We will prove that \(\sqrt2\) is irrational by contradiction:

Assume that \(\sqrt2\) is rational. Then:

\[\sqrt2=\frac{a}{b}\]

where a and b are integers and \(\frac{a}{b}\) is a fraction in its simplest form.

\[b\sqrt2=a\]

\[2b^2=a^2\]

From this point onwards there are two methods (although it could be argued that in essence, they are the same).

## Method 1

\[2b^2=a^2\]

The prime factorisation of any square number has even powers. Thus the power of 2 of the prime factorisation of the RHS is even (and possibly 0).

The power of 2 in \(b^2\) is also even, thus \(2b^2\) must have an odd power of 2.

This is a contradiction, thus the original assumption that \(\sqrt2\) is rational is false, and thus \(\sqrt2\) is irrational.

## Method 2

\[2b^2=a^2\]

\(2b^2\) is even, therefore \(a^2\) is also even, and thus a is even.

Let \(a=2k\),

\[2b^2=(2k)^2\]

\[2b^2=4k^2\]

\[b^2=2k^2\]

Since \(2k^2\) is even, \(b^2\) is even, and thus b is even.

As a and b are both even, this violates the assumption that \(\frac{a}{b}\) is in its simplest form, and thus \(\sqrt2\) is irrational.