A Proof that the Square Root of 2 is Irational

We shall proof that the square root of 2 is irational.

Let's presume the contrary:
$ \sqrt{2} $ is rational.

Then:
$ \sqrt{2}=\frac{m}{n} $ with m, n integer numbers and m and n relatively prime ($ \frac{m}{n} $ is an irreducible fraction).

so:
$ \sqrt{2}n=m $
$ 2n^2=m^2 $

So $ m^2 $ is even. This means that m is even. So there is an k with $ m=2k $.

$ 2n^2=(2k)^2 $
$ 2n^2=4k^2 $
$ n^2=2k^2 $

So $ n^2 $ is even. This means that n is even too.

Contradiction because m and n are relatively prime.

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