A Proof that the Square Root of 2 is Irational

Posted September 25th, 2007 by Isoscel

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.

Instant start