$\sqrt{2} \notin \mathbb{Q}$ and Galois Cohomology 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.


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*$\sqrt{2}\notin\mathbb{Q}$ but ...
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.
 A: Nothing to say about cohomology - I know too little about that. But I may have a positive proof of $\sqrt{2}$ not being rational, without using a contradiction.
A property of any (positive) rational is, that if you take a square with sides equal to the rational value $\frac{p}{q}$, and then duplicate the square horizontally and vertically $q$ times, then the total figure is a square with area equal to $p^2$, an integer square. So there exists at least one integer $q$ for which this all will work.
Now starting with a square with sides $\sqrt{2}$ long, and an area of 2, then multiplying by any (any!) integer $q$ as above gives a square with an area equal to $2q^2$, which is not an integer square for any integer $q$, as then 2 would be an integer square, which I think you agree it is not. Hence $\sqrt{2}$ does not have this property and hence is not a rational.
Try find the use of a contradiction in this story :-).
EDIT
promoting my remark below into this post as it is important enough to emphasize.
The proof of $\sqrt{2}$ being irrational does not generally make use of proof by contradiction. It simply states '$\sqrt{2}$ is rational' and then derives false, thereby proving the negation of the statement, which is '$\sqrt{2}$ is not rational'. See http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/.
