# 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, ${\displaystyle {\sqrt {2}}\notin \mathbb {Q} }$

## Proof

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

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

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

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

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

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

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

## History

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