Why is this proof that $\sqrt{2}$ is irrational titled as "Proof by infinite descent"?

Question

I am reading this wikipedea article on the proof of irrationality of $\sqrt{2}$. It uses the principle of infinite descent. I understand it as:

1. We assume $\sqrt{2}=\dfrac pq$, where $p$ and $q$ are some positive integers.
2. We have $2q^2 = p^2 \implies$ $p$ is even, i.e. $p=2r$ for some positive integer $r$.
3. Now we have $2q^2=(2r)^2 \implies q$ is also even, i.e. $q=2s$ for some positive integer $s$.

From steps 1,2 and 3 we conclude that both $p$ and $q$ have $2$ as their factor at least one time. So we can say $\sqrt{2}=\dfrac pq = \dfrac rs$.

Now we can repeat the steps 1,2 and 3 with $\sqrt{2}= \dfrac rs$. This will eventually imply that intezers $r$ and $s$ also have 2 as their factor at least one time, or $r=2r_1$ and $s=2s_2$. Recalling $p=2r$ and $q=2s$ implies that $p$ and $q$ have $2$ as their factor at least two number of times.

The notable fact is that we can repeat steps 1,2 and 3 infinite number times, which implies that $p$ and $q$ have 2 as their factor infinite number of times, but this can't be for any finite $p$ and $q$, that is to say their does not exist any intezer which can have 2 as its factor infinite number of times. So we conclude that there doesn't exist any $p$ and $q$ which satisfies $\sqrt 2 = \dfrac pq$, hence $\sqrt 2$ is irrational.

Question 1: Do I understand the proof correctly?

Question 2: I noted that this proof on wikipedea is titled as Proof by infinite descent, where it does not use the concept of infinite descent at all. So why is it titled as Proof by infinite descent?

Answer 1

Answer to Question 1: Yes, that is one version of the most commonly given ("Pythagorean") proof of the irrationality of $\sqrt 2$.

Answer to Question 2: Using the proof you give, you are constructing two infinite, descending sequences $p_i,q_i$ of natural numbers by repeatedly extracting their common factor $2$. Given that there are only finitely many natural numbers less than a given natural number (in particular, less than your original $p$ and $q$), you have established an impossible infinite descent, so the classification of the proof as a proof by infinite descent is entirely correct.

Answer 2

It would seem that the first proof you linked to, and transcribed in the question, is a proper proof by infinite descent.

The proof in the second link is not a proper proof by infinite descent; it is more or less formulated as a proof of the non-existence of a minimal counterexample. An not even quite that, since the hypothesis used is not that the pair $(a,b)$ is minimal, but that it is reduced (has no common factors); as stated it depends on the preliminary result that all rational numbers can be represented by a reduced fraction. But that dependency could be avoided by rephrasing it to assume a counterexample $(a,b)$ with (say) $a+b$ minimal; clearly this implies easily that $a/b$ is reduced. Also note the reducedness is not used anywhere except at the end to detonate the contradiction; minimality would have done quite as well for that purpose.

Although close in spirit, a proof by infinite descent is not the same as a proof by non-existence of a minimal counterexample. The former establishes from any counterexample an (impossible) infinite descending chain of counterexamples. Clearly this implies that a minimal counterexample cannot exist. But a proof of non-existence of a minimal counterexample just needs to derive a contradiction from the supposed existence of a minima counterexample. The contradiction may, but does not have to, derive from the construction of an even smaller counterexample. Only if that is the case, and moreover the minimality is not used in the construction itself, can one convert the proof into one by infinite descent. Since therefore proofs by infinite descent are no easier than by non-existence of a minimal counterexample, possibly harder, and in any case the invariably end with the not very elegant "one can repeat this construction indefinitely, leading to an infinite descent, which is impossible", they are not very popular nowadays.

Proofs by non-existence of a minimal counterexample are of course very close to proofs by induction as well, more precisely of a proof by strong induction of a statement "for any $n$ there are no counterexamples up to stage $n$". But that was not really the question here.