## Square Root of 2 is Irrational Proof

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.