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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.

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cactus314
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    Infinite descent can very much happen in $\Bbb Z$. In $\Bbb N$, on the other hand... – Arthur Feb 08 '17 at 06:34
  • @Arthur $\mathbb{Z}^3$ will have an $\mathbb{N}$-valued height. If have two or three numbers or more, the height is the max of the absolute values e.g. $ \mathrm{ht}(2, -3, -4 ) = \max{ |2|, |-3|, |-4| } = 4$. And then, by the well-ordering principle, there can be no infinite descent on $\mathbb{N}$, it has to stop at $0$. – cactus314 Feb 15 '17 at 16:58
  • That's true. I was just referring to the sentence "an infinitely decreasing sequence (which can never happen in $\Bbb Z$)." It can happen in $\Bbb Z$. You probably just meant to say $\Bbb N$. – Arthur Feb 15 '17 at 17:07
  • Comment on your comment : the proof of contradiction is bad when we proof the existence of something using contradiction without providing a construction strategy, there is nothing wrong with proving the nonexistence by contradiction, after all : the word “irrational” means “not equal to $p/q$ for any pair of integers $(p,q)$“. But you can see a more detailed article in Gowers's Blog – Elaqqad Feb 16 '17 at 20:21
  • @elaqqad I apologize if my argument is not picture-perfect. mainly my question is about the use of descent and why that is an instance of cohomology – cactus314 Feb 16 '17 at 21:33
  • I remember the proof of $\sqrt{2}$ not being rational making use of the unique prime factorization: $2b^2$ has an odd number of factors 2, where $a^2$ has an even number - hence they can never be equal. For Galois cohomology and infinite descent I advise you to have a look at https://www-fourier.ujf-grenoble.fr/~berhuy/fichiers/NTUcourse.pdf . – Maestro13 Feb 20 '17 at 19:04
  • Selmer groups..? – Ben Feb 20 '17 at 21:20
  • @Maestro13 What you have just done is a proof by contradiction. You want to prove that $\sqrt 2$ is irrational, you assume that it's rationnal hence there must be some integers $a$ and $b$ such that $a^2 = 2 b^2$ but this is impossible because $2b^2$ has an odd number of factors $2$, hence $\sqrt 2$ is irrational. The argument here uses the proof by contradiction . As Terence Tao said in the link in Gowers Blog you can not escape proof by contradiction for this problem ! – Elaqqad Feb 21 '17 at 20:48
  • Ah wait you want a positive proof of $\sqrt{2}$ being irrational. Well I think I have one for you: $\sqrt{2}$ is a root of $x^2-2$ but that is an Eisenstein polynomial which cannot have any rational roots. – Maestro13 Feb 21 '17 at 21:29
  • Hm maybe the proof for Eisenstein polynomials uses proof by contradiction also. Man, you should take up studying intuitionism if you hate that kind of proof so much. Brouwer reconstructed a large part of mathematics without it - you will probably love it. – Maestro13 Feb 21 '17 at 21:48

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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/.

Maestro13
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  • how you read the Wikipedia article on infinite descent? I at least want you to see I'm not making things up – cactus314 Feb 21 '17 at 22:36
  • I have and I know a little about Galois theory, but not about cohomologies. And I don't fancy reading up on it now. Note that I submitted an answer only because the text was too long for a comment, not for trying to snitch the bounty. – Maestro13 Feb 21 '17 at 22:42
  • "Try find the use of a contradiction in this story" — Easy: "[…] 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." – celtschk Feb 21 '17 at 22:49
  • Let me reformulate. If $2q^2 = b^2$ for two integers $q$ and $b$, then apply prime factorization left and right. Now the argument of odd/even number of prime two's leads to concluding that the $q$ we tried does not work, so let's try the next one. But none will work. Of course there is an infinity argument in this, and there might be a use of contradiction in there, that much I admit: "if no $q$ works, then it must be an irrational" looks suspiciously like a proof by contradiction. – Maestro13 Feb 21 '17 at 23:02
  • I just read a nice article, see http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/ . It also relates to the remark of Elaqqad above. The point is we have to prove negation but that is not the same as supplying a proof by contradiction. – Maestro13 Feb 22 '17 at 00:04