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I got to wondering if a non-zero integral multiple of any irrational number is guaranteed to be irrational? This seems intuitive but I can't prove it to myself.

There is an answer on this site with regard to factors, but not multiples of irrationals.

If someone could sketch a quick proof, my day would be golden. Thanks!

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Chris Bedford
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    Let $\alpha$ ba an irrational number, $m$ a nonzero integer. Suppose for contradiction that $m \alpha$ were rational, Then we would have $$m \alpha = \frac{p}{q}$$ for some integers $p$ and $q \neq 0$, giving $$\alpha = \frac{p}{mq},$$ which contradicts the fact that $\alpha$ is irrational. – Jordan Green Jun 21 '15 at 03:12

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Yes, it is guaranteed. Suppose for the sake of contradiction that $\alpha$ is irrational, and that $n\alpha$ is rational for some non-zero integer $n$. In other words, suppose $$n\alpha=\frac{a}{b}$$ for some non-zero integer $n$ and rational number $\frac{a}{b}$. Then this implies $$\alpha=\frac{a}{bn}$$ contradicting the definition of $\alpha$ being irrational.

Zev Chonoles
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  • You would’t necessarily have an integer on the RHS of the first equation; you’d have to assume a quotient of integers. – Jordan Green Jun 21 '15 at 03:15
  • Just misread the question, edited. – Zev Chonoles Jun 21 '15 at 03:15
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    You never actually use the assumption that $\alpha$ is irrational, so you could just as well take it out.. Just assume $n\alpha$ is rational and show that $\alpha$ is rational; then you've proven the contrapositive. – mathmandan Jun 21 '15 at 04:42
  • Wow. easy when you say it ;^) Thanks, Zev. answer accepted. (thanks to Jordan too). – Chris Bedford Jun 21 '15 at 06:19
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    @mathmandan: see https://gowers.wordpress.com/2010/03/28/when-is-proof-by-contradiction-necessary/ for some discussion, in particular which argues that any proof of irrationality must be by contradiction in some sense. – Ben Millwood Jun 21 '15 at 12:26
  • @BenMillwood, thanks, and I think my comment is pretty much in agreement with Gowers when he says "proof by contradiction is a very useful tool, but try not to use it unless you really need it." Anyway, all I was saying was: (1) when you can remove an assumption from an argument and leave the rest unchanged, you probably might as well do it, and (2) mathematicians can and do distinguish between contradiction and contraposition. But these things are better debated at http://math.stackexchange.com/questions/262828/proof-by-contradiction-vs-prove-the-contrapositive/705291#705291 – mathmandan Jun 21 '15 at 17:52