User:Paxinum/Proof styles

The square rooth of 2 is irrational 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.