# $\sqrt{2}$ is irrational Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Problem

Fact: The square root of $2$, $\sqrt{2}$, is irrational.

Sort the statements below into a proof of this fact. Can you explain how the ordered statements prove this result?

This is a contradiction, so our original assumption that $\sqrt{2}$ is rational must be wrong.

In the prime factorisations of $m^2$ and $n^2$, $2$ occurs to an even power.

Multiply across to get $2n^2 = m^2$.

But prime factorisations are unique, so $2$ should appear to the same power in both $2n^2$ and $m^2$.

Suppose, for a contradiction, that $\sqrt{2}$ is rational.

That is, we can write $\sqrt{2} = \frac{m}{n}$ where $m$ and $n$ are integers and where $n \neq 0$.

In the prime factorisation of $2n^2$, $2$ occurs to an odd power.

So $\sqrt{2}$ is irrational.

Squaring, we have $2 = \frac{m^2}{n^2}$.

Can you adapt this argument to show that $\sqrt{3}$ is irrational?

For which positive integers $d$ is $\sqrt{d}$ irrational? Justify your answer.

Can you show that $1 + \sqrt{2}$ is irrational? Is $\sqrt{2} + \sqrt{3}$ rational or irrational? Justify your answers.