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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 + = ⋅ + ⋅ . bd 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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