Below is a simple proof of irrationality of square-roots that I discovered as a teenager (inspired by a proof of Dedekind). It employs the Bezout identity for the gcd, i.e. $\rm\,\gcd(a,b)\,$ is an integral linear combination of the integers $\rm\,a,b,\,$ i.e. $\rm\,\gcd(a,b)\, =\, a d - b c\,$ for some integers $\,\rm c,d$.

**Theorem** $\ \ \ \rm r = \sqrt{n}\;\;$ is integral if rational, $\:$ for $\:\rm n\in\mathbb{N}$

**Proof** $\ \ $ Note that $\rm\ r = a/b,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad\!-\!bc \,=\, \color{#C00}{\bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\ $ by Bezout.

$\rm\color{#C00}{That\,}$ and $\rm\: r^2\! = \color{#0a0}{ n}\:\Rightarrow\ 0 \,=\, (a\!-\!br)\, (c\!+\!dr) \, =\, ac\!-\!bd\color{#0a0}{ n} \:+\: \color{#c00}{\bf 1}\cdot r,\ $ thus $\ r \in \mathbb{Z}$

This generalizes to roots of *monic* quadratic polynomials (and to higher degree, see here).

**Theorem** $\,\ $ If $\rm\,\ r^2 =\, \color{#0A0}{m\ r + n}\ \,$ for $\rm\ m,n\in\mathbb Z\ $ then $\rm\ r\in \mathbb Q\ \Rightarrow\ r\in\mathbb Z$

**Proof** $\quad \rm r = a/b\in \mathbb Q,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad\!-\!bc \,=\, \color{#C00}{\bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\ $ by Bezout.

So $\rm\,\ 0\, =\, (a\!-\!br)\: (c\!+\!dr)\, =\, ac\! -\! bd(\color{#0A0}{m\:r\!+\!n})+\color{#C00}{\bf 1}\cdot r \, =\, ac\!-\!adm\!-\!bdn + r \ \Rightarrow\ r \in \mathbb{Z}$

Alternatively denominator descent can be done by the Division Algorithm instead of Bezout.

**Theorem** $\, \ $ If $\rm \ n\in\Bbb Z_{\phantom{\frac{i}.}} $ and $\rm \ r = \sqrt{n}\in \Bbb Q\ $ then $\rm \ r\in \Bbb Z$

$\rm \begin{align}\\[-2em]
& {\bf Proof}\,\ \ {\rm Deny,\ \,so}\,\ \ \smash[t]{\overbrace{ \rm r = {\small \frac{a}b}}^{\!\!\!\rm\large\color{#c00}{\ b\:\!r\ =\ a}}},\ \ {\rm and}\,\ \ \rm \color{#b0f}{b\nmid a}\,\ {\rm by}\,\ r\not\in\Bbb Z. {\rm\, wlog}\,\ b = \color{#0a0}{\rm least}\ {\rm denom}. \\[.1em]
&\rm \color{#c00}ar\, =\, (\color{#c00}{br})r\, =\, bn.\ \ a = qb+\bar a,\,\ \color{#b0f}{0 < \bar a} < b\ \ \text{via $\rm \,a\div b\, $ (Division Algorithm)}\\[.1em]
&\rm\!\!\vphantom{\dfrac{}{|_{|_{|_|}}}}\smash[b]{\underbrace{\bar a r\! =\! a\color{#c00}r\!-\!q\color{#c00}{br}}_{\rm\Large (\bar a \ =\ a\, -\, q b)\:\!r}}\!\! = bn\!-\!q\color{#c00}a := j\in\Bbb Z\,\Rightarrow\, r = \frac{j}{\color{#b0f}{\bar a} },\,\ \text{contra }\, to\, \ b = \text{ $\rm \color{#0a0}{\rm least}\,$ denom}.\end{align}$

**Remark** $\ $ It is instructive to compare the various denominator *descents* employed.

The denom descent in the first proof is $\rm\,a,b\,$ denom $\rm\,\Rightarrow\, \gcd(a,b)\,$ [$= 1$] $ $ denom.

In the last proof the denom descent is: $\rm\,a,b\,$ denom $\rm\,\Rightarrow\, a\ {\rm mod}\ b\,=:\, \bar a\:$ denom.

In this answer, the denom descent is: $\rm\, a\!>\!b\,$ denom $\rm\,\Rightarrow\ a\:\!-\:\!b\ $ denom.

These are all essentially special cases of proofs that the ideal $\,a\Bbb Z + b\Bbb Z = \Bbb Z\,$ using fast or slow descent based on the (Euclidean) Division algorithm (with remainder), i.e. specializations of a proof that ideals are principal in a Euclidean domain (compare posts on *denominator ideals*).

That the numerator $\rm\,a = br\,$ is also a denom of $\rm\,r\,$ generalizes to any algebraic integer $\rm\,r,\,$ most efficiently by using Dedekind's **conductor ideal**. This generalizes the above proofs into a slick one-line proof that PIDs are integrally-closed, as I explained at length elsewhere. It beautifully abstracts the denominator descent that governs ad-hoc "elementary" irrationality proofs.