Irrationality of the square root of two

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

[edit] Argument

Assert for any $n, m \in \mathbb{Z}$ , 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,

$\mathrm{GCF}(m,n) \geq 2$ ,

$\mathrm{GCF}(m,n) = 1$ ,

there is a contradiction.

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