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I came across a nice-looking proof that $\sqrt{2}$ is irrational here. It somehow seems to good to be true. What are the assumptions being made in the proof and if this proof is indeed correct, why is not as popular as the classic proof? Is the latter burdened with more trivial assumptions? I've always found that part about $p$ and $q$ being in lowest terms slightly lacking in rigor.

Martin Sleziak
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Simeon
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    This 'new' proof also suggests a fraction in lowest terms. – anak Jun 19 '15 at 19:25
  • It seems fine. Maybe it is not as popular because it is more general. Regarding the lowest term rigor, its rigor is usually skipped over when learning the proof. https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent – Jonathan Hebert Jun 19 '15 at 19:26
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    What is unrigorous about "lowest terms"? - it is equivalent to the assertion that the highest common factor of numerator and denominator is $1$. – Mark Bennet Jun 19 '15 at 19:26
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    Here is a proof by Fermat. Suppose that $\sqrt{2}$ is rational. Then the right-angled triangle with sides $(1,1,\sqrt{2})$ has rational sides length and area $1$. Hence $1$ is a congruent number, i.e., $x^4+y^4=z^4$ has a nontrivial integral solution. Contradiction. – Dietrich Burde Jun 19 '15 at 19:30
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    It's not unrigorous, but it does rely on facts about factorization of integers. Specifically, that rationals can be reduced to lowest terms, such that if $B/A$ is in lowest terms and $B/A = b/a$, then there is an integer $c$ such that $b=cB$ and $a=cA$. It's true, but it has to be proved. – Robert Israel Jun 19 '15 at 19:30
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    @DietrichBurde Nice, adding to the archive. (Although showing from the integer axioms up that $x^4 + y^4 = z^4$ has no solutions is probably more demanding that the standard proof! ;-) ) – Simon S Jun 19 '15 at 19:40
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    @SimonS yes, of course. This is not a "very simple proof", but a nice one. – Dietrich Burde Jun 19 '15 at 19:43
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    I find it just a complicated way to say that if $\sqrt{N}=A/B$, then $N=\sqrt{A^2}{B^2}$. – egreg Jun 19 '15 at 19:50
  • I remember the sense of "unrigor" that the op hints at. It seems that we declare the fraction to be in lowest terms, and then our contradiction that we arrive at is precisely a contradiction of that declaration that we didn't need to make. What's subtly missing in this line of thought is that although we didn't need to make that declaration, it is done without a loss of generality. – Jonathan Hebert Jun 19 '15 at 20:12
  • See here for a more conceptual presentation, and see here for another form using fractional parts. – Bill Dubuque Jun 20 '15 at 05:10
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    Here is another short proof that I don't quite remember the origin of. If $a/b = \sqrt{2}$ then $a^2 = 2b^2$, which is impossible, as the parity of the power of 2 in the prime factorization of both sides is different. – Lynn Jun 27 '15 at 13:21
  • Of course, "unique prime factorization" is another thing more advanced than the usual proof based on "even/odd". – GEdgar Jun 27 '15 at 14:04
  • @GEdgar But said proof doesn't require Existence & Uniqueness of prime factorizations; rather it needs only E & U of factorizations of the form $,2^k n,,$ for $,n,$ odd, which is much more elementary. – Bill Dubuque Jun 27 '15 at 19:45

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