Theodor Estermann proved the irrationality of $\sqrt2$ without rely...

Exactly why is it sufficient to prove sqrt(m) - n is irrational? Th...

Note that $q$ is by assumption the smallest integer we can use as a...

In fact it is even possible to generalize: for positive integers $m...

Note that $q$ is by assumption the smallest integer we can use as a denominator in $\alpha$. Since $m,n,q$ and $p$ are integers $r=(m-n^2)q-2np$ is also an integer and so we were able to rewrite $\alpha$ as a quotient of 2 integers but this time with $p$ in the denominator. As $p<q$ we have reached a contradiction and $\sqrt m$ must be irrational.
You can remove the first $k$ digits in the decimal expansion of any irrational number without changing its irrationality. So therefore a number is irrational iff its fractional part is also irrational. Since $n < \sqrt{m} < n+1$, $\sqrt{m} - n$ is precisely this fractional part.
Exactly why is it sufficient to prove sqrt(m) - n is irrational? This stopped me cold.
rational + irrational = irrational (if rational + irrational = rational we would get irrational = rational - rational, but rationals are closed under addition and subtraction). Hence (sqrt(m) - n) + n = sqrt(m) is irrational (irrational + rational = irrational)
This sort of argument is common for properties that are closed under some operation (like rationality with addition).
In fact it is even possible to generalize: for positive integers $m$ and $k$, $^{k}\sqrt m$ is rational only when $m$ is a perfect $k^{th}$ power.
To prove this let us suppose that $^{k}\sqrt m=\frac{a}{b}$ where $a,b>0$ and are integers and the Greatest Common Divisor between $a$ and $b$ is 1. Now
$$
^{k}\sqrt m=\frac{a}{b} \equiv m = \frac{a^k}{b^k} \equiv mb^k = a^k
$$
If $b$ has a prime factor $c$ that does not
divide $a$, then $c$ does not divide any positive power of $a$. If $b>1$, $a^k$ can be written in terms of powers of $b$ ($a^k=mb^k$) and consequently contradicts the initial assumption, since $a$ and $b$ don't have any common divisors apart from 1. The only option we have is $b=1$ and in that case $m=a^k$ is a perfect power.
Theodor Estermann proved the irrationality of $\sqrt2$ without relying on the prime factorization of $m$. He claimed that his proof was even easier than [Pythagoras' proof](https://en.wikipedia.org/wiki/Square_root_of_2#Pythagorean_theorem_proof).
Estermann started by defining a set $S$ of all natural numbers $n$ such that $n\sqrt2$ is an integer. If the set $S$ was not empty, it would have a least element $k$, for instance. Now if we consider:
$$
(\sqrt2 -1)k\sqrt 2=2k -k\sqrt2
$$
Since $k\in S$, both $(\sqrt2 -1)k$ and $2k -k\sqrt2$ are naturals. So by definition $(\sqrt2 -1)k \in S$. However $(\sqrt2 -1)k < k$, contradicting the initial assumption that k is the least element of S. This way we conclude that $S$ is empty and $\sqrt2$ must be irrational.
Flanders generalizes the proof of Estermann by replacing 2 by $m$ and replacing 1 by the greatest integer less than $\sqrt m$.