Square Root of 2 is Irrational

Theorem:
$$\sqrt{2} \,\!$$ is irrational.

Proof by Contradiction:
Assume that $$\sqrt{2} \,\!$$ is rational; so $$\sqrt{2}={\frac{p}{q}}$$ for some $$p,q \in \mathbb{Z}$$ and $$gcd(p,q)=1\,\!$$.

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

Therefore, $$2|p^2 \Rightarrow 2|p \,\!$$; that is, $$p \,\!$$ is an even number. So, $$p = 2k \,\!$$ for some $$k \in \mathbb{Z}$$.

Thus, $$2q^2 = p^2 = (2k)^2 = 4k^2 \Rightarrow q^2 = 2k^2 \,\!$$, so by the same reasoning, $$2|q^2 \Rightarrow 2|q \,\!$$, contradicting our assumption that $$gcd(p,q)=1\,\!$$, since $$2|p,q \,\!$$.

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

Q.E.D.

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