Rational and Irrational Numbers
A rational number can be written as fraction, so not surprisingly an irrational number
can not be written as a fraction.
3
 6
 23 
1.5 is rational because it can be written as
as is 6    and 0.23  

2
 1
 100 
All integers are rational.
An irrational number will be a decimal which goes on forever (non-terminating)
without ever repeating.
The main examples of irrational numbers are surds (see later) and multiples of and π.
We can express irrational numbers in terms of π and surds
e.g. Write down an irrational number between 7.2 and 7.3
π = 3.141………
so π + 4.1 will be irrational and between 7.2 and 7.3
61  7.81...
61  0.6 will be irrational and lie between 7.2 and 7.3
Writing a Non-Terminating Recurring Decimal as a Fraction
If a decimal goes on forever (non-terminating) but repeats, then it is rational i.e. it can
1
be written as a fraction. The obvious example is 0.333333333… which is
3
Some numbers are not quite so obvious but can still be written as fractions

e.g. Write 0. 4 as a fraction
Let x = 0.444444….. so 10x = 4.4444444…..
10x – x = 4 = 9x
If 9x = 4 then x =

4
but as x = 0. 4
9

0. 4 =
4
9

e.g. Write 3. 4 as a fraction


4
4
31
We know that 0. 4 = so 3. 4 must equal 3 or
9
9
9
 
e.g. Write 0.2 3 as a fraction
 
0.23  0.23232323.....

Let x= 0.23 so 100x = 23.232323…….
100x – x = 23.23232323…. – 0.23232323….. = 23 = 99x
23
If 99x = 23 then x =
but x = 0.23232323…..
99
 
23
So 0.2 3 =
99
After a while you can just recognise the answers if the recurring part of the decimal is
next to the decimal point
 
 
152
27 3
0.15 2 
0.2 7 

(in simplest form)
999
99 11
However you must remember the ‘long-winded’ method as you may be asked to
 
23
prove that, for example, 0.2 3 =
99
If the recurring part is not next to the decimal point then we have to split the recurring
and non-recurring parts and use a fancy bit of division/multiplication to put the
recurring part next to a decimal point. This is probably going to appear on the Higher
paper.

e.g. Write 0.2 5 as a fraction


0.2 5 = 0.2 + 0.0 5
Split into recurring and non-recurring parts

= 0.2 + 0. 5 ÷ 10
2 5
   10
10 9
2 5
 
10 90
18 5


90 90
23

90
Move recurring part next to decimal point
Write as fractions
Perform the division
Find common denominator and re-write
using equivalent fractions
Add up
 
e.g. Write 0.41 53 as a fraction
 
 0.41  0.00 53
 
 0.41  0.53  100
41 53
  100
100 99
 
41
53
0.41 53 

100 9900
4059
53


9900 9900
4112

9900
1028

2475

You can check your answer by converting the fraction to a decimal by dividing and
1028
 1028  2475  0.415353....
seeing if you get the starting value e.g.
2475
Surds
Surds are the square root of a prime number (or a multiple thereof).
They are always irrational.
It is more accurate to write your answer as a square root rather than as a decimal if
you are dealing with surds – particularly useful in situations using Pythagoras’
Theorem
There are two laws of surds which are often used to simplify expressions:
Law 1
ab  a b
Law 2
e.g. Simplify
(and vice versa)
a
a

b
b
48
48  16 3
4 3
( 16  4)
The trick is to look for the largest square number (the square root of which will
always be an integer) which will divide into your expression.
e.g. Simplify
3 8  18  4 2
3 8  18  4 2  3  4 2  9 2  4 2
 3 2 2  3 2  4 2
 6 2 3 2 4 2
 13 2
Rationalising a Fraction
If we rationalise a fraction containing a surd, we write it so that there are no surds in
the denominator (bottom). There will probably be some surds left in the numerator.
Rationalising does not mean ‘get rid of all the surds’.
The secret to rationalising is to use equivalent fractions and multiply by one which
will not change the value of our expression
(we just write one in a very clever way)
e.g. Rationalise
2
5
 5

 1

 5

2
2
5


5
5
5


e.g. Rationalise
2 5
5 5
2 5
5
2 3
5
2 3 2 3
5


5
5
5



5 2 3

5 5
5 2 3


5
Higher Level
If you have a more complicated denominator then you need to be aware of the
difference of 2 squares i.e. a2 – b2 = (a – b)(a + b)
1 2
there is no point multiplying by
5 1
as when you multiply out the denominators you will get:
( 5  1)( 5  1)  5 5  1 5  1 5  1
e.g. If you have to rationalise
5 1
5 1
 25  2 5  1
 26  2 5
This means we will still have a surd in our denominator so it has not been rationalised
The difference of 2 squares would suggest that we multiply by
5 1
5 1
1 2 1 2
5 1


5 1
5 1
5 1

1  2 
 5  1

1 5 1 2 5 1 2
5 5 1 5 1 5 1


5  1
5 1
5  10  2  1
5 1
5  10  2  1
4
There are no surds in the denominator so it has been rationalised.
You may prefer to write it without expanding the numerator i.e.




5 1
1 2 1 2

4
5 1
Past Paper Questions
18 18
2


2
2
2

18 2
2 2
18 2
2
9 2

12


75  48  4 3


25 3  16 3
 2 3 5 34 3

 10 3 3  8 3 3
 10  3  8  3
 30  24
6
(as required )
