Irrationality of the square root of two

We want to show that $\sqrt{2}$ is irrational. We will proceed by contradiction.

[edit] Argument

Assert for any $n, m \in \Z$ , if $m \neq 0$ ,

n

and

m

are relatively prime, and

$\sqrt{2} = n/m$ then

$m \cdot \sqrt{2} = m \cdot (n/m)$ ,

$m \cdot \sqrt{2} = n$ ,

$(m \cdot \sqrt{2})^2 = n^2$ ,

$m^2 \cdot \sqrt{2}^2 = n^2$ ,

$m^2 \cdot 2 = n^2$ ,

$n^2 = m^2 \cdot 2$ ,

$n^2 = 2 \cdot m^2$ ,

n 2

is even,

n

is even, and

$n^2 = (2 \cdot (n/2))^2$ ,

$n^2 = 2^2 \cdot ((n/2)^2)$ ,

$n^2 = 4 \cdot ((n/2)^2)$ ,

$2 \cdot m^2 = 4 \cdot (n/2)^2$ ,

$m^2 = 2 \cdot (n/2)^2$ ,

m 2

is even,

m

is even,

$GCF(m,n) \geq 2$ ,

G C F (m, n) = 1

,

there is a contradiction.