Square Root of 2 is Irrational/Classic Proof

Proof
First we note that, from Parity of Integer equals Parity of its Square, if an integer is even, its square root, if an integer, is also even.

Thus it follows that:
 * $(1): \quad 2 \mathrel \backslash p^2 \implies 2 \mathrel \backslash p$

where $2 \mathrel \backslash p$ indicates that $2$ is a divisor of $p$.

Now, assume that $\sqrt 2$ is rational.

So:
 * $\sqrt 2 = \dfrac p q$

for some $p, q \in \Z$, and:
 * $\gcd \left({p, q}\right) = 1$

Squaring both sides yields:
 * $2 = \dfrac {p^2} {q^2} \iff p^2 = 2q^2$

Therefore from $(1)$:
 * $2 \mathrel \backslash p^2 \implies 2 \mathrel \backslash p$

That is, $p$ is an even integer.

So $p = 2 k$ for some $k \in \Z$.

Thus:
 * $2 q^2 = p^2 = \left({2 k}\right)^2 = 4 k^2 \implies q^2 = 2k^2$

so by the same reasoning:
 * $2 \mathrel \backslash q^2 \implies 2 \mathrel \backslash q$

This contradicts our assumption that $\gcd \left({p, q}\right) = 1$, since $2 \mathrel \backslash p, q$.

Therefore, from Proof by Contradiction, $\sqrt 2$ cannot be rational.

Historical Note
This proof is attributed to [, or to a student of his.

The ancient Greeks prior to Pythagoras, following, believed that irrational numbers did not exist in the real world.

However, from the Pythagorean Theorem, a square with sides of length $1$ has a diagonal of length $\sqrt 2$.