Shorter proof of irrationality of $\sqrt{2}$?
Here is one of my favorite proofs for the irrationality of $\sqrt{2}$.
Suppose $\sqrt{2}\in\Bbb Q$. Then there exist an integer $n>0$ such that $n\sqrt{2}\in\Bbb Z$.
Now, let $n$ be the smallest one with this property and take $m=n(\sqrt{2}-1)$. We observe that $m$ has the following properties:
- $m\in\Bbb Z$
- $m>0$
- $m\sqrt{2}\in\Bbb Z$
- $m<n$
and so we come to a contradiction!
Yes, that is the point - the simple parity-based proof does not require existence and uniqueness of prime factorizations so it can be used before those much deeper results are developed.
It is illuminating to note that the proof based on comparing the parity of powers of two does not need the full uniqueness result but only a much simpler case. Namely, it is easy to prove that every natural can be written uniquely in the form $\, 2^i n\,$ for odd $\,n.\,$ Thus $\, A= 2^{\large J} a,\ B = 2^{\large K} b\,$ for $\,a,b\,$ odd. Therefore $\, A^2\! = 2 B^2$ $\Rightarrow$ $\,2^{\color{#c00}{\large 2J}} a^2 = 2^{\color{#0a0}{\large 1+2K}} b^2.\ $ But $\,a,b\,$ odd $\,\color{#f2f}\Rightarrow$ $\,a^2,b^2\,$ odd too, which contradicts uniqueness (LHS has $\rm\color{#c00}{even}$ vs. RHS $\rm\color{#0a0}{odd}$ power of $2).$
To generalize that proof from $\,p = 2\,$ to an arbitrary prime $\,p\,$ would require generalizing the $\,\rm odd^2 = odd\,$ step that we used above, $ $ i.e. we would need to prove $\ p\nmid a,b\,\color{#f2f}\Rightarrow\, p\nmid a^2,b^2.\ $ Just as for the case $\,p=2,\,$ for any fixed prime $\,p\,$ we could prove it by a brute-force case analysis. But to prove the result for all primes $\,p\,$ we need something much deeper, e.g. Euclid's lemma $\ p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\ $ (or closely related results).
Here is a funny proof of the fact that $\sqrt{m}$ is irrational for non-square $m$. I have seen in it in American Monthly. It is a very cute overkill.
Let $m$ be a non-square integer and suppose that $\sqrt{m}$ is rational, that is, $\sqrt{m}=a/b$ for some coprime integers $a,b$. Well, $mb^2 = a^2$, so $a^2$ is in the ideal generated by $b^2$. However $b^2\neq 1$, so the ideal $b^2\mathbb{Z}$ is contained in some maximal ideal $p\mathbb{Z}$, where $p$ is a prime. Thus, $p|a^2$ and $p|b^2$, that is to say, $p|a$ and $p|b$; a contradiction.