How is this a proof of the irrationality of $\sqrt2$

Question

Proof. Suppose for the sake of contradiction that $\sqrt2$ is rational, and choose the least integer, $q \gt 0$, such that $(\sqrt2 − 1)q$ is a non negative integer. Let $q':=(\sqrt2 − 1)q$. Clearly $0 \lt q' \lt q$. But an easy computation shows that $(\sqrt2 − 1)q'$ is a nonnega­tive integer, contradicting the minimality of $q$.

Now here's my question. If $(\sqrt2-1)q$ is a non-negative integer, and $q$ is an integer, then $(\sqrt2-1)$ must be an integer. So the contradiction about the minimality of $q$ only proves that $(\sqrt2-1)$ cannot be an integer. How does it prove the irrationality of $\sqrt2$?

Answer 1

The proof has the following form: $\,\sqrt{2}\in\Bbb Q\,\Rightarrow\,\sqrt{2}-1\in \Bbb Q\,\Rightarrow\,\sqrt{2}-1 = q'/q\, \ldots = n/q',$ which, since $\,q'< q,\,$ contradicts the choice of $q$ as the least denominator of $\,\sqrt{2}-1.$

That contradiction implies that the initial hypothesis $\,\sqrt{2}\in\Bbb Q$ is false, which completes the proof.

One way to view the proof is this: $\sqrt{2}\in\Bbb Q\iff\sqrt{2}-1\in \Bbb Q\,$ so it suffices to show $\,\sqrt{2}-1\not \in \Bbb Q$

Note $\ \ \overbrace{\sqrt{2}\!-\!1}^{\large q'/q}\, =\, \color{#0a0}{\sqrt{2}\!+\!1} - 2\, =\,\color{#c00}{\underbrace{\smash[t]{\color{black}{\overbrace{\color{#0a0}{\dfrac{1}{\sqrt{2}\!-\!1}}}^{\large \color{#c00}{q/q'}}}}\color{#c00}{ -2}}}\ \ $ by $\,\ \color{#0a0}{(\sqrt{2}-1)(\sqrt{2}+1)} = 2-1 =\color{#0a0} 1.$

So we obtain $\smash{\ \sqrt{2}-1 \, =\, \dfrac{q'}{q} \ =\ \color{#c00}{\dfrac{q-2q'}{q'}}}\ $ with smaller denominator $\ q' < q,\ $ contradiction.

Answer 2

By the way of Contradiction, assume that $\sqrt{2}$ = $\frac{p}{q}$ where p and q are integers with no common factor ( fraction reduced to lowest terms).

Then $\frac{p^2}{q^2}$ = 2

Since $p$ and $q$ are integers so $p$$^2$ = 2 $q$$^2$ $\implies$ $p$$^2$ is even $\implies$ $p$ is even.

So $p$ can be expressed as; $p$ = $2$$k$ for some integer $k$.

Then from $\frac{p^2}{q^2}$ = 2 $\implies$ $\frac{(2k)^2}{q^2}$ = 2 $\implies$ $4k^2$ = 2 $q$$^2$ $\implies$ $q$$^2$ = 2 $k$$^2$ $\implies$ $q^2$ is even $\implies$ $q$ is even.

Which contradicts the fact that $p$ and $q$ do not have common factor (contrdicts that fraction $\frac{p}{q}$ is in lowest terms).

So we can not express $\sqrt{2}$ in the form of fraction, $i.$$e.$ $\sqrt{2}$ is an irrational number.