# User:Paxinum/Proof styles

## The square rooth of 2 is irrational theorem

This result uses the following: [hide] Definition of rational number. Definition of prime and coprime. Definition of square rooth. Gödels incompleteness theorem =)

The square rooth of 2 is irrational, $\sqrt{2} \notin \mathbb{Q}$

## Proof

This is a proof by contradiction, so we assumes that $\sqrt{2} \in \mathbb{Q}$ and hence $\sqrt{2} = a/b$ for some a, b that are coprime.

This implies that $2 = \frac{a^2}{b^2}$. Rewriting this gives $2b^2 = a^2 \!\,$.

Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., $2 | a^2$. Since 2 is prime, we must have that $2 | a$.

So we may substitute a with $2a'$, and we have that $2b^2 = 4a^2 \!\,$.

Dividing both sides with 2 yields $b^2 = 2a^2 \!\,$, and using similar arguments as above, we conclude that $2 | b$.

Here we have a contradiction; we assumed that a and b were coprime, but we have that $2 | a$ and $2 | b$.

Hence, the assumption were false, and $\sqrt{2}$ cannot be written as a rational number. Hence, it is irrational.

## History

Some nice history about the one that first proved this theorem.