Square Root of 2 is Irrational

Theorem

 * $$\sqrt{2} \,\!$$ is irrational.

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

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

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

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

Squaring both sides yields $$2 = \frac{p^2}{q^2} \iff p^2 = 2q^2\,\!$$.

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

That is, $$p \,\!$$ is an even number. So, $$p = 2k \,\!$$ for some $$k \in \Z$$.

Thus, $$2 q^2 = p^2 = (2k)^2 = 4k^2 \implies q^2 = 2k^2 \,\!$$, so by the same reasoning, $$2 \backslash q^2 \implies 2 \backslash q \,\!$$.

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

Therefore,$$\sqrt{2} \,\!$$ can not be rational.

Note: this is a special case of the result that the square root of any prime is irrational.