Square Root of 2 is Irrational/Classic Proof

Theorem

 * $\sqrt 2$ is irrational.

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

Thus it follows that:
 * $(A) \qquad 2 \mathop \backslash p^2 \implies 2 \mathop \backslash p$

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

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

So $\displaystyle \sqrt 2 = \frac p q$ for some $p, q \in \Z$ and $\gcd \left({p, q}\right) = 1$.

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

Therefore, $2 \mathop \backslash p^2 \implies 2 \mathop \backslash p$ (see $(A)$ above).

That is, $p$ is an even integer.

So $p = 2k$ 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
 * $2 \mathop \backslash q^2 \implies 2 \mathop \backslash q$

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

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

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

The ancient Greeks prior to Pythagoras, following Eudoxus of Cnidus, 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 $\sqrt 2$.