) without much conscious thought. Often two or more of the
properties are ‘combined’ into a single step. This can and often does lead to
mistakes, which in turn result in the ‘solution’ being incorrect. Combining or
‘skipping’ steps is therefore discouraged. We will use the statement-reason
format for solving all equations whereby the property used to go from one
step to the next is listed next to the transformed equation at each and
every step. Listing all steps has the benefit of providing a ‘trace-back’ once
the solution is found. This allows us to easily find where any mistakes were
made during the solution process.

Example. Find the solution for the equation 2x + 4 = -6.

2x + 4 = -6                  Given
2x + 4 +(-4) = -6 + (-4)     Addition Property of Equality
2x = -10                     Substitution Property
 (½)2x = -10(½)              Multiplication Property of Equality
x = -5                       Substitution Property

As mentioned at the beginning of our discussion, real-world problems (as well
as many exercises in school) are presented or given in descriptive format i.e.
as word problems. In the process of solving these problems we will often

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            25


need to apply ‘algebraic thinking’ and solve an equation to obtain the solution.
To do this we use the four step problem solving procedure we discussed in
the first lecture. The following example from the text illustrates this
process.

Example. Bruno has five books overdue at the library. The fine is 10¢ per
day per book. He remembers that he checked out an astronomy book a week
before he checked out the other four books. If his total fine is $8.70, how
long was each book overdue?

1. Understanding the Problem. From the problem we have the following
information:

         5 books are due: the astronomy book plus four others.
         10¢ per day per book fine
         $8.70 total is due for all 5 five books
         4 books checked out the same length of time
         1 book (the astronomy book) is checked out 1 week longer

We need to find how long the astronomy book is overdue and how long the
other four books are over due. Common knowledge required: numbers of days
in a week, number of cents in a dollar.

2. The Plan. We want to relate the five pieces of information together to
obtain an equation for the number of days the books are overdue. Since we
are being asked to find the length of time the books are overdue this is our
unknown. Since the unknown is a length of time in days, we let d be our
variable and let it represent the number of days the 4 books that were
checked out at the same time are overdue.

We now need to relate all the other information to our variable to obtain an
equation. Now we know that the astronomy book was checked out one week
or 7 days longer than the other four. Therefore the number of days this
book is overdue is:

                                      d+7




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                            26


The amount of the fine for a book is 10¢ times the number of days it was
overdue. We therefore have:

Fine for the astronomy book = 10(d + 7)
Fine of the other 4 books = 4 • 10d

The total fine is the sum of the fines for each book or 10(d + 7) + 40d. Since
we know that this is equal to $8.70 we have:



                            10(d + 7) + 40d = 870

Where we have converted $8.70 to 870¢ using the common knowledge that
there are 100¢ in $1.

3. Carrying Out the Plan. We have our equation and we must solve it using
the properties and rules of algebra.



10(d + 7) + 40d = 870                   Given
10d + 70 + 40d = 870                    Substitution Property
50d + 70 = 870                          Substitution Property
50d + 70 +(-70) = 870 + (-70)           Addition Property of Equality
50d = 800                               Substitution Property
 1         1                        Multiplication Property of Equality
  50d =   800
 50        50 
d = 16                                  Substitution Property

4. Looking Back. We want to check our answer and make sure that it fits
with the information of the problem. According to our solution the four
books are each overdue by 16 days. The astronomy book is therefore
overdue by 16 + 7 = 23 days. The fine for the astronomy book is therefore
$2.30 and the fine for each of the other books is $1.60. The total fine is
therefore: $2.30 + 4 • $1.60 = $2.30 + $6.40 = $8.70. This checks with the
value of the fine given in the problem, so our answer fits with the data and is
therefore correct.




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                  27


Logic

Logic is the method used by mathematics for constructing formal proofs
(see the proof to Gauss’s problem in the fist lecture). In logic, a statement
is a sentence which is either true or false, but not both. That is a sentence
whose truth can be determined absolutely as either true or false i.e. not
ambiguous. Examples of sentences that are not statements are as follows:

                 1.   She has blue eyes. Who does she refer to?
                 2.   x + 7 = 18.        What is x?
                 3.   2 + 3.             No assertion is made here.
                 4.   How are you?       This is a question not an assertion.
                 5.   Carter was the best president. What is meant by ‘best’?

The following are examples of sentences that are statements:

    1.   If he is over 6ft tall then he is over 70 inches tall.
    2.   It rained in Los Angeles on November 20, 2006.
    3.   Hillary Clinton has blue eyes.
    4.   2(x + y) = 2x + 2y
    5.   All men have black hair.

Negation

Given a logical statement, a new statement having opposite truth value can
be formed by negation. If the original statement is true, the negated
statement is false and vice versa. Examples:

               Statement                                 Negated Statement
It is raining now.                              It is not raining now.
2+3=5                                           2+3≠5
A hexagon has six sides                         A hexagon does not have six sides

Care must be taken when forming the negation of statement. The
statements ‘My shirt is blue’ and ‘My shirt is red’ are statements but they
are not negations of each other since if my shirt is actually green, both
these statements are false. The statements ‘My shirt is yellow’ and ‘My shirt



Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                           28


is not yellow’ are negations of each other because they have opposite truth
values regardless of the actual color of my shirt.

Some statements involving quantifiers can be difficult to negate. The text
makes the following classification of quantifiers:

Universal Quantifiers: all, every, and no. These refer to each other and
every element in a particular set.

Existential Quantifiers: some and there exists at least one. These refer to
one or more, or possibly all elements in a set.

Consider the following statement: Some professors at Pierce College have
blue eyes. This statement means that at least one professor at Pierce has
blue eyes. It does not rule out the possibility that all professors at the
college might have blue eyes or that some professors do not have blue eyes.
Since the negation of this statement must have the opposite truth value of
the original statement the following statements are not negations to this
statement:

         Some professors at Pierce College do not have blue eyes
         All professors at Pierce College have blue eyes.


A correct negation is:

         No professors at Pierce College have blue eyes.

The following list gives the negation of general quantified statements:



             Statement                                     Negation
Same a are b.                            No a is b.
Same a are not b.                        All a are b.




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                           29



Notation and Symbolic Logic

Logic, like algebra, has its own notation and rules for manipulating
statements and symbols defined by the notation. Formally, the letters p and
q are used to represent or act as placeholders for logical statements similar
to the way x and y are used to act as placeholders for numbers in algebra.
The symbol ‘~’ is used to denote negation of a statement. The negation of
the statement p is written as ~p.

Statements can be combined to form compound statements using the
connectives and and or, represented symbolically by ∧ and ∨, respectively.
They are analogous the arithmetic operators used in algebra. For example,
if we let p represent the statement ‘it is snowing’ and q represent the
statement ‘the ski run is open’ then the compound statement ‘it is snowing
and the ski run is open’ is written p ∧ q.

Logical statements p and q and also be linked or combined conditionally. That
is we can say: ‘If it is snowing, then the ski run is open’. We use the arrow
symbol → to represent this conditional statement and write p → q (note: the
‘if’ and ‘then’ are not part of statements p and q, but are indicated by the
arrow). In conditional statements the ‘if’ part is called the hypothesis and
the ‘then’ part is called the implication. Statements can also be linked bi-
conditionally, that is we can say: ‘It is snowing if and only if the ski run is
open’. This statement is written symbolically as p ↔ q and is logically
equivalent to the compound conditional statement (p → q) ∧ (q → p).

Two statements are said to be logically equivalent if and only if they have
the same truth value. If the statements p and q are logically equivalent we
write p ≡ q. Note: this does not mean that the statements p and q say the
same thing; rather it means that if statement p is true, then statement q is
also true or if statement p is false, statement q is also false.

Truth Tables

Truth tables are used to show all possible truth values for logical
statements. These include compound statements, conditionals, and bi-
conditionals. For p and ~p we have the following truth table:

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            30



   p                ~p   The table shows that when statement p is true, then its
True             False   negation ~p is false and vice versa. Since a statement
False            True    must be either true or false these are the only
                         possibilities.

The truth value for a compound statement is derived from it constituent
simple statements. Each constituent statement in the compound statement
may be true or false independent of the other statements. Compound
statements formed using ‘and’ are called conjunctions and are true only if
both the constituent statements are true. Compound statements formed
using ‘or’ are called disjunctions and are false only if both the constituent
statements are false. Note: there are actually two types of ‘or’ statements:
the inclusive or and the exclusive or. In spoken language (as well as
mathematics) the ‘default’ meaning of the word ‘or’ is inclusive. For example:
‘A triangle is a polygon with three sides or a polygon with three vertices’.
This statement is taken to be true if either statement is true and if both
statements are true. Now consider the following statement: 'You can follow
the rules or be disqualified'. In this case the usage of the word ‘or’ is
exclusive. That is, the statement is true if only one of the conditions is true,
but not both conditions. The symbol ∨ refers to the inclusive or. Symbols
for exclusive or vary with the ‘+’ being one of the more popular. The following
truth table shows the truth values for these compound statements.



                     p        q       p ∧ q      p ∨ q        p + q
                 true     true      true      true         false
                 true     false     false     true         true
                 false    true      false     true         true
                 false    false     false     false        false



Since two statements are logically equivalent if they have the same truth
values, their truth tables will also be the same. This means that we can use
truth tables to show whether or not two statements are equivalent. If they
are, their truth tables will be the same, if not their truth tables will be
different.



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                            31


Example. Show that ~(p ∧ q) ≡ ~p ∨ ~q.

The truth tables for both statements are as follows:

                             p          q     ~(p ∧ q)
                       true         true      false
                       true         false     true
                       false        true      true
                       false        false     true



                             p          q     ~p ∨ ~q
                       true         true      false
                       true         false     true
                       false        true      true
                       false        false     true

Since the truth tables for both statements are the same, they are logically
equivalent.

Example. Show that p + q ≡ (p ∧ ~q) ∨ (~p ∧ q). We construct a truth table
for each component of this statement:


          p              q       (~p ∧ q) (p ∧ ~q)     (p ∧ ~q) ∨ (~p ∧ q)
    true             true        false    false      false
    true             false       false    true       true
    false            true        true     false      true
    false            false       false    false      false

Comparing with the truth table for p + q we see the tables are the same.



The truth table for a conditional is derived by considering the implication as
a ‘promise’ that is conditioned on the hypothesis. The text gives the
following example:

                 If Betty gets a raise, then she will take Susan to dinner.

Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                   32



In this case the hypothesis is ‘Betty gets a raise’ and the implication is ‘she
will take Susan to dinner.’ If Betty keeps her promise, then the implication is
true, if not the implication is false. Now there are four possibilities:

                 1.   Betty   gets the raise and takes Susan to dinner.
                 2.   Betty   gets the raise and does not take Susan to dinner.
                 3.   Betty   does not get the raise and takes Susan to dinner.
                 4.   Betty   does not get the raise and does not take Susan to dinner.

In the first case the hypothesis is true i.e. Betty gets the raise, and the
implication is also true i.e. she follows through on her promise and takes
Susan to dinner. In this case the conditional is true. In the second case,
Betty gets the raise but does not take Susan dinner i.e. she breaks her
promise and the conditional is therefore false. For the third case, Betty
does not get the raise, but she takes Susan to dinner anyways. Again the
conditional is true. In the last fourth case, Betty does not get the raise and
does not take Susan to dinner. Since she does not break her promise in this
case, the conditional is true. We therefore have the following truth table:




                                p          q       p → q
                         true          true       true
                         true          false      false
                         false         true       true
                         false         false      true

The conditional p → q has there related statements:

Converse:         q→ p
Inverse:         ~p → ~q
Contra-positive: ~q → ~p

Example: If I am in Los Angeles, then I am in California.

Converse:                If I am in California, then I am in Los Angeles.
Inverse:                 If I am not in Los Angeles, then I am not in California.

Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                           33


Contra-positive: If I am not in California, then I am not in Los Angeles.

This example shows that if the conditional is true then the contra-positive is
true, but the converse and inverse are not necessarily true. In general we
can show that:

 A conditional statement and its contra-positive statement have the same
truth table and are therefore logically equivalent.

Valid Reasoning

Logic along with its notation and rules is used to determine if the reasoning
used in an argument or proof is valid. That is, the reasoning is said to be
valid if the conclusion (implication) follows unavoidably from the hypothesis.
Note: the validity of the reasoning is independent of the truth of the
hypothesis. That is the reasoning can be valid even if the hypothesis is false.
Valid reasoning asserts that the conclusion must follow i.e. is unavoidable
provided that the hypothesis is true. If the hypothesis is false, the
reasoning is still valid, however the conclusion does not necessarily result
since it is conditional on the hypothesis being true.



Example. Consider the following example:

Hypothesis: All roses are red.

Argument: The flower is a rose, therefore it is red.

Conclusion: The flower is red.

This reasoning in this argument is valid since the conclusion follows
unavoidably from the hypothesis. However since the hypothesis is false in
this case, so is the conclusion.

In mathematics, theorems and their proofs follow this pattern. The theorem
typically makes a statement such as if A (is true), then B (follows). In order
to justify the theorem, a proof is given showing that if the hypothesis is



Pierce College                                                              MAP
Math 215 Principles of Mathematics                                               34


true, then the conclusion logically follows. The reasoning (or argument) used
in the proof must always be logically valid.

Rules for Logical Reasoning

 In order to determine if the reasoning used in an argument is valid, it is
examined or analyzed to see if it follows or uses the rules of logic. We will
discuss three of these rules:

1. Modus Pones or direct reasoning. Modus Pones is also known as the law
of detachment states that:

If the statement ‘if p, then q’ is true and if p is true, then q must also be
true.

For example, if the statement ‘if the sun is shining, we shall take a trip’ is
true and the statement ‘the sun is shining’ is true, then we can conclude that
the statement ‘we shall take a trip’ is true.

2. Modus Tollens or indirect reasoning. This rule states that:

If the statement ‘if p, then q’ is true, and the conclusion, q, is false, then the
hypothesis, p, is false.

For example, if the statement ‘if a figure is a square, then it is a rectangle’
is true, then if the statement ‘the figure is not a square’ is true, the
conclusion is false as is the hypothesis and we can conclude that statement
‘the figure is not a square’ is true.

3. The Chain Rule. The chain rule says:

If ‘if p, then q’ and if q, then r’ are true, then ‘if p, then r’ is true.

For example, if the statements ‘if I save, then I will retire early’ and ‘if I
retire early, I will play golf’ are both true, then we can conclude that the
statement ‘if I save, then I will play golf’ is also true.




Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                          35


Using these rules we can ‘symbolize’ an argument i.e. assign letters p, q, r,
etc to logical statements, and then see if we can derive the conclusion from
the hypothesis using the rules of logical reasoning. If we can, then the
argument or reasoning is said to be valid, if not it is invalid.

Example. If x > 2, then x2 > 4.

Let p be the statement x > 2 and q the statement x2 > 4. Then we have:

p → q – original statement
p     - hypothesis
q     - Modus Pones

Thus the conclusion follows logically from the hypothesis and if the
hypothesis is true, the conclusion will be true.

Example. Consider the following true statement: ‘Healthy people eat
oatmeal’. Are the following conclusions valid:

    1. If a person eats oatmeal, then the person is healthy.
    2. If a person is not healthy, the person does not eat oatmeal.

Let p be the statement ‘a person is healthy’ and q the statement ‘a person
eats oatmeal’. The original statement is then p → q. The fist conclusion is
q → p which is the converse of the original and the second is ~p → ~q which
is the inverse of the original. As we saw above, neither the converse nor the
inverse have the same truth table as the original and therefore they are not
logically equivalent. Hence neither is a valid conclusion.




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                               36



Lecture 3

Sections 2.1 and 2.2 – Sets

Definition of a Set

Set theory was developed in the late 1800’s by Cantor to place
number theory and calculus on a more solid theoretical
footing. Set theory is used to define the concept of whole
numbers, addition, subtraction, less-than and greater-than.
We start by defining what we mean by a ‘set’. A set is defined
as any collection of objects. ‘Objects’ is used in a broad and                     Georg Ferdinand
                                                                                 Ludwig Philipp Cantor
generic sense and is meant to include numbers, letters, people,                      1845 - 1918
animals, and just about any physical or abstract entity or item.
The only ‘limitation’ placed on our objects is that they are
discrete and distinct items.

We also require a set to be well defined. This means that if we are given a
set and some object we must be able to state definitively whether or not
the object belongs to the set. Thus using the example in the text, if we are
given the set of letters in the English alphabet:

         β = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}

then the letter ‘h’ is definitely a member of the set while the number ‘3’ is
not a member of the set. Note that some care must be made in making this
determination. For example a distinction is made between upper i.e. capital
letters and lower case letters. Thus the upper case letter ‘H’ is not a
member of this set even though the lower case letter ‘h’ is. In general, this
‘case-sensitivity’ applies everywhere in mathematics. Symbols are unique and
may not be replaced or substituted with other ‘similar’ symbols without
explicitly stating their equivalence.

The objects comprising or making up the set are called the members or
elements of the set. We thus say that the letter ‘h’ is a member or element
of the set of letters in the English alphabet. The order in which the



Pierce College                                                                                  MAP
Math 215 Principles of Mathematics                                                             37


elements of the set are listed are unimportant and each element is listed
only once. Thus we could write:

          β = {u, v, w, x, y, z, o, p, q, r, s, t, g, h, i, j, k, l, m, n, a, b, c, d, e, f}



for the set of letters in the English alphabet.

Set Notation

As indicated in our example above curly-brackets ‘{}’ are used to designate a
set and the elements of the set are placed inside these brackets, typically
separated by commas. The set is given a name or designator, usually a capital
letter or another distinctive name. We have used the Greek letter beta, β,
to represent the name for the set of letters in the English alphabet.

The symbol ∈ is used to indicate that an element is a member of the set.
We write h ∈ β (read h is a member of β) to state that the letter h is a
member of the set. The symbol ∉ indicates that an element is not a member
of the set. Thus, for example, we write 3 ∉ β to state that the number 3 is
not a member of the set β.

In some cases, a set may have an infinite number of elements, like the set of
Natural Numbers, N. In this case listing all the elements of N is impossible,
so instead we list the first few members followed by the ellipsis symbol ‘. . .’
to indicate that the members continue indefinitely in the same manner. Thus
for example we write the set of natural numbers as:

                                     N = {1, 2, 3, 4, 5, . . .}

An alternative method is to use the set builder notation. For example, let
the set C = {1, 2, 3, 4}. Instead of writing C by listing all its members we can
write it using set builder notation as:

                                   C = {x | x ∈ N and x < 5}

In words we read this as: ‘C is equal to the set of all elements x, such that x
is a natural number and x is less than 5’. Breaking this down we read the

Pierce College                                                                                 MAP
Math 215 Principles of Mathematics                                                  38


curly brackets as ‘the set of’, the vertical bar as ‘such that’ and ‘x’ as ‘all
elements x’:

                 C   =       {        x          |      x∈N       and   x < 5}
             Set C   is      the      all        such   x is a    and   x is less
                     equal   set      elements   that   natural         than five
                     to      of       x                 number




Set builder notation is used when it is inconvenient or impossible to list the
elements of a set. For example, suppose D is the set of all real numbers R,
between zero and one. In this case, listing all the elements of the set is
impossible and ellipsis will not help us either. However we can use set
builder notation to define the set:

                                   D = {x | x ∈ R and 0 < x < 1}

In general, the most convenient notation is used when defining a set. In some
cases, either notation may be used (as with the set C), and the choice is left
to the judgment of the person defining the set.

Set Properties

We now discuss the general properties of sets. These include equality, one
to one correspondence, equivalency, and cardinal number.



1. Equality

Two sets are said to be equal if and only if they contain exactly the same
elements.

Note again, that the order of the members may be different but if they are
equal they will both contain exactly the same members. Thus the sets
A = {1, 2, 3} and B = {3, 1, 2} are equal but the sets C = {1, 2, 3, 4} and
D = {1, 2} are not equal. If two sets A and B are equal we write A = B. If two
sets C and D are not equal we write C ≠ D.

Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                            39




2. One-to-One Correspondence

If the elements of two sets A and B can be paired such that for each
element of A there is exactly one element of B and for each element of B
there is exactly one element of A, then the two sets are said to be in one-
to-one correspondence.

The idea here is that if we have two sets with the same number of elements
we can establish an exact pairing between them. The text provides the
following example. Suppose we have three swimmers: Tom, Ellen, and Marie
and a swimming pool with three lanes. We can form two sets:

                 S = {Tom, Ellen, Marie} and L = {Lane1, Lane2, Lane3}

We can then assign each swimmer to a lane, and since the number of
swimmers and the number of lanes is the same, the pairing of swimmers to
lanes is one-to-one. It should be noted that this pairing is not unique. That
is, there is more than one way to assign the swimmers to a lane as the
following illustrates:

1. Tom – Lane 1             2. Tom – Lane 1           3. Tom – Lane 2
   Ellen – Lane 2              Ellen – Lane 3            Ellen – Lane 1
   Marie – Lane 3              Marie – Lane 2            Marie – Lane 3

4. Tom – Lane 2            5. Tom – Lane 3           6. Tom – Lane 3
   Ellen – Lane 3             Ellen – Lane 1            Ellen – Lane 2
   Marie – Lane 1             Marie – Lane 2            Marie – Lane 1

Each of these parings defines a one-to-one correspondence between the
swimmers and the lanes in the pool. Note: if there were only two lanes in the
pool or only two swimmers then it is not possible to have a one-to-one
correspondence between the sets. We would either have an extra swimmer
or an extra (unused) lane in the pool.




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                            40


3. Equivalent Sets

Two sets A and B are said to be equivalent if and only if there exists a one-
to-one correspondence between the sets. We write A ~ B to indicate that
the sets are equivalent.

In the previous example of the swimmers and pool lanes, the sets S and L
are equivalent and we can write S ~ L. Note: equivalent sets and equal sets
are different. For two sets to be equal they must contain the same
elements. For two sets to be equivalent, they must be in one-to-one
correspondence, but their elements may be different.



4. Cardinal Numbers

The cardinal number for a set A, denoted n(A) is the number of elements in
A.

Two sets that are in one-to-one correspondence will have the same cardinal
number. Thus in the example of the swimmers and the pool lanes we have
n(S) = n(L). Note: we do not necessarily have to know (i.e. have counted) the
number of elements to know that two sets have the same cardinal number.
Instead, if we place them in one-to-one correspondence, then we can say
that their cardinal numbers are the same, without performing a tally of their
members.

Special Sets

There are four categories of sets worth special mention: the empty set,
finite sets, infinite sets, and the universal set.

1. Empty Set

A set that contains no elements or equivalently whose cardinal number is
zero is called the empty set and is denoted by the symbol ∅. Note that ∅ ={}
and should not be confused with the set {0} which contains one element i.e. 0
and is not empty. Also note that it is not correct to write {∅} since this
indicates a set which contains the empty set, and is therefore not empty.

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                            41


2. Finite Sets

A set is finite if the cardinal number of the set is zero or a natural number.
Finite sets can be large e.g. the set of all women on earth over 5 feet tall
but there will always exist some natural number m, such that the cardinal
number of the finite set is less than or equal to m.

3. Infinite Sets

Put simply, an infinite set is a set that is not finite. That is, there is no
natural number, m, such that the cardinal number of the set is less than or
equal to m. Examples of infinite sets are the set of natural numbers and the
set of whole numbers.

4. Universal Set

The universal set or the universe, denoted U, is the set that contains all the
elements being considered. The universal set will vary from problem to
problem and it is important to know exactly what the universal set is for the
problem being considered. Examples of universal sets are: all the people
living in California, all the students attending Pierce College, all the cell
phones in the classroom, etc. In each case, the universal set is defined by
the problem under consideration.

Venn Diagrams

Venn diagrams provide a pictorial view of the relationship
between sets. Consider the following sets: the set of all people in
California, which is designated as the universe, and the set of all
females in California. Using a Venn diagram these sets are
represented as follows:
                                                                        John Venn
                                                                       1834 - 1923
                               U

                   F




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                            42


The universal set (the set of all people in California) is drawn as a rectangle
and represents the ‘bounds’ of the problem. That is, all other sets and
elements of sets must be part of this set (by definition) and are drawn
inside it. The set of all females in California is represented by a circle which
is drawn inside the rectangle representing the universe.

Using Venn diagrams we can illustrate the relationships between various sets
and the universe as well as the relationships between different sets inside
the universe.

Set Relationships

Given a universe and a set in that universe we can define three important
relationships: the complement of a set, the subset of a set, and the proper
subset of a set.

Complement

The complement of a set A, written Ā, is the set of all elements in the
universal set U that are not in A; that is Ā = {x | x ∉ A}.

If we consider U as the set of all people in California and F as the set of all
females in California, then the complement of F is depicted as follows using a
Venn diagram:



                                            U

                                F




where the compliment of F is the shaded region in the figure.

Subset

B is a subset of A, written B ⊆ A, if and only if every element of B is an
element of A.



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             43


Note that the definition above allows B to be equal to A, that is a set is a
subset of itself.

Proper Subset

If a subset B of A is not equal to A, then B is called a proper subset of A
and we write B ⊂ A.

Note: for B to be a proper subset of A, all the elements of B must be in A,
however there must be at least one element of A that is not in B.



We can use Venn diagrams to show the relationship of two sets to each
other. Consider the following diagrams:



                     U                        U                        U

                                  A   B                   A
          A      B                                            B




In the left most diagram, neither A nor B is a subset of the other and they
share no common elements. In the middle diagram, the situation is the same,
however A and B do have some elements in common. In the right most
diagram B is a subset of A.

Finally we note that:

The empty set is a subset of itself and a proper subset of any other set
other than itself.

To see this consider some non-empty set A. Then either ∅ ⊆ A or ∅ is not a
subset of A. However, if ∅ is not a subset of A then by definition there
must be some element of ∅ that is not in A. But since ∅ contains no
elements, this cannot be and we must therefore have that ∅ ⊆ A.




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                               44


Number of Subsets in a Set

We now ask the question: ‘Given a finite set A, how many distinct subsets of
A are there?’ To answer this question we follow the strategy used in the
text and start with the simplest case and work towards more complex cases.
Since the simplest case is a set having no elements, we start with the empty
set.

Since the empty set has no elements, we can only form one subset, and that
subset is the empty set itself (remember a set is a subset of itself by
definition). Therefore, the empty set has one subset.

The next case is a set having only one element, say A1 = {a1}. Here we can
form two subsets: ∅ and A1 where we note again, that a set is always a
subset of itself and that the empty set is a subset of every set. Thus for
the case of a set having one element, we can form two subsets.

For a set having two elements, say A2 = {a1, a2} we can form the following
subsets: ∅, {a1}, {a2}, {a1, a2}. We therefore have that four subsets can be
formed from a set with two elements.

If A3 = {a1, a2, a2}, then we can form the following subsets: ∅, {a1}, {a2}, {a3},
{a1, a2}, {a1, a3}, {a2, a3}, {a1, a2, a2}. Thus a set with three elements has 8
subsets.

At this point we stop and make a table of our results:

                 Number of Element in Set   Number of Subsets
                            0                       1
                            1                      2
                            2                      4
                            3                      8

We observe that the number of subsets is always a power of 2. Specifically,
number of subsets = 2n where n is the number of elements in the set. We
therefore hypothesize that number of subsets = 2n. In order to state that
this is true for finite sets having more than three elements we need to



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                          45


provide a formal proof. This is beyond the scope of the course, but in fact,
this hypothesis can be proved. We therefore state (without proof):

The number of subsets that can be formed from a finite set A is 2n(A).

Inequalities

We can use our definition of one-to-one correspondence to define the
concepts of less-than and greater-than. Given two finite sets A and B we
can try and establish a one-to-one correspondence between them. One of
three possible outcomes will result:

    1. A one-to-one correspondence cannot be established because A has
       fewer elements than B.
    2. A one-to-one correspondence can be established since A and B have
       the same number of elements.
    3. A one-to-one correspondence cannot be established since A has more
       elements than B.

For the first outcome we can write n(A) < n(B) meaning that the number of
elements in A is less-than the number of elements in B. For the second
outcome we write that n(A) = n(B) meaning that the number of elements in A
is equal to the number of elements in B. For the last outcome we write n(A) >
n(B) meaning that the number of elements in A is greater than the number
of elements in B.

It is important to note here that we do not need to actually count the
elements in A and B to make this determination. All that is required is to try
and establish a one-to-one correspondence between them.

Set Operations

Up to this point we have defined the concept of set, subset, and the
notation used for describing sets. We now discuss operations on sets; that is
forming a set from two other sets. This is analogous to using addition or
multiplication to ‘form’ a number given two other numbers. We discuss four
fundamental operations on sets: intersection, union, difference, and
Cartesian products.

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                          46


Intersection

The intersection of two sets A and B, written A ∩ B, is the set of all
elements common to both A and B. A ∩ B = {x | x ∈ A and x ∈ B}.

The intersection of two sets is associated with the word ‘and’. Recalling what
we discussed about truth tables in logic, if two statements are linked by the
word ‘and’ common usage as well as mathematical usage of this word implies
that both statements must be true for the compound statement to be true.

If two sets have no elements in common they are called disjoint sets and
their intersection is the empty set i.e. A ∩ B = ∅. We also note that the
intersection of two sets is a set.

Example

Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6, 7}, and C = {6, 7}. Then:

A ∩ B = {3, 4}
A∩C=∅
B ∩ C = {6, 7} = C

We can also use a Venn diagram to represent the intersection of two sets.
In the example above we can draw a Venn diagram for A ∩ B:


                                                     U


                                A        B
                                    2
                            2                3




The diagram shows the number elements in the intersection (shaded region),
and the number of elements in A and B that are not in the intersection.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                                 47


Union

The union of two sets A and B, written A ∪ B, is the set of all elements in A
or in B. A ∪ B = {x | x ∈ A or x ∈ B}.

The union of two sets is associated with the word ‘or’. Recalling what we
discussed about truth tables in logic, if two statements are linked by the
word ‘or’ common (inclusive) usage as well as mathematical usage of this word
implies that the compound statement is true if one or both of the
statements is true. We also note that the union of two sets is a set.



Example

Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6, 7}, and C = {6, 7}. Then:

A ∪ B = {1, 2, 3, 4, 5, 6, 7}
A ∪ C = {1, 2, 3, 4, 6, 7},
B ∪ C = {3, 4, 5, 6, 7} = B

Note: if an element appears in both sets e.g. 3 in A and B in the example
above, it is listed only once in the union. That is, ‘duplicates’ or ‘copies’ of
such an element do not appear in the union.

We can also use a Venn diagram to represent the union of two sets. In the
example above we can draw a Venn diagram for A ∪ B:


                                                     U


                                A         B




The shaded region in the diagram shows the union of the two sets. Note: the
sets need not overlap to have a union.




Pierce College                                                                     MAP
Math 215 Principles of Mathematics                                                    48


Difference

The difference or complement of set A relative to set B, written B – A, is
the set of all elements in B that are not in A. B – A = {x | x ∉ A and x ∈ B}.

Note B – A is not read as B minus A. We note that B – A is a set and the
minus sign does not represent an operation on numbers but rather an
operation on sets.

We can use a Venn diagram to represent the difference of two sets. The
difference of A and B i.e. B – A is shown in the following Venn diagram:


                                                       U


                                 A         B




The shaded region in the diagram shows the difference of the two sets.
Note: the difference B – A is different than A – B. We also note that B – A
= B ∩ Ā.

Example

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4}, B = {3, 4, 5, 6, 7}, and
C = {6, 7}. Then:

A – B = {1, 2}
B – A = {5, 6, 7}
B – C = {3, 4, 5}
C–B=∅




Pierce College                                                                        MAP
Math 215 Principles of Mathematics                                                 49


Cartesian Product

For any two sets A and B, the Cartesian Product of A and B, written A × B, is
the set of all ordered pairs such that the first component of each pair is an
element of set A and the second component of each pair is an element of set
B. A × B = {(x, y) | x ∈ A and y ∈ B}.

Note A × B is not read as A times B. We note that A × B is a set and the
multiplication sign does not represent an operation on numbers but rather an
operation on sets. Sometimes the multiplication sign is read as ‘cross’.

We also note that since the empty set contains no elements, we cannot form
ordered pairs with elements of another set. We therefore define the
Cartesian product of the empty set and any other set, A, as the empty set:



                                   A×∅=∅×A=∅

Example

Let A = {a, b, c} and B = {1, 2, 3}. Then:

A × B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3), (c, 1), (c, 2), (c, 3)}
B × A = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}



Properties of Set Operations

Because the order of elements in a set is not important, A ∪ B = B ∪ A. Thus
sets are commutative with respect to the union operation. Likewise sets are
also commutative with respect to the intersection operation: A ∩ B = B ∩ A.
In addition the union operator is also associative as is the intersection
operator. Note: union and intersection operators are not associative if mixed
together:

                               A ∩ (B ∪ C) ≠ (A ∩ B) ∪ C



Pierce College                                                                     MAP
Math 215 Principles of Mathematics                                                    50


To see this let:

A = {a, b, c, d}, B = {c, d, e}, and C = {d, e, f, g}. Then:

A ∩ (B ∪ C) = {a, b, c, d} ∩ ({c, d, e} ∪ {d, e, f, g}) = {a, b, c, d} ∩ {c, d, e, f, g}
            = {c, d}



(A ∩ B) ∪ C = {a, b, c, d} ∩ ({c, d, e} ∪ {d, e, f, g}) = {c, d} ∪ {d, e, f, g}
            = {c, d, e, f, g}



Although these operators are not associative when mixed, they are
distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). We therefore have the
following results:

If A, B, and C are three sets, then:

Commutative Property: A ∪ B = B ∪ A and A ∩ B = B ∩ A.

Associative Property: A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C

Distributive Property: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)



Example

The following problem shows how we use sets, Venn Diagrams, and set
properties to find a solution. Consider the following:

In a survey of 110 college freshman that investigated their high school
backgrounds, the following information was gathered:

25 took physics                         6 took physics and biology
45 took biology                         5 took all thee subjects
48 took math
10 took physics and math
8 took biology and math

Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                           51


 How many students took biology but neither physics nor math? How many
took physics, biology, or math? How many did not take any of these three
subjects?

We solve this problem by making a Venn diagram to represent the sets
involved. In this case there are three subjects, we have three sets (of
students): those that took physics, set P, those that took math, set M, and
those that took biology, set B. Since, from the given information, we know
that 5 students took all three subjects, we know that all three of these sets
intersect. We therefore have:


                                                 U
                                P       B




                                    M




From the information given we have:

n(U) = 110       n(M ∩ P) = 10
n(P) = 25        n(B ∩ M) = 8
n(B) = 45        n(B ∩ P) = 6
n(M) = 48        n(B ∩ M ∩ P) = 5



The first thing we are asked to find is the number of students that took
biology but neither math nor physics. The phrase ‘neither math nor physics’
means all the students that did not take math or physics. The set of
students taking math or physics is M ∪ P. The set of students taking biology
but not math or physics is therefore B - (M ∪ P), where we have used the
definition of set difference.

The second question asks how many took math, biology, or physics. This set
of students is B ∪ M ∪ P. Again, we remember that the word ‘or’ is
associated with the union operator.



Pierce College                                                             MAP
Math 215 Principles of Mathematics                                          52


The last question asks how may did not take any of these subjects. The set
of students taking all three subjects is just the union of the students taking
biology, math, and physics: B ∪ M ∪ P. Using the definition of set
difference, the set of students that did not take any of these subjects is U
– (B ∪ M ∪ P).

The answers to the three questions therefore are n(B - (M ∪ P)), n(B ∪ M ∪
P), and n(U – (B ∪ M ∪ P))

In order to find the answers to these questions, we start by adding the
intersection information given in the problem statement to our Venn diagram
for the problem. Let’s deconstruct the intersections so we can see where
each piece of information goes:


            P        B                    B               P
                 6                                8           10

                                                      M            M




                                  P                   B

                                                  5



                                                  M




In the bottom figure, we can find the number of students in the
intersections shaded by the lines, by subtracting 5 from the numbers in the
other intersections. We therefore have:



                          P           1       B

                                  5
                              5           3



                                  M




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                               53


We can now add the number of students in each subject that are not in the
areas of intersection (white areas in figure above) by subtracting the
number of students in the intersections from the number of students in the
subject. Thus for example, 5 + 5 + 3 = 13 students intersect with math and
since 48 students took math, there are 48 – 13 = 35 students in the white
area of the math set. Doing this for physics and biology we have:



                                P    14               1       B
                                                                   36
                                                     5
                                             5                3

                                                     35
                                                     M



Finally we add back the universe and compute the number of students in the
universe not in the three sets. To find out the number of students in all
three sets, we just add the number in each area in the diagram above: 14 +
36 + 35 + 1 + 5 + 5 + 3 = 99. Therefore there are 110 – 99 = 11 students who
did not take any of these subjects. We therefore have:


                                                                             U

                            P       14           1        B
                                                                  36
                                                 5
                                         5                3

                                                 35
                                                                        11
                                                 M




With our diagram complete, we can now answer the questions posed by this
example. For the first question we need to find n(B - (M ∪ P)). Now B - (M ∪
P) is the number of students who took biology but not physics or math i.e.
the number of students who just took biology. From the figure we see this
is number is 36.

For the second question we need to find n(B ∪ M ∪ P). We can find this by
summing all the numbers inside the ‘circle’ in our diagram. Therefore n(B ∪ M



Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                           54


∪ P) = 24 + 1 + 5 + 5 + 3 + 35 + 36 = 99. Thus the number of students that
took one or more of these three subjects is 99.

Finally for the last question we need to find n(U – (B ∪ M ∪ P)). Since the
universe has 110 students and we know that 99 of these took at least one of
the three subjects, the number who did not take any of these three
subjects is 110 – 99 = 11.




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           55



Lecture 4

Sections 2.3 and 2.4 – Addition, Subtraction, Multiplication, and Division

We shall discuss the basic models or underlying principles governing the four
canonical operations of arithmetic. In all four cases, our work with set
theory will provide the principle model for each operation.

Addition

For addition two models are given by the text: the set model and the
number-line model. As the name implies, the set model uses the ideas we
developed in the previous lecture to construct a model for addition of whole
numbers.

Set Model

The idea here is that given two sets, say of blocks, if we form the union of
the two (disjoint) sets i.e. combine all the blocks into a single set, how many
blocks will we have in the union? Further more how is this number related to
the number of blocks in the two original sets?



                 A                                         B

                     a   b                                     e       f

                     c   d                                         g




                         A∪B
                                       a           b

                               c           d           e

                                   f           g

Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           56


Using the figure above, we have two sets A and B. Set A has 4 blocks and set
B has 3 blocks. If we combine the blocks together in to a single set, then the
resulting set has 7 blocks. In this case, we simply counted the number of
blocks in the union to get our answer. We can relate this answer to the
number of blocks in the original sets or better yet use it to define what we
mean by addition. If we use the strategy the text calls ‘counting on’, we can
start by counting the blocks in set A and then ‘counting-on’ as we count the
blocks in set B. Thus block ‘e’ in set B is counted as 5 (instead of 1) and the
result gives us the total number of blacks. We can therefore define
addition in terms of the union of these sets:

                             n(A) + n(B) = n(A ∪ B)



Note: it is important that sets A and B are disjoint, otherwise the
relationship between the cardinal numbers of the three sets will not equate
as stated above. We therefore have the following result:

Definition of Addition of Whole Numbers. Let A and B be two disjoint
finite sets and let a = n(A) and b = n(B). Then a + b = n(A ∪ B). The numbers a
and b are called the addends and a + b is called the sum.

Number Line Model

The number line model uses (not surprisingly) the number line to compute
the sum of two numbers. The number line is just a way of graphically
ordering the whole numbers. Each number on the number line is placed the
same fixed distance away from its immediate neighbors on the left and the
right. The number zero ‘starts’ the line and does not have a neighbor to its
left.




                 0   1   2    3    4     5     6      7   8     9   10

Numbers are placed on the line, starting with zero, in increasing order. A
number to the left of another number on the line is less-than that number
e.g. 2 is less than 5 because it is to the left of 5. Conversely a number to the

Pierce College                                                               MAP
Math 215 Principles of Mathematics                                                                         57


right of another number is greater-than that number e.g. 7 is greater than 4
because it is to the right of 4.

We can use the number line to compute the sum of two numbers, by
representing each number i.e. addend as a line segment beginning at zero and
ending that the number in question. Thus for the number 4, we represent
this number as a line segment starting as zero and ending at 4, as shown
below.




            0        1       2           3        4      5           6       7       8       9       10

Likewise the number 3 is represented as a segment starting at zero and
ending at three:




           0         1       2           3       4       5           6       7       8       9       10


If we take our two segments and lay them end to end, with the beginning (no
arrow end) of the first segment at zero and be beginning of the second
segment at the end (arrow end) of the second segment, then the end of the
second segment lies at the sum of the two numbers:

                                                 4+3
                                     4                           3


                 0       1       2           3       4       5           6       7       8       9    10


From the number line and the definitions of greater-than and less-than, we
see that since the number 4 is less than 7, this means that there is some
other number i.e. 3 such that 4 + 3 = 7. If we generalize this concept we
have:

Pierce College                                                                                             MAP
Math 215 Principles of Mathematics                                            58


Missing Addend Property. For any two numbers a and b, if a is less-than b
i.e. a < b, then there exists a natural number c such that a + c = b.

We will use the missing added property when we discuss subtraction models.

Addition Properties

Addition of whole numbers follows or has certain properties. In particular
addition is commutative, associative, closed, and has and identity element.

Closure

This property is often ‘overlooked’ because it is so obvious. However its
importance cannot be over emphasized since it provides the very foundation
of arithmetic. Closure means that if we add two whole numbers we will obtain
a whole number as the result. Formally stated we have:

Closure Property of Whole Numbers. If a and b are whole numbers, then
a + b is a whole number.

Note: the closure property says that the sum both exists and is unique. If
this was not true, then the sum of two whole number might not exist or
worse, might not be a whole number.

Commutative Property

The commutative property states that the order in which we add two whole
numbers is unimportant. That is, if we reverse the order of the addends we
will still obtain the same sum:

Commutative Property of Addition of Whole Numbers. If a and b are two
whole numbers then a + b = b + a.

Associative Property

The associative property states that it is unimportant how we group three
(or more) whole numbers together when we compute their sum. The idea
here is that we will be adding the numbers pair wise to compute the sum. The

Pierce College                                                                MAP
Math 215 Principles of Mathematics                                          59


sum is independent of our choice of grouping, as for example, was
demonstrated by Gauss’s problem.

Associative Property of Addition of Whole Numbers. If a, b, and c are
three whole numbers then (a + b) + c = a + (b + c).

Identity Property

The identity property states that there exists a unique whole number such
that when this whole number is added to any other whole number, the result
is just the whole number in question. In particular:

Identity Property of Addition of Whole Numbers. There is a unique whole
number 0, called the additive identity, such that for any whole number a, a +
0 = 0 + a = a.

The identity property is important since it allows us to define inverses. In
particular, when integers are discussed we define the additive inverse of the
whole number a as the number b such that a + b = 0.

Addition Facts

These are basic strategies for performing elementary addition. The text
discusses four such strategies: counting-on, doubles, making ten, and
counting back.

    1. Counting-on. This technique or strategy starts with the larger of the
       two addends and ‘counts-on’ by the remaining addend. See the
       discussion of the number line above. This method is clearly inefficient
       and practical only if one of the addends is small i.e. less than 5 say.
    2. Doubles. By knowing the doubles i.e. 1 + 1, 2 + 2, 3 + 3, 4 + 4, 5 + 5,
       6 + 6, 7 + 7, etc, other sums can be derived by expressing them in
       terms of the doubles. For example 6 + 7 = (6 + 6) + 1.
    3. Making Ten. Since our number system is base ten, breaking one of the
       addends into two smaller numbers so that one of these numbers plus
       the other addend forms ten, allows sums to be computed using the
       resulting simplification of adding with ten. For example 8 + 5 = 8 + (2
       + 3) = (8 + 2) + 3 = 10 + 3 = 13.

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                              60


    4. Counting Back. In this strategy a number is added to the larger of the
       two addends to form ten, then the second addend is added and the
       number that was added is subtracted off. For example 8 + 5 = [(8 +
       2) + 5] – 2 = =[10 + 5] – 2 = 15 - 2 = 13.

Subtraction

There are four models discussed by the text for subtraction: the set model
(take-away model), the missing addend model, the comparison model, and the
line number model.

Set Model

In this case we start with a set of objects and remove a subset of these
objects to form two sets. In this case, we ask how many blocks are left in
the original set?

                                                                A

        C                     B
                                                                    f       h

         a       b                a       b   f                         g
                                                   A

         c       d                c       d   g


                 e                    e       h




In the figure above, we start with set B containing 8 elements and form a
subset A with 3 elements. We then take-away the elements in the subset A,
leaving us with set C, which is set B without the elements in subset A. We
can define subtraction in terms of the set difference since the elements in
C are all the elements in B that are not in A.




Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                       61


Definition of Subtraction of Whole Numbers. Let B be a finite set and A
be a subset of B and let a = n(A) and b = n(B). Then b - a = n(B ∩ Ā). The
number a is called the subtrahend and the number b is called minuend, and b
- a is called the difference.

Missing-Addend Model

We discussed this model above as the missing addend property. The idea
here is that we can ‘convert’ our subtraction problem into an addition
problem by asking what number needs to be added to the subtrahend so that
it equals the minuend. Thus for example, if we have:

                                   8–3=

we can convert it to the following problem involving addition:

                                  3+     =8

Thus instead of being asked to find the difference of 8 and 3 we are being
asked to find the number, that when added to 3 gives 8.

Comparison Model

This model uses the concept of one-to-one correspondence from sets to
determine the difference by seeing what’s left over after a one-to-one
correspondence is attempted between the minuend and subtrahend.




                                        difference




In the example of figure, the difference of 8 – 3 is represented as what’s
left over when a one-to-one correspondence is attempted between 8 blocks
and 3 blocks.



Pierce College                                                           MAP
Math 215 Principles of Mathematics                                               62


Line Number Model

This is similar to what we did for addition. In this case, the segments are
the subtrahend and minuend. The segment for the subtrahend is reversed
(i.e. arrow points left instead of right) and the beginning of the subtrahend
segment is placed at the end of the minuend segment. The end of the
subtrahend segment then ‘points’ to the difference, as shown in the figure
below.


                             Minuend = 8


                 0      1     2         3   4       5   6       7   8   9   10

                     Subtrahend = 3


                 0      1     2         3   4       5   6       7   8   9   10


                                  8-3                       3

                                                8


                 0      1     2         3   4       5   6       7   8   9   10

This model is useful when presenting negative numbers. The revering of the
arrow on the subtrahend corresponds to a change of sign e.g. making 3, -3.

Properties of Subtraction

Subtraction does not enjoy the same properties as addition does. In
particular, subtraction is not a closed operation under the whole numbers,
since, if the minuend is smaller than the subtrahend the result is not a whole
number e.g. 3 – 5 is not a whole number. In addition subtraction is not
commutative or associative and although zero might serve as an identity
element i.e. 5 – 0 = 5, 0 - 5 is not defined so there is no identity element
either.




Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                            63


Multiplication

For multiplication four models are given by the text: the set model
(Cartesian Product), repeated addition, line number, and the array and area
models. The set model uses the Cartesian product to determine the number
of possible pairings that can be created from the elements of two sets.

Set Model

We consider the following problem. Suppose a person has three shirts
(white, blue, and red) and two pairs of pants (black and brown). We ask the
question ‘how may outfits (i.e. pant-shirt combinations) can be created from
these items?’ We can answer this question using our knowledge about sets.
If we let set A be the two pairs of pants and set B be the three shirts, then
the Cartesian product of A and B gives us all possible pairings and therefore
all possible outfits. We therefore have:

Definition of Multiplication of Whole Numbers. For finite sets A and B, if
a = n(A) and b = n(B), then a·b = n(A×B). The numbers a and b are called the
factors and a·b is called the product.

Note: unlike addition, the sets A and B do not have to be disjoint, though in
the example given above they are.

Array and Area Model

If we go back and consider our pant-shirt outfit example, we can find all
possible outfits by constructing an array. We form our array like a table
with the column header being the shirts and the row header being the pants:

                       White Shirt            Blue Shirt          Red Shirt
                     st                  nd                  rd
Black Pants         1 Outfit            2     Outfit        3     Outfit
Brown Pants         4th Outfit          5th   Outfit        6th   Outfit

We can then count the number of cells in our array (table) to determine the
number of possible outfits. Since the cells have the shape of rectangles or
squares, we can also view this arrangement as giving an area for the array or
table.

Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                   64



Repeated Addition Model

Suppose we have an arrangement of 5 rows of chairs, with each row
containing 4 chairs. We can find the total number of chairs by adding the
together the chairs in each row:

                         Total number of chairs = 4 + 4 + 4 + 4 + 4 = 4·5 = 20

That is, adding 4 together five times to obtain 20 chairs.

Line Number Model

In this model, we create a line segment corresponding to one of the factors.
We then duplicate this segment so that the number of segments is equal to
the other factor. We then place the segments end to end on the number
line with the end of the last segment giving us the product.

Consider the product 4·5.

1. Create a line segment corresponding to the factor 4.



        0            1      2      3   4       5         6    7   8        9   10

2. Duplicate this segment 5 times.




      0          1         2       3   4       5     6        7   8        9   10


3. Lay the segments end to end on the number line.
                                                   4·5


          0                    4           8             12           16        20
Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                             65


Properties of Multiplication

Like addition, multiplication has commutative, associative, and closure
properties. In addition, it has an identity property and zero multiplication
property.

Properties of Multiplication of Whole Numbers.

Closure property of multiplication of whole numbers. For whole numbers a
and b, a·b is a unique whole number.

Commutative property of multiplication of whole numbers. For whole numbers
a and b, a·b = b·a.

Associative property of multiplication of whole numbers. For whole numbers
a, b and c, (a·b) ·c = a·(b·c).

Identity property of multiplication of whole numbers. There is a unique
whole number 1 such that for any whole number a, a·1 = 1·a = a.

Zero property of multiplication of whole numbers. There is a unique whole
number 0 such that for any whole number a, a·0 = 0·a = 0.



In addition to these fundamental properties, multiplication is distributive
over addition. This means that if we have three whole numbers a, b and c
and we add b and c and multiply by a we get the same result as if we first
multiply b by a and c by a and then add these results together.

Distributive property of multiplication over addition for whole numbers.
For any whole numbers a, b, and c: a·(b + c) = a·b + a·c.
It should be noted that both the associative and distributive properties can
be generalized to any (finite) number of terms e.g.:

                             a·(b + c +d) = a·b + a·c + a·d

                 ((a·b) ·c)·d = (a·(b ·c))·d = (a·b) ·(c·d) = a·((b ·c) ·d)


Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                               66


Division

The text presents three models for division: the set model, the missing
factor model, and the repeated subtraction model. We can also construct a
line-number model for division.

Set Model

In this model we have a set A which we wish to partition i.e. ‘divide-up’ into
equal parts. That is we wish to form some number of subsets from the
elements of A each having the same number of members.


                                                    B




                 A
                        B     C     D

                                                    C




                                                    D




In the example of the figure we wish to partition set A into subsets each
having 4 elements. After partitioning A, we break it up into 3 sets, each
having four elements.




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                              67



Definition of Division of Whole Numbers. For finite sets A and B, if a =
n(A) and b = n(B) ≠ 0, then a÷b = c if and only if c is a unique whole number
such that when we construct a set C with c = n(C), then n(A) = n(B×C)
i.e. b·c = a. The number a is called the dividend and the number b is called
the divisor and a÷b is called the quotient.

Missing Factor Model

The idea here is that we convert our division problem into a multiplication
problem. This is similar to what we did with subtraction when we converted
the subtraction problem into an addition problem (see Missing Addend
Model, above). In this case we have:

                                   12 ÷ 4 =

And we convert it to the equivalent multiplication problem:

                                   4·   = 12

We must now find the number, that when multiplied by 4 gives 12.

Repeated Subtraction Model

This is again similar to what we did with multiplication (see Repeated
Addition Model, above). The idea here is that we will repeatedly subtract
the divisor from the dividend until we obtain zero. Counting the number of
times we performed the subtraction yields the quotient. As an example, we
could think of doling out cookies to a group. Suppose we have a dozen
cookies. A person comes by and we give them four cookies, leaving use with
8. When the next person comes by, we give him/her 4 cookies, leaving us
with 4. When another person comes by we give away our last 4 cookies, so
that we have none (i.e. zero) remaining. Since we gave out cookies 3 times,
we know that 12 ÷ 4 = 3.




Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                                           68


Line Number Model

This model is included for completeness and is similar to what we did for
multiplication. In this case we construct a segment equal to that of the
dividend and the divisor.

                                       divisor
                                                             dividend


       0         1       2     3   4      5         6        7      8   9     10   11   12

We then make copies of the divisor and lay them end to end until the end of
the last copy is coincident with the end of the segment for the dividend:


                     divisor                  1st copy                  2nd copy




       0         1       2     3   4      5         6        7      8   9     10   11   12

We now count the total number of divisor segments (in this case 3) and that
is the quotient.
                                                 3 divisor
                                                 segments




       0         1       2     3   4      5         6        7      8   9     10   11   12




Dividing by Zero and One

From the identity property of multiplication we have that for any whole
number a, a·1 = a. From our definition of division, we therefore have that for
any whole number a, a÷1 = a. Thus any whole number divided by one is equal to
itself.

For zero, we have a similar result. From the zero property of multiplication
we have 0·a = 0 for any whole number a ≠ 0. Using the definition of division

Pierce College                                                                               MAP
Math 215 Principles of Mathematics                                           69


we have that: 0÷a = 0 for any whole number a ≠ 0. That is, if we divide 0 by
any whole number that is not zero, then the result is zero.

We now discuss division by zero. If division by zero were possible, this
means that for every whole number a, there would exist a unique whole
number c, such that a÷0 = c. But this means (by our definition of division)
that 0·c = a. But 0·c = 0 for every whole number c, so (unless possibly a = 0),
no such whole number c exists. Now if a is zero (special case) we have 0÷0 =
c or equivalently 0·c = 0. Now this is true for every whole number c and
therefore c is not unique, as required by our definition. We therefore have
that division by zero is undefined for every whole number, including zero.

Properties of Division

Like subtraction, division is not closed for the whole numbers. That is, it is
possible to find two whole numbers such that their quotient is not a whole
number e.g. 27÷5 is not a whole number. However, if we introduce the
concept of remainder we can ‘patch things up’ so to speak. Let’s go back to
the repeated subtraction model for division and our cookie example. Suppose
instead of handing out 4 cookies at a time, we hand out five. This
corresponds to the problem 12÷5. In this case, the first two people that
come by get 5 cookies each. However when the third person comes by, there
are only 2 cookies left. This means that we cannot give out the same share
of cookies to this person. In effect we will leave with two cookies remaining
from our original batch of 12 cookies.

Mathematically we can divide 12 by 5 with the result that we get 2, with 2
remaining. We therefore obtain the following result:

Division ‘Algorithm’. Given any whole numbers a and b with b ≠ 0, there exist
unique whole numbers q (called the quotient) and r (called the remainder)
such that a = bq + r with 0 ≤ r < b.

We also note that division is not commutative or associative and that 1 is not
the identity element for division.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            70



Order of Operations

Order of operations refers to the order in which addition, subtraction,
multiplication, and division operations are performed when these operations
are mixed together. Consider the following example:

                                  2 + 3·5 = ?

There are two ways to perform this calculation. In the first, we add 2 and 3
and then multiply the result by 5 to obtain 25. In the second, we multiply 3
and 5 and add 2 to the result to obtain 17. The result is different and
therefore dependant on the order in which the operations are performed.

To obtain an unambiguous result, a formal convention has been adopted that
specifics that multiplication and division are performed (from left to right)
before addition and subtraction. In our example, this means that the second
method should be used and the correct answer is 17, not 25. If we want to
add 2 to 3 before multiplying by 5, then we need to add parenthesis to the
problem:
                                  (2 + 3)·5 = ?

Without the parenthesis we must multiply 3 and 5 before adding 2. Note: as
the text points out, calculators will not necessarily follow this convention, so
care should be exercised when using machines.

Relating Multiplication and Division

The text points out that addition and subtraction are ‘inverse’ operations
and that multiplication and division are also ‘inverse’ operations. All four
operations can be related as shown in the following figure using the various
models we have discussed.

With the expansion of the whole numbers to the real numbers, subtraction
and division become non-operations or ‘special’ cases of addition and
multiplication. We can view subtraction as addition with the additive inverse
(requires negative numbers) and division as just multiplication by the
reciprocal (requires rational numbers) of the number.

Pierce College                                                               MAP
Math 215 Principles of Mathematics                              71




                                 Inverse
                            +    Operations
                                              -


                 Repeated                         Repeated
                 Addition                         Subtraction



                                Inverse
                            ×                 ÷
                                Operations




Pierce College                                                  MAP
Math 215 Principles of Mathematics                                              72



Lecture 5

Sections 3.1 and 3.2 – Numeration Systems and Algorithms for Addition and
Subtraction of Whole Numbers

The number system used in the United States, Europe, and most of the
world is the Hindu-Arabic system. This system uses 10 cardinal numbers (0,
1, 2, 3, 4, 5, 6, 7, 8, 9) or numerals to represent all possible numbers. The
text defines a numeration system as a collection of properties and symbols
used to represent numbers systematically. Other systems used in the past
and even today are shown in the following table:

Babylonian                                                                 <
Egyptian                                                                   ∩
Mayan                ●    ●●    ●●●   ●●●●       ●     ●●    ●●●    ●●●●
Greek                α    β     γ     δ      ε   φ     ζ     η      υ      ι
Roman                I    II    III IV       V   VI    VII VIII IX         X
Hindu-Arabic 0       1    2     3   4        5   6     7   8    9          10

Hindu-Arabic System

The Hindu-Arabic system is the one we use today; it was developed by the
Hindus (who are credited with the ‘invention’ of zero) and transported to
Europe by the Arabs. This system used 10 digits (i.e. 0 - 9) and is a place
value system based on powers of 10 i.e. base-10 system.

Place value assigns a value to each digit in a number based on its place i.e.
position in the number. To find the value of a digit in a number we multiply
its face value (i.e. the digit itself) by its place value. The place value is
found by counting the position of the digit in the number, starting from the
right, till the digit is reached. Multiplying 10 by itself this number of times,
gives the place value. Thus for example, consider the number 5984. The
value of 5 in this number is 5 times 10 × 10 × 10 = 1000. Thus the digit 5 has
a value of 5000. Likewise, 9 has a value of 9 times 10 × 10 = 100. The value
of the digit nine is therefore 900. In a similar manner the value of the digit
8 is 80 and the value of 4, is just 4.


Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                      73


Doing this for each digit in a number allows us to write a number in expanded
notation. Thus for 5984 we can write:

                           5984 = 5 × 103 + 9 × 102 + 8 × 10 + 4



Where we have used exponential notation to write the factors (powers) of
10 e.g. 103 = 10 × 10 × 10 = 1000. In particular, we note:

Exponential Notation. For any number a and any whole number n:

                                      an = a⋅ a⋅ a⋅ . . . ⋅a
                                                n − tim e s



For n = 0, and a ≠ 0, we define a0 = 1.

Tally System

This system uses strokes or vertical lines, called tally marks, to represent a
number. As the name implies, such a system could be used in counting or
tallying objects. The ‘numbers’ 1 to 10 in this system are:

            |, ||, |||, ||||, |||||, ||||||, |||||||, ||||||||, |||||||||, ||||||||||

In this system, there is a one-to-one correspondence between the tally
marks and the number or count being represented. Although this system is
simple it has several disadvantages:

    1. As numbers become larger they are harder to read.
    2. The amount of space required to write a number is proportional to the
       ‘size’ of the number. The larger the number, the more space required.
    3. It is error prone in both reading and writing a number.
    4. It is inefficient from both space and time required to record a
       number.




Pierce College                                                                          MAP
Math 215 Principles of Mathematics                                         74


Egyptian System

The Egyptian system is a modified tally system. It modifies the tally system
to make it more efficient and easier to read and write by assigning a second
symbol to represent groups of tally marks. For this reason it is called a
grouping system. The Egyptians used the symbol ∩ to represent 10 tally
marks. Other symbols were introduced to represent the various powers of
10:



       Egyptian Numeral          Description               Hindu-Arabic
                                                            Equivalent
                          Vertical Staff              1
                          Heel Bone                   10
                          Scroll                      100
                          Lotus Flower                1,000
                          Pointing Finger             10,000
                          Polliwog or burbot          100,000
                          Astonished Man              1,000,000

The Egyptian system was not place value, but additive. That is the value of
the number is the sum of its digits. For example the number:




has a value of 100,322 in our system. This is found by summing the digits as
follows:

                          represents           100,000
                          represents              300 (100 + 100 + 100)
                          represents                20 (10 + 10)
                          represents                 2 (1 + 1)
                          represents           100,322

Although this system is more efficient than the tally system and addresses
many its disadvantages it still somewhat inefficient in its spatial usage of
symbols (e.g. the number 22 requires 4 symbols) and it requires some
calculation to determine the value of the number.

Pierce College                                                            MAP
Math 215 Principles of Mathematics                                           75


Babylonian System

The Babylonia system uses the symbols and < in the same manner the
Egyptians used the staff and heel bone symbols. The numbers 1 to 59 were
constructed using these symbols in an additive manner. However for
numbers greater than 59, a place value system was used with spaces
between the symbols used to represent powers of 60. The Babylonian system
was therefore base 60, not base 10 like the Hindu-Arabic system we use
today. The following table shows some examples of numbers and their
modern day equivalents.

                          represents         22   ( 10 + 10 + 1 + 1)
                          represents        140   (2⋅60 + 10 + 10)
                          represents      1,201    (20⋅60 + 1)
                          represents     40,261   (11⋅602 + 11⋅60 + 1)
                          represents    256,261   (1⋅603 + 11⋅602 + 11⋅60 + 1)

Although this system has modern features, it is still additive and the use of
spaces requires care to avoid mistakes in reading numbers. Apparently the
Babylonian’s introduced a third symbol (as mentioned in the text) to replace
the space. This symbol was used much like we use zero today in writing our
numbers.

Mayan System

The Mayan system was similar to the Babylonian system except that it was
base 20 and contained a symbol for zero. Three symbols were used in this
system:



                 Mayan Numeral            Hindu-Arabic
                                           Equivalent
                                                1
                                               5
                                               0




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                                    76


Numbers are written vertically instead of horizontally with the lower place
values at the bottom. The following provide several examples of numbers in
both the Mayan and modern systems:

     1.             = 3 fives plus 4 ones = 19.

     2.             = 20. In this case, the   stands for one group of 20.

     3.             = (bottom) 2⋅5 + 1 = 11 plus (top) 2⋅5 + 3 = 13 times 20 = 271.

     4.             = (bottom) is 0 plus (top) 3⋅5 + 1 = 16 times 20 = 320.



As the text mentions, due to reasons having to due with the calendar, the
Mayan system does not increase strictly by powers of 20, but rather by 18
times the powers of 20. Thus the place values increase as follows:

                 1, 20, 18⋅20 = 360, 18⋅202 = 7200, 18⋅203 = 144,000, etc.

The following examples demonstrate this:



     5.              = (top)     1⋅5 + 1 = 6 times 360 = 2160 plus
                       (middle) 2⋅5 + 2 = 12 times 20 = 240 plus
                       (bottom) 1⋅5 + 4 = 9           = 2409.



Although this system has modern features, it is still additive and uses
spaces (see text) between the horizontal bars representing 5 to indicate
100 = 5⋅20 thereby requiring care to avoid mistakes in reading numbers.

Roman System

The Roman system was used in Europe from about 300 B.C. It is still used
today for special applications: cornerstones, title pages in books, the faces
of clocks, dates of release/production in films, etc. This system is very
similar to the Egyptian system, with letters denoting various larger numbers,
as the following table shows:

Pierce College                                                                        MAP
Math 215 Principles of Mathematics                                          77


                 Roman Numeral           Hindu-Arabic
                                          Equivalent
                       I                        1
                       V                       5
                       X                       10
                       L                       50
                       C                      100
                       D                     500
                       M                     1000

Like the Egyptian system, the Roman system is additive. Thus, for example
MDCLXVI = 1000 + 500 + 100 + 50 + 10 + 5 + 1 = 1666. The Roman system has
a subtractive property (apparently introduced in the Middle Ages) to avoid
repeating I more than three times. Thus the number 4 is written IV (five
minus one) instead of IIII. Likewise 9 is written IX (ten minus one) instead
of VIIII. This subtractive property was also extended to the ten symbol, X
when used with the 100 symbol. Thus 90 is written as XC instead of LXXXX.

In order to represent larger numbers, (again in the Middle Ages) bars are
placed over letters to indicate multiplication by 1000. This is the
multiplicative property of the Roman system. Thus for example, V is 5000
and CDX is 410⋅1000 = 410,000. Two bars over these letters indicate
multiplication by 1,000,000. Thus CXI is 111⋅10002 = 111,000,000.

The Roman system has all the disadvantages of the Egyptian system with
added complexity of the subtractive and multiplicative properties.

Bases

In our discussion of the Hindu-Arabic, Mayan, and Babylonian systems, we
saw that these systems use various bases to represent numbers using a place
value system. In the case of the Hindu-Arabic system, base-10 is used while
the Mayan system is base-20, and the Babylonian, base-60. The text
provides other examples of various bases. For example the Luo of Kenya use
a base-5 system and the duodecimal system is base-12.




Pierce College                                                           MAP
Math 215 Principles of Mathematics                                           78


Of particular importance is the binary (base-2) and hexadecimal (base-16)
systems which are widely used in computing. We will discuss these systems
now.

Base-16 or Hexadecimal

In this case there are 16 digits defined for the system. These are shown in
the following table:

                       Digit              Value
                        0                   0
                         1                  1
                        2                   2
                        3                   3
                        4                   4
                        5                   5
                        6                   6
                        7                   7
                        8                   8
                        9                   9
                        A                  10
                        B                  11
                        C                  12
                        D                  13
                        E                  14
                        F                  15

The letters A, B, C, D, E, and F stand for the numbers 10, 11, 12, 13, 14, and
15, respectively. Thus for example, if we wish to write 12 in hexadecimal we
would write C. Since our base is 16, each place in a number represents a
power of 16, starting with 160 = 1 (ones place) and increasing to 161 = 16
(‘tens’ – place), 162 = 256 (‘hundreds’-place), and so on.

In hexadecimal the number 2AF7 is expanded as follows:

                 2AF7 = 2 × 163 + A × 162 + F × 16 + 7 × 160
                     = 2 × 4096 + 10 × 256 + 15 × 16 + 7 = 10,999


Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          79


When writing numbers in bases other than 10, a subscript is added to the
end of the number specifying the base. Thus we should write 2AF716 instead
of 2AF7. Converting a number from another base is straight-forward: write
the number in expanded notation and then perform the necessary
multiplications and additions as we did in the example above. Thus to
convert 1345616 to base 10:



                 1345616 = 1 × 164 + 3 × 163 + 4 × 162 + 5 × 16 + 6 × 160
                         = 1 × 65536 + 3 × 4096 + 4 × 256 + 5 × 16 + 6
                         = 78,934

We can also go the other way, converting a number from base-10 to base-16
(or any other base). This process is more involved, since we must divide by
the successive powers of the base, starting with the largest power that is
less than or equal to the number. Thus to convert 160,533 to hexadecimal we
precede as follows, using the following table of powers of 16:



                       160              1
                       161              16
                       162              256
                       163              4,096
                       164              65,536
                       165              1,048,576

Since our number is less than 1,048,576, we start with 164, dividing our
number by 65,536:

                                     2 R 29461
                              65,536 160,533


Thus 65,536 goes into 160,533 twice with a remainder of 29,461. This means
that 160,533 = 2 × 164 + 29,461. We now repeat this process with the
remainder. Since 29,461 is larger than 4,096, we divide by 4,096:

                                       7 R 789
                                4, 096 29, 461

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          80


Thus 4,096 goes into 29,461 seven times with a remainder of 789. We
therefore have:

                         160,533 = 2 × 164 + 7 × 163 + 789.

Repeating again, 789 is larger than 256, so we divide by 256:



                                         3 R 21
                                     256 789


So we now have:

                     160,533 = 2 × 164 + 7 × 163 + 3 × 162 + 21.

Since 21 is greater than 16, we divide by 16:



                                         1 R 5
                                      16 21


And we have:

                 160,533 = 2 × 164 + 7 × 163 + 3 × 162 + 1 × 161 + 5.

Since there are no remaining powers of 16 left, we are done. We can remove
the powers of 16 from the right hand side of the expression above to obtain
our number in base 16:

                                160,533 = 27,31516

This process requires long division and can be labor intensive if done by
hand. Some calculators will provide keys and buttons to automatically
convert between various bases.




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                                   81


Base-2 or Binary

This base is widely used in computing due to the design of modern digital
computers. Digital computers use transistors as their basic building blocks.
Transistors are similar to switches, having two positions (or states) ‘on’
(electrical current is flowing through them) and ‘off’ (current is not flowing
through them). Since a transistor can be in only one of two states, a zero is
used to represent the off state and a one the on state. Information (and in
particular numbers) can be stored using the on/off state of one or more
transistors to represent the data. Thus for example, if we wish to represent
the numbers 0 and 1, we can use a transistor and place it in the off state to
represent 0 and in the on state to represent 1.

Larger numbers such as 2, 3, 10 or 100 require more than one transistor.
Most computers use 32 transistors to store whole numbers. Each transistor
represents a power of 2, starting with 20 and increasing to 231. A number
such as 100 is represented as a series of ones and zeros corresponding to
the various powers of 2 required to represent it. In the computer hardware,
this sequence of ones and zeros correspond to the on/off state of the 32
transistors used to store the number. Note: in the language of computing,
each one or zero is called a bit and a sequence of 32 such bits is called a
word.

Since only two numbers are used in base-2 (i.e. 0 and 1), it is called a binary
system. Just like with hexadecimal and other place value representations
each digit in the number corresponds to a power of two based on the
position of the digit in the number. Thus for example:

                 1100100 = 1 × 26 + 1 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0
                         = 1 × 64 + 1 × 32 + 0 × 16 + 0 × 8 + 1 × 4 + 0 × 2 + 0
                         = 64 + 32 + 4
                         = 100

Conversion from base-10 to base-2 proceeds in the same manner as with
hexadecimal except we divide by powers of 2 instead of powers of 16. Thus
consider the following example:




Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                                   82


Convert 1,055 to binary representation. In this case we need a table of
powers of 2:



                                20                1
                                21                2
                                22                4
                                23                8
                                24                16
                                25                32
                                26                64
                                27                128
                                28                256
                                29                512
                                210               1024
                                211               2048
                                212               4096

Since 1,055 is less than 2048, we start our division with 210 = 1024:



                                                1 R 31
                                          1, 024 1, 055


We therefore have:

                                      1,055 = 1 × 210 + 31

In this case we can skip division by 29, 28, 27, 26, and 25 since our remainder
is less any of these numbers. However, we need to add them into our
expanded representation of our number with zero as a multiplying factor,
otherwise when we ‘contract’ our number i.e. remove the powers of two, at
the end, we will be missing these required zeros:

                 1,055 = 1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 31

We can now continue by dividing our remainder, 31, by 16:


Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                                                       83


                                                1 R 15
                                             16     31
We now have:

        1,055 = 1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 15

Next we divide by 8:

                                                 1 R 7
                                               8    15


And we have:

   1,055 = 1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 1 × 23 + 7

Dividing by 4:

                                                 1 R 3
                                               4     7

       1,055 = 1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 1 × 23+ 1 × 22 + 3


Finally we divide by 2:

                                                 1 R 1
                                               2     3


And we obtain:

   1,055 = 1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 1 × 23+ 1 × 22 + 1 × 21 + 1


Removing the powers of 2, we get:

                                      1,055 = 100000111112




Pierce College                                                                                      MAP
Math 215 Principles of Mathematics                                           84


With a 32-bit word, the largest whole number that can be stored is 232-1 =
4,294,967,295 (slightly over 4 billion). As technology progresses computers
will be made using 64-bit words and this number will increase to
18,446,744,073,709,551,615 (over 18 quintillion).



Algorithms for Addition

We now discuss several methods for performing whole number addition. The
text presents four methods: the Standard Algorithm, Left-to-Right
Algorithm, Lattice Algorithm, and Scratch Algorithm.

Standard Algorithm

In this algorithm single digit addition is performed from right to left. The
numbers are lined up vertically, a horizontal line (called the summation line)
is drawn under the lowest number, and the result is placed blow this line. If
the sum of the digits in a column is greater than 9, the ‘excess digits’ are
carried into the next column where they become part of the sum of the
digits in that column. The text notes that the term ‘carry’ has been
‘replaced’ by the terms regroup or trade. The idea here is that if we write
our numbers in expanded form, then we ‘regroup’ or ‘trade’ them using the
associative law after adding the digits corresponding to the various powers
of 10. We consider the following example:

                                   376
                                 + 459

If write these numbers in expanded notation we have:

                            3⋅102 + 7⋅10 + 6
                          + 4⋅102 + 5⋅10 + 9




The digits of our number are now grouped by powers of 10, and we add the
digits in each column:



Pierce College                                                              MAP
Math 215 Principles of Mathematics                                                              85


                              3⋅102 + 7⋅10 + 6
                            + 4⋅102 + 5⋅10 + 9
                             7⋅102 + 12⋅10 + 15



Because the numbers in the ones and tens place are greater than nine, we
must expand them and then regroup or trade:

7⋅102 + 12⋅10 + 15 = 7⋅102 + (10 + 2) ⋅10 + (10 + 5)       – result after adding

                   = 7⋅102 + 1⋅102 + (2 + 1)⋅10 + 5        - break 12 into 10 +2 and 15 into 10 + 5

                   = (7 + 1) ⋅102 + (2 + 1)⋅10 + 5         - use distributive property and regroup

                   = 8⋅102 + 3⋅10 + 5                      - our number is now in expanded form

                   = 835                                   - convert to standard form



This step by step break-down provides a ‘behind the scenes’ examination of
what is happening when we carry. This procedure is therefore not meant to
actually performed, but to serve as an illustration of what is happening
‘behind the scenes’ when we carry (or trade/regroup). In practice, the
addition proceeds in the familiar fashion starting with the right most column
of numbers (digits):

                               1
                              376
                            + 459
                                 5-   add 9 and 6 to get 15, place the
                                      five in the column under these digits
                                      and carry (trade/regroup) the one
                                      (actually 10) into the next (tens) column.



We now repeat this with the next (middle, in this case) column:

                              11
                              376
                            + 459
                               35 -   add the 1 (carried over), the 7, and the 5
                                      To get 13. The 3 is placed directly below the
                                      5 and he one (actually 100) is again carried
                                      into the next (hundreds) column.




Pierce College                                                                                 MAP
Math 215 Principles of Mathematics                                               86


Finally we add the digits in the last (left most) column. In this case (since we
are in the last column), if the result is greater than 10, we do not carry over,
but write all the digits of the result below the summation line:



                            11
                            376
                          + 459
                            835 -   add the 1 (carried over), the 3, and the 4
                                    To get 8.



The text also discusses a variation on the Standard Algorithm, which they
term the Expanded Algorithm. In this case, carrying is replaced by writing
the result of the summation for each column on a separate line under the
summation line, and then adding these once the summation of all the columns
is complete:



                                   376
                                 + 459
                                    15     - result from adding 6 and 9

                                   120     – result from adding 70 and 50

                                   700    – result from adding 300 and 400



Note: when we add the second and third columns we must note that we are
adding 70 and 50 and 300 and 400, not 7 and 5 and 3 and 4. Once this ‘first
pass’ is done, we must now add the resulting three numbers we obtained:



                                   376
                                 + 459
                                    15
                                   120
                                   700
                                   835 –    result from adding 15 + 120 + 700




Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                                                  87


Left-to-Right Algorithm

 This is a variation on the Expanded Algorithm of the previous section,
where addition of the columns precedes from left to right (reading order) as
opposed to from right to left. Consider the following example:



                                            568
                                         + 757
                                          1200 – result from adding 5 and 7 from left most column
                                            110 - result from adding 6 and 5 in middle column
                                             15 - result from adding 8 and 7 in right column
                                           1325 – result from adding 1200 + 110 + 15

Note: When adding 5 and 7 we are really adding 500 and 700 and this is
reflected by the 1200 in the fist line below the summation line. Likewise in
adding 6 and 5 we are really adding 60 and 50 to obtain 110, which must be
aligned correctly under the 1200 above it.

Lattice Algorithm

This is yet another version of the expanded algorithm. In this case, the
results from adding the digits in each column are placed horizontally instead
over vertically in a ‘lattice’. The diagonals in adjacent portions of the lattice
are added to give the final results. The following example illustrates this
procedure:

                  3          5       6          7
                 +5          6       7          8
                 0       1       1          1
                                                               lattice
                     8       1       3          5


                                                              8 + 7 = 15


                                                              6 + 7 = 13


                                                             5 + 6 = 11


                                                             3+5=8
Pierce College                                                                                  MAP
Math 215 Principles of Mathematics                                                         88



First we add the digits in each column and place the results in the lattice
below the summation line as shown in the figure above. No carrying is
performed. If the result is larger than 9, both digits are written in the
lattice, as shown above.

After digits in each column are added and placed into the lattice, we sum the
adjacent diagonals in the lattice to obtain the final result:


                  3          5       6       7
                 +5          6       7       8
                 0       1       1       1
                     8       1       3       5


                 9       2           4       5
                                                               Nothing to sum, just bring 5 down


                                                               1+3=4


                                                              1+1=2


                                                             1 + 8 = 9; no need to bring 0 down




Scratch Algorithm

This algorithm is similar to the standard algorithm, except it keeps track of
the sum of the digits in each column using a ‘scratch’ when ever the sum
exceeds 10. The following example from the text illustrates this procedure:

                                                 Start at the top with the first (right most)
                              8 7                column and work down.
                              6 52
                             +4 9                Any time the sum is greater than 10, scratch
                                                 out the current number and replace it with
                                                 the sum to that point minus 10. Thus, since
                                                 7 + 5 = 12, we scratch out the 5 and replace
                                                 it with 2 (= 12 – 10).
Pierce College                                                                            MAP
Math 215 Principles of Mathematics                                              89



                         8 7         We repeat this process as we continue down.
                         6 52        Since the sum of 9 and 2 is 11, we scratch
                        +4 9 1       the 9 and replace it with 1 (= 11 – 10).




                         2           After we reach the last digit in the column,
                         8 7         we place the sum below the summation line
                         6 52        and the count the scratches.
                        +4 9 1
                                     In this case there are 2. We place the two
                           1         at the top of the next column, and repeat
                                     the process with that column.




                         2           Since 2 + 8 = 10, we scratch the 8 and
                                     replace it with 0.
                         80 7
                         6 52        0 + 6 = 6, so we don’t need to scratch here,
                        +40 9 1      but our sum is 6 at this point.
                            1
                                     6 + 4 = 10, so we scratch the 4 and replace it
                                     with zero.




                        2
                        80 7
                        6 52
                       +40 9 1
                       20 1            There are no further digits after the 4, so
                                       we write the current sum below the
                                       summation line (zero in this case) and then
                                       count the number of scratches. There are
                                       two again, however, this is the last column,
                                       so we write the 2 below the summation line
                                       and we are done.




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                      90


Subtraction Algorithms

We now discuss several methods for performing whole number subtraction.
The text presents two methods: the Standard Algorithm and the Equal
Addends Algorithm.

Standard Algorithm

The text makes use of the same terminology about ‘trading’ when
subtraction is performed. This is essentially just another name for
borrowing. Just like addition, the process begins with the right most column
of digits:

                          243
                         - 61
                            2 – since 3 is greater than 1, we just
                                   subtract, placing the result below
                                   the difference line.



Proceeding to the middle column, we note that since 4 is less than 6, we need
to borrow from the next column to the left. Note that the 4 is really 40 and
the 6 is really 60. We borrow 1 from the 2. This is really a regrouping
(associative property) of these numbers since we have 200 + 40 = 100 + 140.
This allows us to subtract 60 from 140 (14 – 6, in place):

                          1
                          2143
                         - 61
                            82 –   borrow ‘1’ from the 2 i.e. regroup; we scratch the
                                   2 and replace it with one, and make the 4 a 14;
                                   We then subtract 6 from 14 and place the result
                                   below the difference line under the 6.




Going to the last (left most) column, we see that there we are essentially
subtracting 0 from 1 (the ‘zero’ is not shown since we do not write numbers
with leading zeros). We therefore bring the one down below the difference
line, completing the computation:




Pierce College                                                                          MAP
Math 215 Principles of Mathematics                                                       91


                           1
                           2143
                          - 61
                            182 –   we bring the one down, since there is nothing to
                                    subtract here, and write it below the difference
                                    line, completing the computation.
                                    .




Equal Addends Algorithm

This algorithm makes use of the property of equality. That is, if we add the
same number to both the minuend and subtrahend, the difference does not
change. For example:

                      5 - 3 = (5 + 2) – (3 + 2) = 7 – 5 = 2

We can also do this ‘in-place’ in the following example:

                                       255
                                       -163



                                                                    Since it’s trivial to subtract zero
                                     255 + 7                        from a number, we add 7 to
                                     -163 + 7                       both numbers, so that the 3 in
                                                                    the subtrahend becomes zero.


                                       262
                                       -170

We can also ‘get rid of’ the 7 in the ‘new’ subtrahend by adding 30 to both
numbers:

                                    262 + 30
                                                               We now add 30 to both numbers
                                    -170+ 30                   so that the subtrahend contains
                                                               2 zeros i.e. eliminate the 7
                                       292
                                       -200

Pierce College                                                                          MAP
Math 215 Principles of Mathematics                                               92


The problem now requires no borrowing and the subtraction is straight
forward:

                                      292
                                     -200
                                       92

Addition and Subtraction in Other Bases

Addition and subtraction in other bases is identical to that in base ten
except when carrying or borrowing, we are adding the value of the base, not
10. Let’s consider some examples from base-16 (see the text for some
examples from base-5). Suppose we wish to add the following base-16
numbers:

                          9816
                         +8916

We can proceed using the standard algorithm and add 8 and 9 to get 17.
Instead of subtracting 10, when we carry the one over to the next column,
we subtract 16 and carry one:

                          1             8 + 9 = 17. But since we are working in
                          9816          base 16, we need to subtract 16 from
                         +8916          17 to get 1, and then carry 1
                                        (representing one-16 instead of one-10)
                            116
                                        into the next column.

                                        We therefore write 1, not 7 below the
                                        summation line.


We now add the 1, 9, and 8 in the next column to get 18. However 18 in base-
16 is one-16 plus 2, or 12.
                                            1 + 9 + 8 = 18. But since we are
                           1                working in base 16, we need to subtract
                           9816             16 from 18 to get 2, and then carry 1
                                            (representing one-16 instead of one-10)
                         +8916
                                            into the next column.
                          12116
                                            We therefore write 2 below the
                                            summation line. Since this is the last
Pierce College
                                            column, we place the one below the MAP
                                            summation line also.
Math 215 Principles of Mathematics                                                93


We can verify our result by converting to base 10 and adding there, and then
converting our answerer from the base-16 addition and comparing. In base-
10, 9816 = 9 × 16 + 8 = 144 + 8 = 152 and 8916 = 8 × 16 + 9 = 128 + 9 = 137. We
therefore have:

                         9816 + 8916 = 152 + 137 = 289

Now 12116 = 1 × 162 + 2 × 16 + 1 = 256 + 32 + 1 = 289. Our answers are
therefore the same.

Subtraction proceeds in the same manner. When we borrow, we are
borrowing a number in that base. Thus is base-16 we are borrowing a 16, not
a ten as the following example illustrates:

                            8               Since 2 is less than 9, we must borrow.
                            91216           We therefore scratch the nine making it
                           -7 916           an 8, and make our 2 in the first column
                                            and 12.




                            8               Note: the 12 is 12 base-16 so it is
                            91216           really 16 + 2 = 18. We therefore
                           -7 916           subtract 9 from 18 to get 9.
                              916

                            8               We now subtract 7 from 8 to get 1,
                            91216           which we place next to the 9 below the
                           -7 916           difference line.
                            1 916

Again, we can verify our answer by converting to base-10, subtracting and
comparing to the base-10 converted answer:

9216 = 9 × 16 + 2 = 144 + 2 = 146 and 7916 = 7 × 16 + 9 = 112 + 9 = 121

Thus: 9216 - 7916 = 146 – 121 = 25 in base-10. Converting our answer to base-
10 we get: 1916 = 1 × 16 + 9 = 25, which is the same.

Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                           94


Lecture 6

Sections 3.3 and 3.4 – Algorithms for Multiplication and Division of Whole
Numbers and Estimation

Algorithms for Multiplication

We present the algorithms used for multiplying whole numbers.
Multiplication is based on the ‘multiplication facts’ which are essentially
single digit multiplication tables and multiplication by 10. The multiplication
tables are listed below, and of course must be mastered i.e. memorized
before multiplying numbers greater than 10. The table itself is derived from
our set definitions of multiplication, or alternatively from the concept of
multiplication being repeated addition.

                             Multiplication Facts
          0      1    2    3     4       5     6      7      8      9     10
0         0      0    0    0     0       0     0      0      0      0     0
1         0      1    2    3     4       5     6      7      8      9     10
2         0      2    4    6     8       10    12     14     16     18    20
3         0      3    6    9     12      15    18     21     24     27    30
4         0      4    8    12    16      20    24     28     32     36    40
5         0      5    10   15    20      25    30     35     40     45    50
6         0      6    12   18    24      30    36     42     48     54    60
7         0      7    14   21    28      35    42     49     56     63    70
8         0      8    16   24    32      40    48     56     64     72    80
9         0      9    18   27    36      45    54     63     72     81    90
10        0      10   20   30    40      50    60     70     80     90    100

We note that the facts-table is symmetrical about the diagonal and the
entries on the diagonal are the ‘squares’ of the whole numbers from zero to
ten i.e. the numbers multiplied by themselves. The other fact required for
multiplication is that any whole number multiplied by 10 annexes zero to the
number. This is a consequence of the number system being base-10. In fact,
we can generalize this fact further by noting that:




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                                           95


Multiplication of any natural number by 10n, where n is a natural number
annexes n zero to the number.

When multiplying powers of 10 e.g. 103⋅104 we use the following property of
exponents:

                                              an⋅am = an+m

where n and m are whole numbers. Thus 103⋅104 = 107. With these ‘facts’ the
standard algorithm for multiplication follows from the distributive property.
Thus for example, with a single digit factor:


                 12 × 4 = (10 + 2) × 4    =       10⋅4 + 2⋅4   =   40 + 8       =   48



                        Write                     Distributive Multiplication   Add to get
                        number in                 property     facts            answer
                        expanded
                        notation


In the standard algorithm, we perform this procedure ‘in-place’ by arranging
our number vertically, instead of horizontally, and ‘skip’ the first step of
writing the numbers in expanded notation. Thus:


                  12                               12                                12
                 ×4                               ×4                                ×4
                        Multiply 2 by 4 and         8     Multiply 1 by 4 and        48
                        write the result                  write the result
                        directly below the                directly below the
                        product line, under the           product line, under the
                        4                                 1




Pierce College                                                                               MAP
Math 215 Principles of Mathematics                                                                 96


With multiple digit factors we use the same procedure as illustrated by the
following example:


         14 × 23 = (10 + 4) × 23          =     10⋅23 + 4⋅23        =   230 + 92       =    322



                   Write number                    Distributive     Multiplication      Add to get answer
                   in expanded                     property         facts
                   notation



In-place, using the standard algorithm, we have:

                                                                             23                     23
                                       23                                   × 14                   × 14
  23
                                     × 14                                    92                     92
× 14
         Multiply 23 by 4 and
                                       92 Multiply 23 by 1 and              23
                                                                                     Add           23
         write the result                     write the result below                 numbers
                                                                                                   322
         directly below the                   92 under the product                   below the
         product line with the ‘2’            line, with the ‘3’ in 23               product
         in 92 below the 3 and 4.             below the 2, 1, and 9.                 line to get
                                              We shift this number to                the
                                              the left by one digit                  answer.
                                              since we are really
                                              multiplying 23 by 10, not
                                              1. The trailing zero is
                                              omitted by convention.




The text also uses an expanded standard notation where they write the
results of the multiplication of each digit directly below the product line.
Thus:

                                            23
                                           × 14
                                             12                    4×3
                                            80                     4 × 20
                                            30                    10 × 3
                                          +200                    10 × 20
                                            322


Pierce College                                                                                     MAP
Math 215 Principles of Mathematics                                                                                     97


In addition to the standard algorithm, there is also a lattice algorithm for
multiplication. In this case, the numbers are arranged on the outside of the
lattice, and the results from multiplying each digit are placed in the lattice.
When this is competed, the numbers in the diagonals are summed to obtain
the answer. The following example illustrates:
                                                                                    This cell holds 4⋅2. The 8
                                                                                                      ⋅
                                                                                    goes below the diagonal with
                                                1               4                   a zero entered above the
                                                                                    diagonal since the result is
                                                            0
                                                                8           2       only a single digit.


                                                            1
                                                                2           3



                                       This cell holds 4⋅3. The 1
                                                        ⋅
                                       goes above the diagonal and
                                       the 2 below.


We now repeat, for the other column:
                                                1               4
  This cell holds 1⋅2. The 2
                    ⋅                       0               0
  goes below the diagonal with
                                                2               8           2
  a zero entered above the
  diagonal since the result is              0               1
  only a single digit.
                                                3               2           3



                                       This cell holds 1⋅3. The 3
                                                        ⋅
                                       goes below the diagonal and a
                                       zero is entered above since
                                       the resut is single digit.


The diagonals are now summed to obtain the answer, which is written along
the left and bottom of the lattice:
                                                        1               4
           3. 0 + 2 + 1                                                                         We scratch the 0 and make
                                                                        1
                                                    0               0                             it one (see step 2).
              Note: The ‘0’ in the upper
                                                        2                   8   2
              right cell is now ‘1’ (see
              step 2). The answer is
                                                    0               1
              322
                                            3           3                   2   3
                                                                                                1. Nothing to sum, just bring
                           2. 3 + 1 + 8           2                         2                      the 2 down.
                              Note: the ‘1’ from
                              the 12 is placed in
                              the next diagonal and
                              summed there.

Pierce College                                                                                                        MAP
Math 215 Principles of Mathematics                                                 98


Division Algorithms

We proceed in the same manner as we did when discussing multiplication
algorithms. We start with a single digit divisor and then look at two-digit
divisors. With both single and double digit divisors we need to estimate or
guess the quotient at each step. If the estimate or guess is incorrect, it
needs to be modified or adjusted accordingly. As with multiplication, it is
important to remember the place-value associated with the face-value of
the digits involved. The following example illustrates the standard (long)
division algorithm:
                                         We proceed from left to right (the opposite
                                         of multiplication). We estimate the quotient
                                         when dividing 7 by 6. Remember the 7 is
                           6 726         really 700, so the quotient needs to be
                                         multiplied by 100.


Since 100·6 < 700 < 200·6 = 1200, the quotient is 100. We write 1 above the
7 on the quotient apparatus or, if doing the expanded form, 100. We then
subtract 600 from 726 to get 126:

                       Standard                  Expanded

                          1                        100
                        6 726                    6 726
                        −600                      − 600
                         126                          126
We now repeat with 126 as the minuend. Since six is greater than one (i.e.
there is no multiple of 100 times 6 that is less than 100), we exhausted the
multiples of 100, so we look at multiples of ten. We therefore want a
multiple of ten (i.e. 10, 20, 30, etc.) such that when multiplied by 6, we get a
number less than or equal to 126. This is clearly 20, since 20·6 = 120. We
therefore have:




Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                       99


                     Standard                Expanded

                                                   20
                        12                     100
                      6 726                  6 726
                       −600                   − 600
                        126                      126
                       − 120                   − 120
                            6                        6
For the standard algorithm, we write the 2 next to the 1 above the quotient
apparatus and subtract 120 from 126. In the expanded version, we write 20
above the 100 and subtract 120 from 126. Finally, we see that 6·1 = 6, the
last quotient is 1. For the standard algorithm we write the 1 next to the 2
above the quotient apparatus. In the expanded version, we write the 1 above
the 20:
                           Standard               Expanded

                                                      1
                                                     20
                           121                     100
                         6 726                   6 726
                          −600                   − 600
                           126                      126
                          − 120                   − 120
                             6                       6
                            −6                      −6
                              0                       0
Pierce College                                                          MAP
Math 215 Principles of Mathematics                                                         100


Since the result of the subtraction is less than 6 i.e. 0, we are done. With
the standard algorithm, the answer is above the quotient apparatus. With
the expanded form, we must add the three numbers above the quotient
apparatus to obtain the answer:

                                            1     Add to obtain the
                                                  answer: 121
                                           20
                                  100
                                6 726
                                 − 600
                                    126
                                  − 120
                                      6
                                     −6
                                           0
There is also a ‘short’ division that can be used with single-digit division. In
this case the work is done ‘in-place’. The following example illustrates:

                5                  5               57                    5 76
             5 2880             5 28 803             3 3
                                                5 28 8 0              5 283830

          Not enough         28÷5 = 5R3.        38÷5 = 7R3.           30÷5 = 6. The
          thousands i.e. 5   The 3 is placed    The 3 is again        remainder is
          is greater than    next to the        placed next to        zero and since
          2, so divide the   eight; the next    the eight; the        zero is less
          hundreds i.e. 5    division will be   next division         than 5 (and this
          into 28            5 into 38          will be 5 into 30     is the last digit,
                                                                      we are done).


Two digit-division proceeds in the same manner. In this case, the mechanics
of the multiplication is more complicated, put the procedure is identical. The
text provides the following example:



Pierce College                                                                             MAP
Math 215 Principles of Mathematics                                                 101



                                 We start by estimating the quotient. Since
                                 1·32 = 32, 10·32 = 320, and 100·32 = 3200, we
                 32 2618         see that the quotient is between 10 and 100.
                                 We therefore multiply 32 by multiples of 10
                                 i.e. 10, 20, 30, 40, etc.



Trying multiples of 10 and 32, yields 80·32 = 2560. Since 90·32 = 2880, we
see that 80 is the correct multiple of 10:



                       8            We must now find the multiple of 32 that is
                                    just less than 58. Since 32·2 = 64, the
                  32 2618           multiple must be 1.


                    −2560
                       58
Since 32·2 = 64, the multiple of 32 that is just less than 58 is 1. Placing the
one next to the 8 above the quotient apparatus, we subtract 32 from 58:


                       81            We must now find the multiple of 32 that is
                                     just less than 58. Since 32·2 = 64, the
                  32 2618            multiple must be 1.


                    −2560
                        58
                      −32
                        26
Since the result of the subtraction is less than 32, we are done. In this case
we have a remainder, which we write next to the quotient above the division
apparatus preceded by an ‘R’.



Pierce College                                                                     MAP
Math 215 Principles of Mathematics                                              102



                        81R 26                We place the remainder next to the quotient.
                                              This is the answer.
                     32 2618
                      −2560
                            58
                           −32
                            26
Multiplication and Division in Different Bases

As with multiplication and division in base-10, the (single-digit) multiplication
facts must be known. We therefore start by constructing a multiplication
table for the base. Division and multiplication proceeds as with base-10,
except that the place values are now powers of the base, not 10. We
illustrate this using the base-5 example in the text. We start by
constructing a multiplication table:




                       0         1       2          3          4
                 0     0         0       0          0          0
                 1     0         1       2          3          4
                 2     0         2       4         11          13
                 3     0         3       11        14          22
                 4     0         4       13        22          31




When multiplying, we must convert the products from base-10 to base-5.
When all the products are formed, we add, remembering our numbers are in
base-5. We use the expanded algorithm since it is easier to annotate:




Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                          103



                            235
                           ×145
                            225            4×3 = 12 = 225

                           1305            4×2×5 = 40 = 1305

                            305            1×5×3 = 15 = 305

                           2005            1×5×5×2 = 50 = 2005

                           4325            12 + 40 + 15 + 50 = 117 =
                                           225 + 1305 + 305 + 2005 = 4325



Division proceeds in the same manner using the multiplication facts for the
base. Again we illustrate using the base-5 example from the text:

                                345
                          435 32415
                            −2345      3×435 = 2345 (see table below)
                             4015      3245 - 2345 = 4055
                            −3325      4×435 = 3325 (see table below)
                              145      4015 - 3325 = 145 (remainder)




Where we have made use of the following base-5 multiples of 435:


                  435           Base-5             Base-10
                    ×0             0                  0
                    ×1            435                23
                    ×2           1415                46
                    ×3           2345                69
                    ×4           3325                92




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                        104


Mental Mathematics

The text defines mental mathematics as the process of producing an answer
to a computation without using computational aids. By ‘aids’ the text means
pencil and paper, not necessarily a calculator. This is accomplished by using
some of the algorithms we discussed earlier as well as taking advantage of
the associative property and the property of equivalence. Five techniques
are given for addition, as illustrated by the following examples:



Adding from the Left


         67      60 + 30 = 90 (add tens)
        +36       7 + 6 = 13 (add ones)
                           103 (add the two sums)



Breaking up and Bridging


         67      67 + 30 = 97 (add the first number to the tens of the second number)
        +36      97 + 6 = 103 (add the sum to the units of the second number)




Trading Off




         67      67 + 3 = 70 (add 3 to make a multiple of ten)
        +36      36 - 3 = 33 (subtract 3 to compensate for the three that was added)
                         103 (add the two sums to obtain the answer)




Pierce College                                                            MAP
Math 215 Principles of Mathematics                                           105


Using Compatible Numbers

Note: the text defines compatible numbers as numbers whose sums are easy
to calculate mentally.

          130                       200
           50
           70                       100
           20
         + 50                       20
                                    320



Making Compatible Numbers


         25       25 + 75 = 100 (25 and 75 add to 100)
        +79      100 + 4 = 104 (add 4 more since this is what’s left after
                                 removing 75 from 79)


For subtraction the text defines three techniques for mental mathematics.
These are similar to the techniques for addition:

Breaking up and Bridging


          67     67 - 30 = 37 (subtract the tens from the second number
                               from the first number)
        -36      37 - 6 = 31 (subtract the units from the second number
                              from the difference)


Trading Off


         71      71 + 1 = 72 (add 1 to first number; note 40 is a multiple of 10)
        -39      39 + 1 = 40 (add one to second number to compensate and make 40)
                          32 (subtract to obtain the answer)



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             106


Drop the Zeros


         8700       87 - 5   = 82 (drop the zeros – note same number of zeros dropped)
        - 500       82 × 100 = 8200 (replace the zeros, in this case multiply by 100)



Similar techniques can also be used for multiplication. The text presents
three which make use of the distributive and associative properties:

Front End Multiplying


         64       60 × 5 = 300 (multiply tens in first number by 5)
        × 5        4 × 5 = 20 (multiply ones in first number by 5)
                            320 (add to obtain the answer)


Using Compatible Numbers


             2 × 9 × 5 × 20 × 5 = 9 × (2 × 5) × (20 × 5)          rearrange to
                                = 9 × 10 × 100                    for products that
                                = 9 × 1000                        are multiples of
                                = 9000                            ten




Thinking Money



             64     Think of the product as 64 nickels which is
            × 5     equivalent to 32 dimes which is 320 cents.




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                   107


Finally for division, the text discusses two methods: breaking up the
dividend and using compatible numbers:

Breaking Up the Dividend


                                               600 + 8                 608
     7 4256      →    7 4200 + 56      →    7 4200 + 56       →     7 4256



                           Break up          7·600 = 4200             Add the
                           the               and 7·8 =56              quotients to
                           dividend                                   obtain the
                           into 4200                                  answer
                           and 56,
                           both of
                           which are
                           multiples
                           of 7




Using Compatible Numbers


                                         30 + 5                35
     3 105       →   3 90 + 15     →   3 90 + 15       →    3 105



                       Break up        3·30 = 90 and        Add the
                       the             3·5 =15              quotients to
                       dividend                             obtain the
                       into 90                              answer
                       and 15,
                       both of
                       which
                       divisible
                       by 3




Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                         108


Computational Estimation

The text defines computational estimation as the process of forming an
approximate answer to a numerical problem. This is useful in both estimating
i.e. obtaining an approximate answer to the problem and checking or verify
answers i.e. making sure they are reasonable. To understand the difference
between estimating and checking/verifying consider the following examples.

Suppose you wish to determine the number of gallons of paint required to
paint 3 rooms in a house. We can measure the walls in these rooms and
compute the total square feet of surface area. Dividing this by the number
of square feet of coverage of a gallon of paint tells us how many gallons of
paint are required. Note that, for this problem, our answer needs to be
correct to within only one gallon (since we can’t buy part of a gallon). In
fact, if we overestimate the number of gallons of paint, this will probably be
acceptable since we might want some paint left over for touchups. In this
case, we need only provide an estimate, since an exact answer is not
necessary.

Now suppose we have three bills that we need to pay and we want to know if
we have enough money to pay all three. We can add the amount of the three
bills to find the total that we must pay and compare this to the money we
have. However, we want to check our answer to make sure it is reasonable,
since if we have made a mistake, there could be penalties or fines imposed
on us. In this case, we can estimate the total to help verify that our answer
is reasonable.

For addition and subtraction, the text provides five methods or techniques
for estimating sums and differences. These methods all use rounding in
various ways to provide an estimate. With rounding, it should be noted that
the higher the place we round our numbers to the less accurate the estimate
is, but easier it is to calculate the estimate. Conversely, the lower the place
we round our numbers to, the higher the accuracy, but the harder the
calculation. Providing ‘good’ estimates therefore requires trading off
between accuracy and ease of computation.




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                         109


Front End

This method rounds the numbers in the problem twice to provide increased
accuracy. It is similar to the Adding from the Left method discussed above.


         423          Step 1: Round each number down to the nearest
         338                  100 (i.e. sum the left most digits): 4 + 3 + 5
       + 561                  = 12.

                     Step 2: The initial estimate is 1200 from step one.

                     Step 3: We can refine this estimate by rounding
                             the digits in the tens place. In this case we
                             use the digits in the one place to decide if
                             we should round up or down. We have: 2
                             (rounded down) + 4 (rounded up) + 6
                             (rounded down) = 12.

                     Step 4: Our estimate from step 3 is 120 and we add
                             this to the estimate from step 2 to get
                             1320 as the final estimate.



Grouping Nice Numbers

This is another variation on using compatible numbers. Here however we are
not interested in the exact sums, but rather estimating sums that are near
10 or 100:


                    23
                    39
  About 100         32          About 100      The sum is therefore about
                    64                         200.
                  + 49




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                          110


Clustering

In the case where all the numbers are close to a particular number we can
compute the sum using multiplication:


            6200          Here all the numbers are near 6000. Since there
            5842          are 5 numbers, we estimate the sum as 5·6000 =
            6512          30,000
            5521
          + 6319



Rounding

In this case we round all the numbers to the same place and add:


              4724         Round to the nearest 1000. We therefore have
            + 3129         5000 + 3000 = 8000.




Using the Range

In this case we wish to know between what range the answer lies. We can
estimate the range by rounding all the numbers down to obtain a lower
estimate and then rounding all the numbers up to obtain an upper estimate.
The exact answer will lie in this range i.e. between the lower and upper
estimates:

     378         We compute the lower estimate or bound by rounding
   + 524         down: 300 + 500 = 800.

                 The upper estimate or bound is found by rounding up:
                 400 + 600 = 1000.

                 The exact answer therefore lies between 800 and
                 1000.

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                        111


Note: most of these techniques can be used with subtraction, especially
rounding.

For multiplication and division the text presents two techniques: Front-End
and Compatible Numbers:

Front End


   524           Multiply 500 by 8 to get 4000. Next multiply 20 by
   × 8           8 to get 160. Add these to obtain 4160 as the
                 estimate.


Compatible Numbers


                                          800                 Perform the
                                                              division to
      5 4163 → 5 4000                → 5 4000
                                                              obtain an
                                                              estimate of
                                                              the quotient.
                    Find a number close
                    to 4163 that is
                    divisible by 5 e.g.
                    4000




Pierce College                                                            MAP
Math 215 Principles of Mathematics                                           112



Lecture 7

Sections 4.1 and 4.2 – Addition, Subtraction, Multiplication, and Division of
Integers

Integers

Historically negative numbers were first used in China and India, but were
resisted by the Europeans until the 16th centaury. However bankers in both
India and Europe found them useful for representing debt. Today negative
numbers are used everywhere: on temperature scales, on maps for
representing altitudes (below sea-level), in accounting (for debt), etc.
Mathematically, negative numbers are required to extend the whole numbers
so that this system is closed under subtraction. As we saw earlier, the
whole numbers are closed under addition and multiplication, but not under
subtraction and division. In particular, expressions such as 4 – 6 are
undefined i.e. there is no whole number n, such that 6 + n = 4.

We can make subtraction a closed operation by extending the whole numbers
with the addition of the negative numbers. The negative numbers combined
with the whole numbers are called the Integers and are denoted by I. From
Lecture 4, we recall that the number line was used as one of our models for
addition and subtraction. If we go back to this model with the problem of
4 - 6, we see that the segment representing the subtraction of 6 from four
points to the left, beyond zero on the number line:


                        4 - 6 using the number line
                        model. The segment
                        representing the subtraction
                        of 6 from 4 extends to the
                        left of zero on the number
                        line.




                  0     1    2     3     4     5       6   7   8   9    10




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                        113


Clearly, if we extend the number line to the left, we can find the solution to
this problem, just as for subtraction problems where the segments
representing the numbers do not extend to the left beyond zero. In
extending the number line, we must decide how to represent the numbers to
the left of zero. In the problem of 4 – 6 we see that the line representing 6
extends 2 units beyond zero, so clearly this number must somehow capture
the concept that it is 2 units from zero, just as the number 2 to the right of
zero does. Furthermore, if we examine other problems like 4 – 5 and 4 – 7
we see that in the first case, the segment will extend one unit to the left of
zero while the latter extends 3 units to the left. From this we can infer
that numbers to the left of zero are ‘reflections’ of the numbers to the
right and should be arranged as follows about zero:



                 5’   4’   3’   2’   1’   0   1   2   3     4    5


where we have temporality used the notation 1’, 2’, 3’, etc to represent these
numbers. Because of the symmetry of the arrangement of these numbers
about zero, we can regard them as opposites of their counterparts to the
right of zero. That is we note that both 4 and 4’ are both four units from
zero. We also note that if we use the number line to find the solution to the
addition problem 4’ + 4 we find:




                 5’   4’   3’   2’   1’   0   1   2   3    4    -5

So that 4’ + 4 = 0. If we look at this same problem with other combinations
such as 3’ and 3 or 11’ and 11 we see that we always get zero because of the
symmetry of these numbers about zero. Thus calling these new numbers
opposites further reflects their mathematical relationship to their
counterparts to the right of zero.

Now as the text points out, we could keep the prime notation for these
numbers or use some other notation such as putting circles around them as
the Hindus did or circles above them as the Arabs did. However, an arguably
more efficient and elegant notation is to use the minus sign:

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                         114




                 -5   -4   -3   -2    -1     0      1   2   3   4   5

Although this notation makes the minus sign do ‘double duty’ it also captures
the symmetry and oppositeness of these numbers. Also, as will be seen later
it provides a compact notation for expressing these numbers in addition and
subtraction problems and elegantly captures the equivalence of these
operations.

We also note that the concept of oppositeness is reciprocal; that is -4 is the
opposite of 4 and 4 is the opposite of -4. We use the minus sign to indicate
this by writing -4 to indicate that -4 is the opposite of 4 and –(-4) = 4 to
indicate that the opposite of -4 is 4.

Finally we note that zero is neither a positive nor a negative number and its
opposite is itself, zero.

Absolute Value

Because any whole number or positive integer is the same number of units
away from zero as its opposite or negative integer counterpart (and vice
versa), this relationship with zero is the same for both numbers and is
represented by a positive number called the magnitude or absolute value.
For the positive integers, the absolute value is just the number itself. For
negative numbers, it’s the opposite of the number. We place vertical bars
about a number to indicate its absolute value. Thus:

For any number x, the absolute value of x, written |x| is defined as follows:

                                |x| = x if x ≥ 0

                                |x| = -x if x < 0




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                       115


Integer Addition

We have already mentioned how an integer and its opposite have a sum of
zero. We already know how to add the positive integers, since this is just
whole number addition. However, in general, how do we add two negative
integers and a negative and positive integer? Remember, that our motivation
for introducing the negative numbers was to make subtraction a closed
operation. In doing so however, we want to keep addition closed with respect
to the integers. Therefore we need the sum of any two integers to be an
integer. The text presents four models for integer addition that keep
addition a closed operation with respect to the integers. These models are
essentially equivalent and give the same result for any problem.

Chip Model

This model uses the one-to-one correspondence from set theory to model
addition of positive and negative numbers. In this model positive numbers
are represented by black chips and negative numbers by red chips. Black and
red chips ‘neutralize’ each other, so if we establish a one-to-one
correspondence between the black and red chips, any excess chips i.e. chips
that are not neutralized provide the answer to the addition problem. Thus,
for example, if we have -4 + 3, then we represent -4 by 4 red chips and 3 by
three black chips:


                                   The one-to-one correspondence
                                   shows that three black chips are
                                   neutralized by three red chips,
                                   leaving one red chip unpaired.
                                   The answer is therefore -1



As we can see from the figure, the three back chips are paired with three
red chips, leaving one red chip. The answer is therefore -1 and we write: -
4 + 3 = -1.




Pierce College                                                           MAP
Math 215 Principles of Mathematics                                                           116


Charged Field Model

This model is essentially identical to the chip model, except that instead of
using red and black chips we use negative and positive signs. Thus, for
example -5 + 3 becomes:


                                                         The one-to-one correspondence
                                                         shows that three +’s are
                 -   -   - - -                           neutralized by three –‘s, leaving
                                                         two –‘s unpaired. The answer is
                                                         therefore -2
                 +   +   +


Pattern Model

This model utilized what we learned about looking for patterns to deduce
the results of addition between negative and positive numbers. For example
suppose we have the problem -6 + 4. We can create a pattern of addition
with 4 as follows:

                                         7+4      = 11
                                         6+4      = 10
                                         5+4      = 9
                                         4+4      = 8
                                         3+4      = 7
                                         2+4      = 6
                                         1+4      =5
                                         0+4      =4

We can then extend this pattern by continuing to decrease the number on
the left:

                             7   +   4   =   11    1+4 =5      -5 + 4 = ?
                             6   +   4   =   10    0+4=4       -6 + 4 = ?
                             5   +   4   =    9   -1 + 4 = ?
                             4   +   4   =    8   -2 + 4 = ?
                             3   +   4   =    7   -3 + 4 = ?
                             2   +   4   =    6   -4 + 4 = 0

Pierce College                                                                               MAP
Math 215 Principles of Mathematics                                                  117



Since -4 and 4 are opposites we know that -4 + 4 = 0. From this pattern we
see that as we go from 7 to 0, the sum decreases by one each time. If we
continue this decrease, we find that the pattern ‘fits’ with -4 + 4 = 0 and we
have:

                           7 + 4 = 11     1+4 =5              -5 + 4 = -1
                           6 + 4 = 10     0+4=4               -6 + 4 = -2
                           5+4= 9        -1 + 4 = 3
                           4+4= 8        -2 + 4 = 2
                           3+4= 7        -3 + 4 = 1
                           2+4= 6        -4 + 4 = 0

Thus we surmise that -6 + 4 = -2. Although this approach works in this case,
it does suffer from the problem of generalizing patterns from limited data
and therefore comes with the usual cautions.



Number Line Model

This model uses the number line in the same manner we did above. Negative
numbers are represented by segments whose length is equal to the absolute
value of the number but that point (i.e. have an arrow head) to the left. The
procedure for adding is the same as discussed in Lecture 4. Thus, for
example -5 + 3 can be solved using the number line by drawing a segment of
length 3 from 0 to 3 and pointing to the right and adding a segment of
length 5 starting at 3 and pointing to the left:

                                                     -5

                                        -5 + 3                  3

                 -5   -4    -3   -2     -1       0        1     2     3     4   5




Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                         118


Properties of Integer Addition

Integer addition has the same properties as addition of whole numbers. In
particular:

Given three integers a, b, and c:

Closure Property of Addition. a + b is a unique integer.
Commutative Property of Addition. a + b = b + a.
Associative Property of Addition. a + (b + c) = (a + b) + c.
Identity Element of Addition. O is a unique integer such that for all
integers a + 0 = 0 + a = a.
Uniqueness Property of the Additive Inverse. For every integer a there
exists a unique integer –a, the additive inverse of a, such that
a + (-a) = -a + a = 0.




The text also notes the following properties for the additive inverse:

For any integers a and b:

-(-a) = a
-a + (-b) = -(a + b)



Integer Subtraction

The text presents the same four models for integer subtraction as for
integer addition.

Chip Model

Again we use red chips to represent negative numbers and black chips to
represent positive numbers. In this case however, subtracting a negative
number corresponds to removing red chips. We start with the black chips
and then add neutralized pairs of red and black chips, i.e. add zero until we


Pierce College                                                              MAP
Math 215 Principles of Mathematics                                           119


have the required number of red chips. Then we remove the red chips. Thus
for example if we have 3 – (-2) we start with 3 black chips:




                                           Start with three black chips




                                        Add two neutralized pairs




                                          Subtract (remove) the two red
                                          chips




                                           We are left with 5 black chips,
                                           which is the answer




Charged Filed Model

This again is identical to the chip model except that we replace the red
chips by minus signs and the black chips by plus signs. Again, for the
example 3 – (2) we have:

Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           120




                    + + +                   Start with three positives




                                          Add two neutralized pairs
                 + + +    + +

                          - -




                  + + +    + +

                                            Subtract (remove) the two
                           - -
                                            negatives




                   + + +    + +              We are left with 5 positives,
                                             which is the answer




Patterns Model

Again, this model utilizes what we learned about looking for patterns to
deduce the results of subtraction between two integers. For example
suppose we have the problem 3 – (-4). We can create a pattern of
subtraction with 3 as follows:




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                            121


                           3   – 3 =0
                           3   - 2 =1
                           3   - 1 =2
                           3   - 0 =3
                           3   – (-1) = ?
                           3   – (-2) = ?
                           3   – (-3) = ?
                           3   – (-4) = ?

We notice that as the subtrahend decreases by one, the difference is
increasing by one. Therefore extending this pattern we get:

                           3   – 3 =0
                           3   - 2 =1
                           3   - 1 =2
                           3   - 0 =3
                           3   – (-1) = 4
                           3   – (-2) = 5
                           3   – (-3) = 6
                           3   – (-4) = 7

Thus we surmise that 3 –(-4) = 7. Although this approach works in this case,
it does suffer from the problem of generalizing patterns from limited data
and therefore comes with the usual cautions.

Number Line Model

The procedure here is the same as for addition except that with subtraction
we reverse the direction of the arrow for the subtrahend. Thus for the
example 3 – (-4):

                                         3 – (-4)
                                     3                       -(-4)

            -3   -2   -1   0     1       2     3     4      5        6   7

                                                    Arrow is reversed since
                                                    we are subtracting.




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                          122


Subtraction as the Inverse of Addition

As the text notes, subtraction of integers, like subtraction of whole
numbers, can be specified in terms of addition:



Subtraction. For integers a and b, a – b is the unique integer n such that a =
b + n.

Note: Perhaps a more ‘mathematical’ way of looking at this is to note that
for any two integers a and b if a = b + n then by the property of equivalence
we can add the same number to each side of this equation:

                              a + (-b) = -b + b + n

but –b + b = 0, so we have:

                              a + (-b) = n = a – b

This means that every subtraction problem can be expressed as an
equivalent addition problem with the subtrahend replaced by its additive
inverse. In fact we can define subtraction as follows:

Definition of Subtraction of Integers. If a and be are two integers, then
a – b = a + (-b).

This means that subtraction is a ‘redundant’ operation; we really don’t ‘need’
it. In essence all we need is addition with integers, since all ‘subtraction
problems’ are really just addition problems involving integers.



Order of Operations

Strictly speaking subtraction is not commutative or associative. That is
5 – 3 ≠ 3 – 5. However if we agree to convert all subtraction problems to
their equivalent addition problem involving integers, then we have:

                          5 – 3 = 5 + (-3) = -3 + 5

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                                  123


which is commutative, since addition is commutative. Likewise if we have
3 – 15 – 8 we can write:

          3 – 15 – 8 = 3 + (-15) + (-8) = 3 + ((-15) + (-8)) = (3 + (-15)) + (-8)

which is associative since addition is associative. In general we agree to
convert all subtraction problems into their equivalent addition problem and
then performing all additions in order from left to right.

Integer Multiplication

The text presents the same four models for multiplication as for addition
and subtraction. However with multiplication we must determine what we
mean when we multiply a positive and negative number together and what we
mean when we multiply two negative numbers together. Let’s first consider
multiplying a negative number by a positive number. Since multiplication is
just repeated addition, we can view this as just adding the negative number
to itself the number of times as specified by the positive number. Thus for
example 3⋅(-2) = -2 + -2 + -2 = -6 by the models we have developed for
addition of negative numbers. We note that in this case the answer is
negative.

For the product of two negative integers, there is no definitive method for
determining what the product is a priori (although we can provide some
justification for a particular choice). Instead we define the product of two
negative numbers as equal the product of their additive inverses. More
generally we say:

Definition of Multiplication of Integers. For any two integers a and b:

                               (-a)⋅b = -(a⋅b)
                            (-a) ⋅(-b) = a⋅b

Note: a and b can be either positive, negative or zero.

Although this might seem arbitrary it gives us a consistent system that is
closed under multiplication and that obeys the familiar commutative and



Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                                  124


associative properties we expect. Note that a direct consequence of this
definition is that for -1 we obtain:

Multiplication by -1. For every integer a, (-1) ⋅a = -a

We can now discuss the four models for multiplication presented by the
text. Since the chip and charged field models are essentially the same we
use them interchangeably and present them together.

Chip and Charged Field Models

For a positive number multiplying a negative number we use the repeated
addition model. Consider the problem 3⋅(-2). We start with two red chips
representing -2 and repeat this pair until we have three pairs:



                                                Start with two red chips
                                                representing -2 and then repeat
                                                until we have 3 pairs. The number
                                                of red chips represents the
                                                answer.




For multiplication involving two negative numbers, we create negative
charges corresponding to the second number then repeat these till we have
the number of groupings corresponding to the absolute value of the first
number. We then add an equivalent number of positive charges and then
remove the negative charges. The removal of the negative charges results
form the sign of the first number which we interpret as being removal. Thus
for example (-2) ⋅(-3):
                        - - -       - - -       -   -   -   + + +
                                    - - -       -   -   -   + + +
                                                +   +   +
                                                +   +   +
                 Start with three   Repeat 2   Add same     Remove
                 negatives          times.     number of    negatives.
                 representing -3.              positives.

Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                        125


Pattern Model

Again, this model establishes a pattern to deduce the results of
multiplication between two integers. For example suppose we have the
problem 3⋅(-4). We can create a pattern of multiplication with 3 as follows:



                         3 ⋅4   =   12
                         3 ⋅3   =    9
                         3 ⋅2   =    6
                         3 ⋅1   =    3
                         3 ⋅0   =    0
                         3⋅(-1) =    ?
                         3⋅(-2) =    ?
                         3⋅(-3) =    ?
                         3⋅(-4) =    ?

We notice that as the second factor decreases by one, the product is
decreasing by three. Therefore extending this pattern we get:

                         3 ⋅4   =    12
                         3 ⋅3   =     9
                         3 ⋅2   =     6
                         3 ⋅1   =     3
                         3 ⋅0   =     0
                         3⋅(-1) =    -3
                         3⋅(-2) =    -6
                         3⋅(-3) =    -9
                         3⋅(-4) =   -12

Thus we surmise that 3⋅(-4) = -12. Although this approach works in this
case, it does suffer from the problem of generalizing patterns from limited
data and therefore comes with the usual cautions.




Pierce College                                                            MAP
Math 215 Principles of Mathematics                                           126


Line Number Model

This is essentially the same as what we did with multiplication of whole
numbers. With integers, the directions of the segments and their
replication on the number line is given by the sign of the numbers. Thus for
example 3⋅(-2) is calculated by creating a segment of 2 units pointing to the
left from zero to -2. This segment is repeated three times, with each
segment added to the arrow end of the previous segment. The result is a
segment from zero to -6:


                             3⋅(-2)
                      (-2)            (-2)        (-2)

                 -6   -5       -4     -3     -2   -1     0   1   2   3   4



                      Create 3 segments from 0 to -2 and
                      lay them end to end.




Properties of Multiplication

Multiplication follows the same properties for integers as for whole
numbers. These are:

Properties of Integer Multiplication

If a, b, and c are integers then:

Closure Property of Multiplication. a⋅b is a unique integer.
Commutative Property of Multiplication. a⋅b = b⋅a.
Associative Property of Multiplication. a⋅(b⋅c) = (a⋅b)⋅c.
Multiplicative Identity Property. 1 is a unique integer such that 1⋅a = a⋅1 = a.
Distributive Property of Multiplication over Addition. a⋅(b + c) = a⋅b + a⋅c.
Zero Multiplication Property. O is a unique number such that a⋅0 = 0⋅a = 0.



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                               127


Integer Division

We define division in terms of multiplication:

Definition of Integer Division. If a and b are two integers and b ≠ 0 then
a÷b is the unique integer c, if it exists, such that a = b⋅c.

Note: The quotient of two negative integers, if it exists, is a positive integer
and the quotient of a positive and a negative integer, if it exists, or a
negative and a positive integer, if it exists, is negative.

Note: division (just as for the whole numbers) is not closed. That is, there is
no integer given by -3÷2. In order to achieve closure for division (ignoring
division by zero), we must extend the integers to the rational numbers.

Order of Operations

These are the same as for the whole numbers: multiplication and division are
performed first, in order from left to right, followed by addition and
subtraction, in order from left to right. If exponents appear, they are
calculated first i.e. before multiplying and dividing.

Thus for example:

24 – 16 ÷ 4 ⋅ 2 + 8 = 16 - 16 ÷ 4 ⋅ 2 + 8 = 16 - 4 ⋅ 2 + 8 = 16 – 8 + 8 = 8 + 8 = 16




Ordering Integers

As with the whole numbers we use the number line to determine order. The
same rules apply: a number to the right of another number on the number
line is greater than the number to its left and a number to the left of
another number is less than the number to its right. Thus for example:




Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                               128


    •    2 is less than 5 because it is to the left of 5
    •    7 is greater than 3 because it is to the right of 3
    •    -2 is less than 0 because it is to the left of zero
    •    -3 is less than 2 because it is to the left of 2
    •    -4 is less than -1 because it is to the left of -1
    •    -2 is greater than -3 because it is to the right of -3

Formally we say:

Definition of Greater Than and Less Than. If a and b are two integers
then a < b (or equivalently b > a) if and only if b – a is equal to a positive
integer; that is b – a is greater than zero.

Equivalently we can say that b > a if and only if there exists a positive
integer k (i.e. k > 0) such that a + k = b.




Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                               129



Lecture 8

Sections 4.3 and 4.4 – Divisibility and Prime and Composite Numbers

Divisibility

Divisibility refers to integer division i.e. dividing one integer by another
integer. We say one integer divides another if the remainder of the division
is zero. Remember, division is not a closed operation for the integers, so in
general one integer will not divide another. More formally we say:

Divisibility. If a and b are any integers and b ≠ 0, then b divides a if and only
if there is a unique integer c such that a = bc. If such an integer exists,
then a is divisible by b.

If b divides a then b is a factor of a and a is a multiple of b. b is also called a
divisor of a. The text uses the notation b|a which is read b divides a and is a
logical statement i.e. is either true or false. The vertical bar is read
‘divides’. As the text points out, this should not be confused with a/b which
is a divided by b and is a number.

The notion of divisibility is part of number theory which deals
with the properties of numbers in general and integers in
particular. Many famous mathematicians have made
contributions to this field, perhaps most notably Fermat who
developed many important theorems and proofs. We will discuss
                                                                     Pierre de Fermat
some of these theorems and present proofs for a few of them.            1601 - 1665


Theorem 1. For any integers a and d ≠ 0, if d divides a and n is any integer,
then d divides a⋅n.

This theorem can also be stated as follows: if d is a factor of a, then d is
also a factor of any multiple of a.

Proof. The proof of this theorem follows immediately from the definition of
divisibility. If d divides a then, by definition there exists an integer c such
that:

Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                               130



                                            a = c⋅d

If n is another integer, then by the general property of equivalence and the
associative property of multiplication:

                                      n⋅a = n⋅c⋅d = (n⋅c)⋅d

Since the integers are closed under multiplication, n⋅c is an integer and by
the definition of divisibility, d divides n⋅a.

The following theorem presents results for the sum and difference of
integers that are divisible by a common factor:

Theorem 2. For any integers a, b, and d ≠ 0:

a.   If   d   divides   a and divides b, then d divides a + b.
b.   If   d   divides   a and does not divide b, then d does not divide a + b.
c.   If   d   divides   a and divides b, then d divides a - b.
d.   If   d   divides   a and does not divide b, then d does not divide a - b.

Proof. If d divides a and b, the by definition there exists integers n and m
such that:

                                      a = md and b = nd
Therefore:

                 a + b = md + nd = (m + n)d by the distributive property
                 a - b = md - nd = (m - n)d by the distributive property

Since the both addition and subtraction are closed for the integers, m + n is
an integer and m – n is an integer. Therefore by the definition of divisibility,
a + b and a – b are divisible by d.

Now suppose d divides a but does not divide b. Then by definition, there is
an integer m such that a = md. However no such integer n exists such that
b = nd. Now suppose that d divides a + b. Then there exists an integer q such
that a + b = qd. However by the general property of equivalence:

Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                           131



a + b = qd
a – a + b = qd – a
b = qd – a = qd – md (using the fact that d divides a)
b = (q – m)d by the distributive property.

However since the integers are closed under subtraction, q – m is an integer,
and by definition d divides b, which contradicts our assumption to the
contrary. Therefore d does not divide a + b. The exact same argument also
shows part d of the theorem.

Divisibility Rules

One way to tell if one integer divides another is to perform long division with
the two numbers and see if there is a non-zero remainder. Although this
method always gives a definitive answer, it often requires a considerable
amount of work. To make matters worse, we might not actually want to know
the result (i.e. quotient) of the division, but only a true/false to the question
‘does a divide b?’ This is often the case when we are trying to factor a
number or determine the least common multiple or greatest common divisor
(see next lecture). To this end, it would be useful to have some simple tests
i.e. tests that require little or no calculation to determine if one integer
divides another. Fortunately such tests exist for the integers 2 through 10.

Divisibility Test for 2

An integer is divisible by 2 if its units digit is 0, 2, 4, 6, or 8.



Divisibility Test for 3

An integer is divisible by 3 if the sum if its digits is divisible by 3.



Divisibility Test for 4

An integer is divisible by 4 if the number given by its tens and units digits is
divisible by 4.

Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           132




Divisibility Test for 5

An integer is divisible by 5 if its units digit is 0 or 5.


Divisibility Test for 6

An integer is divisible by 6 if it is divisible by both 2 and 3.


Divisibility Test for 8

An integer is divisible by 8 if the number given by its hundreds, tens and
units digits is divisible by 8.



Divisibility Test for 9

An integer is divisible by 9 if the sum if its digits is divisible by 9.



Divisibility Test for 10

An integer is divisible by 10 if its units digit is 0.




Note: the text also gives a test for 11, but this might be more bother than
its worth, just like the test for 7.




Examples.

    1. The number 120.
         a. Is divisible by 2, since it ends in 0.


Pierce College                                                               MAP
Math 215 Principles of Mathematics                                                 133


                 b. Is divisible by 3, since the sum of the digits is 3.
                 c. Is divisible by 4, the tens and ones digits give 20 and 4 divides
                    20.
                 d. Is divisible by 5, since it ends in 0.
                 e. Is divisible by 6 since it is divisible by both 2 and 3.
                 f. Is divisible by 8, since 8 divides 120 (test not useful here).
                 g. Is not divisible by 9, since the sum of the digits is 3, which is
                    not divisible by 9.
                 h. Is divisible by 10, since it ends in 0.

    2. The number 111,111,111.
         a. Is not divisible by 2, since it does not end in 0, 2, 4, 6, or 8.
         b. Is divisible by 3, since the sum of the digits is 9.
         c. Is not divisible by 4, the tens and ones digits give 11 and 4 does
             not divide 11.
         d. Is not divisible by 5, since it does not end in 0 or 5.
         e. Is not divisible by 6 since it is not divisible by 2.
         f. Is not divisible by 8, since 8 does not divides 111.
         g. Is divisible by 9, since the sum of the digits is 9.
         h. Is not divisible by 10, since it does not end in 0.
    3. The number 97,128
         a. Is divisible by 2, since it ends in 8.
         b. Is divisible by 3, since the sum of the digits is 27.
         c. Is divisible by 4, the tens and ones digits give 28 and 4 divides
             28.
         d. Is not divisible by 5, since it does not end in 0 or 5.
         e. Is divisible by 6 since it is divisible by both 2 and 3.
         f. Is divisible by 8, since 8 divides 128.
         g. Is divisible by 9, since the sum of the digits is 27, which is
             divisible by 9.
         h. Is not divisible by 10, since it does not end in 0.




Example: Inventory Problem. The manger of a warehouse is informed that
there are 11,368 cans of juice in inventory. All cans are packed in boxes
holding either 6 or 24 cans. Is the inventory correct?



Pierce College                                                                     MAP
Math 215 Principles of Mathematics                                            134


Understanding the Problem. We must determine if the number 11,368
correctly gives the inventory of cans of juice in the warehouse. We are told
that all these cans are contained in boxes holding either 6 or 24 cans.

Devising a Plan. We must relate the total number of cans to the number of
cans in boxes and see if this provides sufficient information to determine if
the inventory is correct. Relating the total number of cans to the cans in
the boxes will require an equation. We let n be the number of boxes holding
6 cans and m the number of boxes holding 24 cans. We can then construct
the following equation:

                               6n + 24m = 11,368

We note that the left side of this equation is divisible by 6 since 6 divides
both itself and 24. This means the right side must also be divisible by six.
If it is, then the inventory might be correct. If not, the inventory is wrong.

Carrying Out the Plan. In order to be divisible by 6, 11,368 must be divisible
by both 2 and 3. Using our divisibility tests for two and three, we see that
11,268 is divisible by 2 since its units digit is 8. However the sum of its
digits is 19 and it is therefore not divisible by 3. This means 11,368 is not
divisible by 6 and hence the inventory must be incorrect.

Looking Back. Note that the plan we developed only yields a definitive answer
if the total number of cans is not divisible by 6. If the inventory had been
11,358 or some other number divisible by 6, then the best we could say is
that the inventory might be correct. In fact, in the case where the
inventory is divisible by 6, there are many possible solutions for n and m. In
order to obtain a definitive answer more information is required.




Prime and Composite Numbers

A positive integer having only two distinct divisors (one and itself) is called a
prime number. Note: one is not a prime number. An integer greater than one

Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                 135


with a positive factor other than one and itself is a composite number.
Examples of prime numbers are:

                    2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37

Examples of composite numbers are:

      4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, and 30

Prime Factorization

 All positive integers greater than one can be expressed as a product of two
or more distinct numbers. For example:

                                 1⋅7 = 7 ad 3⋅4 = 12

Such expressions are called factorizations. A factorization expresses a
number as the product of two or more factors. We note that prime numbers
have only one factorization. However for composite numbers, several
factorizations are possible. For example:

                                     1⋅12 = 12
                                      2⋅6 = 12
                                      3⋅4 = 12
                                     2⋅2⋅3 = 12

We notice however, in this example, that one of these factorization i.e. the
last one, involves only prime numbers and that there is only one such
factorization for 12. This result happens to be true for all positive integers
greater than one and is known as the fundamental theorem of arithmetic:

Theorem 3: Fundamental Theorem of Arithmetic. Every positive composite
integer greater than one can be written uniquely as a product of prime
numbers.

Note: the word ‘uniquely’ in the theorem means that there is only one such
factorization (the order of the factors is unimportant; that is rearranging
the numbers in the factorization is not considered as a new or different

Pierce College                                                                     MAP
Math 215 Principles of Mathematics                                             136


factorization). Thus the fundamental theorem of arithmetic assures us that
once we find a prime factorization of a number, a different prime
factorization cannot be found for the same number.

The prime factorization of any number can be found by starting with the
smallest prime 2, and seeing if it divides the given number. If not, we go to
the next largest prime number and see if it divides the given number. We
repeat this process until we find a prime number that divides the given
number. If we can’t find such prime number then the given number is prime.
If we do, then we repeat this process with the quotient. The entire process
is repeated until we find a quotient that is prime. Thus for example consider
the number 260:

    1.   260 is divisible by 2 with quotient 130.
    2.   We start over with 130, which is also divisible by 2 with quotient 65.
    3.   65 is not divisible by 2 or 3 but it is divisible by 5 with quotient 13.
    4.   Since 13 is prime we are done. Collecting our factors we have:
    5.   260 = 2⋅2⋅5⋅13 which is the prime factorization.

When writing a number’s prime factorization it is customary to order the
factors from least to greatest and use exponential notation for repeated
factors. Thus for 260 we would write: 260 = 22⋅5⋅13.

Number of Divisors

Another question we can ask is: ‘How many divisors does a number have?’
Note this question asks for all divisors, not just prime ones. One way to
determine this is to list all possible pairs of factors starting with one and
the number in question. If we place these numbers so that the factors form
two columns, the fist column starting with one, and the second starting with
the given number, then we can create this list by seeing if 2, then 3, and so
on divides the given number. Whenever we find a factor we place it in the
first column and the quotient in the second column. We are done when the
next factor is a number that is greater than or equal to a number in the
second column. Thus for example consider finding all the factors of 24. We
start with 1 and 24:
                                1          24



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                                137


Since 2 divides 24 with quotient 12, we add 2 to the first column and 12 to
the second:
                              1         24
                              2          12

Three also divides 24, so we add it to the first column and 8 to the second
column:

                                1               24
                                2               12
                                3               8

Four divides 24 as well, so again we add it to the first column and 6 to the
second column:

                                1               24
                                2               12
                                3               8
                                4               6

Five does not divide 24, so we proceed to 6. Six however appears in the
second column, so we are done and the two columns list all 8 factors of 24.

We can also determine the number of factors for a given number by finding
its prime factorization. In general, this factorization will be of the form:

                                                  n
                                          n
                                    p1n1 p2 2    pmm

where p1, p2, . . ., pm are the prime factors and n1, n2, . . . nm are positive
integers. The number of factors is then given by the formula:

                                (n1+1) (n2+1)⋅⋅⋅ (nm+1)

Thus for example, for 260, its prime factorization is:

                                       24 = 23⋅3



Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                           138


 so that m = 2 and p1 = 2 and p2 = 3 and n1 = 3 and n2 = 1. Therefore the
number of factors is:

                             (3 + 1)(1 + 1) = 4⋅2= 8

which is the same result we obtained above.

Determining if a Number is Prime

In principle if we want to determine if a number is prime we must see if
there are any numbers between 1 and the number itself that divide the given
number. Thus for example, if we wish to check if 97 is prime we must check
all numbers between 1 and 97. If we use the fundamental theorem of
arithmetic, we can reduce this task to checking only prime numbers between
2 and 97, since if 97 is not prime, by the theorem, it can be expressed as a
product of primes.

With a little thought we can reduce this task even further by noting, as the
text does, that we need only check prime numbers p, such that p2 ≤ n. To see
this we develop some theorems that prove this result.



                                                                            n
Theorem 4. If n and d are positive integers and d is a divisor of n, then
                                                                            d
is also a divisor of n.

Proof. By definition, since d is a divisor of n, there exists a unique integer m,
such that n = m⋅d. If we divide both sides of this equation by d, we have: m =
 n                     n                                          n
   , which shows that     is an integer. We therefore have n = ⋅d and since d
 d                     d                                          d
                              n
is an integer, by definition     divides n.
                              d




Theorem 5. If n is a composite number, then n has a prime factor p such
that p2 ≤ n.


Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             139



Proof. Since n is composite, then by the fundamental theorem of arithmetic
it can be expressed as a product of two or more prime numbers. Consider
the smallest of these prime factors and call it p. Let q be any other prime
factor of n. Then, by our choice of p we have p ≤ q and therefore:

                                  p2 = p⋅p ≤ p⋅q ≤ n

since p⋅q × (any remaining prime factors) = n.



Theorem 6. If n is an integer greater than 1 and not divisible by any prime
number p2 ≤ n, then n is prime.

Proof. We will assume the opposite of the theorem and show this leads to a
contradiction. Suppose that n is not divisible by any prime p2 ≤ n and that n is
composite. Then by the fundamental theorem of arithmetic n can be
factored in to a product of prime numbers. Let q be the smallest such
                                                               n
factor. By hypothesis we must have q2 > n. From theorem 4         is also a
                                                               q
divisor of n and:

                                         n
                                    q⋅     = n < q2
                                         q
Dividing both sides by q gives:

                                          n
                                            <q
                                          q
                                                            n
However this gives us an immediate contradiction since if     is prime then q
                                                            q
                                        n
is not the small prime factor of n and if is not prime we can (by the
                                        q
fundamental theorem of arithmetic) express it as a product of primes.
However these prime must all be less than q, which again contradicts our
hypothesis.

Another way of finding all the prime numbers less than a given
number is to use the Sieve of Eratosthenes, named after the

Pierce College                                                                 MAP


                                                                    Eratosthenes of Cyrene
                                                                       276 BC - 194 BC
Math 215 Principles of Mathematics                                                                                                                  140


Greek mathematician Eratosthenes of Cyrene who is credited
as one of the first people to measure the circumference of the
Earth. In this scheme all the numbers between 1 and the
number of interest, say 100 for example, are listed in order.
The number 1 is crossed out since it is neither prime nor
composite. The next number is two, which is prime. All multiples
of two are then crossed out.

We then proceed to three, which is also prime, and cross out all multiples of
three. Since four will have been crossed out as a multiple of 2, we then
proceed to 5, which is prime and cross out all multiples of 5. We continue
this process until we reach 100 (or whatever number we are interested in).
All the numbers that are not crossed out will be prime.

Example. Find all the prime numbers between 1 and 100.


     1. List numbers from 1 to 100                                         2. Cross out all multiples of 2.
     and cross out 1.
1     2    3       4        5        6       7    8   9    10         1     2    3    4     5       6        7        8         9         10
11 12 13 14 15 16 17 18 19                                 20         11 12      13   14    15      16       17       18        19        20
21 22 23 24 25 26 27 28 29                                 30         21 22      23   24    25      26       27       28        29        30
31 32 33 34 35 36 37 38 39                                 40         31 32      33   34    35      36       37       38        39        40
41 42 43 44 45 46 47 48 49                                 50         41 42      43   44    45      46       47       48        49        50
51 52 53 54 55 56 57 58 59                                 60         51 52      53   54    55      56       57       58        59        60
61 62 63 64 65 66 67 68 69                                 70         61 62      63   64    65      66       67       68        69        70
71 72 73 74 75 76 77 78 79                                 80         71 72      73   74    75      76       77       78        79        80
81 82 83 84 85 86 87 88 89                                 90         81 82      83   84    85      86       87       88        89 90
91 92 93 94 95 96 97 98 99 100                                        91 92      93   94    95      96       97       98        99 100


     3. Cross out all multiples of 3.                                      4. Cross out all multiples of 5.
 1     2       3       4        5        6       7    8    9    10    1      2    3    4        5        6        7        8         9         10
11    12   13          14       15       16      17   18   19   20    11    12   13   14     15      16          17       18         19        20
21    22   23          24       25       26      27   28   29   30    21    22   23    24    25      26          27        28        29        30
31    32   33          34       35       36      37   38   39   40    31    32   33    34    35      36          37        38        39        40
41    42   43          44       45       46      47   48   49   50    41    42   43    44    45      46          47        48        49        50
51    52   53          54       55       56      57   58   59   60    51    52   53    54    55      56          57        58        59        60
61    62   63          64       65       66      67   68   69   70    61    62   63    64    65      66          67        68        69        70
71    72   73          74       75       76      77   78   79   80    71    72   73    74    75      76          77        78        79        80
81    82   83          84       85       86      87   88   89    90   81    82   83    84    85      86          87        88        89     90
91    92   93          94       95       96      97   98   99   100   91    92   93    94    95      96          97        98        99    100




Pierce College                                                                                                                                      MAP
Math 215 Principles of Mathematics                                                                       141



       5. Cross out all multiples of 7.                       6. Cross out all multiples of 11 (none).

   1    2        3   4    5    6    7    8    9    10    1      2   3    4    5    6    7    8    9    10
  11    12   13      14   15   16   17   18   19   20    11    12   13   14   15   16   17   18   19   20
  21    22   23      24   25   26   27   28   29   30    21    22   23   24   25   26   27   28   29   30
  31    32   33      34   35   36   37   38   39   40    31    32   33   34   35   36   37   38   39   40
  41    42   43      44   45   46   47   48   49   50    41    42   43   44   45   46   47   48   49   50
  51    52   53      54   55   56   57   58   59   60    51    52   53   54   55   56   57   58   59   60
  61    62   63      64   65   66   67   68   69   70    61    62   63   64   65   66   67   68   69   70
  71    72   73      74   75   76   77   78   79   80    71    72   73   74   75   76   77   78   79   80
  81    82   83      84   85   86   87   88   89    90   81    82   83   84   85   86   87   88   89    90
  91    92   93      94   95   96   97   98   99   100   91    92   93   94   95   96   97   98   99   100




We can continue this process, but since 112 > 100 we know by Theorem 6 that
there are no more primes. Note: strictly speaking, with this method we would
continue to check until we reached 100.

As the text notes, there are infinitely many prime numbers, but no known
formula for finding or generating them. An active area of mathematical as
well as computation research is finding large prime numbers. Currently the
largest known prime number is 232,582,657-1. It has 9,808,358 digits and was
found in 2006. Besides being a mathematical curiosity and contest for
computer scientists primes have practical applications, especially in
cryptography where they are used to generate difficult to break
encryptions of data.




Pierce College                                                                                           MAP
Math 215 Principles of Mathematics                                         142




Lecture 9

Sections 4.5 and 4.6 – Greatest Common Divisor, Least Common Multiple, and
Clock and Modular Arithmetic

As we have noted previously, addition, multiplication, and subtraction are
closed operations for the integers. Division however is not closed since in
general dividing one integer by another does not yield an integer. In order to
bring division closer to being a closed operation we must extend the number
system (just like we did for subtraction by adding the integers). This
extension will be to add rational numbers i.e. fractions to the number
system, which we will do in subsequent lectures. However before doing this
we will need some tools (mathematical infrastructure) to help us work with
fractions. In particular, we will find it useful to compute the Greatest
Common Divisor and Least Common Multiple for two or more numbers.

Greatest Common Divisor

Given two (or more) integers we know that by the fundamental theorem of
arithmetic that we can factor them into a product of prime numbers. We can
then compare these factors and find all possible factors (prime and
composite) that these numbers share. Since this list is finite, there will be a
common factor (which can be one) that is greatest. This factor is known as
the greatest common divisor. More formally:

The greatest common divisor (GCD) of two integers a and b is the greatest
integer that divides both a and b.

There are three methods discussed by the text for finding the GCD: the
intersection of sets method, the prime factorization method, and the
Euclidean Algorithm method. In addition the text discusses the colored rods
method which is a way of visualizing the GCD for two numbers and the
calculator method which is programming procedure that uses a
computer/calculator to find the GCD. We will not discuss these two
‘methods’ and the text may be consulted for more information.


Pierce College                                                             MAP
Math 215 Principles of Mathematics                                           143


Before discussing each of these methods we present the following theorem,
which is used in the Euclidean Algorithm:

Theorem 7. If a and b are any whole numbers greater than zero, and a ≥ b,
then the GCD of a and b is equal to the GCD of r and b, where r is the
remainder when a is divided by b.

Proof. By Theorem 2, every divisor of a and b is also a divisor of a – b.
Conversely (also by Theorem 2) every divisor of b and a-b is also a divisor of
b and (a-b)+b = a. Thus these two pairs of number b and a-b and b and a
share the exact same set of divisors, and hence the GCD of one pair is also
the GCD of the other pair. W e can therefore reduce the size of the
numbers involved by subtracting the smaller number from the larger and
finding the GCD for this pair instead. We need not stop with this pair, since
by the same argument the GCD of b and a -2b will be the same as the GCD of
b and a, and so on. In particular, if n is the quotient resulting from dividing a
by b and r the remainder, then we can form n such pairs, with the final pair
being r and b, which will therefore have the same GCD as b and a.

Intersection Sets Method

This is a ‘brute-force’ method where we list all the divisors of both numbers
in two sets and then find the intersection of these sets. The largest
element in the intersection is the GCD. Thus suppose we want to find the
GCD of 20 and 32. First we must find all the divisors of both numbers:

                           1     20          1      32
                           2     10          2      16
                           4      5          4       8
Let A be the set of all divisors of 20 and B be the set of all divisors of 32.
The using the results above we have:

                 A = {1, 2, 4, 5, 10, 20}    B = {1, 2, 4, 8, 16, 32}

                                  A ∩ B = {1, 2, 4}

Therefore 4 is the GCD since it’s the largest element of A ∩ B.


Pierce College                                                               MAP
Math 215 Principles of Mathematics                                       144



Prime Factorization Method

Although the set intersection method always works, unless the numbers
involved are relatively small and have few factors, this method requires
significant work. A somewhat more efficient method is the prime
factorization method. In this case we factor both numbers into a product
of primes. We then extract all the common factors. Multiplying these
common factors together gives the GCD. Thus for example, suppose we want
the GCD of 168 and 180. First we factor both these numbers:

                               168 = 2⋅2⋅2⋅3⋅7

                               180 = 2⋅2⋅3⋅3⋅5

Next we find all the common factors:

                               168 = 2⋅2⋅2⋅3⋅7

                               180 = 2⋅2⋅3⋅3⋅5
Finally we multiply the common factors together to get the GCD:

                              GCD = 2⋅2⋅3 = 12

 Note: if the two numbers do not have any prime factors in common, then
their GCD is one e.g. 4 and 9. Numbers whose GCD is one are called relatively
prime.

Euclidean Algorithm Method

Both the intersection of sets and prime factorization methods require
factoring the numbers in question. For larger numbers, this may require a
significant amount of effort. The Euclidean method reduces this effort by
reducing the size of the numbers involved using Theorem 7. However we
need not be content with stopping with the remainder, we can apply Theorem
7 repeatedly until the remainder is zero. When this happens, the non-zero
number in the pair is the GCD. The following example illustrates:



Pierce College                                                            MAP
Math 215 Principles of Mathematics                                           145



Find the GCD of 2300 and 10,764. We divide 2300 into 10,764. Performing
this division reveals that 2300 goes into 10,764 four times with a remainder
of 1564. Our new pair is therefore (1564, 2300). We now divide 1564 into
2300. The quotient in this case is one, and the remainder is 736. We now
have the pair (736, 1564). Dividing again yields a quotient of 2 and a
remainder of 92. The new pair is therefore (92, 736). Since 92 divides 736
exactly 8 times, the final pair is (0, 92) and the GCD is 92.

Finally we note that finding the GCD is not confined to a pair of numbers.
Often we wish to know the GCD for three, four, or even more numbers.
Since a common divisor of three numbers, say, is also a common divisor of
any two, the GCD of three numbers cannot be greater than the GCD of any
two. We can therefore find the GCD of three numbers by finding the GCD
of any two. The two we choose is up to us, so we choose the pair that
results in the least effort. Consider the following example.

Find the GCD of 6339, 6341, and 13,791. We work with the pair (6339,
6341) since they are the smallest two numbers and also because they are
close to each other compared with the third number. We get:

                 (6339, 6341) → (6341-6339, 6339) = (2, 6339) → (1, 2)

Thus the GCD is 1 in this case. Note: if the number we found was greater
than one, we would need to test all devisors less than this number to find
the GCD for the triple of numbers.

Least Common Multiple

Given two (or more) integers we can ask the question: ‘what is the smallest
number that both these integers divide?’ This number is known as the least
common multiple. More formally:

Suppose a and b are positive integers. The least common multiple (LCM) of a
and b is the least positive integer that is a multiple of both a and b.




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           146


Clearly such a common multiple exists, one example being the product of the
two numbers. However this may not be the smallest such number. The text
presents four methods for finding the LCM: the intersection of sets
method, the prime factorization method, the Euclidean Algorithm method
and the Division-by-Primes method. Again the text discusses the colored
rods method which is a way of visualizing the LCM for two numbers. Refer to
the text for details.

Before discussing the details of these methods, we present the following
theorem which will be used in the Euclidean Algorithm:

Theorem 8. For any two natural numbers a and b, the product of the GCD of
a and b and the LCM of a and b is equal to the product of a and b.

Proof. The proof of this theorem follows directly from the fundamental
theorem of arithmetic and the prime factorization methods for finding both
the GCD and LCM. By the fundamental theorem of arithmetic we can factor
both a and b into a product of prime numbers:

                          a = q1q2⋅⋅⋅qn and b = p1p2⋅⋅⋅pm

where q1, q2,⋅⋅⋅ ,qn and p1, p2,⋅⋅⋅ ,pm are prime numbers. The product of a and b
is therefore:

                              a⋅b = q1q2⋅⋅⋅qn p1p2⋅⋅⋅pm

We now divide the prime factors of a into two groups. The first group
contains all the q’s that are equal to one of the p’s. The second group
contains the remaining q’s. Without loss of generality, we assume that the
first k q’s are in the first group, and the remaining n-k q’s are in the second
group. We therefore have:

                        a⋅b = (q1q2⋅⋅⋅qk)( qk+1⋅⋅⋅qn p1p2⋅⋅⋅pm)

Now from the prime factorization method for constructing the LCM (see
below) we have LCM = qk+1⋅⋅⋅qn p1p2⋅⋅⋅pm. Also by the prime factorization
method for constructing the GCD we note that GCD = q1q2⋅⋅⋅qk.



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             147




Intersection-of-Sets Method

For this method we create a set containing all the positive multiplies of the
first number and another set containing all the positive multiples of the
second number. We then find the intersection of these two sets. The
smallest element in this set is the LCM. Note: as a practical matter, the set
of all positive multiples of a given number contains an infinite number of
elements, so in practice we can only list a finite number. If too few elements
are listed the intersection of the two sets containing the multiples of the
given numbers may have zero elements. If this is the case, we must (go back
and) list more elements in these sets and continuing repeating this process
until the intersection of the sets contains at least one element. The
following example illustrates this procedure.

Find the LCM of 8 and 12. We start by creating two sets, A and B containing
the multiples of 8 and 12, respectively:

             A = { 8, 16, 24, 32, 40, 48,…}   B = {12, 24, 36, 48, 60, 72,…}

                                  A ∩ B = {24, 48,…}

The first (and smallest) element in the set A ∩ B is 24 and this is the LCM.

Prime Factorization Method

The intersection method requires that we list a sufficient number of
multiples of each number so that a common multiple appears in each list/set.
Since the number of multiples that must be listed before finding a common
multiple is not known a priori, we are forced to guess at the number
required. As a result, more multiples will be listed than necessary, resulting
in extra work. In addition, just generating these multiples requires effort,
and since we really only care about one of these i.e. the LCM, the work
required to generate all the others is ‘wasted’ effort. Although this method
works, it is clearly inefficient. The prime factorization method, in general, is
more efficient. We start by factoring each number into it prime factors. We
then select one of these factorizations, typically that of the largest

Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                         148


number, and then add the factors in the other number that are ‘missing’. The
resulting number will be the LCM. The following example demonstrates.

Find the LCM of 12 and 40. We start by finding the prime factorization of
each number:

                           12 = 2⋅2⋅3    40 = 2⋅2⋅2⋅5

Starting with the factorization for 40, we see that the only factor from the
factorization for 12 that does not appear is 3. We therefore add 3 to this
factorization which gives us the LCM:

                             LCM = 2⋅2⋅2⋅5⋅3 = 120

Euclidean Algorithm Method

This method is useful when it is difficult to find the prime factorizations of
the numbers whose LCM we want. In this case, we find their GCD instead
using the Euclidean method and then compute the LCM by dividing their
product by their GCD. Thus for example:

Find the LCM of 731 and 952. We start by finding the GCD of 731 and 953.
Using the Euclidian method:

 1.   Divide 952 by 731. The remainder is 221.
 2.   Divide 731 by 221, the remainder is 68.
 3.   Divide 221 by 68 to obtain the remainder 17
 4.   Dividing 68 by 17 we obtain a remainder of zero.
 5.   The GCD is therefore 17.

 We therefore have 17⋅LCM = 731⋅ 952 = 695,912. Dividing by 17, we find
 that the LCM = 40,936.

 Division by Primes Method

 Another method for finding the LCM of several natural numbers is division
 by primes. This method is presented in the following example: Find the LCM



Pierce College                                                             MAP
Math 215 Principles of Mathematics                                         149


 of 12, 75, and 120. We start with the least prime that divides at least one
 of the numbers, 2 in this case, and divide as follows:




                             2 12 75 120
                                6 75 60

 Because 2 does not divide 75, we simply bring down 75. For 12 and 120, we
 divide by 2 and bring down the quotients. We continue this process until the
 bottom row contains only relatively prime numbers:


                               2 12 75 120
                               2    6 75 60
                               2    3 75 30
                               3    3 75 15
                               5    1 25 5
                                   1 5 1

 The LCM is the product of the numbers on the left side and the bottom:
 LCM = 2⋅2⋅2⋅3⋅5⋅1⋅5⋅1 = 23⋅3⋅52 = 600.

 Clock Arithmetic

 ‘Clock’ arithmetic is a specific example of a more general type of arithmetic
 known as circular or modulo arithmetic. In this case, our ‘arithmetic’ is
 confined to the numbers 1 to 12 that appear on the face of a standard wall
 clock. If we start at midnight and watch the hands on the clock move we will
 see them progress from one o’clock to two o’clock and so on, until noon when
 they will return to their original position at 12. If we continue to watch the
 clock, the process will repeat over and over again, indefinitely. The behavior
 of the clock marks the passage of time in half-day increments. It answers
 the ‘local’ question of ‘what time is it?’ When we ask this question we are

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          150


 not interested in the ‘absolute’ time with respect to some established point
 of reference. That is, if we ask what time it is, we expect an answer such
 as 7 o’clock, not April 23, 2007 at 7PM, GMT. The clock therefore answers
 the question ‘what time is it?’ within a particular day without regard to what
 day it is or even what half day it is. The user of the clock is assumed to
 either not care about the day, month, or year or to have this information
 from another source.

 If we look at the clock from this perspective, then its circular repeating
 behavior defines a type of arithmetic which the text calls clock arithmetic.
 If we know the current time, then we can ask the question, what time it will
 be 5 hours from now or what time it was 9 hours ago. We can calculate
 these times (hours) by adding or subtracting from the current time. Thus
 for example, if the time is 4 o’clock and we agree to meet in 5 hours, then
 we can compute this time by adding 4 and 5 to get 9 o’clock. We see in this
 case that the normal rules for arithmetic apply.

 Now consider the case where it is 9 o’clock and we agree to meet again in 5
 hours. In this case simply adding 5 to nine gives us the ‘nonsense’ answer of
 14 o’clock. The reason this is a ‘nonsense’ answer, of course, is that 14
 o’clock does not correspond with any hour on the clock, where all times must
 be specified by a number between 1 and 12. So what is going on here? The
 answer is that the 12 on the clock acts as ‘zero’ or reference point (just like
 the zero on the number line). All times are measured relative to this time
 just like all numbers on the number line are ‘measured’ relative to the zero
 on the line.

 On the number line, if we have two numbers, say 9 and 14, and we want to
 know the ‘distance’ between them, we subtract 9 from 14. This works
 because they are both referenced to the same point on the line i.e. zero.
 Conversely if we add 5 to 9, we can use the ‘count-on’ strategy and start at
 9 and count-on by 5 which takes us to 14. If we do this on the clock, then
 we will find ourselves at 2 o’clock (not 14 o’clock). The reason is that as we
 count-on from 9, when we reach 12, the next count brings us to 1 o’clock. As
 we pass the 12 in our count we pass the ‘reference point’ for the clock and
 the hours are ‘reset’ and start ‘repeating.’




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                                   151



                                                       To find 5 hours from 9
                                   12                  o’clock we count-on by 5,
                              11           1
                                                       starting at 9. As we pass
                         10                    2       12, the hours are ‘reset’,
                                                       and we end up at 2 o’clock.
                     9                             3



                         8                     4

                              7            5
                                   6




 This ‘resetting’ tells us that we have crossed a half-day period, and as we
 recall, the clock is designed to provide the time only within half-day periods.
 The crossing of the 12, tells us that we agree to meet at 2 o’clock in the
 next half-day period. Now suppose its 9 o’clock and we agree to meet in 12
 hours. If we use the count-on method for computing the hour we find
 ourselves back at 9:


                                                       To find 12 hours from 9
                                   12                  o’clock we count-on by 12,
                              11           1
                                                       starting at 9. As we pass
                         10                    2       12, the hours are ‘reset’,
                                                       and we end up back at 9
                                                       o’clock.
                     9                             3



                         8                     4

                              7            5
                                   6




 We therefore have that 9 ⊕ 12 = 9 in our clock arithmetic. Note we use the
 symbol ‘⊕’ instead of ‘+’ to indicate clock addition instead of integer
 addition. We also note that if we use any multiple of 12 e.g. agree to meet
 36 hours from 9 o’clock we also end up back at 9 o’clock. The only
 difference is that in counting on we will move past 12 multiple times (3
 times in the case of 36 hours). We therefore have that:

                                       9 ⊕ 12n = 9

Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                             152



 where n is any whole number. In fact we find that this is the case no matter
 what the starting time is. Thus if t is any number between 1 and 12 we have:

                                     t ⊕ 12n = t

 Going back to our original problem of meeting 5 hours from 9 o’clock we
 note that:

                                      9⊕5=2

 Now we note that 9 + 5 = 14 = 2 + 12·1. That is 2 is what is left over after
 dividing 14 by 12. Likewise if we add 12 to 9 we get 21 and we note that:

                                    21 = 12·1 + 9

Thus 9 is the remainder of dividing 21 by 12. Similarly if we add 24 to 9 we
get 31 and:

                                    31 = 12·2 + 9

From this we can derive the rule for clock addition i.e. what we mean by ⊕.
We see that if we add any whole number to the time, the resulting time
(hour on the clock) is the remainder from dividing the regular sum by 12. To
illustrate consider the following examples:

    •    11 ⊕ 8 = 7 since 11 + 8 = 19 and 7 is the remainder of dividing 19 by 12
         i.e. 19 = 12·1 + 7.

    •    12 ⊕ 8 = 8 since 12 + 8 = 20 and 8 is the remainder of dividing 20 by
         12 i.e. 20 = 12·1 + 7.

         5 ⊕ 6 = 11 since 5 + 6 = 11 and 11 is the remainder of dividing 11 by 12
         i.e. 11 = 12·0 + 11.

Now suppose its 2 o’clock and we ask the question ‘where were you 5 hours
ago?’ To find this answer we can go back to the clock and count-back:



Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                   153




                                                            To find 5 hours before 2
                                         12                 o’clock we count-back by 5,
                                    11          1
                                                            starting at 2. As we pass
                             10                     2       12, the hours are ‘reset’,
                                                            and we end up counting
                                                            back from 12 o’clock.
                         9                              3



                             8                      4

                                    7           5
                                          6



 We therefore have that 2 Θ 5 = 9 where the symbol Θ stands for clock
 subtraction. Since 9 ⊕ 5 = 2, we can define clock subtraction in term of
 clock addition, just as we did for subtraction of whole numbers. Thus we
 define 2 Θ 5 as the number n, such that n ⊕ 5 = 2. Thus:



    •    4 Θ 4 = 12 since 12 ⊕ 4 = 4

    •    4 Θ 8 = 8 since 8 ⊕ 8 = 4

    •    11 Θ 7 = 4 since 4 ⊕ 7 = 11

Clock multiplication i.e. ⊗ can be defined in terms of clock addition as
repeated clock addition just like multiplication of whole numbers can be
defined as repeated addition of whole numbers:

                                 3 ⊗ 5 = 5 ⊕ 5 ⊕ 5 = 10 ⊕ 5 = 3

Clock division can in turn be defined in terms of multiplication (see text).
However with respect to clock arithmetic these operations are not as
common and we will not peruse them further. Finally we note that adding or
subtracting 12 on the clock always gives the same result: the original time.
Thus 12 acts as the additive identity for clock arithmetic i.e. as its ‘zero’.
The text also notes (see page 287) that clock division by 12 is undefined,
just dividing by zero is undefined with the whole numbers.



Pierce College                                                                       MAP
Math 215 Principles of Mathematics                                           154


Modular Arithmetic

Clock arithmetic is a specific example of circular or modulo arithmetic. In
particular, there is nothing special (from a mathematical perspective) about
the number 12. We could conceive of 10 hour clocks or 16 hour clocks or
even 7 hour clocks. The arithmetic of such clocks would be exactly the same
as for the 12 hour clock except that when adding we divide by 10 or 16 or 7
and keep the remainder of that division. Thus for example:

    •    On a 7 hour clock 4 ⊕ 5 = 2 since 4 + 5 = 9 and 2 is the remainder of
         dividing 9 by 7 i.e. 9 = 7·1 + 2.
    •    On a 10 hour clock 9 ⊕ 5 = 4 since 9 + 5 = 14 and 4 is the remainder of
         dividing 14 by 10 i.e. 14 = 10·1 + 4.
    •    On a 16 hour clock 10 ⊕ 7 = 1 since 10 + 7 = 17 and 1 is the remainder
         of dividing 17 by 16 i.e. 17 = 16·1 + 1.

In fact we can do this sort of arithmetic with any integer number, n greater
than one. In order to avoid confusion (or overly complex notation) with the
addition symbol ⊕ we adopt a more general notation. We define the modulo
‘operator’, mod n, as follows:

If b is an integer and n is any integer greater than one, then b mod n is the
remainder resulting from dividing b by n.

Thus for example:

    •    23 mod 10 = 3
    •    23 mod 16 = 7
    •    23 mod 7 = 2

From these examples we see that there is a relationship between these
reminders and 23. With respect to arithmetic modulo ten, say, the numbers
3 and 23 are essentially equivalent. That is, on a 10-hour clock, they both
represent the same hour: 3. Likewise on a 16 hour clock, 7 and 23 are
equivalent in the same manner, as is 2 and 23 on a 7 hour clock. Formally we
say 3 and 23 are congruent modulo 10:




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                             155


Modular Congruence. For integers a and b, a is congruent to b modulo m,
written a ≡ b mod m, if and only if, a – b is a multiple of m, where m is a
positive integer greater than one.

If we examine our previous examples we see that:

    •    23 – 3 = 20, which is a multiple of 10, therefore 23 ≡ 3 mod 10.
    •    23 – 7 = 16, which is a multiple of 16, therefore 23 ≡ 7mod 16.
    •    23 -2 = 21, which is a multiple of 7, therefore 23 ≡ 2 mod 7.

In another example, we note that 18 and 25 are congruent modulo 7, and
that each number is leaves the same remainder, 4, upon division by 7:

                           18 = 2·7 + 4 and 25 = 3·7 + 4

In general, we can say that two numbers are congruent modulo m, if and only
if their remainders on division by m are the same.

Modular arithmetic has many important applications beyond clocks. Consider
the following:

    •    The week with seven days, provides a modulo sequencing of days
         throughout the year. Thus if Monday, say falls on the fourth of the
         month, then the 11th, 18th, and the 25th are also Mondays.

    •    Location on the Earth is measured in degrees of latitude and
         longitude. Since a circle had 360 degrees, longitudinal position is
         congruent modulo 360. This has important consequences for navigation
         on and above the Earth.

    •    Music uses an arithmetic that is modulo 12 in the consideration of the
         system of twelve-tone equal temperament, where octave and
         enharmonic equivalency occurs.

    •    The method of casting out nines offers a quick check of decimal
         arithmetic computations performed by hand. It is based on modular
         arithmetic modulo 9, and specifically on the crucial property that 10 ≡
         1 (mod 9).

Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                         156




Lecture 10

Sections 5.1 and 5.2 – Rational Numbers and Addition and Subtraction of
Rational Numbers

Rational Numbers

As we noted in previous lectures, the negative numbers were ‘added’ to the
whole numbers to form the integers so that subtraction would be a closed
operation. Without the negative numbers, expressions like 3 – 5 were
undefined. Likewise, division is also not a closed operation with respect to
the whole numbers, and for that matter the integers. Expressions such as
3÷5 or -12÷7 are undefined. We were able to ‘patch things up’ by introducing
the concept of a remainder when dividing two whole numbers (or integers),
however, the result of the division is still not a whole number or integer. To
solve this problem, we again add ‘new’ numbers to the integers so that (for
the most part) division becomes a closed operation.

We note, that unlike subtraction, division can never be truly closed, since
division by zero will always remain undefined. However, we can ‘limit’ this
exception to just the number zero. The ‘new’ numbers we add are, of course,
the rational numbers. The set of rational numbers is denoted by Q and
defined as follows:

                          a
                   Q={      | a and b are integers and b ≠ 0}
                          b
                                                              a
Note: at this point we have not defined what we mean by , other than this
                                                              b
‘new’ number is composed (in some manner) of two integers, one of which i.e.
b cannot be zero. Our next task therefore is to define exactly what we mean
    a
by     and then to specify how the four canonical operations of arithmetic i.e.
    b
addition, subtraction, multiplication, and division ‘work’ with these new
numbers. We also expect, that the integers will be a subset of the rational
numbers i.e. I ⊂ Q and that addition, multiplication, and subtraction will
remain closed operations. We also point out that the term fraction is often


Pierce College                                                             MAP
Math 215 Principles of Mathematics                                            157


used describe the rational numbers, however the term fraction refers to any
                      a
number of the form      where b ≠ 0. That is, where a and b are not
                      b
necessarily integers.

To begin our discussion we start, as usual, with definitions and notation. For
                   a
a rational number , we call the number a the numerator and the number b
                   b
the denominator. The word numerator comes from a Latin word meaning
‘numberer’ and the word denominator comes from a Latin word meaning
‘namer’. The word fraction is derived from the Latin word fractus, meaning
                                                                      a
to break. The terms are reflective of what we mean by the number . Taken
                                                                      b
             a
together as , the numbers a and b represent part or a portion of a given
             b
set. The number b represents the number of elements in the set, and the
number a represents the number of elements in a portion of the set i.e. in
                           a
some subset. The number is the ratio of the number of elements in the
                           b
subset to the total number of elements in the (complete) set. That is, is
specifies what part or portion of the whole (total number of elements in the
set) are contained in the subset. Consider the following examples:

    •    Suppose we have a set with five elements and we form a subset with
                                                                    2
         two elements. The subset contains two of the 5 elements or of the
                                                                    5
         elements in the set.

                                                  Two of five
                                                  elements




    •    We can form a ‘set’ from a whole by breaking or dividing the whole
         into pieces. Consider a circle divided into eight pieces:




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                            158



          We can then form a ‘subset’ containing three of the eight pieces:


                                                    Three of the eight
                                                    pieces from dividing
                                                    the circle. These
                                                    represent 3/8 of
                                                    the original circle.




          We note that this division of the circle can be ‘virtual’. That is, we
          need not actually break it apart. Thus the three ‘pieces’ in the subset
                                                                             3
          represent 3 equal parts of 8 constituting the original circle or of
                                                                             8
          the circle.

    •    Suppose we have a set of six pencils to be distributed equally among 3
         people. We can give each person 2 pencils which is an equal share of
                                                                2
         the original set of 6. Each person therefore receives of the original
                                                               6
         set of 6 pencils.

                                                 a
We can therefore regard the rational number         as representing a
                                                 b
partitioning of a set (or some object(s) e.g. cake, pizza, pencils, sheets of
paper, etc) into b equal parts and then taking a of these equal parts. The
         a
number     represents the amount of the whole i.e. the fraction that was
         b
taken. It is important to note that when applied to physical objects, the
                                                                         3
same fraction may represent different amounts. Thus, for example, of a
                                                                         4
                                3
large pizza is more pizza than of a small pizza even though the
                                4
        3
faction accurately represents the relationship of the part to the whole in
        4
each case. It is therefore the ‘whole’ being considered that represents the
                                                                       a
absolute amount. We thus note that to understand the meaning of in the
                                                                       b
absolute sense we must know or understand:


Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                  159



    1. The whole being considered.
    2. The number of b equal-size parts into which the whole is divided.
    3. The number of a parts of the whole that are selected.

We can also represent rational numbers on the number line as follows. Mark
the line with zero as reference and one to establish scale. We can now assign
a rational number to a point on the line by dividing the unit segment
(segment between zero and one) into parts as specified by the denominator
of the rational number and then counting the number of these parts as
specified in the numerator. Thus for example, to find the point representing
 3
   on the number line, we divide the unit segment into 4 equal parts, and
 4
starting from zero we count 3 of these parts and mark the corresponding
point on the line:
                                                Unit Segment




                 -1                   0                         1


                                                              Divide unit segment
                                                              into 4 equal parts




                 -1                   0                         1


                                 Count 3 parts starting
                                 at zero. ¾ goes here.




                 -1                   0                   ¾     1




Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                            160



In a similar manner we can place      −5 ,   -½, 9 , ⅔, and ⅓:
                                      4          8



                                                      9
                       −5
                                 -½          ⅓ ⅔
                 -2    4    -1        0              18          2


In the examples above we note that in some cases |a| ≥ |b|. Such factions
are called improper fractions. If |a| < |b|, then the fraction is called a
proper fraction.



Equivalent Fractions

Suppose we divide two circles (of the same radius) into equal parts. The
first circle we divide into 8 parts and the second into 16, as shown in the
figure below.




Now consider the shaded regions in the figure. In the circle we divided into
eight parts we shaded one of the parts, and for the circle we divided into 16
parts, we shaded two of the regions. If we remove the shaded regions, we
have two wedge shaped portions of the circles, as shown below.




If we compare these two wedges, we see that they are exactly the same
size and shape. That is, they are equal. If we represent each wedge as a
rational number then:




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                         161




                                   1                           2
                                   8                          16

and since these wedges are exactly equal we must have that:

                                    1 2
                                     =
                                    8 16

        1       2
That is    and    must represent the same number despite the fact that
        8      16
their numerators and denominators are different. Such fractions are called
equivalent fractions or equal fractions since they represent the same
number. Examining these fractions more closely we note that:

                                 1 2 2 ⋅1
                                  =  =
                                 8 16 2 ⋅ 8
                        2                        1
That is, we can view      as being obtained from   by multiplying both the
                       16                        8
numerator and denominator by 2. In general, if we take two (or more)
objects of the same size (circles, squares, rectangles, etc.) and divide them
into equal parts of different sizes and then compare the various parts we
will discover similar relationships between the fractions representing the
parts. Thus for example suppose we take a rectangle and two identical size
copies and divide them respectively into 3, 6 and 12 parts, as shown below.




If we take the same size portion of each rectangle, as shown in the figure,
and draw a box around it, and count the number of parts inside the box, we
find that:

                                  1 2 4
                                   = =
                                  3 6 12




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                          162


                   1 2 2 ⋅1 4 2 ⋅ 2 4 ⋅1
And we note that:    = =      =    =      =     . We therefore make the
                   3 6 2 ⋅ 3 12 2 ⋅ 6 4 ⋅ 3
following generalization: The value of (i.e. the number represented by) a
fraction does not change if its numerator and denominator are multiplied by
the same nonzero whole number. Thus we have:

                                       a
Fundamental Law of Fractions. Let        be any fraction and n a nonzero
                                       b
                 a n⋅a
number, then      =     .
                 b n ⋅b

Example

                               7
                                 =
                                     ( −1) ⋅ 7 = −7
                              −15 ( −1) ⋅ ( −15 ) 15

           a −a           −a
In general:   =     where     is preferred i.e. negative sign is associated
           −b b            b
with number in the numerator.

Simplifying Fractions

                            a
Given a particular fraction   , we can generate as many equivalent fractions
                            b
as we wish by multiplying the numerator and denominator by nonzero
integers. Likewise, given a fraction we can generate an equivalent fraction
by removing a common factor from the numerator and denominator:

                                   4   4 ⋅1 1
                                     =     =
                                  12 4 ⋅ 3 3

This process of multiplying and canceling common factors from the
numerator and denominator changes the form of the fraction, but not its
value. Thus there are many equivalent ways to represent the value of a
fraction. However we note that for some representations, the numerator
and denominator do not contain any common factors. That is, the numerator
and denominator are relatively prime. Such fractions are said to be in
simplest form:


Pierce College                                                                MAP
Math 215 Principles of Mathematics                                            163


                                 a
Simplest Form. A rational number   is in simplest form if a and b have no
                                 b
common factor greater than one, that is if a and b are relatively prime.

If a fraction is not in simplest form, we can generate the equivalent fraction
that is in simplest form by canceling the greatest common divisor from the
numerator and denominator. Thus for example:

                                  60 30 ⋅ 2 2
                                     =      =
                                  210 30 ⋅ 7 7

We note that 30 is the greatest common divisor of 60 and 210. In practice
is may be more efficient to remove two or three common factors to find the
simplest form for a fraction than to find the greatest common divisor.
However, finding the greatest common divisor insures that the fraction will
be in simplest form after it is canceled from the numerator and
denominator. In general, it is mandatory to place a fraction in simplest form
when the fraction is being supplied as the solution to a problem (unless
specifically specified otherwise).

Equality of Fractions

Since equivalent fractions can be represented in a variety of ways, given two
fractions, we can ask the question: ‘are they equal?’ The text presents three
methods for answering this question:

    1. Simplify both fractions so they are in simplest form. Then compare
       the numerators and denominators. If they are the same, the fractions
       are equal. For example:



                                12 2    10 2
                                  = and   =
                                42 7    35 7

         Therefore these fractions are equal. Note: geometrically, this is
         equivalent to finding a (larger) common piece of the whole that is
         exactly equal to pieces specified by (in the numerators of) the
         fractions.


Pierce College                                                                MAP
Math 215 Principles of Mathematics                                             164




    2. Compute the least common multiple of the numbers in the
       denominators of the two fractions and then multiply the numerator
       and denominator of each fraction by the least common multiple
       divided by the denominator and compare the numerators. If they are
       equal, then the fractions are equal. The idea here is that the two
       fractions each represent a partition or division of the whole into
       difference size pieces. If we can make the size of the pieces the
       same (by sub-dividing them further), we can just count the number of
       (now equal size) pieces and see if we have the same number. If we do
                                                                    10
       the fractions are equal. For example consider the fractions      and
                                                                     35
       12
           :
        42

         The least common multiple of 35 and 42 is 210. Since 210 ÷ 35 = 6, we
                                                    10           12
         multiply the numerator and denominator of     by 6. For    , we
                                                    35           42
         multiply the numerator and denominator by 210 ÷ 42 = 5:



                        10 6 ⋅10 60       12 5 ⋅12 60
                          =      =    and   =      =
                        35 6 ⋅ 35 210     42 5 ⋅ 42 210

         Since the numerator in both cases is equal to 60, the fractions are
         equal. Geometrically this is equivalent to sub-dividing the pieces
         represented (by the numerators) in each fraction so that all the
         pieces are the same size and then comparing the number (of equal
         size) pieces. In the case of the example, we slice each of the 10
                                 10
         pieces represented by       into six pieces for a total of 60 pieces. For
                                 35


Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                             165


         12
             , we divide in each of the 12 pieces into 5 pieces for a total of 60
         42
         pieces. Since in both cases we get the same number of equal size
          1 
               pieces, the fractions are equal.
          210 




                 Divide each                       Divide each
                 piece into 6                      piece into 5
                 pieces.                           pieces.




    3. This is similar to the second method, except instead of finding the
       least common multiple, we multiple the numerator and denominator of
       the first fraction by the denominator of the second fraction and the
       numerator and denominator of the second fraction by the denominator
       of the first fraction. Since the denominators will be the same, we
       compare the numerators, if they are the same, the fractions are
       equal. Using are previous example:

                       10 42 ⋅10 420       12 35 ⋅12 420
                         =       =     and   =       =
                       35 42 ⋅ 35 1470     42 35 ⋅ 42 1470

         Since the numerator in both cases is 420, the fractions are equal. .
         Geometrically this is again equivalent to sub-dividing the pieces
         represented (by the numerators) in each fraction so that all the
         pieces are the same size and then comparing the number (of equal
         size) pieces. In this case, the size of the sub-divisions is determined
         by the denominator of the other fraction. For the example, we slice
                                                  10
         each of the 10 pieces represented by        into 42 pieces for a total of
                                                  35
                            12
         4200 pieces. For      , we divide in each of the 12 pieces into 35 pieces
                            42
         for a total of 420 pieces. Since in both cases we get the same number

Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                            166


                        1 
         of equal size        pieces, the fractions are equal. Note: in practice
                        1470 
         we need not actually compute the denominators (since we know a priori
         they will be the same) because we only need to compare the
         numerators. We can summarize this result as follows:

                                   a      c
Equality Property. Two fractions      and   are equal if and only if ad = bc.
                                   b      d
This is some called cross multiplying.

Ordering Rational Numbers

Ordering refers to placement of rational numbers on the number line. In
particular, we recall that if a number appears to the right of another
number on the number line, then that number is greater than the other
number. Conversely, if a number appears to the left of another number on
the number line, then that number is less than the other number.
Numerically, we can use methods (2) and (3) above for determining equality
of rational numbers to determine their order i.e. which one is greater or less
than the other.

From our discussion of equality, we saw that in order to compare two rational
numbers, we had to find a common multiple (methods 2 and 3) of both
numbers. By finding a common multiple, we could replace each rational
number by an equivalent rational number such that the two ‘replacements’
had the same (i.e. common) denominator. We then compared the numerators.
If they were equal, then the two numbers were equal, otherwise they were
not. Finding a common denominator is the numerical equivalent of
partitioning or subdividing the portion of the whole represented by each
number into the same size pieces. We can then count the number of pieces
i.e. look at the numerators, and determine equality. In the case where the
numerators are not equal, then the number with the larger numerator is the
larger number, or conversely, the number with the smaller numerator is the
smaller number. We therefore have the following result:

                                              a c
If a, b, and c are integers and b > 0, then    > if and only if a > c.
                                              b b



Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                         167


Note: we must always insure that the denominators are the same before
comparing the numerators. If the denominators are not the same, then we
are effectively comparing pieces of different sizes giving us an erroneous
result. Consider the following example. Suppose we have two identical cakes.
One cake we cut into eight equal size pieces and the other into six equal size
pieces:




Suppose 3 pieces of the first cake are eaten and 2 of the second are eaten.
              5                       4
We then have of the first cake and of the second cake left.
              8                       6




Which of the remaining cakes is larger? Since the pieces are of different
sizes, we cannot simple count the number of pieces remaining in each cake
and compare. Instead, we must first make the sizes of each piece of cake
the same. We therefore cut each piece in the first cake into three pieces
and each piece in the second cake into four pieces:




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                168




                 Cut each piece           Cut each piece
                 into three pieces.       into four pieces.

We now have equal size pieces in both cakes. Counting the pieces in the first
cake we find that there are 15 pieces while a count of the pieces in the
second cake shows that there are 16 pieces. Thus there are more pieces in
                                      5 4
the second cake and we conclude that < .
                                      8 6

Mathematically we can reduce the amount of work if we are just interested
in determining the order of the two rational numbers. Just as in method 3
for determining equality, all we need to do is compare the numerators for a
common denominator. The actual calculation of the common dominator can be
avoided by cross multiplying. We therefore have the following result:

                                                              a c
If a, b, c, and d are integers and b > 0 and d > 0, then       > if and only if
                                                              b d
ad > bc.



Denseness of Rational Numbers

                               1     2
Consider the rational numbers    and . We can find a rational number
                               2     3
between these two numbers as follows. First we rewrite both rational
                                                      1 3        2 4
numbers so they have a common denominator. We have = and = .
                                                      2 6        3 6
Looking at the numerators we note that is no whole number between 3 and 4.
We therefore rewrite our two rational numbers again, using a larger common
                                     1 6        2 8
denominator, say 12. Do so, we obtain =     and = . Comparing the
                                     2 12       3 12

Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                        169


                                                                    7
numerators again, we note that 7 lies between 6 and 8. Therefore      lies
                                                                   12
          1      2                                     1 7        7 2
between      and . We can verify this by noting that < and          < . In
          2      3                                     2 12      12 3
general, it can be shown that for any two (distinct) rational numbers we can
find another rational number between them. Note: this property is not true
for the natural numbers, the whole numbers, or the integers. We therefore
have the following property:

Denseness Property of Rational Numbers. Given any two distinct i.e. not
                       a    c
equal rational numbers   and , there is another rational number between
                       b    d
these two numbers.

                                                               a      c
We can now ask the question, given two distinct rational numbers and ,
                                                               b      d
how do we find a rational number between them? One technique is to use the
following theorem:

    a     c
Let   and   be any two distinct rational numbers with positive
    b     d
                    a c           a a+c c
denominators, where < . Then <             < .
                    b d           b b+d d

              a c
Proof. Since   < and b > 0 and d > 0, we can multiply both sides of the
              b d
inequality by bd:

                                        a          c
                               ( bd )     < ( bd )
                                        b          d

Simplifying we get:

                                     ad < bc

Adding ab to both sides we obtain:

                              ad + ab < bc + ab



Pierce College                                                            MAP
Math 215 Principles of Mathematics                                          170


                               a(d + b) < b(c + a)

Since b > 0 and d > 0, we can divide both sides by b(d + b) to get:

                                    a a+c
                                     <
                                    b b+d

Now, from above:

                                    ad < bc

Adding cd to both sides we get:



                               ad + cd < bc + cd

                               d(a + c) < c(b + d)

Since d > 0 and b > 0, we can divide both sides by d(b + d) to get:

                                   a+c c
                                      <
                                   b+d d

Finally we note that the rational number that is exactly half way between
                      a      c     ad + bc
the rational numbers and       is          .
                      b      d      2bd



Addition and Subtraction of Rational Numbers

Now that we have defined the rational numbers we must determine how to
compute the sum and differences (and eventually the products and
quotients) of two rational numbers. We also want the rational numbers to be
closed under these operations. For addition, we first consider the case
where both rational numbers have the same denominator. Consider the
following example:

                                    1 2
                                     + =?
                                    5 5

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          171


If we consider these rational numbers as representing parts of circles we
have divided into five equal pieces then, geometrically we have:




                         +                        =



Where the circle on the right represents all of the pieces placed together
i.e. summed (recall set model for addition). Counting the pieces in this circle
shows that:

                                    1 2 3
                                     + =
                                    5 5 5

Thus when all the pieces are the same size, we can just count the number of
pieces to obtain the sum or more generally, the sum of two rational numbers
is just the sum of their numerators. We therefore have the following result:

     a     c                               a c a+c
If     and   are two rational numbers, then + =    .
     b     b                               b b  b

We will use this result to define addition for the general case i.e. where the
denominators of the rational numbers are not the same. To do this, we use
the same strategy we employed for determining order. Consider the
                    2 1
following example:    + = ? . In this case, the denominators are not the
                    3 4
same, so we cannot use the result from above directly. However if we were
to write these numbers in an equivalent form such that they had a common
denominator, then we could use the result from above. Geometrically we
represent our rational numbers as parts of circles:




                                +                         =     ?


Pierce College                                                               MAP
Math 215 Principles of Mathematics                                       172


Finding a common denominator is geometrically equivalent to subdividing each
piece into smaller, equal size pieces. Once we have subdivided each piece, we
can count them to obtain the result. This works, because all the subdivided
pieces are the same size. Since 12 is the LCM of 3 and 4, we subdivide each
piece in the first circle into 4 equal pieces and each piece in the second
circle into 3 equal pieces:




                     +                        =


     Divide each            Divide each            Combine and
     piece into 4           piece into 3           count each
     pieces.                pieces.                piece.


Combining and counting the equal sized pieces yields:

                                   2 1 11
                                    + =
                                   3 4 12

Mathematically we can cross multiply and then add the numerators to
compute the sum. This gives the following result:

     a     c                               a c ad + bc
If     and   are two rational numbers, then + =        .
     b     d                               b d   bd

Mixed Numbers

Although our discussion of rational numbers has been completely general, we
have explicitly avoided discussing the case where the numerator of the
                                                                          a
rational number is greater than the denominator i.e. a rational number ,
                                                                          b
where a > b. Such rational numbers are called improper fractions. Consider
                        11
the following example:     . Geometrically (using circles again) this number
                         4
                                                       1
represents 11 pieces of a circle where each piece is     of a circle:
                                                       4

Pierce College                                                            MAP
Math 215 Principles of Mathematics                                               173




We can ‘assemble’ these pieces back into circles and parts of circles by
rearranging the pieces:




                                                                3
We see that in this case, we end up with 2 whole circles plus     of a circle.
                                                                4
We therefore must have that:



                                   11      3
                                      = 2+
                                    4      4

That is, these two quantities are equivalent, representing the same number.
We therefore see that when the numerator is greater than the denominator
in a rational number, the rational number represents one or more ‘whole
                                                                              11
parts’ plus a ‘fractional part’. Instead of expressing a rational number like
                                                                               4
       3
as 2 + , we drop the ‘+’ sign between these numbers and write them as a
       4
                  3
single number, 2 . This form of the number is called a mixed numeral or
                  4
mixed number.

                               11       3
It should be noted that although  and 2 equivalent i.e. represent the
                                4       4
same number, the mixed number format is preferred since it provides us

Pierce College                                                                   MAP
Math 215 Principles of Mathematics                                          174


with ‘additional; information’. In particular it tells us that the number
represents two whole parts plus a fractional part. Thus for example, saying
that we have eleven-fourths of a cake does not provide us with quite the
same information as saying we have two whole cakes plus three-quarters of a
cake. For this reason, we ‘require’ that all rational numbers having numerator
greater than their denominator be expressed as mixed numerals.

Caution must be exercised about the meaning of mixed numbers. It is
                            3             3         3
important to recall that 2 means 2 + , not 2 × . Mathematically if we
                            4             4         4
                    3
wish to write 2 × we must explicitly use the multiplication sign or
                    4
                                                                    3      3
parenthesis. The following forms are standard and mean 2 times : 2 × ,
                                                                    4      4
    3         3
 2 ⋅ , and 2   . If the multiplication sign or parenthesis are absent then
    4        4
   3                               3
 2 is always interpreted as 2 + .
   4                               4

                                                                     3
Care must also be used for negative mixed numbers. The number −2       is
                                                                     4
                          3           3
always interpreted as −2 −  , not −2 + . That is there is an implicit set of
                          4           4
                                           3     3        3      3
parenthesis around the mixed number: −2 = −  2  = −  2 +  = −2 − . The
                                           4     4        4      4
                                                                      3
parenthesis are omitted as a notational ‘simplification’. However −2 is
                                                                      4
                          3
always interpreted as −2 − .
                          4

Converting Between Mix Numbers and Improper Fractions

                     3     11
Since the numbers 2    and    represent the same number, we can convert
                     4      4
between these two forms. To convert from a mixed number to an improper
fraction we note that any whole number can be written as a rational number
                                                 1     2     3
by placing the whole number over one. That is 1 = , 2 = , 3 = , and so on.
                                                 1     1     1
Using this fact we have:


Pierce College                                                              MAP
Math 215 Principles of Mathematics                                             175


                         3     3 2 3 4 2 3 8 3 11
                     2     = 2+ = + = ⋅ + = + =
                         4     4 1 4 4 1 4 4 4 4

                                    a
In general if A is a whole number and is a rational number, then the mixed
                                    b
        a                                                            Ab + a
number A is equivalent to the improper fraction i.e. rational number        .
        b                                                              b

To convert an improper fraction to a mixed numeral in the example above
                                              11
where we represented the rational number         as 11 pieces of a circle, where
                                               4
each piece was one-quarter of a circle. In this case we repeatedly removed
(i.e. subtracted) 4 pieces of from the 11 to form 2 whole circles with 3
pieces remaining. Referring back to our discussion of division of whole
numbers, we see that this process exactly represents dividing the numerator
by the denominator, with the remainder represented as a fraction of the
whole circle that is ‘left over’. Generalizing this process we see that we can
convert a rational number (improper faction) into a mixed number by dividing
the numerator by the denominator. Thus:

                                            2 R3
                            11                       3
                               → 4 11 → 4     11 → 2
                             4                       4



Properties of Addition of Rational Numbers

Rational numbers have the same properties as integers with respect to
addition: closure, commutative, associative, additive identity, and additive
inverse properties.

                                                         a
Additive Inverse Property. For any rational number         , there exists a unique
                                                         b
        a                         a
number − , the additive inverse of , such that:
        b                         b
                          a  a         a a
                            +−  = 0 = − +
                          b  b         b b




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                         176


                     a     −a
Note the numbers −     and    are equivalent. This can be shown by
                     b     b
considering the following:

                             a  a
                              +−  = 0
                             b  b
                             a − a a + ( −a )
                              +   =           =0
                             b b       b

                                          a a
Just as for integers we also note that −  −  = . Rational numbers also
                                          b b
obey the addition property of equality:



                               a     c
Addition Property of Equality. Ifand are any two rational numbers
                               b     d
         a c        e                          a e c e
such that = , and if is a rational number, then + = + .
         b d        f                          b f d f



Subtraction of Rational Numbers

Subtraction of rational numbers is similar to addition. Mechanically we find a
common denominator (just like for addition) but subtract, instead of add,
the numerators. Conceptually we can model subtraction using the ‘take-away’
                                                          5
model or the number line. Thus, for example, if we have of a pizza and we
                                                          8
            3                               2
take away     of the original pizza we have left. More specially, if we cut a
            8                               8
pizza into eight pieces (slices) and at some point have 5 slices left, and we
                                                   2
remove another 3 slices we have 2 slices left, or of the original pizza.
                                                   8

With the number line model, we create a number line, but we make our unit
        1
spacing where b is the (common) denominator of our faction(s). Thus with
        b
                                                    1
the pizza problem, we would mark our number line at spacing and then draw
                                                    8
our segments and proceed as we did with whole numbers.

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                    177


We can also define subtraction in terms of addition just a we did for whole
numbers:



                                                  a      c
Subtraction of Rational Numbers (via Addition). If  and are any two
                                                  b      d
                     a c                               e
rational numbers then − is the unique rational number     such that
                     b d                               f
a c e
  = + .
b d f


As with integers, we can define subtraction of rational numbers in terms of
addition with the additive inverse of the subtrahend:



                                                                              a    c
Subtraction of Rational Numbers (via Additive Inverses). If                     and are
                                                                              b    d
                                    a c a −c
any two rational numbers then        − = +   .
                                    b d b d

Finally, we can mechanize subtraction using cross multiplication to compute a
common denominator:



                                                                               a    c
Subtraction of Rational Numbers (via Cross Multiplication). If                   and are
                                                                               b    d
                                    a c ad − bc
any two rational numbers then        − =        .
                                    b d   bd



Note: when dealing with mixed numerals we can first convert them to
improper fractions and then subtract, or we can deal with them directly
(which still requires finding a common denominator and possibly borrowing).
Thus for example:

                  1  3 16 11 4 16 3 11 64 − 33 31                         7
                 5 −2 = − = ⋅ − ⋅ =           =    =                 2
                  3  4 3 4 4 3 3 4       12     12                       12

                       Convert to      Find a        Subtract   Convert to
                       improper        common                   mixed
                       fractions       denominator              numeral
Pierce College                                                                        MAP
Math 215 Principles of Mathematics                                            178


Although this works, it requires converting back and forth between mixed
numerals and improper factions. We can avoid this by working directly with
the rational numbers as mixed numerals:




                    1                4                16                 16
                  5                   5             4                  4
                    3               12                12                 12
                    3                9                 9                  9
                 −2              −2                −2                 −2
                    4               12                12                 12
                                                                          7
                                                                       2
                        Find a            Borrow           Subtract      12
                        common
                        denominator



Estimation

In Lecture 6 we talked about estimation with whole numbers. The same
concepts and principles apply with rational numbers. In some cases we may
                                                                    1 1 1
wish to round to our rational numbers to ‘convenient’ fractions e.g. , , ,
                                                                    2 3 4
1 2 3
  , , , or 1. Thus for example, if a student answers 59 out of 80
5 3 4
                                                             59
questions correctly on an exam, then the student’s score is     which is
                                                             80
               60     3                                                  3
approximately     or . We estimate that the student answered about
               80     4                                                  4
of the questions correctly.

As with whole numbers we can use the same techniques to estimate sums and
differences of rational numbers. Thus for example we can estimate the sum
  9    3 11
3 +2 +       as 4 + 2 + 1 = 7. In this case we have rounded each rational
 10    8 12
number to the nearest whole number. If we wanted a more accurate
                                          1     1
estimate we could round to the nearest       or .
                                          2     4




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                            179


Lecture 11

Sections 5.3 and 5.4 – Multiplication and Division of Rational Numbers and
Proportional Reasoning

We now consider multiplication of rational numbers. Again we want the
rational numbers to be closed under multiplications and they should also have
all the properties of integers and whole numbers. To determine what we
mean by multiplication of rational numbers we use the repeated addition and
area models for multiplication. First, consider the special case of 3×¾ where
the rational number 3 is also a whole number. Using the repeated addition
model for multiplication we have:

                               3× ¾ = ¾ + ¾ + ¾

Graphically we can represent ¾ as 3 quarter sections of a circle. The
repeated addition of ¾ is then:




                                        =




                                                 9
From the figure we see that 3×¾ = ¾ + ¾ + ¾ =      = 2¼. We can also obtain
                                                 4
                                                                6 3 9
this result using addition since 3×¾ = ¾ + ¾ + ¾ = (¾ + ¾) + ¾ = + = . If
                                                                4 4 4
multiplication is commutative, then ¾ × 3 = 3×¾ = 2¼.



Now we consider the case when neither factor in the product is a whole
                                                   1 3
number or integer. Consider the following example: × . If we identify
                                                   2 5
these rational numbers with areas, like the text does, we can use the area
                                                3
model to determine the product. Thus suppose represents the amount of
                                               5



Pierce College                                                                MAP
Math 215 Principles of Mathematics                                          180


                                                   1
land on the Earth originally covered by forest and   represents the amount
                                                   2
of remaining forest. We can ask, how much of the Earth’s land surface is
currently covered by forest? If use a rectangle to represent the land
                                       3
surface of the Earth, then originally,   was covered by forest, and we
                                       5
                 3
therefore shade of the rectangle green:
                 5




Since half of the forests remain we divide the shaded portion of the figure
in half:




The three green squared represent the current amount of forest; we now
want to know, in relation to the entire amount of land surface, how large i.e.
how much area, does each of these squares represent. We can easily
determine this by dividing the remaining two rectangles in half, so that the
figure is composed of equal sized rectangles:




We count the total number of squares, and see that there are 10. Thus
 3
   of the Earth’s land surface is now covered by forest. This means that:
10

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                             181



                                       1 3 3
                                        × =
                                       2 5 10

Finally we note that in both cases, the product is the result of multiplying
numerator by numerator and denominator by denominator:




                               3 3 3 3 ⋅1 9
                             3× = × =     =
                               4 1 4 1⋅ 4 4
                             1 3 1⋅ 3 3
                              × =     =
                             2 5 2 ⋅ 5 10

We can generalize this result to obtain the following definition for
multiplication of rational numbers:



                                            a    c
Multiplication of Rational Numbers. If        and are any rational numbers,
                                            b    d
    a c a⋅c
then × =    .
    b d b⋅d



Multiplication of rational numbers has the same properties as the whole
numbers and integers:

Multiplicative Identity. The number 1 is the unique number such that for
                       a
every rational number :
                       b

                                       a a     a
                                  1⋅    = ⋅1 =
                                       b b     b




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                     182


                                                                  a b
Multiplicative Inverse. For any non-zero rational number           ,  is the unique
                                                                  b a
                                 a b     b a             b
rational number such that         ⋅ = 1 = ⋅ . The number   is the multiplicative
                                 b a     a b             a
                 a
inverse of         and is also called the reciprocal.
                 b


                                 a c      e
Distributive Property. If         , , and   are any rational numbers then:
                                 b d      f

a c e        a c  a e 
  +          =  ⋅ + ⋅ .
bd f         b d  b f 

                       a      c                                   a c
Equality Property. If    and    are any rational numbers such that =
                       b     d                                    b d
    e                              a e c e
and   is any rational number, then ⋅ = ⋅ .
    f                              b f d f

                         a                            a         a
Zero Property. If          is any rational number then ⋅ 0 = 0 ⋅ = 0 .
                         b                            b         b




                                    −2            4
Examples. Find the inverse of 4 and    . Since 4 = , the reciprocal (and
                                    5             1
                      1      −2                                5 −5
hence inverse of 4 is . For     we take the reciprocal to get     =    . We
                      4      5                                 −2 2
             1        −2 −5 10
note that 4 ⋅ = 1 and   ⋅  =    =1.
             4         5 2 10

                                       3
Example. A bicycle is on sale at         of its original price. If the sale price is
                                       4
$330, what is the original price?

                                                             3
Solution. Let x be the original price. Then the sale price is  x = 330 . We can
                                                             4
solve by using the multiplication property of equality and multiplying both
                            3
sides by the reciprocal of :
                            4
Pierce College                                                                         MAP
Math 215 Principles of Mathematics                                          183



                                4 3     4
                                  ⋅ x = ⋅ 330
                                3 4     3
                                x = 440



Multiplication with Mixed Numbers

Remember that mixed numbers are another way of writing rational numbers
when the numerator is greater than the denominator. If we wish to multiply
                         1       3
two mixed numbers, say 2 and 3 then we have:
                         2       4

                      1 3  1  3          1    3      Definition of mixed
                     2 ⋅3 =  2 ⋅3  =  2 + ⋅3 +       number
                      2 4  2  4          2    4
                                    3 1    3             Distributive
                            = 2⋅3 +  + ⋅3 + 
                                    4 2    4             Property

                                  6 3 3                      Distributive
                            = 6+ + +                         Property
                                  4 2 8
                                  12 12 3      27
                            = 6+ + + = 6+                    Equality Property
                                   8 8 8       8
                                    3                        Equality Property
                            = 6 +3+                          (Mixed Number)
                                    8
                                  3   3                      Definition of mixed
                            =9+ =9                           number
                                  8   8

In order to multiply we need to make use of the definition of mixed numbers
to express our mixed number as a sum of the whole and fractional parts,
then use the distributive property to multiply, and then convert the result
back into a mixed numeral. We can simplify this process by converting our
numbers in to fractional form, multiplying, then converting the result back to
a mixed numeral:




                         1 3  5   15   75  3
                        2 ⋅3 =  ⋅  =   = 9
                         2 4 2  4   8      8




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                          184


Division of Rational Numbers

Recall, that for whole numbers 6 ÷ 2 means breaking-up or ‘dividing’ a set of 6
                                                                    1
elements into sets each having 2 elements. Using this concept 6 ÷ means
                                                                    2
breaking up a set of six elements such that the resulting sets each have half
an element. Graphically we have:




Counting the number of elements after we divide the elements of the set,
we see that we have 12 ‘pieces’ or 12 sets each with ½ of an element from
the original set. Thus we have that:



                                          1
                                   6÷       = 12
                                          2

If we us the definition that division is the inverse of multiplication then
    1
 6 ÷ is the number such that when this number is multiplied by ½ the result
    2
is 6:



                                          1
                                   12 ⋅     =6
                                          2




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            185


We could also use the repeated subtraction model and subtract ½ from 6
and count the number of times (12) we need to do this until we reach zero.
Next we consider the problem of dividing two rational numbers, say
 3 1                                              3
  ÷   . In this case we wish to divide or break-up (of a circle) into
 4 8                                              4
1
  pieces:
8




                   3                  1
After dividing the   of a circle into   pieces and counting the number of
                   4                  8
pieces we see that we have 6 pieces. Therefore:

                                  3 1
                                  ÷  = 6
                                  4 8

                      1   3
Again we note that   ⋅ 6 = . Generalizing we get the following result for
                      8   4
division of rational numbers:



                               a     c                               c
Division of Rational Numbers. If and are any rational numbers and
                               b     d                              d
                a c e               e
is not zero then ÷ =  if and only if is the unique rational number such
                b d f               f
      c e a
that ⋅ =
      d f b




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                186



Although the above definition provides us with what we mean by division of
rational numbers, it provides little help in finding the quotient. To do this we
must develop a procedure or algorithm for computing the quotient of two
rational numbers. To do this we use the definition above and note that:

                                      a
                                        = a ÷b
                                      b

                           a
since the rational number    represents dividing or breaking-up a set into b
                           b
pieces and then taking a of those pieces (see previous lecture). Then:

                                     a      1 a 1
                            a ÷b =     = a ⋅ =  ⋅ 
                                     b      b 1 b

This means that dividing a by b is equivalent to multiplying a by the
                       1
reciprocal of b i.e. by . With rational numbers as part of our system of
                       b
numbers we can ‘replace’ division by multiplication by the reciprocal of the
divisor. If we generalize this result to the rational numbers themselves then
we have a practical procedure i.e. an algorithm for multiplying rational
numbers:

                                                         a     c
Algorithm for Division of Rational Numbers. If             and   are any rational
                                                         b     d
                 c                  a c a d a⋅d
numbers and        is not zero, then ÷ = ⋅ =    .
                 d                  b d b c b⋅c



Example. There are 35½ yards of material available to make towels. Each
                3
towel requires yard of material. How many towels can be made? How much
                8
material is left over?

                                                                            3
Solution. To find the number of towels we can make we divide 35½ by           :
                                                                            8




Pierce College                                                                    MAP
Math 215 Principles of Mathematics                                                  187


                                   1 3 71 3 71 8 568      2
                                 35 ÷ = ÷ = ⋅ =      = 94
                                   2 8 2 8 2 3    6       3

                     2
Since we can’t make    of a towel, we can make 94 towels and we will have
                     3
                                    2
enough material left over to make     of a towel. The amount of material left
                                    3
                  2 3 6 1
over is therefore ⋅ =       = yard.
                  3 8 24 4

Estimation and Mental Math with Rational Numbers

Estimation and mental math strategies developed for whole number can also
be used with rational numbers. We demonstrate by example.

                                             1   1         4
Example. Use mental math to find (12 ⋅ 25 ) ⋅ , 5 ⋅12 , and ⋅ 20 .
                                             4   6         5

                      1                                1
Solution. For (12 ⋅ 25 ) ⋅
                        we note that multiplication by is equivalent to
                      4                                4
diving by 4. Using commutative and associative properties we have:

               1               1           1
(12 ⋅ 25 ) ⋅     = ( 25 ⋅12 ) ⋅ = 25 ⋅ 12 ⋅  = 25 ⋅ (12 ÷ 4 ) = 25 ⋅ 3 = 75 .
               4               4           4

     1
For 5 ⋅12 we use the definition of mixed numbers and the distributive and
     6
commutative properties:

 1          1              1
5 ⋅12 =  5 +  ⋅12 = 5 ⋅12 + ⋅12 = 60 + 12 ÷ 6 = 60 + 2 = 62
 6          6              6

    4
For   ⋅ 20 we can break up the fraction as a product of the numerator and
    5
denominator, use the associative and commutative properties:

4         1               1                 1
  ⋅ 20 =  4 ⋅  ⋅ 20 = 4 ⋅  ⋅ 20  = 4 ⋅  20 ⋅  = 4 ⋅ ( 20 ÷ 5 ) = 4 ⋅ 4 = 16
5         5               5                 5



Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                         188


                   1 8        5   1
Example. Estimate 3 ⋅ 7 and 24 ÷ 4 .
                   4 9        7   8

     1 8                        1                        8
For 3 ⋅ 7 we round noting that 3 is approximately 3 and 7 is
     4 9                        4                        9
                       1 8
approximately 8. Thus 3 ⋅ 7 is approximately 3×8 = 24.
                       4 9

      5   1                                         5   1
For 24 ÷ 4 we use compatible numbers and estimate 24 ÷ 4 as 24 ÷ 4 = 6 .
      7   8                                         7   8

Exponents

Recall that in Lecture 5 we defined an as a multiplied by itself n times. The
number a was taken to be a natural, whole, or integer number and n was a
natural number. We can extend what we mean by ‘number’ to include the
rational numbers and our previous definition applies to all numbers, including
rational numbers.

Exponential Notation. For any number a and any whole number n:

                                an = a⋅ a⋅ a⋅ . . . ⋅a
                                          n − tim e s



For n = 0, and a ≠ 0, we define a0 = 1.



With the addition on rational numbers we can also extend our definition and
allow n to be negative. We define:

Integer Exponent. Let a be any non zero number and n any integer. Then:

If n > 0                       an = a⋅ a⋅ a⋅ . . . ⋅a
                                          n − tim e s



If n = 0                        a0 = 1
                                           −n
                                     1
If n < 0                        an =  
                                     a


Pierce College                                                             MAP
Math 215 Principles of Mathematics                                              189


Note when n is less than zero, -n is greater than zero and therefore for a
negative exponent, we multiply the inverse of a by itself –n times. Using the
definition of exponents we can derive or show the following properties:

Properties of Exponents. Let a and b be any rational numbers and n and m be
integers, Then:

1. a n ⋅ a m = a n + m
    an
2. m = a n − m , a ≠ 0
   a
          m
3. ( a n ) = a n⋅m
         n
   a  an
4.   = n , b ≠ 0
   b  b
              n
5. ( a ⋅ b ) = a n ⋅ b n
         −n            n
   a              b
6.              =   , a ≠ 0 and b ≠ 0
   b              a



Proportional Reasoning

Numbers characterizing two quantities are often combined so as to quantify
the relationship in terms of a single number. The characterizing numbers are
combined using division in fractional form with one number forming the
denominator and the other forming the numerator. The result is called a
ratio and is often expressed in raw form i.e. not simplified or expressed as a
mixed numeral. Although the term ratio is used to express the relationship
between any two quantities in fractional form, if the quantities have
different units, the ratio is called a rate. Ratios are also written using a
colon ‘:’ to separate the numbers. Thus for example:

    •    A ratio of 1:3 for boys to girls in a glass means that the number of
                 1
         boys is that of girls; that is there is 1 boy for every 3 girls;
                 3
    •    There us a 2:3 ratio of Democrats to Republicans on a legislative
         committee meaning that there are 2 Democrats for every 3
         Republicans on the committee.



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                             190


    •    A person is diving her car at 65 miles per hour will travel 65 miles in
         one hour; in this case the number 65 is not expressed in fractional
         form but as a single number. This ratio is an example of a rate since
         the quantities in question (distance and time) have different units.

Proportions

Two ratios are said to be proportional if and only if the fractions
representing them are equal. The equal ratios are said to form a proportion.
In general we have:

Proportions. If a, b, c, and d are all real numbers and b≠ 0 and d ≠ 0, the the
           a c
proportion = is true if and only if ad = bc.
           b d



                 2      8                       2 8
Example.           and    are proportional since =   or 2·12 = 24 = 3 ·8.
                 3     12                       3 12

Proportions can be use to solve problems where the relation between the
parameters involved are multiplicative. Consider the following problem: Carl
types 8 pages for every 4 pages that Dan can type. If Dan has typed 12
pages, how may pages had Carl type? In this case we can express the number
                                                     8
of pages that Carl types versus Dan as a ratio: 8:4 =  . This means that
                                                     4
Carl types twice as fast as Dan. Thus at any time we expect that Carl will
have typed twice as many pages as Dan. We can therefore establish a
proportion between the pages typed:

                                          8 x
                                           =
                                          4 12

with x representing the (unknown) number of pages Carl has typed. Using
algebraic properties we can solve for x to obtain 24 pages. Care should be
taken when making this determination. If the relationship between the
problem parameters is not multiplicative we cannot establish a proportion to
help us solve the problem. Consider the following example: Allie and Betty
type at the same speed. Allie started first. When Allie had typed 8 pages,
Betty had typed 4 pages. When Betty had typed 10 pages, how many pages

Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                              191


had Allie typed? Note: in this case Betty and Allie type at the same speed.
This means that since Betty starts after Allie she will always remain behind
by the number of pages Allie had type before she started. Thus from the
problem, Allie is four pages ahead of Betty when we are given the first count
of their pages. She will therefore remain 4 pages ahead throughout since
both type at the same speed. We therefore have: x = 4 + 10 where x is the
number of pages Allie has typed. Again we can solve this to obtain x = 6. In
this case the relationship between the parameters is additive, not
multiplicative and therefore we cannot use proportions and ratios to help us
solve this problem.



In practical applications the ratio of two quantities is often a constant. For
example the ratio of the miles (distance) traveled in a car and the number of
gallons of gasoline used by the car in traveling that distance is a constant,
say 28:

                                     m
                                       = 28
                                     g

where m represent the miles travels and g the amount of gasoline used in
gallons. This ratio is independent of both the miles traveled and the number
of gallons used. That is, if we keep track of the miles we drive in the car
and the number of gallons of gasoline used to travel those miles, the ratio
will always be 28 i.e. be constant. When two quantities are related in this
way we say they vary proportionally with the constant of proportionality
given by their (constant) ratio. Thus in our example the constant of
proportionality is 28 and m varies proportionally with g. We therefore make
the following definition:

Proportionality. If the variables x and y are related in the equality y = kx,
    y
 k =  , then y is said to be proportional to x and k is the constant of
    x
proportionality between x and y.

It is important to note that the numbers involved in ratios do not need to be
                                                    7
integers (and often are not). Thus for example, if    of the population
                                                   10

Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                              192


                                     3
exercises regularly and the other      do not, the ratio of the population that
                                    10
                                            7 3
exercises regular to those that do not is     : . When expressed as a
                                           10 10
fraction this result yields non-integer numbers in both numerator and
denominator.

Units are also important when using proportions to solve problems. For
example, if a turtle travels 5 inches every 10 seconds and we ask how long
will it take for the turtle to travel 15 feet, we need to use a consistent set
of units. Thus we need to either convert 5 inches to feet, or 15 feet to
inches when establishing the proportion used to solve the problem:

                                     5 180
                                       =
                                    10   x

where we have chosen to convert 15 feet to 180 inches. It will therefore
take the turtle 360 seconds or 6 minutes to travel 15 feet.

Other problems involving proportions can be solved by finding the constant
of proportionality. The text calls this the scaling strategy. Thus consider
the following problem: Is it better to by 12 tickets for $15.00 or 20 tickets
for $23.00? If the price of tickets is the same, then the ratio of cost to
number of tickets purchased will be constant. If there is a discount for
buying more tickets then this ratio will not be constant. In this case, the
‘constant’ of proportionality is the cost per ticket:

15 5      1                                            23    3
   = = 1 dollars per ticket for the 12 tickets and        = 1 dollars per
12 4      4                                            20    20
                                  1 5
ticket for the 20 tickets. Since =      , the cost per ticket is smaller i.e.
                                  4 20
 3    5
   <    if 20 tickets are purchased and therefore the ratio of cost to
20 20
number of tickets purchased is not constant. A discount is applied for
purchasing more tickets. We say that buying 20 tickets is a ‘better deal’
because we get more tickets for our money. The ratio of cost to amount
purchased is called the unit rate. In general, the lower the unit rate, the
better the deal.



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                          193


Example. Kerry, Paul, and Judy made $2520 for painting a house. Kerry
worked 30 hours, Paul 50 hours, and Judy 60 hours. If they divide the money
in proportion to the hours worked, how much did each earn?

Solution. We assume a constant unit rate. That is, each person makes the
same amount per hour. If x represents this rate, then we must have:

                          30x + 50x + 60x = 2520

or

                                140x = 2520.

Solving we get that x = 18. This means that each person earns 18 dollars per
hour, so Kerry gets 30·18 = $540, Paul gets 50·18 = $900, and Judy gets
60·18 = $1080.

Proportional relations can also be established in more than one way. For
example in the turtle problem we could have placed the time in the
numerator and distance in the denominator when writing the proportional
relationship between time and distance:

                                  10   x
                                     =
                                   5 180

Note that we still get x = 6 minutes for the answer, so this choice is
arbitrary or more importantly these proportions are equivalent. We also
note that we could have also established a proportion by forming ratios of
distance to distance and time to time:

                                   x 180
                                     =
                                  10   5

Again we find that x = 6 minutes and that the proportions are equivalent.
Generalizing we get the following results:




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                        194


                                      a       c                      a c
Equivalency. For any rational numbers    and    with a ≠ 0 and c ≠ 0, = ,
                                      b       d                      b d
              b d                                             a c
if and only if = . Likewise, if b ≠ 0, c ≠ 0, and d ≠ 0 and if = if and
              a c                                             b d
        a b
only if = .
        c d

Scale Drawings

An import application of ratios and proportions are in scale drawings and
maps. For example, if the scale is 1:300, then the length of 1 cm in such a
drawing represents 300 cm or 3m in true or actual size. The scale is the
ratio of the size of the drawing to the size of the actual object. In maps
different units are used for the scale. For example one inch on the map may
represent 10 miles. Two determine the actual distance between two points
(e.g. cites) we measure the distance on the map in inches and multiply by the
scale. Thus for example if the distance is 5½ inches on the map, the actual
distance is 5½ × 10 = 55 miles.




Pierce College                                                            MAP
Math 215 Principles of Mathematics                                            195



Lecture 12

Sections 6.1, 6.2 and 6.3 – Decimals and Operations and Decimals; Non-
terminating Decimals

Decimal comes from the Latin word decem, meaning ten. Mathematically, it
refers to a base-ten method for representing rational (and irrational)
numbers by extending the place value system used to write natural and whole
numbers. The most ubiquitous application, and perhaps the first encountered
by most people, is its use associated with money. The place value system can
be extended to represent rational numbers in a procedure that is analogous
to extending the number line to accommodate negative numbers. In this
case however, a marker analogous to zero on the number line is needed to
separate left from right. This marker is the decimal point which is
symbolically denoted using a period ‘.’. Note: in Great Britain and other
countries, a comma is used instead and the period is used like a comma is
used in the United States, to separate groups of three digits.

Digits to the left of the decimal point have the same place values as with
whole numbers, with the first (or digit directly next and to the left of the
decimal point) in the ones place, and subsequent digits in the tens, hundreds,
thousands, etc. places. We note as we move to the left, the place value
increases by a power of ten. Or conversely if we move to the right, each
place value decreases by a power of ten. If we continue this decrease as we
move past the ones place and the decimal point, then the first digit to the
                                                           1
right of the decimal point has a place value of 10-1 or       . The next place to
                                                          10
                                                          1
the right will therefore have a place value of 10-2 or        , and so on. The
                                                        100
table below shows several place values on both the left and right of the
decimal point. When reading a decimal number the decimal point is read as
the word ‘and’ and it is therefore given this ‘value’ in the table. Place values
to the right of the decimal point have the same names as their
corresponding place value to the right of the decimal point with ‘ths’
appended to the name. Thus for example, the first place to the right is the
                                                            1
tenths-place corresponding to the place value 10-1 or          .
                                                           10


Pierce College                                                                MAP
Math 215 Principles of Mathematics                                                          196




                             Place Value Chart with Decimal Places
       4              3     2
  10             10       10    101 100 Decimal 10- 10-2 10-3            10-4       10-5
                                                     1

10,000 1000               100   10   1     ‘and’     1     1      1        1         1
                                                    10    100   1000    10, 000   100, 000
    T            T        H     T    O       .      T      H      T        T            H
    e            h        u     e    n              e      u      h        e            u
    n            o        n     n    e              n      n      o        n            n
                 u        d     s    s              t      d      u                     d
    T            s        r                         h      r      s        T            r
    h            a        e                         s      e      a        h            e
    o            n        d                                d      n        o            d
    u            d        s                                t      d        u
    s            s                                         h      t        s            T
    a                                                      s      h        a            h
    n                                                             s        n            o
    d                                                                      d            u
    s                                                                      t            s
                                                                           h            a
                                                                           s            n
                                                                                        d
                                                                                        t
                                                                                        h
                                                                                        s

As the table implies, decimals can be written in expanded form using place
values with negative exponents for the powers of ten. This is the logical
extension of expanded notation for whole numbers. Thus for example
12.618437 becomes:

           1 ×101 + 2 ×100 + 6 ×10-1 + 1 ×10-2+ 8 ×10-3 + 4 ×10-4 + 3 ×10-5 + 7 ×10-6




Pierce College                                                                              MAP
Math 215 Principles of Mathematics                                                              197


Note that by using the fractional representation for the powers of ten with
negative exponents we can also represent this decimal number in fractional
notation. Thus:

    12.618437 = 1× 101 + 2 × 100 + 6 ×10 −1 + 1×10 −2 + 8 ×10 −3 + 4 × 10−4 + 3 × 10−5 + 7 × 10−6
                                 6   1    8      4       3          7
                   = 10 + 2 +      +   +     +       +        +
                                10 100 1, 000 10, 000 100, 000 1, 000, 000



Since 1,000,000 is a common denominator for all the fractions in this sum,
we can add using this denominator to obtain:

                          600, 000     10, 000       8000        400          30          7
12.618437 = 12 +                     +          +           +           +           +
                         1, 000, 000 1, 000, 000 1, 000, 000 1, 000, 000 1, 000, 000 1, 000, 000
                          618, 437
                 = 12 +
                         1, 000, 000
                       618, 437
                 = 12
                      1, 000, 000

This shows that the decimal number 12.618437 and the mixed numeral
    618, 437
12             represent the same number, or in more formal terms, are
   1, 000, 000
equivalent. In particular any decimal number can be converted to its
equivalent rational number by writing it in expanded notation, replacing
negative powers of ten by their equivalent fractional forms, find a common
denominator, and then adding.

This also suggest that any fraction whose denominator is a power of ten can
be converted using the reverse process. Thus for example:

                   56 50 + 6 50    6   5  6
                      =     =    +   = +     = 5 ×10 −1 + 5 ×10 −2 = 0.56
                  100   100   100 100 10 100

            205     200 + 5   200     5     2     5
                  =         =     +       =   +        = 2 × 10−2 + 5 × 10−4
           10, 000 10, 000 10, 000 10, 000 100 10, 000
                      = 0 × 10−1 + 2 × 10 −2 + 0 × 10−3 + 5 × 10−4
                      = 0.0205


Pierce College                                                                                  MAP
Math 215 Principles of Mathematics                                       198


Thus we can convert any decimal number into a rational number and any
rational number whose denominator is a power of ten into a decimal number.
This further suggests that decimal numbers and rational numbers are
equivalent. That is we can convert any decimal number into a rational number
and any rational number (regardless if its denominator is a power of ten or
not) into a decimal number. We already know the former is true, therefore
we must show the later is also true to have equivalency. We will discuss this
topic more fully in the next lecture. For now we note that in the case where
the prime factors of the denominator are only powers of 2 and 5, we can
convert the denominator to an equivalent fraction whose denominator is a
power of ten, and proceed as above. We therefore have the following result:

                              a
Theorem. A rational number      in simplest form can be written as a
                              b
terminating decimal if and only if the prime factorization of the denominator
contains no primes other than 2 or 5.



We note that terminating decimals are decimal numbers that can be written
with a finite number of digits to the right of the decimal point. Thus for
example consider:

                          3 3 2 6
                           = ⋅ = = 0.6
                          5 5 2 10

                          7 7 125 875
                           = ⋅   =     = 0.875
                          8 8 125 1000

                          11   11 4   44
                             =    ⋅ =    = 0.044
                          250 250 4 1000

These examples satisfy the theorem and therefore can be written as
decimal numbers. It should be noted that not all rational numbers satisfy
                                 2
the theorem. Thus for example       cannot be expressed as an equivalent
                                11
fraction with a denominator that is a power of 10 and we therefore cannot
convert it to a decimal number using the same method as shown in the
examples above.


Pierce College                                                            MAP
Math 215 Principles of Mathematics                                         199



It should be noted that all terminating decimals can be easily located on the
number line. Recall that by placing a number on the number line we determine
its order relative to other numbers. That is whether the number is less than
or greater than other numbers. Placement of terminating decimals on the
number line is accomplished by converting them to rational numbers and
using what we know about the placement of rational numbers on the number
                                                       56
line to place these numbers. Thus for example: 0.56 =      and we place it as
                                                      100
follows:




             0    1     2   3          55    56     57         99   1
                                 ··                      ··
                 100   100 100        100   100    100        100


                                            0.56

                           50   55   56   57   60
From this we see that         <    <    <    <    . Thus we have that
                          100 100 100 100 100
 50    56    60
    <      <     or 0.5 < 0.56 < 0.6. This result suggest that we can order
100 100 100
decimal numbers without first converting them to rational numbers. We
note that 0.5 < 0.56 < 0.6 is equivalent to 0.50 < 0.56 < 0.60. Comparing the
digits in each number starting in the tenths place we note that 0.60 > 0.56
since 6 > 5. Likewise 0.50 < 0.56 since comparing digits we note in the tenths
place both numbers contain a 5. However in the hundredths place 6 > 0 and
therefore 0.56 > 0.50. We therefore make the following generalization:

In order to compare two decimal numbers:

    1. Line up the numbers by place value.
    2. Start at the left and find the first place where the face values of the
       digits are different.
    3. Compare these digits. The number containing the greater face value in
       this place is the greater of the two numbers.



Pierce College                                                             MAP
Math 215 Principles of Mathematics                                         200


Thus for example determine the order of 0.532 and 0.54. First we line up
the two numbers:



                                      0.532
                                      0.54

Starting at the left and moving towards the right, the first digits that are
different in both numbers occurs in the hundredths place:

                                      0.532
                                      0.54

Since 4 > 3 we have that 0.54 > 0.532 or 0.532 < 0.54.

Multiplying Decimals

We can multiply two terminating decimal numbers by first converting them
to fractions, multiplying, and then converting them back. Thus for example:



                                      462 24 11, 088
                       4.62 × 2.4 =      ×  =        = 11.088
                                      100 10 1, 000

The denominator of the result is a power of ten and is related to the
denominators of the factors as follows:

10n × 10m = 10n+m where n is the power of ten in the denominator of the first
factor, m is the power of ten in the denominator of the second factor and
n+m is the power of ten of denominator of the product. In the example
above, n = 2, m = 1, and n+m = 3. This means that the placement of the
decimal point in the product is always determined by this relationship, since
dividing by a power of ten is equivalent to moving the decimal point to the
left by the number of places represented by the power of ten. We
therefore have the following result which provides a ‘short-cut’ method for
multiplying two terminating decimal numbers:




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                       201


If there are n digits to the right of the decimal point in one number and m
digits to the right of the decimal point in a second number, multiply the two
numbers ignoring the decimals and then place the decimal point so that there
are n+m digits to the right of the decimal point in the product.

Thus for example:

                             1.43    (2 digits after the decimal point)
                            × 6.2    (1 digit after the decimal point)
                         ____________
                              286     (multiply ignoring decimal points)
                            858
                         ___________
                            8.866     (add and place decimal point so
                                       there are 3 digits to the right of
                                       decimal point)

Scientific Notation

Scientific notation is an alternate method or notation for representing
decimal numbers. It was developed as an efficient method for writing either
very small or very large numbers and is especially useful when both very
large and very small numbers appear together in formulas, equations, and
computations.

In scientific notation, a positive number is written as the product of a
number greater than or equal to one and less than ten and an integer power
of 10.

The following are examples of numbers written in scientific notation:

                                  8.3 X 108

                                 1.2 X 1010

                                 7.84 X 10-6




Pierce College                                                           MAP
Math 215 Principles of Mathematics                                             202


Note that the numbers 0.43 X 109 and 12.3 3 X 10-5 are not considered to be
in proper scientific notation since 0.43 is less than one and 12.3 is greater
than 10. To write a number such as 934.5 in scientific notation we divide by
the power of ten than makes the number fall in the range between 1 and 10
and then multiply by this same power of ten:

                                 934.5
                       934.5 =         × 100 = 9.345 × 100 = 9.345 × 102
                                  100

In this case, the division by 100 moves the decimal point two places to the
left and multiplying by 100 insures that the value of the number remains
unchanged. We then re-write 100 as a power of 10 i.e. 102 to complete the
‘conversion’. If the number is less than one, such as 0.000078, we multiply
by the power of ten that makes the number greater than one but less than
10, and then divide by this same power of ten:

                              0.000078 × 100, 000     7.8    7.8
                 0.000078 =                       =         = 5 = 7.8 × 10−5
                                   100, 000         100, 000 10

This amounts to moving the decimal place 5 places to the left and multiplying
by 10-5. Numbers in scientific notation are easy to manipulate using the laws
of exponents we discussed earlier in the course. For example, the
acceleration, g, of Earth’s gravity can be calculated using Newton’s Law of
Gravity:

                                                  M
                                           g =G
                                                  R2

Where G is the universal gravitation constant, M is the mass of the Earth,
and R is the radius of the Earth. The value of G is 6.67 × 10−11 N m2 kg−2 and
the values of M and R are 5.9742 × 1024 kg and 6.3781 × 106 m, respectively.
Substituting these numbers into the above equation gives:




Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                                    203


                                     -11    5.9742 × 1024
                     g = 6.67 × 10                           2
                                           ( 6.3781 × 10 )
                                                       6


                                            5.9742 × 10 24
                       = 6.67 × 10-11
                                        4.068015961 × 1013
                       = ( 6.67 × 10-11 )( 5.9742 × 1024 )( 2.4582 × 10-14 )
                       = 6.67 ×5.9742×2.4582×10-11× 1024 × 10-14
                       = 97.954×10-1
                       = 9.7954

Dividing Decimals

As we did with multiplication we can divide a terminating decimal number by
an integer by first converting both numbers to fractions, dividing, and then
converting them back. Thus for example:

                                    7545 3 7545 1 2515
                      75.45 ÷ 3 =       ÷ =    × =     = 25.15
                                    100 1 100 3 100

By writing the dividend as a fraction whose denominator is a power of 10 and
using the associative property of multiplication, the division is changed into a
division of whole numbers (7545 ÷ 3) times a power of 10 (10-2) .

We can also view this process as one of converting the decimal number into a
whole number by multiplying by the appropriate power of ten and then
dividing by the same power of ten after performing the (now whole number)
division. Thus using the previous example:



                                       100                   1   2515
                 75.45 ÷ 3 = 75.45 ×       ÷ 3 = 7545 ÷ 3 ×    =      = 25.15
                                       100                  100 100



When the divisor is not an integer we can convert it to an integer by
multiplying and dividing by the appropriate power of ten. Thus for example:

                                  100             32            100            1
1.2032 ÷ 0.32 = 1.2032 ÷ 0.32 ×       = 1.2032 ÷     = 1.2032 ×     = 120.32 × = 120.32 ÷ 32
                                  100            100             32           32


Pierce College                                                                        MAP
Math 215 Principles of Mathematics                                                  204


We can then use the method for dividing a terminating decimal number by an
integer to carry out the division and obtain the quotient. If we examine this
procedure we see that we can obtain the following generalization:

To divide two terminating decimal numbers, convert the divisor to a whole
number by moving the decimal point the required number of places to the
right so that it is converted to an integer. Move the decimal point in the
dividend the same number of places to the right. Then divide the two
numbers to obtain the quotient.

We can divide using the familiar division apparatus placing the divisor on the
left and the dividend inside the apparatus. The quotient is placed on top, its
decimal point aligned with the decimal point of the dividend underneath. Like
with multiplication, the actual division is performed ignoring the decimal
point. Thus for example:

                                                          .        101.3
                     13.169 ÷ 0.13 → 0.13 13.169 → 13 1316.9 → 13 1316.9
                                                                      13
                                                                           16
                                                                           13
                                                                            39
                                                                           39
                                                                                0
Mental Computation

Some of the tools used for mental computation with whole numbers can be
used to perform mental computation with decimals. Consider the following
examples:

1. Breaking and bridging
                                                                Add 9.2 and 0.48
                                    Add 4.5 and 0.7

          1.5+3.7+4.48 = 4.5+0.7+4.48 = 5.2+4.48 = 9.2 + 0.48 = 9.68


                 Combine 3 in 3.7
                                                  Combine 4 in 4.48
                 with 1 in 1.5
                                                  with 5 in 5.2


Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                     205




2. Using compatible numbers

   Decimal numbers are compatible when they add up to whole numbers:


             7.91
                                             12

             3.85

             4.09

        + 0.15
                                            + 4
       _________
                                             16




3. Making compatible numbers



 9.27 = 9.25 + 0.02
+3.79 = 3.75 + 0.04
_________________
        13.00 + 0.06 = 13.06



4. Balancing with decimals in subtraction



  4.63 = 4.63 + 0.03 = 4.66
- 1.97 = -(1.97 + 0.03) = - 2.00
___________________________________
                            2.66




Pierce College                                                         MAP
Math 215 Principles of Mathematics                                          206



5. Balancing with decimals in division



8 ÷ 0.25 = 8 ÷[ 0.25 × (4 ÷ 4)] = 8 ÷ (1 ÷ 4) = (8 × 4) ÷ 1 = 32 ÷ 1 = 32



Rounding Decimals

As with whole numbers and integers, it is often not necessary to know the
exact numerical answer to a problem involving decimal numbers. For example
if we want to know the distance to the moon or the population of New York
City, approximate answers will often suffice. The particulars of the problem
determine how to round the numbers involved. The rules for rounding remain
the same as with whole numbers and integers; however the place values are
now extended to include tenths, hundredths, thousandths, and so on. The
following examples illustrate rounding with decimal numbers:

7.   456   to    the   nearest hundredth is 7.46
7.   456   to    the   nearest tenth is 7.5
7.   456   to    the   nearest one is 7
7.   456   to    the   nearest ten is 10
7.   456   to    the   nearest hundred is 0

Estimating Using Rounding

Rounding is used to estimate an approximate answer to a problem. As noted
above, often an exact answer is not desired or we might want to obtain an
approximate answer either before or after solving a problem as a method
for checking the reasonableness of our answer. In each case we can round to
obtain an estimate for the actual answer. Consider the following example:

Karly goes to the grocery store to buy items that cost the following
amounts. She estimates the total cost by rounding each amount to the
nearest dollar and adding the rounded numbers:




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                        207


                            $2.39    →      2.00
                             0.89    →       1.00
                             6.13    →      6.00
                            4.75     →       5.00
                           + 5.05    →    + 5.00
                           _____          _______
                                            19.00



Round-off Errors

Round-off errors are typically compounded when computers are involved. For
example, if two distances are 42.6 miles and 22.4 miles and are rounded to
the nearest tenth, then the sum is 65 miles. To the nearest hundredth, the
distances might have been more accurately reported as 42.55 and 22.35.
The sum of these distances is 64.9 miles. Alternatively, the original
distances may have been rounded from 42.64 and 22.44. The sum is now
65.08 or 65.1, rounded to the nearest tenth. The original sum of 65 miles is
between 64.9 and 65.1, but the simple rounding process used is not precise.

Similar errors arise in other arithmetic operations. When computations are
done with approximate numbers, the final result should not be reported
using more decimal places then the number with the fewest decimal places.
In other words an answer cannot be more accurate than the least accurate
number used to find it.

Non-terminating Decimals

We have discussed the equivalency of decimal and rational numbers and
showed how to convert a decimal number into a rational number and how to
convert rational numbers into decimal numbers for the case where the
denominator is a power of ten or where the rational number can be
expressed as an equivalent fraction with a denominator that is a power of
ten. We now want to show that any rational number can be expressed as a
decimal number, even if its denominator is not a power or ten or it cannot be
expressed as an equivalent rational number whose denominator is a power of
ten.



Pierce College                                                            MAP
Math 215 Principles of Mathematics                                          208


To do this we go back to the concept of a rational number as division of the
numerator by the denominator (see Lecture 10). In this case we use decimal
division to convert the rational number to a decimal number. Thus for
example:

                                             0.875
                             7
                               → 8 7.000 → 8 7.000
                             8
                                             64
                                               60
                                               56
                                                40
                                                40
                                                    0



      7
Thus    is equivalent to 0.875. We can verify this by converting 0.875 back
      8
to a rational number:

                                       875 7 ⋅125 7
                            0.875 =       =      =
                                      1000 8 ⋅125 8

Repeating Decimals

Although the above technique always work for converting a rational number
into a decimal number, there is no ‘guarantee’ that the division will terminate
i.e. reach a point where the ‘remainder’ is zero. Thus consider the following
example:




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                         209


                                              0.181818
                         2
                           → 11 2.000000 → 11 2.000000
                        11
                                              11
                                               90
                                               88
                                                20
                                                11
                                                    90
                                                    88
                                                     20
                                                     11
                                                         90
                                                         88
                                                          2

In this case we see that the division does not terminate but rather repeats
with the original dividend ‘2’ appearing over and over again. The quotient
reflects this repetition with the digits 18 occurring over and over again.
       2
Thus     is equivalent to the decimal number 0.181818… where the ellipsis ‘…’
      11
indicates that the digits repeat in the indicated pattern ‘forever’. Such
numbers are called repeating decimals or non-terminating decimals. These
numbers can also be written using the ‘bar’ notation:

                                    2
                                      = 0.18
                                   11

Where the digits under the bar are repeated. We note that with repeating
decimals, that whenever a remainder reoccurs, the process repeats itself. In
                                a
general for any rational number it will take at most b-1 steps before a
                                b
repeating remainder occurs. Therefore a rational number may always be
represented either as a terminating decimal or as a repeating decimal.

Converting Repeating Decimals Back to Rational Numbers

At this point we have showed how to convert rational numbers to decimal
number (both terminating and repeating) and how to convert terminating

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                    210


decimals to rational numbers. To complete this cycle and show equivalency we
must also show how to convert repeating decimals to rational numbers.

We recall that to convert a terming decimal such as 0.55 to a rational
number, we expanded this number in powers of ten, computed the common
denominator, and wrote the number as a rational number:

                                                 5   5   50   5   55
                 0.55 = 5 ×10 −1 + 5 × 10−2 =      +   =    +   =
                                                10 100 100 100 100

We cannot use this approach with repeating decimals since there are not a
finite number of terms in the expansion in the first step of the procedure.
In order to eliminate this problem we need to write a repeating decimal
number in an equivalent form such that it contains a finite number of digits.
Consider the following example:

Let n = 0.5 then we note that: 9n = (10 − 1) n = 10n − n = 5.5 − 0.5 = 5 , where we
have used the distributive property. If we solve this ‘equation’ for n, then we
have:

                                          9n = 5
                                                5
                                            n=
                                                9
                                                5
                                          0.5 =
                                                9

Note: we can check our answer by dividing 5 by 9 to obtain 0.5 . Multiplying
by 10 and then subtracting 0.5 , eliminates the repeating part of the decimal
number leaving us with an integer result. Using the general property of
equivalence then allows us to find the rational number equivalent. Another
approach is to use following result for the summation of a geometric
sequence of numbers:

Sum of a Geometric Sequence. If S = a + ar + ar 2 + ar 3 + ... where 0 < r < 1,
          a
then S =      .
         1− r




Pierce College                                                                        MAP
Math 215 Principles of Mathematics                                                              211


Poof. We can prove this using a method similar to what we did for
converting 0.5 to rational form. Thus:



                         S = a + ar + ar 2 + ar 3 + ...
                   S − rS = ( a + ar + ar 2 + ar 3 + ...) − r ( a + ar + ar 2 + ar 3 + ...)
                           = ( a + ar + ar 2 + ar 3 + ...) − ( ar + ar 2 + ar 3 + ar 4 + ...)
                           =a
                 S (1 − r ) = a
                               a
                         S=
                              1− r

Using this result, we can convert 0.5 to rational form. We note that:

                         0.5 = 0.5 + 0.05 + 0.005 + 0.0005 + ...
                              = 0.5 + 0.5 × 10−1 + 0.5 ×10 −2 + 0.5 × 10−3 + ...

Letting S = 0.5 , a = 0.5, and r = 10-1, we have:

                                             0.5        0.5    0.5 5
                                  0.5 =          −1
                                                    =        =    =
                                          1 − 10      1 − 0.1 0.9 9

Now consider the repeating decimal 0.235 . In this case the digits 235
repeat over and over. We can still use the techniques above but note that we
must multiply by 1000 using the first technique and that r = 10-3 for the
second technique. Thus letting n = 1000 and S = 0.235 we have:

                    999n = (1000 − 1) n = 1000n − n = 235.235 − 0.235 = 235
                         235
                    n=       = 0.235
                         999

                                0.235        0.235    0.235 235
                          S=          −3
                                         =          =      =    = 0.235
                               1 − 10      1 − 0.001 0.999 999

Using these results we make the following generalization:




Pierce College                                                                                  MAP
Math 215 Principles of Mathematics                                           212


If the repetend (part of the number that repeats) is immediately to the
right of the decimal point, the rational number equivalent is the repetend
(written as an integer) over 10m-1 where m is the number of digits in the
repetend.

Now if the repeating part of the number does not immediately follow the
decimal such as with the number 2.345 , we convert this number to a form
where the repeating part does immediately follow the decimal point and use
the techniques above. Thus we note:

                                 1          1             23 1
                      2.345 =
                                10
                                  (     )
                                   23.45 =
                                           10
                                             (       )
                                              23 + 0.45 =   +
                                                          10 10
                                                               (0.45   )
We can now use either of our two techniques or the generalization above, to
                                               45
express 0.45 as a rational number. Thus 0.45 =    . Substituting into the
                                               99
result above, we have:



                   23 1          23 1 45 23 ⋅ 99 + 45 2277 + 45 2322
            2.345 =   +
                   10 10
                          (
                          0.45 =  )+ ⋅ =
                                 10 10 99     990
                                                     =
                                                        990
                                                               =
                                                                 990
                   1161 129     19
                 =      =    =2
                    495 55      55

Ordering Repeating Decimals

Since any repeating decimal can be written as a rational number in the form
 a
   where b ≠ 0, it can be placed on the number line in the manner we
 b
discussed earlier with terminating decimals. We can also compare two
repeating decimals directly to determine their order relative to each other.
Thus consider the numbers 1.3478 and 1.347821 . To see which number is
larger, write the decimals one under the other in their equivalent form
without the bars and line them up vertically at the decimal point. Thus:

                                       1.34783478…
                                       1.34782178…



Pierce College                                                               MAP
Math 215 Principles of Mathematics                                          213


If we compare the digits in each number starting on the left, we see that
the first place where the digits are not the same is in the hundred-
thousandths place where the top number has the digit 3 and the bottom
number has the digit 2. Since 3 > 2, this means that 1.3478 > 1.347821 .




Pierce College                                                              MAP
Math 215 Principles of Mathematics                                              214



Lecture 13

Sections 6.4, 6.5 and 2.5 – Real Numbers, Percents and Functions


Real Numbers

With the addition of the rational numbers, our number system is seemingly
complete with all the canonical operations closed (with the exception of
division by zero). With respect to the rational numbers and their
equivalents, the decimal numbers, this means that every number can be
expressed as either a terminating or repeating decimal number. However
this is not the case. That is, there are numbers, called the irrational
numbers, that are not of this form. That is, they cannot be expressed as
the ratio of two integers or equivalently, cannot be expressed as either a
terminating or repeating decimal number.

Historically the ancient Greeks are credited with the ‘discovery’
of the irrational numbers. In particular the philosopher/
mathematician Pythagoras is credited with their discovery. His
school or society believed that whole numbers provided the
foundation for everything and, as a consequence, irrational
                                                                  Pythagoras of Samos
number could not exist. As a result, they tried to suppress         569 BC - 475 BC
their discovery, and as legend goes, even drowned one of their
members, Hippasus, for revealing this ‘secret’.

Pythagoras’s discovery of the irrational numbers was related to square roots
and the Pythagorean Theorem. This theorem states that if a and b are the
lengths of the legs of a right triangle and c is the length of the hypotenuse
of the triangle then:
                                   a2 + b2 = c2

For triangles with legs of lengths 3 and 4, say, things work out with only
whole numbers:

                            32 + 42 = 9 + 16 = 25 = 52



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                        215



That is, the sum of the squares of the legs is 25 and there exists a whole
number 5 such that 52 is 25. The number 5 is called the principal square root
of 25 and is denoted as 25 . In particular, for any number we have the
following definition:

Principal Square Root. If a is any nonnegative number, the principal square
root of a (denoted a ) is the nonnegative number b such that b2 = a.

Note that any nonnegative number has two square roots, each the additive
inverse of the other. That is -5 and 5 are both square roots of 25 since (-
5)2 = 52 = 25. However, by convention, the positive square root is designated
as the principal square root.

Although the Pythagoras’s equation relating the lengths of the sides of a
right triangle can be satisfied using whole numbers for many right triangles,
this is not true in general. Consider the case where a = b = 1. In this case
we have that

                            12 + 12 = 1 + 1 = 2 = c2



Or c = 2 i.e. c multiplied by itself must be two. We note that c is clearly
not a whole number since 12 = 1 and 22 = 4 and since there are no whole
numbers between 1 and 2 there cannot be a whole number such that its
square is two. This means that if 2 is a rational number it must be the
                                              a
ratio of two whole numbers i.e. of the form , with b ≠ 0. We can use
                                              b
indirect reasoning to show that this is not the case. That is, suppose there
                                                       2
                                                a
are two whole numbers a and b, b ≠ 0, such that   = 2 . Then:
                                                b
                                     2
                                 a
                                   =2
                                 b
                                   a2
                                       =2
                                   b2
                                   a 2 = 2b 2
                                  a ⋅ a = 2⋅b ⋅b


Pierce College                                                            MAP
Math 215 Principles of Mathematics                                             216


Now by the Fundamental Theorem of arithmetic, the prime factors of a2 and
2b2 are the same. In particular, the prime number 2 must appear the same
number of times in the prime factorization of a2 as in 2b2. But since b2 = b·b,
if two appears n times in the prime factorization of b, then it will appear 2n
times in the prime factorization of b2 and therefore 2n+1 times in the prime
factorization of 2b2. That is, 2 appears an odd number of times in the prime
factorization of 2b2. However, if 2 appears m times in the prime
factorization of a, then it will appear 2m times in the prime factorization of
a2. This means that 2 appears an even number of time in the prime
factorization of a2. Since 2 appears an even number of times in the prime
factorization of a2 and an odd number of times in the prime factorization of
2b2 we have contradiction since the factorization of these two numbers is
not the same. This contradiction means that our original hypothesis that two
such whole numbers a and be could be found such that the square of their
ratio was 2 is incorrect. Hence 2 cannot be a rational number.

Other Roots

While on the subject of square roots, we can also extend this concept to
other roots of higher order. In particular we can define the nth root of a
number a, denoted n a as the number b such that bn = a. The number n is
called the index of the root and is omitted from the notation n a when n = 2.

We note that when the index is even a must be a positive number. If n is
odd then there is only one root if a is negative and that root is a negative
number. Thus for example we have:
                                        9 = ±3
                                      3
                                          −27 = −3
                                          4
                                              16 = ±2
                                  5
                                      1024 = 4
Estimating Square Roots

Suppose we wish to estimate the 2 . We know that 1 < 2 < 2 since 12 = 1 <
2 < 4 = 22. This gives us the value of the 2 to this nearest whole number or
units-place. We can refine or increase the accuracy of our estimate by
determining 2 to the nearest tenth. To do this we guess that 2 lies
halfway between 1 and 2 or near 1.5. Now (1.5)2 = 2.25 and therefore we

Pierce College                                                                 MAP
Math 215 Principles of Mathematics                                        217


have 1 < 2 < 1.5. We can refine this further by noting that 1.3 is
approximately half way between 1 and 1.5. Since (1.3)2 = 1.69 we have that
1.3 < 2 < 1.5. This can be refined further by noting that 1.4 is half way
between 1.3 and 1.5 and that (1.4)2 = 1.96. Thus 1.4 < 2 < 1.5. We have
therefore estimated 2 to the nearest tenth.

We can refine our answer further by estimating 2 to the nearest
hundredth. To do this we note that 1.45 is half between 1.4 and 1.5 and that
(1.45)2 = 2.1025. Thus 1.40 < 2 < 1.45. Since 1.43 is approximately half way
between 1.40 and 1.45 and (1.43)2 = 2.0449 we have Thus 1.40 < 2 < 1.43.
We can now try either 1.41 or 1.42. We note that (1.41)2 = 1.9881 and (1.42)2
= 2.0164 so that 1.41 < 2 < 1.42.

If we wish we can continue this process by estimating 2 to the nearest
thousandths, and so on. In particular we can estimate 2 to as many
decimal places as desired, although the computation effort increases with
the precision.

Real Numbers

If we include the irrational numbers into our number system we obtain the
set of real numbers, denoted R. Real numbers can be terminating, repeating,
or non-terminating and non-repeating. Every integer is a rational number as
well as a real number. Every rational number is a real number, but not every
real number is rational or an integer. We can construct the following ‘family
tree’ for the real numbers showing the relationship of all its subsets.


                                    Real Numbers


                    Irrational                    Rational

                                    Integers



                 Negative Numbers              Whole Numbers


                                 Zero             Natural Numbers

Pierce College                                                              MAP
Math 215 Principles of Mathematics                                            218


As mentioned earlier, the concept of fraction includes the ratio of any two
numbers where the denominator is nonzero. By number we mean real number.
                     1
                      
               , and   are all fractions. The canonical operations of
        5 1.7         3
Thus      ,
       3 6.2         2
                      
                     7
addition, subtraction, multiplication, and division are all defined for the set
of real numbers such that these operations have the properties of closure,
associatively, and so on that we have discussed for subsets such as the
integers. In particular we note that the following properties hold for the
real numbers:



                       Properties of the Real Numbers

Closure. For real numbers a and b, a + b and ab are unique real numbers.

Commutative. For real numbers a and b a + b = b + a and ab = ba.

Associative. For real numbers a, b, and c a + (b + c) = (a + b) + c and a(bc) =
(ab)c.

Identity. The number 0 is the unique additive identity and 1 is the unique
multiplicative identity such that for any real number a, a + 0 = 0 + a = a and
1·a = a·1 = a.

Inverse. For every real number a, there exists a unique additive inverse, -a,
such that a + (-a) = -a + a = 0. For every nonzero real number a, there exists
                       1                1 1
a unique real number , such that a ⋅   =   ⋅ a = 1 .
                       a                a a

Distributive. For any real numbers a, b, and c a(b+c) = ab + ac.

Denseness. For any real numbers a and b, a < b, there exists a real number c,
such that a < c < b.




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                         219


Rational Exponents

We can use the real numbers to extend the concept of exponents to include
square root and nth roots. In particular we define:

                                1
                               x =nx
                                n

                                      1           m

                               ( xm ) n = x n = n xm
Properties of Exponents

Using the definitions above we can show that the following properties hold
for rational exponents:

                                                 1
                                    x−n =
                                                 xn
                                             n
                                    ( xy )       = xn yn
                                             n
                                    x   xn
                                       = n
                                     y  y
                                       n m
                                    (x )         = x nm


Note n and m are real number here, not necessarily integers or whole
numbers.

Percents

Percents are used to present and summarize data and information. Common
applications include interest rates which are expressed in percent, statistics
and demographic data presentation, etc. The word percent comes from the
Latin phase per centum, meaning per hundred. The number one hundred
provides the basis for percents. An interest rate of 6 percent for example,
              6
means that       of the amount lent is paid each year as a fee for borrowing
             100
the money. The use of 100 as a common denominator for expressing
fractional amounts provides a standardized methodology for specifying and
relating information to the whole. In the case of interest rates, the fee
changed to the lender for borrowing a given amount of money, is related to

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                              220


the total amount borrowed. In general the following definition and notation
is used:

                      n
Percent. n% =            . The symbol ‘%’ indicates percent.
                     100

                                n                               1
Thus n% of a given quantity is     of the quantity. Thus 1% is     of the
                               100                             100
whole and 100% represents the entire quantity. The number n is not confined
                                                 200
to be less than 100. Thus for example 200% is        or 2 times the amount or
                                                 100
quantity in question. The number n need not be a whole number either. It
can be given as either a decimal number (terminating or repeating) or as a
rational number. To convert a number to percent, we use the above
definition, expressing the number in equivalent form as a rational number
with denominator 100. Thus for example:

                             3 3 ⋅ 25 75
                               =       =    = 75%
                             4 4 ⋅ 25 100
                                    0.03 ( 0.03) ⋅100    3
                             0.03 =      =            =     = 3%
                                     1      1 ⋅100      100


                 0.3 =
                            ( )
                       0.3 0.3 ⋅100 33.3
                           =             =       = 33.3%
                        1      1 ⋅100      100
                       1.2 (1.2 ) ⋅100 120
                 1.2 =     =             =      = 120%
                        1      1 ⋅100      100
                     1 1 ⋅100 100
                 1= =          =       = 100%
                     1 1 ⋅100 100
                      3 3
                 3 5 5    ⋅100 60
                   =   =             =      = 60%
                 5     1      1 ⋅100      100
                      2 2
                 2 3 3    ⋅100 66.6
                   =   =             =       = 66.6%
                 3     1      1 ⋅100       100
                        1   15              1500 
                        2    ⋅100                
                 2 =
                           7  7 
                                            =
                   1                               7        2
                              =                         = 214 % = 214.285714%
                   7      1        1 ⋅100        100         7



Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                       221


A number can also be converted to percent using a proportion. Thus for
example:

                                   3   n
                                     =
                                   5 100

We can solve for n, which gives the desired percent:

                               3    n
                                 =
                               5 100
                                       3
                               n = 100   = 60
                                       5

Thus:

                                   3
                                     = 60%
                                   5

Another method for converting to percent is to recall that 1 = 100%. Thus,
                      3
for example to convert :
                      4

                       3 3    3
                        = ⋅1 = ⋅100% = 3 ⋅ 25% = 75%
                       4 4    4

It should be noted that the % symbol is crucial when expressing a number as
a percent. Leaving off or omitting the symbol changes the value of the
                            1       1
number. Thus for example,     and % are different numbers:
                            2       2
                                1
                                  = 50%
                                2
                                      1
                                       
                                  %=  =
                                1      2   1
                                2     100 200

Likewise 0.01 = 1% is different that 0.01% = 0.0001.

So far we have showed how to convert a number to a percent. We can also
convert a percent to a number. This is important since information such as

Pierce College                                                           MAP
Math 215 Principles of Mathematics                                          222


interest rates which are usually given as percents, must be converted to
numbers before they can be used in calculations. To convert a number from a
percent to a number, we again use the definition and some simplification.
Thus for example:

                                 5      1
                           5% =      =     = 0.05
                                100 20
                                  6.3
                           6.3% =      = 0.063
                                  100
                                  100
                           100% =      =1
                                  100
                                   250
                           250% =       = 2.5
                                   200
                                       1
                                   33
                              1            33.3
                           33 % = 3 =           = 0.3
                              3     100 100

                                                                  1
We can also convert a percent to a number by noting that 1% is       = 0.01 .
                                                                 100
Thus for example:

                             5% = 5·(0.01) = 0.05
Applications

Applications or problems involving percent usually take one of the following
forms:

    1. Finding a percent of a given number.
    2. Finding what percent one number is of another.
    3. Finding a number when a percent of that number is known.

We note that the word ‘of’ appears in all the statements above. We recall
that to find a fraction ‘of’ a number means that we multiply the fraction by
                                  2             2
the number. Thus for example of 70 means ⋅ 70 . Similarly 40% of 70
                                  3             3
        40
means      ⋅ 70 = 28 .
       100




Pierce College                                                                  MAP
Math 215 Principles of Mathematics                                                  223


Example 1. A house sells for $92,000 requires a 20% down payment. What
is the amount of the down payment?

                                                       20
Solution. The down payment is 20% of $92,000 or           ⋅ $92, 000 = $18, 400 .
                                                      100
Hence the down payment is $18,400.

Example 2. Alberto has 45 correct answers on an 80-question test. What
percent of his answers are correct?

                     45
Solution. Alberto had   of the answers correct. To find the percent of
                     80
                            45
correct answers we convert     to percent. We can do this by any of the
                            80
methods discussed above. In this case, we multiply by 100%:

                         45 45   45
                           = ⋅1 = ⋅100% = 56.25%
                         80 80   80



Example 3. Forty-two percent of the parents of the school children in the
Paxton School District are employed at Di Paloma University. If the number
of parents employed by the University is 168, how many parents are in the
school district?

Solution. Let n be the number of parents in the school district. Then 42% of
n is 168. We translate this statement into an equation by identifying with
word ‘of’ with multiplication:

                               42% of n is 168
                                42
                                    ⋅ n = 168
                               100
                                         100
                               n = 168 ⋅      = 400
                                          42

Example 4. Kelly bought a bicycle and a year later sold it for 20% less than
what she paid for it. If she sold the bike for $144, what did she pay for it?




Pierce College                                                                      MAP
Math 215 Principles of Mathematics                                         224


Solution. Let P be the original price Kelly paid for the bicycle. Since the
$144 selling price includes the 20% loss, we can write the following equation:

                                   $144 = P – loss

Now Kelly’s loss is 20% of the original price or P. Thus we have:

                              $144 = P – 20%·P
                     $144 = P – 0.2·P = (1 – 0.2)P = 0.8·P
                                   $144
                               P=        = $180
                                    0.8
Thus Kelly paid $180 for the bike.

Example 5. Westerner’s Clothing Store advertised a suit for 10% off, for a
savings of $15. Later the manager marked the suit at 30% off the original
price. What is the amount of the current discount?

Solution. A 10% discount amounts to $15 savings. We set a sub-goal of
finding the original price from this information. Once we know the original
price we can find the amount of the 30% discount. If P is the original price
then 10% of P is $15. Thus we have:

                                      10%·P = $15
                                       0.1·P = $15
                             10·0.1·P = P = $15·10 = $150

Now that we know P we can find 30% of P:

                 30% of P = 30% of $150 = 30%·$150 = 0.3$150 = $45

Thus the current discount is $45.

Mental Math

Again, we can apply the techniques we learned in previous lectures to
problems involving percents. We consider the following cases:




Pierce College                                                             MAP
Math 215 Principles of Mathematics                                           225


    1. Using fraction equivalents. Knowing fraction equivalents for some
       percents can make some computations faster. The following table
       contains fractions equivalents for some common percents.

           Percent      25%     50%     75%       1            2     10%    1%
                                                33 %      66     %
                                                  3            3
           Fraction      1        1      3        1            2      1      1
           Equivalent    4        2      4        3            3     10     100

         These equivalents can be used in such computations as in the following:

                       1
         50% of $80 is   ⋅ 80 is $40
                       2
            2           2
          66 % of 90 is ⋅ 90 is 60
            3           3

    2. Using a known percent. Frequently we may not know a percent of
       something, but we know a close percent of it. For example, to find
       55% of 62 we might do the following:

                        1
         50% of 62 is     of 62 is 31
                        2
                    1                  1
         5% of 62 is   of 10% of 62 is    of 6.2 is 3.1
                    2                  2
         Adding we get 55% of 62 is 31 + 3.1 is 34.1

         This technique is useful in tipping, especially when tipping 15% of the
         amount. Thus for example, if the amount is $46, we can compute the
         15% tip by noting that 10% of $46 is $4.60 (divide by 10) and 5% of
         $46 is half of 10% i.e. half of $4.60 or $2.30. Adding we get $4.60 +
         $2.30 is $6.90.



Estimations Involving Percents

Again, estimations can be used to determine if answers are reasonable or to
provide an approximate answer. Examples:




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                           226


    1. To estimate 27% of 598 note that 27% of 598 is a little more than
       25% of 598 and that 598 is a little less than 600. Thus we estimate
                                      1
       27% of 598 as 25% of 600 or      of 600 which is 150.
                                      4
    2. To estimate 148% of 500 we note that 148% is close to 150% and that
       150% is equivalent to 1½. Thus we multiply 500 by 1.5 to get 750.




Functions

A (mathematical) function is a deterministic relationship between pairs of
quantities. The first quantity in the pair is the ‘input’ or independent variable
while the second is the ‘output’ or dependant variable. Since the relationship
is deterministic, the input or independent variable determines a unique
output or value of the dependent variable. Although the relationship
between the input and output is deterministic, it can take a variety of forms.
Functions can be expressed as rules, equations, machines, tables, graphs,
diagrams, and sequences. Formally we define a function as follows:

Definition of Function. A function from set A to set B is a correspondence
from A to B in which each element of A is paired with one, and only one,
element of B.

Note that is many cases the sets A and B contain numbers, but this need not
be the case (see Functions as Machines, below for example). Also note that a
function associates exactly one output with each input. If an element x is
used as the input and some output y is obtained, then every time x is used as
the input, the same output y is obtained.

The set A in the definition is the set of all allowable inputs and is called the
domain of the function. The set of all allowable outputs is B and is called the
range. In many cases the domain and range of a function are not explicitly
given. In this case, the domain is assumed to be all inputs for which the
function is meaningful. The range is then the corresponding outputs for
these inputs.




Pierce College                                                               MAP
Math 215 Principles of Mathematics                                         227


Functions as Rules

In this case the output of the function is determined from the input by
some rule. Examples include “take the original number and add 3” or “take
the input, multiply by 5 and subtract 2”. In some cases, a table of input and
outputs are given for the function, and the rule is determined from the
table. For example:




                        Input            Output
                          2                5
                          3                7
                          5                11
                         10                21

In this case the rule is “multiply the input by 2 and add 1”. Rule based
functions are often generated from empirical data collected by researchers.
The rule is developed from the data to provide a model for the underlying
process.

Functions as Machines

In this case the function is modeled as a ‘back-box’ or machine. The input is
put into the machine, which then generates the output, as illustrated below.

                                 INPUT




                                         OUTPUT


An example of function as a machine is a television. The inputs to the
‘machine’ are channel numbers and the outputs are stations. The following is
table lists the inputs and outputs of the television function.




Pierce College                                                             MAP
Math 215 Principles of Mathematics              228


                     Channel          Station
                     (Input)         (Output)
                        2               CBS
                        4               NBC
                        5              KTLA
                        7               ABC
                        8               PBS
                        9              KCAL
                        10             KMEX
                        11             KTTV
                        13             KCOP
                        26             DISC
                        27              TLC
                        28             ARTS
                        29              USA
                        32             ESPN
                        38              TBS
                        39             HIST
                        43             NICK
                        49               FX
                        50               E!
                        51              BET
                        52              MTV
                        53              VH1
                        55            BRAVO
                        56              TNT
                        57             AMC
                        63              CNN
                        71            SCIFI
                        73             LIFE
                        74             HGTV
                        75            FOOD
                       103              TBS
                       104              TNT
                       105              USA
                       106               FX
                       108               E!

Pierce College                                  MAP
Math 215 Principles of Mathematics                                         229


                         110            BRAVO
                         112             ARTS
                         126            FOXM
                         127             TCM
                         152             SOAP
                         172              DYI




Functions as Equations

In this case an equation is used to represent the function. The notation f(x)
is used to represent the output of the function given the input x. The
following are examples of functions as equations:

                                  f(x) = x + 3

                               f(x) = x2 + 3x + 2

We can tabulate the inputs and corresponding outputs by selecting values
for x and using the function equation to compute the outputs. Thus:

                          x           f(x) = x+ 3
                          0                3
                          1                4
                          2                5
                          3                6
                          4                7
                          5                8
                          6                9

Functions as Diagrams

Arrow diagrams can be used to represent a correspondence between two
sets that define a function. It should be noted that not all such diagrams
represent a function. The definition of a function must be satisfied by the

Pierce College                                                             MAP
Math 215 Principles of Mathematics                                                      230


arrow diagram for the representation to be that of a function. The
following illustrate this concept.
          A                       B               A                       B

          1                           a           1                           a

          2                                       2

          3                           b           3                           b



          This arrow diagram represents           This arrow diagram does not
          a function since each element           represent a function since each
          of A is mapped to a single              element of A is not mapped to a
          element of B.                           single element of B. Element 1 is
                                                  mapped twice to different elements
                                                  of B and element 2 is not mapped at
                                                  all.
Functions as Tables and Ordered Pairs

As we have seen above, functions can be described using a table. Consider
the table below relating the amount spent on advertising and the resulting
sales. This table describes the amount of dollars in Advertising and the
amount of dollars in sales.

                             Advertising       Sales
                             (in $1000s)   (in $1000s)
                                  0              1
                                  1              3
                                  2              6
                                  3              8
                                  4             10

We can also describe the function in terms of the two sets A = {0,1,2,3,4}
and S = {1,3,6,8,10}. The table describes a function from A to S, where A
represents thousands of dollars in advertising and S represents thousands
of dollars in sales.

We can also create a set of ordered pairs of numbers, taking one number
from each row to generate the ordered pairs: {(0,1), (1,3), (2,6), (3,8),
(4,10)}. The first number in the pair is in the function’s domain while the
second number in the pair is in the function’s range. Again we note that a set

Pierce College                                                                          MAP
Math 215 Principles of Mathematics                                            231


of ordered pairs does not necessarily represent a function. Consider the
following sets of ordered pairs: {(1,2),(1,3), (2,3),(3,4)} and
{(1,0),(2,0),(3,0),(4,4)}. The first set is not a function since the number 1 in
the domain is mapped to both 2 and 3 in the range, which violates the
definition of a function. The second set is (or can) define a function, since
the domain consists of the set {1, 2, 3,4} and these numbers are all paired
with one and only one element in the range which is the set {0,4}.

Data collected by measurement, survey, etc. are often tabulated to generate
empirical functions such as the function shown in the advertising/sales table
above. Such functions do not necessarily have a defining equation associated
with them, but are none-the-less valid and useful functions.

Functions as Graphs

Functions that are given by tables or ordered pairs are often graphed to
provide a visual representation of the function. Graphing not only provides a
visual representation of the function but facilitates identification of trends
and other interesting or important behavior of the function such as local
maximums or minimums or an increasing or decreasing trend. If the data in
the advertising/sales table are plotted, the following graph results.




Pierce College                                                                MAP
Math 215 Principles of Mathematics                                        232


This graph shows that as spending on advertising increases, sales increase.
Although this is the expected or hoped for result, graphing the function
shows this at a glance.

Sequences as Functions

Arithmetic, geometric, and other sequences introduced in the first part of
the course can be thought of as functions whose inputs or domain are the
natural numbers and whose output or range is the sequence in question. For
example the sequence 2, 4, 6, 8,… whose nth term is 2n can be described as a
function from N (natural numbers) to the set E (even natural numbers) using
the function f(n) = 2n where n is an element of N.

Composition of Functions

If we go back to the concept of functions as machines we can think of
chaining two machines together such that the output of the first function
(machine) is the input of the second function (machine).

                  INPUT1


                                         INPUT2 = OUTPUT1
                           OUTPUT1




                                              OUTPUT2



If for example the first machine represents the function f(x) = x + 4 and
the second machine the function g(x) = 2x, then if we input x = 2 into the
first machine we get f(2) = 2 + 4 = 6. If we take this output and use it as
the input to the second function we have g(6) = 2·6 = 12. The input 2
therefore yields the output 12 when we combine these two functions in this
manner. Combining functions in this manner, with the output of one function
providing the input to another function is known as composition of functions.
The range of the first function becomes the domain of the second. The
notation g◦f = (g◦f)(x) = f(g(x)) is used to denote the composition of the
functions f and g. In the above example we write (g◦f)(2) = 12.

Pierce College                                                            MAP

				
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