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.