Square Root of 2 is Irrational/Proof 4

Proof
that $\sqrt 2$ is rational.

Let $n$ be the smallest positive integer so that


 * $\sqrt 2 = \dfrac m n$ for some $m \in \Z_{>0}$.

Then $m=n\sqrt2>n$, so


 * (1) $m-n>0$.

We also have $m = n\sqrt2 < 2n$, so


 * $m<2n$

and therefore


 * (2) $m-n<n$.

Finally, we have

It follows that $\frac{2n-m}{m-n} = \sqrt2$.

By (1), the denominator of $\frac{2n-m}{m-n}$ is positive.

By (2), the denominator of $\frac{2n-m}{m-n}$ is less than $n$.

We have thus written $\sqrt2$ as a fraction with a smaller denominator than $n$, which is a contradiction.