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I'm talking specifically about a proof that I've found. I don't seem to get some parts of it. It states that if you take: $$\sqrt{n}=\frac{p}{q} \:\: \;\;p,q \in \mathbb{Z} $$ where $p$ and $q$ share no common prime factors. Now $p^2=nq^2$ so that every prime factor of $q$ divides $p^2$, and hence (by the fundamental theorem of arithmetic) it also divides $p$. We deduce that $q$ has no prime factor; thus $q=1$ and $n=p^2$.


I get up to the part where it says that every prime factor of $q$ divides $p^2$ since $p^2/q=nq\in \mathbb{Z}$. It's the next part where it says that hence it also divides $p$. Since it seems that would imply: $$ q|p^2 \implies q|p $$ which is false ($q=4$ and $p=6$ for example, it seems that if $p$ is square it's false, but true otherwise. Though I can't seem to prove that either...). It's probably just me misreading it but I can't seem to make the proof clear; I'm just wondering if anyone can explain that last bit in a little more detail.

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  • $\begingroup$ "p and q share no common prime factors." is important as $q=4$ and $p=6$ would be a case where 2 is a common prime factor and thus doesn't count. $\endgroup$ – JB King Aug 25 '15 at 22:48
  • $\begingroup$ Here's another proof that doesn't use divisibility: math.stackexchange.com/questions/263864/… It starts with $m^2 < n < (m+1)^2$ and concludes that $\sqrt{n}$ is irrational. $\endgroup$ – marty cohen Aug 25 '15 at 23:08
  • $\begingroup$ just for future reference it seems that the thing I missed is Euclid's Lemma. $\endgroup$ – Jay Aug 26 '15 at 10:46
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$r \mid a^2 \implies r \mid a$ if $r$ is prime. That is the key.

Compliments on your showing attention to detail and trying to find counterexamples. This will serve you well in mathematics.

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    $\begingroup$ Thanks! Initially I thought you meant the $q$ in the question which doesn't have to be prime but then realised that you could decompose $q$ into its prime factors and then use your implication on each one, hence why the proof had "every prime factor". Therefore $q$ must divide $p$. Again thanks for the help! Just a quick side question, do you know if it's easy to prove $q|p^2 \implies q|p$ if $q$ isn't a square (if indeed it's true); it seems simple but I don't want to take too much time trying to prove it if it's false! $\endgroup$ – Jay Aug 25 '15 at 22:44
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    $\begingroup$ @jay The condition is that $q$ should be square-free (i.e. contain no repeated prime factors), and then it is true. $q$ not being a square is not enough. Of course $p$ doesn't have to be a prime here. Try factoring $p$ and note that if $a$ is square-free, $a$ divides $b^2$ iff the prime factors of $b$ (without repetition) contain the prime factors of $a$. But that implies that $a \mid b$. Note that the implication I mention in the answer is a special condition of this more general one! $\endgroup$ – Eric Auld Aug 25 '15 at 22:53
  • $\begingroup$ That's make intuitive sense now, since obviously if there was repetition in the prime factorisation of $a$, say $p_k$, then it could be the case that $b$ may only have $p_k$ in its factorisation once and thus $a|b^2$ but clearly it would be false that $a|b$ since $p_k$ would feature twice in $a$ but only once in $b$. $\endgroup$ – Jay Aug 25 '15 at 23:19
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    $\begingroup$ Again thanks for the help! It makes sense of the seemingly odd pattern that Mathematica outputted when I tried to find counter examples, it's clear now they were all square-free! $\endgroup$ – Jay Aug 25 '15 at 23:24
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    $\begingroup$ @jay Yes, you are correct. The "(without repetition)" comment is unnecessary. :) $\endgroup$ – Eric Auld Aug 30 '15 at 22:21
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Alternative.

You can write every numbers $n$ as

$$ \prod_{p \in \mathbb{P}} p^{n_p}. $$

So $a^2 = n b^2$ is written as

$$ \prod_{p \in \mathbb{P}} p^{ 2 a_p} = \prod_{p \in \mathbb{P}} p^{ n_p} \prod_{p \in \mathbb{P}} p^{ 2 b_p} $$

whence

$$ n_p = 2(a_p - b_p), $$

thus $n_p$ is even.

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  • $\begingroup$ Thanks! I've seen this proof whilst researching the question, I think I prefer it the one in the question $\endgroup$ – Jay Aug 25 '15 at 22:47
  • $\begingroup$ Yep. This answer contains 'alternative' so you can look at it. Sometimes that could be helpfull. Good luck with your quest and questions! $\endgroup$ – johannesvalks Aug 25 '15 at 22:51

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