**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.