Square Root of 2 is Irrational Proof

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

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


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



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

Method 1


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


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

Let \(a=2k\),




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.

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