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I just guess that the following statement is true:

$\sqrt{d}\in{\mathbb Z}_+$ or $\sqrt{d}\in {\mathbb R}\setminus{\mathbb Q}$ for positive integer $d$?

But I don't see a way to deal with it. I think the point is to discuss $dn^2=m^2$ where $n$ and $m$ are positive integers. If I can conclude that $n|m$ then the proof will be done. Help?

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    It follows from unique factorization. Better to assume $m,n$ are relatively prime and show that $dn^2=m^2$ implies $n=1$. – Thomas Andrews Jul 26 '13 at 20:04
  • http://math.stackexchange.com/questions/4467/sqrt-a-is-either-an-integer-or-an-irrational-number http://math.stackexchange.com/questions/324724/p1-n-is-irrational-if-p-is-prime-and-n1 http://math.stackexchange.com/questions/448172/what-rational-numbers-have-rational-square-roots – Baby Dragon Jul 26 '13 at 20:38
  • Suppose to the contrary $\sqrt{d} = a/b$ with $(a,b) = 1$ and $b > 1$. Then $b^2 d = a^2$. Hence $b | a^2$, but $b > 1$ and $(a,b) = 1$ so we got a contradiction! – blabler Jul 26 '13 at 20:43

3 Answers3

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Lemma: If $a|bc$ and $(a,b)=1$ , then $a|c$.

Lemma: If $(m,n)=1$ then $(m^2,n^2)=1$.

Theorem: If $(m,n)=1$ and $dn^2=m^2$, then $d=m^2$.

Proof of theorem only: If $dn^2=m^2$, then $m^2|dn^2$ and therefore $m^2|d$ by lemmas. So $m^2|d$ and $d|m^2$ implies $d=\pm m^2$. But clearly $d$ is positive.

Both of the lemmas can be proven using the fact that if $(a,b)=1$ then $ax+by=1$ has a solution for $x,y\in\mathbb Z$.

Thomas Andrews
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  • Regarding your comment about, can I just count the prime factors on both side to reach a conclusion? –  Jul 26 '13 at 20:15
  • Nice presentation! :) – Kunnysan Jul 26 '13 at 20:15
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    @Jack Sure, that way works just as fine. This is essentially a coded version of that counting argument. Arnold Ross insisted that the first Lemma is what really ought to be called the "fundamental theorem of arithmetic," and a lot of unique factorization proofs can be quickly represented via that lemma. – Thomas Andrews Jul 26 '13 at 20:16
  • @ThomasAndrews Fair enough. I was just about to ask how you would write the proof neatly according to your comment. Thanks! –  Jul 26 '13 at 20:18
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$dn^2=m^2 $. If $p|d$ prime such that $p^{2n+1}||d$ (such a prime exists) then $p^{2n+1}||m^2$ , which is clearly impossible.

Kunnysan
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More directly, without assuming $m,n$ are coprime, you can try the following steps:

  1. $n\vert m$ if and only if for any prime $p$ and natural number $k$, if $p^k$ divides $n$, then it divides $m$ (by prime factorization).
  2. Choose arbitrary $p^k$ dividing $n$. Then $p^{2k}$ divides $n^2$ and consequently $m^2$.
  3. If $p^k$ did not divide $m$, then by uniqueness of factorization, $p^{2k}$ could not divide $m^2$, so $p^k$ divides $m$ and we're done.

In fact, you can prove a similar theorem for arbitrary UFD in the same way: any element of an UFD $R$ has a square root in $R$ if and only if it has one in the quotient field. This is a special case of a more general fact that a unique factorization domain is always integrally closed.

tomasz
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