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So the proof is as follows:

For a contradiction, suppose that $\sqrt{2} = \dfrac{m}{n}$ is a ratio of integers. Squaring both sides and cross-multiplying gives $2n^{2} = m^{2}$. Note that 2 divides the left-hand side an odd number of times while 2 divides the right-hand side an even number of times. Thus we have an odd number equal to an even number, a contradiction.

I'm not quite sure how 2 divides the left-hand side an odd number of times and divides the right-hand side an even number of times. Can someone explain this please? I feel like its obvious, but I just can't wrap my head around it.

Nostradamus
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  • If $2$ divides $n$ $k$ times, then it divides $n^2$ $2k$ times. – eyeballfrog Jul 19 '22 at 18:09
  • By existence & uniqueness of prime factorizations, $2$ occurs to even power in the squares $n^2,m^2$ so odd power in $2n^2$. More simply, it suffices to prove existence and uniqueness of factorizations only for the prime $2$, i.e. factorizations the form $2^j k$ for $k$ odd, see this answer in the linked dupe., – Bill Dubuque Jul 19 '22 at 18:15
  • @BillDubuque While I don't feel strongly enough to vote to reopen, I don't really think this is a duplicate - this question is asking specifically about the justification of a particular step, not for any proof of the irrationality of $\sqrt{2}$. Certainly this doesn't deserve a downvote (although that may not have been you)! – Noah Schweber Jul 19 '22 at 18:24
  • @Noah The proper justification is already supplied in the linked dupe (and many other answers). There is no reason to duplicate these arguments ad infinitum (and it greatly damages site health by doing so). In fact your answer did not originally give a rigorous argument until after I linked to the dupe and posted the prior comment (and it's still not completely rigorous). – Bill Dubuque Jul 19 '22 at 18:38
  • @BillDubuque I also don't see how its a duplicate question, it really doesn't help when I'm trying to learn and my question is being shut down. Now I'm not even able to post another question for 2 days. I wasn't asking for a proof, I was just asking for the justification of a step. – Nostradamus Jul 19 '22 at 18:48
  • @Nostradamus If you have any questions about the justification given in the linked proof then please post them in comments there. This is the way we improve content on the site (with the hope of eventually reaching "proofs from the book" through after enough feedback) – Bill Dubuque Jul 19 '22 at 18:52
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    To be honest I think this proof did fail to justify the step @Nostradamus had trouble with. A simple justification is if we assume that $2$ must divide into $m$ a certain number of times (call it $k_m$) and divide into $n$ a certain number of times (call it $j_m$) then $2$ divides into $2n^2$ a total of $2k_n + 1$ times (which is an odd number) and into $m^2$ a total of $2j_m$ times (which is an even number). Um, where is this proof from. My opinion of the proof will vary depending on whether it was a fellow student doing the best she can (it's good) or a professor trying to rush it by. – fleablood Jul 19 '22 at 21:57
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    FWIW I feel a lot of proofs try to ram through UNIQUE FACTORIZATION (this proof more than most) or that all rationals can be reduced to lowest terms (which is an assumption of UNIQUE FACTORIZATION) when I don't think UNIQUE FACTORIZATION had been proven yet. Euclid's original proof didn't assume UF but relied on infinite descent... which has the advantage of being more basic. But the disadvantage of being tedious and abstract and liable to lose the students in the hand-waving of why we can't have it go on forever. – fleablood Jul 19 '22 at 22:05
  • @fleablood It was my professor, during a lecture he showed different proofs for the the irrationality of $\sqrt{2}$ with varying levels of detail and complexity, this was just one of them. Actually the proof he presented after this touched upon the fundamental theorem of arithmetic and unique factorizations. Thanks for your comment, it really does make sense now! Appreciate it! – Nostradamus Jul 20 '22 at 12:12
  • FWIW I'd say 90% of the proofs I see on this like to brush everything aside with a "Let $r =\frac mn$ where $\frac mn$ is in lowest terms" or "Let $r = \frac mn$ where at most one of $m$ and $n$ are even and at least one is odd". It's usually acceptable to students this is possible (although without unique factorization it isn't really) and then the proof goes on to show that if $2n^2 = m^2$ then both $m$ and $n$ are even which is a contradiction. (without UF though you can do INFINITE DESCENT if $n,m,\frac n2, \frac m2,\frac n4, ...etc$ must all be even but these numbers get smaller so – fleablood Jul 20 '22 at 16:58

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Think about factorization. We can always write a natural number $k$ as $2^ma$ for some odd $a$. But this means that $k^2=2^{2m}a^2$, and if $a$ is odd $a^2$ will also be odd. This means that if $2$ divides $k$ exactly $m$ times, then $2$ divides $k^2$ exactly $2m$ times.

More snappily, we've just proved the following:

For any $k$, the largest power of $2$ dividing $k^2$ is even.

This handles the right hand side. Do you see why it also handles the left hand side? (Think about how many times $2$ divides into $n^2$ ...)

Note that this argument is actually using some nontrivial assumptions about how factorization works. Most obviously, I'm assuming that every number has a prime factorization. But that's not all! There's another assumption I'm making, which is more subtle and harder to prove ...

I need prime factorizations to be unique: in particular, we need to rule out the possibility of $2^uw=2^ls$ for $w,s$ odd and $u\not=l$.

Noah Schweber
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