Square Root of 2 is Irrational/Proof 4

Theorem

$\sqrt 2$ is irrational.

Aiming for a contradiction, suppose 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

$\ds m^2$

$=$

$\ds 2 n^2$

$\ds \leadsto \ \ $

$\ds m^2 - n m$

$=$

$\ds 2 n^2 - n m$

$\ds \leadsto \ \ $

$\ds m \paren {m - n}$

$=$

$\ds \paren {2 n - m} n$

$\ds \leadsto \ \ $

$\ds \dfrac {2 n - m} {m - n}$

$=$

$\ds \dfrac m n$

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.

$\blacksquare$