I mean the argument that uses the fundamental theorem of arithmetic. Doesn't this theorem assume that p, if we are proving √p is irrational, is prime? However I see that proofs online of any other number that's not a perfect square use the same logic for proof (e.g √6).
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3It works for all squarefree numbers. – TravorLZH Feb 17 '23 at 13:42
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2It works for all nonsquare numbers – lhf Feb 17 '23 at 13:43
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It is not necessary to $m$ be a prime to do the argument on $\sqrt{m}$, just to not be a perfect square, as in that case it is rational.
As we know every natural has a prime factorization so you can take $p$ the smallest prime that divides $m$ and its power is not even in the factorization, and do kind of the same argumentation.
- the argumentation
let's suppose $\sqrt{m} = \frac{a}{b}$ with $a$ and $b$ naturals and coprimes, then we have that $ m = \frac{a^2}{b^2} \iff mb^2 =a^2$ then as $p \mid m \implies p \mid mb^2 = a^2 \implies p\mid a$ , lets call $a = a_1p$ then we have that $mb^2 = a_1^2p^2$ in here we have two options $p^2 \mid m$ or not if it doesn´t we conclude the same way. but if it does you have that at least $p^3\mid m$ because the power of $p$ in $m$ is not even, and dividing by $p^2$ you arrive to the beginning but with a $m_1 = \frac{m}{p^2}$ so this algorithm will end in a finite number of cases, then you conclude that $a_n$ and $b$ are not coprime and that is a contradiction.
J. W. Tanner
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 17 '23 at 15:30