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                   Revision Topic: Powers, Surds and Rational Numbers
The objectives of this unit are to:
* recap the basic rules for indices;
* to evaluate fractional and negative powers of numbers;
* to simplify surds and expressions involving surds;
* to solve problems involving surds;
* to convert a recurring decimal to a fraction.

Brief recap of Grade B and C material:
Rules of indices:

Zero power:                Anything to the power 0 is 1
                           So
                                  70  1
                                   a0  1
                                   ( xy ) 0  1

Multiplying indices: Add the powers
                     So
                            a m  a n  a mn
                                   83  84  87
                                   52  7  54  73  (52  54 )  (7  73 )  56  7 4

Dividing indices:          Subtract the powers
                           So
                                   y m  y n  y mn
                                   511  54  57
                                   r4  r5 r9
                                        3
                                             3  r6
                                      r       r

Power of a power:          Multiply the powers together
                           So
                                  ( a m ) n  a mn
                                   (93 ) 2  96

Examination style question
Simplify each of the following by writing the answer as a single power of 7:

        a)       74  72

        b)       78  73

        c)       7 
                    2 6




                  78  7 2  7
        d)
                      74

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Negative powers
A power of -1 represents the reciprocal of a number.
So:
              1
       a 1 
              a
              1
       7 1 
              7
You find the reciprocal of a fraction by swapping the numerator and denominator over.
Therefore
             1
       a   b
         
       b   a
            1
       1  3
          3
       3  1
             1
       3   4
         
       4   3

General negative powers are handled as follows:
              1
       an  n
             a
So,
              1 1
       23  3 
             2     8
              1    1
       52  2 
             5     25
Likewise with fractions:
             2          2
       2         5   25
                        (square top and bottom)
       5         2   4
             3          3
       2  3   27
           
       3  2   8

Examination style question

Evaluate (i.e. work out)

       a)         80

       b)         32

       c)         24

                        3
                  5
       d)          
                  2




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Fractional Powers
Fractional powers correspond to roots of numbers:
         1
       a2  a
         1
       a3  3 a
         1
       a4  4 a
Therefore
         1
       42  4  2
                  1
       125 3  3 125  5
         1     1      1   1
       81 4  1         
              814
                     4
                       81 3
                                  1

        27  3 3 27    3
                                                            (cube root top and bottom)
        1000    1000 10
                              1                        1
                          
        1                   2      25  2 1

                                    25 2  5
        25                         1 

More complex fractional powers are handled as follows:
                           a
         m                                m
       an                    n


So
                        4   2  32
         5                                5
       42                                             5



                        27   3  9
              2                                2
                                  3                        2
       27 3
                      3                            3
        9 2  9   3 
                              3
                                  27
              4  2  8
                        
       4           
              2     1          1      1    1
       1000 3          2
                           3      2
                                      2 
                   1000 3 ( 1000)     10   100


Examination question
Evaluate

       (i)                        52
                                      2

       (ii)                       83

                                          1
       (iii)                      49       2




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Examination question
Find the value of

        (i)        640
                        1
        (ii)       64 2

                        2
        (iii)      64    3




Surds
                                   6 5
Expressions like            2, 5 3  1 and
                                         are all examples of surds as they are expressed in terms of
                                     7
a root. In general, surds are numbers that are left in a form involving a root (typically a square
root).

Surds are often used when it is important to give an exact answer.
For example, suppose we wished to calculate the length of the hypotenuse in this triangle:

                                                    Using Pythagoras:
                                  x cm
                                                           x 2  42  62
     4 cm
                                                           x 2  16  36
                                                           x 2  52
                            6cm
Although we can use a calculator to evaluate the length of the hypotenuse, giving x = 7.21…cm, the
decimal would have to be rounded (giving an answer which is not exact). If an exact answer was
required, the hypotenuse could be expressed in surd form – the exact length is x = 52 cm.


Simplifying Surds
Some surds can be simplified. This is done by finding an equivalent expression that involves the
square root of a smaller number.

The surd        n can be simplified if n is divisible by a square number (bigger that 1).
To simplify a surd, we use the result:          ab  a  b




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Example:
Simplify 52

Solution:
As 52 is divisible by a square number (4 goes into 52),   52 can be simplified:

         52 =    4  13  4  13  2 13


Example 2:
Simplify 72

Solution:
We know that 72 can be simplified, as 72 can be divided by several square numbers (for example
9 goes into it 8 times).

         72  9  8  9  8  3 8

However, this is not the final answer as   8 can also be simplified:

       3 8  3 4  2  3 4  2  3 2  2  6 2

So:   72  6 2

Note: This simplified answer could have been obtained directly in one step if 72 had been split up
as 36 × 2.

Examination style question
Simplify 18




Examination style question
Simplify 50




Examination style question
Simplify 54




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Calculating with surds
Addition and subtraction of surds

The methods involved are similar to those used when simplifying algebraic expressions.

Example 1:
     4 2 3 3 2 2  3  6 2 4 3                     (collect together similar roots)

Example 2:
     5 7 3 5 2 7 2 5 3 7 5 5

However, it is important to make sure that all surds are simplified as much as possible before trying
to collect together like terms.

Example 3:
Simplify        7 2  3 8  4 18

Solution: Begin by simplifying the surds:      8  4 2  2 2
                                               18  9  2  3 2
Therefore,
       7 2  3 8  4 18 = 7 2  3  2 2  4  3 2  7 2  6 2  12 2  2

Example 4:
Simplify        5 2  3 3  2 12  50

Solution:
First we simplify the surds:      12  4  3  2 3
                                  50  25  2  5 2
So:
        5 2  3 3  2 12  50 = 5 2  3 3  2  2 3  5 2  7 3

Worked examination style question
2 72  32  k 2 . Find the value of k.

Solution:
Simplify the surds:       72  36  2  6 2
                          32  16  2  4 2
Therefore,
       2 72  32  2  6 2  4 2  8 2
So k = 8.

Worked examination style question
 27  3n . Find the value of n.

Solution:
  27  9  3  3  3  3  31/ 2  33 / 2
       So, n = 3/2


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Multiplication of surds

We will need to use the following result:
         a  b  ab

Example:
     4 3  2 3  4  2  3  3 = 4  2  3  24

Note: Here we used the fact that         3  3  3 (by definition of the square root of 3). We can think
of this as 3  3  9  3

Example 2:
     6 2  3 18  6  3  2  18  18  36  18  6  108

Example 3:
Expand and simplify:        
                          2 3 32 2          
Solution: Use the usual rules for expanding brackets to get:
                2 3 3  2 2 2
               3 6 2 4
                3 6  2 2
               3 6 4

Example 4:
Expand and simplify: (3 5  2)(3 5  2)

Solution:
We use any usual method for expanding double brackets:

       (3 5  2)(3 5  2) = 9 5  5  6 5  6 5  4
                          = 95  4
                          = 41

Example 5:
Expand and simplify:   5           
                            2  1 2 2 1         
Solution:
       5                 
            2  1 2 2  1  10 4  5 2  2 2  1
                                 20  3 2  1
                                 19  3 2




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Worked examination question
Calculate the area of the triangle shown.


Solution:
                  1                                                                  5 3 cm
Area of triangle =   base×height
                  2
                 1
               =  4 3 5 3
                 2
               = 2 3 5 3                                              4 3 cm
               = 10  3  3
               = 30 cm2


Examination question 1:
                              
                                   2
Work out the value of 5  3 . Give your answer in the form a  b 3 where a and b are
integers.

Remember: To square a bracket, multiply it by itself.




Examination style question :
The length of a rectangle is (5  3) cm. The width of the rectangle is (6  3) cm.
Work out:
a) the perimeter of the rectangle;
b) the area of the rectangle.




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Examination style question
a) Simplify 75  27
b) Simplify      75  27      75  27   




Examination question (Edexcel)
a) Find the value of n in the equation 2n  8 .             A

Triangle ABC has an area of 32 cm².
b) Calculate the value of k.                             8 cm


                                                            B   k      C
                                                                2 cm




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Dividing Surds

                                                                   a       a
When dividing, we sometimes make use of the result                          .
                                                                   b       b

                      40       40
Example:                          42
                      10       10


Rationalising a denominator

In mathematics, it is considered untidy to leave a surd in the denominator of a fraction.
If there is a surd in the denominator, you should try to find an equivalent answer which only has
surds on the top of the fraction. This process is called rationalising the denominator.

                                              a
To rationalise the denominator in                 , you multiply top and bottom by   b:
                                              b
         a        a        b       a b
             =                
         b        b        b        b

Note: Because you multiply the top and the bottom by the same thing, you haven’t changed the
value of the number.

                                                       6
Example: Rationalise the denominator in                     .
                                                        2

Solution:
Multiply top and bottom by 2 :
        6      6     2 6 2
            =             3 2
          2     2    2    2

                                                        8
Example: Rationalise the denominator in
                                                      3 6

Solution:
Multiply top and bottom by              6:
         8            8        6       8 6
              =                   
        3 6     3 6            6       3 6
                8 6
              =
                 18
                4 6
              =                         (dividing top and bottom by 2)
                  9




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Examination style question
                                 40
Rationalise the denominator of
                                  5




Examination style question:
        12
Express    in the form a b where a and b are integers.
         3




Examination question
           6
a) Express    in the form a b where a and b are positive integers.
            2

The diagram shows a right-angled isosceles triangle.
                                                                             6
                                         6                                      cm
The length of each of its equal sides is    cm.                               2
                                          2
b) Find the area of the triangle. Give your answer as an integer.

                                                                     6
                                                                        cm
                                                                      2




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Changing recurring decimals to fractions
                                                 .    .    .        . .
Recurring decimals, such as 0.5 , 0.217 and 0.142 , all can be written as fractions.


          This is short for the                      This is short for the       This is short for the
          decimal                                    decimal                     decimal
             0.555555…                                  0.217217217…                0.142424242….

Note: The dots go over each end of the set of repeating digits.

Example 1:
                .
Change 0.5 to a fraction.

                                       .
Step 1:                       Write 0.5 as x:             x = 0.55555555…..         (1)
Step 2:                       Find 10x:                   10x = 5.555555555….       (2)
Step 3:                       Subtract (2) – (1):          9x = 5
                                                                 5
Step 4:                       Find the fraction:             x
                                                                 9

Example 2:
            .       .                       a
Write 0.217 in the form                       , where a and b are integers.
                                            b

                                       .    .
Step 1:                       Write 0.217 as x:               x = 0.217217217…              (1)
Step 2:                       Find 1000x:                 1000x = 217.217217…               (2)
Step 3:                       Subtract (2) – (1):          999x = 217
                                                                217
Step 4:                       Find the fraction:            x
                                                                999

Example 3:
                        . .
Change 0.142 to a fraction.

                                           . .
Step 1:                       Write 0.142 as x:               x = 0.142424242…              (1)
Step 2:                       Find 100x:                   100x = 14.24242424…              (2)
Step 3:                       Subtract (2) – (1):           99x = 14.1
                                                                                             14.24242424...
                                                                                              0.14242424...
                                                                                             14.10000000...
                                                            14.1
Step 4:                       Find the fraction:               x
                                                             99
                              To change this to a fraction with integers top and bottom, multiply through by 10:
                                                            141
                                                        x
                                                            990
                                                       47
                              This simplifies to
                                                      330

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Examination question
Change to a single fraction
                              . .
a) the recurring decimal 0.13
                                . .
b) the recurring decimal 0.513




Examination question
            3
a) Change      to a decimal. (Hint: divide the numerator by the denominator)
           11
                                      . .   13
b) Prove that the recurring decimal 0.39 
                                            33




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