$P$ and $Q$ are two distinct prime numbers. How can I prove that $\sqrt{PQ}$ is an irrational number?
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3I assume you want $P$ and $Q$ to be two distinct prime numbers. – Hagen von Eitzen Oct 23 '14 at 15:39
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More generally: If $X^2-n$ has a rational root, then it has an interger root. – Hagen von Eitzen Oct 23 '14 at 15:41
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1You should be able to adapt the proof that $\sqrt 2$ is irrational. Do you know that proof? That seems to be a logical place to start. – MJD Oct 23 '14 at 15:42
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I've seen the proof that $\sqrt 2$ is irrational. I am very fresh in Infi and don't feel confident in this yet. – Dean Oct 23 '14 at 16:19
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If $$ \sqrt{pq}=\frac{m}{n}, \quad (m,n)=1, $$ then $$ n^2pq=m^2, \tag{$\star$} $$ which means that $p\mid m^2$ and hence $p\mid m$. Thus $m=pm_1$, and $(\star)$ becomes $$ n^2q=pm_1^2. $$ But this means that $p\mid qn^2$, and as $p\ne q$ and hence $p\not\mid q$, then $p\mid n^2$, and thus $p\mid n$. Therefore, $n=pn_1$.
This is a contradiction, since $p\mid m$ and $p\mid n$, and we had assumed that $(m,n)=1$.
peter.petrov
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Yiorgos S. Smyrlis
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Could you please clarify what (m,n) = 1 means? We usually assume (m,n) exist in Z. – Dean Oct 23 '14 at 16:01
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2$(m,n)=1$ means that $m$ and $n$ are relatively prime - i.e., they do not have any common divisor other than 1. – Yiorgos S. Smyrlis Oct 23 '14 at 16:26
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Once you get to $n^2pq=m^2$ you're actually done - because a prime number has to appear an even number of times in the prime decomposition of a number squared, and in this case since $p ≠ q$ both will appear an odd number of times in the prime decomposition of $m^2$, since either they appear an even number of times in $n^2$, or they don't at all and either way it's an odd number of appearances in the left side of the equation - which is a contradiction. – Elad Avron Mar 26 '15 at 17:19
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Proof: Assume, to the contrary, that $\sqrt{pq}$ is rational. Then $\sqrt{pq}=\frac{x}{y}$ for two integers $x$ and $y$ and we further assume that $gcd(x,y)=1$. Observe that $pqy^2=(qy^2)p=x^2$. Since $qy^2$ is an integer, $p\mid x^2$ and by Euclid's Lemma, $p\mid x$. Thus $x=pd$ for some integer $d$ and so $pqy^2=x^2=(pd)^2$ and so $qy^2=p(d^2)$. Hence, $p\mid qy^2$. Since $p,q$ are distinct primes, $p \neq q$ and so $p \nmid q$, which means $p\mid y^2$. Thus, $p\mid y$. This contradicts our assumption that $gcd(x,y)=1$.
Alexander Berliner
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If you follow through the usual proof that $\sqrt 2 $ is irrational, it goes through in this case as well. One of them (for $\sqrt 3$) is here, but a search for irrational+sqrt will find many choices
Ross Millikan
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