Square Root of 2 is Irrational/Proof 4

Proof
that $\sqrt 2$ is rational.

Let $n$ be the smallest positive integer such that:
 * $\sqrt 2 = \dfrac m n$

for some $m \in \Z_{>0}$

Then:
 * $m = n \sqrt2 > n$

so:
 * $(1): \quad m - n > 0$

We also have:
 * $m = n \sqrt2 < 2 n$

so:
 * $m < 2 n$

and therefore:


 * $(2): \quad m - n < n$

Finally, we have

It follows that:
 * $\dfrac {2 n - m} {m - n} = \sqrt2$

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

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

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