Revision Topic 21: 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  r6
3
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  7 3
c)
7 
d)
78  7 2  7
74
2 6
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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
b
a
  
a
b
1
3
1
   3
1
3
3
 
4
1

4
3
General negative powers are handled as follows:
1
an  n
a
So,
1 1
23  3 
8
2
1
1
52  2 
25
5
Likewise with fractions:
2
 
5
2
2
 
3
3
2
25
5
  
4
2
(square top and bottom)
3
27
3
  
8
2
Examination style question
Evaluate (i.e. work out)
a)
80
b)
3 2
c)
2 4
d)
5
 
2
3
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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  4

81 3
814
1
3
 27  3 3 27


 
1000 10
 1000 
 1 
 
 25 

1
2
(cube root top and bottom)
1
1
 25  2
    25 2  5
 1 
More complex fractional powers are handled as follows:
m
an 
So
5
 a
n
 4   2  32
  27   3  9
5
42 
2
27 3
m
5
2
3
3
2
3
3
27
 9 2  9   3 
    
   
8
4
 4  2
1
1
1
1
2
1000 3 
 3
 2 
2
2
100
10
1000 3 ( 1000)
Examination question (Edexcel June 2001)
Evaluate
(i)
5 2
(ii)
83
(iii)
49
2
 12
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Examination question (Edexcel June 2005)
Find the value of
(i)
640
(ii)
64 2
(iii)
64
1
 23
Surds
6 5
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).
Expressions like
2, 5 3  1 and
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
4 cm
x 2  42  62
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 =
52 can be simplified:
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
of this as 3  3  9  3
3  3  3 (by definition of the square root of 3). We can think
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:
Solution:
5

5


2 1 2 2 1

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
 base×height
2
1
=  4 3 5 3
2
= 2 3 5 3
5 3 cm
Area of triangle =
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 .
Triangle ABC has an area of 32 cm².
b) Calculate the value of k.
A
8 cm
B
k
2 cm
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C
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Dividing Surds
When dividing, we sometimes make use of the result
40
Example:
10

a
b

a
.
b
40
 42
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.
To rationalise the denominator in
a
b
=
a
b

b
b

a
b
, you multiply top and bottom by
b:
a b
b
Note: Because you multiply the top and the bottom by the same thing, you haven’t changed the
value of the number.
Example: Rationalise the denominator in
6
2
.
Solution:
Multiply top and bottom by 2 :
6
2 6 2
6
=


3 2
2
2
2
2
Example: Rationalise the denominator in
Solution:
Multiply top and bottom by
8
3 6
=
8
3 6
8 6
=
18
4 6
=
9

6
6

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8
3 6
6:
8 6
3 6
(dividing top and bottom by 2)
10
Examination style question
Rationalise the denominator of
40
5
Examination style question:
12
Express
in the form a b where a and b are integers.
3
Examination question (Edexcel November 2004)
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
The length of each of its equal sides is
cm.
2
b) Find the area of the triangle. Give your answer as an integer.
6
cm
2
6
cm
2
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Changing recurring decimals to fractions
.
.
.
. .
Recurring decimals, such as 0.5 , 0.217 and 0.14 2 , all can be written as fractions.
This is short for the
decimal
0.555555…
This is short for the
decimal
0.217217217…
This is short for the
decimal
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:
Step 2:
Step 3:
Write 0.5 as x:
Find 10x:
Subtract (2) – (1):
Step 4:
Find the fraction:
x = 0.55555555…..
10x = 5.555555555….
9x = 5
5
x
9
(1)
(2)
Example 2:
.
.
Write 0.217 in the form
.
a
, where a and b are integers.
b
.
Step 1:
Step 2:
Step 3:
Write 0.217 as x:
Find 1000x:
Subtract (2) – (1):
Step 4:
Find the fraction:
x = 0.217217217…
1000x = 217.217217…
999x = 217
217
x
999
(1)
(2)
x = 0.142424242…
100x = 14.24242424…
99x = 14.1
(1)
(2)
Example 3:
. .
Change 0.14 2 to a fraction.
. .
Step 1:
Step 2:
Step 3:
Write 0.14 2 as x:
Find 100x:
Subtract (2) – (1):
14.24242424...
0.14242424...
14.10000000...
Step 4:
14.1
99
To change this to a fraction with integers top and bottom, multiply through by 10:
141
x
990
47
This simplifies to
330
Find the fraction:
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x
12
Examination question (March 2002 Edexcel)
Change to a single fraction
. .
a) the recurring decimal 0.13
. .
b) the recurring decimal 0.513
Examination question (Edexcel June 2005)
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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