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.

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    $\begingroup$ This 'new' proof also suggests a fraction in lowest terms. $\endgroup$ – anakhro Jun 19 '15 at 19:25
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    $\begingroup$ What is unrigorous about "lowest terms"? - it is equivalent to the assertion that the highest common factor of numerator and denominator is $1$. $\endgroup$ – Mark Bennet Jun 19 '15 at 19:26
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    $\begingroup$ 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. $\endgroup$ – Dietrich Burde Jun 19 '15 at 19:30
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Jun 19 '15 at 19:30
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    $\begingroup$ @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! ;-) ) $\endgroup$ – Simon S Jun 19 '15 at 19:40

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