If I look up infinite descent on Wikipedia we get the sample of proving that $\sqrt{2} \notin \mathbb{Q}$ -- it is irrational.
$$\sqrt{2} = \frac{a}{b} \to 2 = \frac{a^2}{b^2} \to 2b^2 = a^2 \to 2b^2 = (2c)^2 = 4c^2 \to b^2 = 2c^2 $$
I apologize for writing in this streamlined way. However we've shown that if $2b^2 = a^2$ has a solution then $b^2 = 2c^2$ has a solution And we can go back and forth forever. Such arguments appear all over this site.
The Wikipedia article suggests that - in between the lines - we have used height function since the numerator and denominator are decreasing:
$$ (|a|, |b|) \to (|b|, |c|) \to \dots $$
forms an infinitely decreasing sequence (which can never happen in $\mathbb{Z}$). Wikipedia has that:
Infinite descent was Pierre de Fermat's classical method for Diophantine equations... Descent is something like division by two in a group of principal homogeneous spaces (often called 'descents', when written out by equations); in more modern terms in a Galois cohomology group which is to be proved finite.
What is the Galois Cohomology group being used here? Was it proved finite? How did it prove the irrationality of the $\sqrt{2}$.
Comment Personally I do not like proof by contradiction. It suggests there's an alternative, constructive proof somehow, but I will not argue it here.