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I'm trying to prove that $2\sqrt{pq}$, with $p$ and $q$ primes and $p \neq q$, is irrational. I have something like this:

Let's suppose that $2\sqrt{pq}$ is rational. Then, we have that $2\sqrt{pq} = \frac{m}{n}$, with $m,n \in \mathbb{Z^{+}}$, and $(m,n) = 1$.

Now, I need to prove that $m \neq 1$ and $n \neq 1$. When I try to prove that $m \neq 1$ I get:

If $m = 1$, then: \begin{align*} 2\sqrt{pq} &= \frac{m}{n}\\ 2n\sqrt{pq} &= 1\\ \end{align*} This implies $2 \vert 1$, wich is false. Thus, $m\neq1$.

But I don't really know how to prove that $n \neq 1$. The most I can get is:

If $n = 1$, then: \begin{align*} 2\sqrt{pq} &= \frac{m}{n}\\ 2\sqrt{pq} &= m\\ \end{align*} And this implies $2\sqrt{pq}$ is an integer.

But I don't know how to continue after this.

Anonymous
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Rararat
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    You have to use somewhere the fact that $p$ and $q$ are distinct primes. Otherwise the statement is false. Nowhere in your proof does this fact appear. – EuYu Oct 30 '20 at 05:09
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    Continuing with what you did, square both sides to get $4pq = m^2$ , do you get the proof now? ($p,q$ are distinct primes) . – Anonymous Oct 30 '20 at 05:11
  • @Anonymous Thanks, I get it now. $(2p)(2q) = m^{2}$ and this is impossible since $p \neq q$, right? – Rararat Oct 30 '20 at 05:15
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    Yeah, it's done. – Anonymous Oct 30 '20 at 05:16
  • @EuYu So, I have that $(2p)(2q) = m^{2}$ and this is impossible since $p\neq q$, right? – Rararat Oct 30 '20 at 05:16
  • @Anonymous Thank you. – Rararat Oct 30 '20 at 05:16
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    @RaulPeña Depending on what this problem is for, I wouldn't say you're done with $(2p)(2q) = m^2$. There is no explicit contradiction reached here yet. If this were for a first course in number theory, I don't think that stopping here is an acceptable solution. You should continue (using Euclid's lemma) until you reach an explicit contradiction by descent somewhere . – EuYu Oct 30 '20 at 05:18
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    I can't believe no one has mentioned it yet, but feel free to drop the factor of $2$. It suffices to show $\sqrt{pq}$ is irrational, for if $2\sqrt{pq}$ were rational, then $\sqrt{pq} = \frac{1}{2} \times 2\sqrt{pq}$ is the product of two rational numbers, and hence is rational. If you can prove this conclusion false, then $2\sqrt{pq}$ is irrational. – user837206 Oct 30 '20 at 05:27

1 Answers1

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If $2\sqrt {pq} = a/b$ with $(a,b) = 1$ and $p,q$ distinct primes, then $4pq = a^2/b^2$ so that $b^24pq = a^2$ which implies that $p | a$, and $a=pk$. Then $4pqb^2 = p^2k^2$. Then $p | 4q$ and if $p > 2$ then $p | q$, a contradiction. You might consider the case where $p = 2$ and $q = 3$ in the following way:

If $2\sqrt6 = a/b$ with $(a,b) = 1$ then $24b^2=a^2$ and $3|a$. Then $24b^2 = 9k^2$ so that $8b^2 = 3k^2$ which implies that $3|b$, a contradiction.

Derek Luna
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