Square Root of 2 is Irrational

Theorem

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

Proof 1: 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:
 * $$(A) \qquad 2 \backslash p^2 \implies 2 \backslash p \,$$.

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

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.

Proof 2
This is a special case of the result that the square root of any prime is irrational.

Decimal Expansion
The decimal expansion of $$\sqrt 2$$ starts:
 * $$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$$