# How do I rationalise the denominator of a fraction?

I am going to assume you know what an irrational number is, given that that should have been taught right at the beginning the topic of surds. When I talk about a “rational” number, here I mean “whole number” or “integer”, although a rational number is a number that is not irrational.

There are two types of problem that you could come across when being asked to rationalise the denominator of a fraction:

The first type is when you are asked to rationalise a fraction which has its whole denominator under a square root. For example 1/sqrt(2) or 5/sqrt(8) etc. To rationalise the denominator here, we somehow have to square the denominator to get a whole number. However, we can’t change the value of the fraction! A great way of doing this is therefore by multiplying the fraction with denominator/denominator as you’ll notice that this is just equal to 1: Let’s say we had the fraction n/d (n for numerator, and d for denominator), if we multiply this by d/d, we get nd/d^{2} (which is still equal to our original fraction). If ‘d’ was irrational, e.g. d=sqrt(2), then d^{2} would be rational, e.g. (sqrt(2))^{2}=2, and therefore we have successfully rationalised the denominator! For example, we could multiply 1/sqrt(2) by sqrt(2)/sqrt(2) to get sqrt(2)/( sqrt(2))^{2} = sqrt(2)/2, or 5/sqrt(8) by sqrt(8)/ sqrt(8) to get 5sqrt(8)/(sqrt(8))^{2} = 5sqrt(8)/8.

The second type is a little trickier as it relies on a concept we call “the difference of two squares”. First, let’s consider (x+y)(x-y) = x^{2}-y^{2}; this result is known as “the difference of two squares”, and it is incredibly useful because if either ‘x’ or ‘y’ were irrational, then we now know that (x+y)(x-y) is rational. For example, if x is irrational (e.g. x=sqrt(2)) and y is rational (e.g. y=3), then (x+y)(x-y)= x^{2}-y^{2} =(sqrt(2))^{2}-3^{2}=2-9=-7, or if x is rational (e.g. x=4) and y is irrational (e.g. y=sqrt(5)), then (x+y)(x-y)= x^{2}-y^{2}=4^{2}-(sqrt(5))^{2}=15-5=11. Now let’s return to the second type of problem: you could be asked to rationalise the denominator of a fraction whose denominator comprises of a rational number added to a whole number, for example 1/(1+sqrt(2)) or 3/(sqrt(3)-5) etc. This is definitely more tricky than the first type of problem, but here is where we use what we’ve just learned about “the difference of two squares” to make life a lot easier for us: The denominators of these fractions look a lot like the examples I gave for (x+y) or (x-y) (so 1+sqrt(2) is just (x+y) where x=1 and y=sqrt(2) etc.) so, we probably want to find a way to multiply the denominator of the fraction by the ‘opposite form’ in order to rationalise it. For example, multiplying (1+sqrt(2)) by (1-sqrt(2)) gives us the difference of two squares: 1^{2}-(sqrt(2))^{2}=1-2=-1, and multiplying (sqrt(3)-5) by (sqrt(3)+5) = (sqrt(3))^{2}-5^{2}=3-25=-22. Now we have a way of turning the denominator into a whole number, we are back to thinking about how we can do this without changing the value of the fraction. Using our trick from the first type, you can see that, if we were given a fraction a/(b+sqrt(c)), then multiplying it by (b-sqrt(c))/ (b-sqrt(c)) wouldn’t change the value ( as (b-sqrt(c))/ (b-sqrt(c))=1) but gives us b^{2}-(sqrt(c))^{2} on the bottom, which is b^{2}-c, which is a whole number! Going back to my earlier examples, 1/(1+sqrt(2)) x(1-sqrt(2))/ (1-sqrt(2)) = (1-sqrt(2))/-1 = sqrt(2)-1, and 3/(sqrt(3)-5) x(sqrt(3)+5)/ (sqrt(3)+5)=3(sqrt(3)+5)/-22 = (3sqrt(3)+15)/-22 =(-3sqrt(3)-15)/22.