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Rationalising denominator
Con il contributo di: Adam Mlynarczyk
2-√3
In this case multiplying the bottom and top by √3wouldn't work as we would still get surd at the bottom
To rationalise denominator we have to multiply the topand the bottom of the fraction by 2+√3.
√3
Rationalise the denominator of 
Example 1:
2-√3
√3
x(2+√3)
x(2+√3)
x√3
x√3
=
=
(2-√3)(2+√3)
(2-√3) x √3
√3 x √3
√3(2+√3)
=
=
4+2√3-2√3-3
2√3-3
3
2√3+3
It's not good as we still
have surd in denominator,
so we have to multiply by
different thing to get rid of
the surd in denominator.
2-√3
√3
=
2√3+3
1+√2
1-√2
In the denominator we have 1+√2, so we have to
multiply the top and the bottom of the fraction by
1-√2.
We always use the two numbers from the bottom, butwe have to change the sign + to - or - to +.
Example 2:
Rationalise the denominator of 
x(1-√2)
x(1-√2)
=
(1+√2)(1-√2)
(1-√2)(1-√2)
=
1-√2+√2-2
1-√2-√2+2
=
3-2√2
1+√2
1-√2
-1
=
-3+2√2
=
√5+1
Rationalise the denominator of 
5-√5+√5-1
3
x(√5-1)
x(√5-1)
=
(√5+1)
 -
3(
=
(
-
√5+1
-
)
3
-
)
=
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