Square root of 2 is not rational

  The Pythagorean school of mathematics originally insisted that only natural and rational numbers exist. The discovery of the irrationality of $\sqrt{2}$, the ratio of the diagonal of a square to its side (around 5th century BC), was a shock to them which they only reluctantly accepted. $^1$
  From the Pythagorean theorem we have that 

 \[ \alpha^2 = \beta^2 + \beta^2 \Leftrightarrow \] \[   \alpha^2 = 2\beta^2  \Leftrightarrow \] \[ \frac{\alpha^2}{\beta^2} = 2 \] and setting $ \frac{\alpha}{\beta} = \rho $ we end in the relation \[ \rho^2 = 2. \] So we have to proove that:

There is no rational number $ \rho$, with $ \rho^2 = 2 $.

Proof




  We have a rational number $\rho = \frac{m}{n}$ with $m, n $ integers which are not both even (if they were, we would simplify with 2 until one of them become odd)(1).
  Then let $\rho$ a rational number with the  property $\rho^2 = 2$. Then we will have \[ \left( \frac{m}{n} \right)^2 = 2   \Leftrightarrow \] \[ \frac{m^2}{n^2} = 2 \Leftrightarrow \] \[  m^2 = 2n^2 \]
  Therefore $ m^2 $ is even (it is a multiple of 2) then $m$ is even. If it was not even, then $ m^2 $ would be odd as the square of an odd number.
  So $m$ is even and $ m = 2k $ for a $ k \in \mathbb{Z} $. So we have: \[ m^2 = 2 n^2 \Leftrightarrow \] \[  (2k)^2 = 2n^2 \Leftrightarrow \] \[  4k^2 = 2n^2 \Leftrightarrow \] \[   2k^2 = n^2. \]
  Like $m$ so $n$ will be even or else $n^2$ would be odd.
  So $m$ and $n$ are even.
  Thus contradiction from (1).
  Thus there is no rational number  $ \rho$, with $ \rho^2 = 2 $. $ \square $
QED

Notes
  1. https://en.wikipedia.org/wiki/Foundations_of_mathematics

References

• Walter Rudin: Principles of Mathematical Analysis, p.2