Square Root of 2 is Irrational

Theorem. The $$\sqrt{2}$$ is irrational.

Proof.

Assume that the $$\sqrt{2}$$ is rational; so $$\sqrt{2}={\frac{p}{q}}$$ where $$p,q \in \mathbb{Z}$$ and $$(p,q)={1}$$.

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

So $$p^{2}={2q^{2}}$$.

By the properties of exponents and the Fundamental Theorem of Arithmetic, $$2|p^{2}$$ and $$2|p$$. So $$p$$ is an even number. Clearly, since $$p$$ is an even number, $$q$$ must also be an even number, that is $$2|q$$.

But this contradicts the assumption that $$(p,q)={1}$$; 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.