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I am wondering what is wrong with my contradiction proof that "The product of two irrational numbers is irrational". I understand that there are examples where this is not true: $\sqrt{2} * \frac{1}{\sqrt{2}}$. I've seen an attempt to prove this on stack exchange (Proving/Disproving Product of two irrational number is irrational), but it doesn't answer why the following contradiction proof doesn't work?

Let p be the statement "m is irrational" q be the statement "n is irrational" and r be the statement "m*n is irrational". Thus, "The product of two irrational numbers is irrational" is the statement:

p $\wedge$ q $\to$ r

Using a contradiction proof of the form p $\wedge$ q $\wedge\neg$ r, I will assume and try to find a contradiction from the following:

m is irrational $\bigwedge$ n is irrational $\bigwedge$ m*n is rational

mn is rational, so mn = $\frac{a}{b}$ where a and b are have no common divisor and b $\neq$ 0. Since both m and n are irrational, neither can be zero. Thus n = $\frac{a}{bm}$ and n is rational. This contradicts the assumption that n is irrational, and so m*n must be irrational.

I'm not sure if there's something wrong with my form in the contradiction proof, or if contradiction proofs do not always work for proving.

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  • $\begingroup$ $n=a/(bm)$ proves nothing about $n$, since $m$ is assumed to be irrational. $\endgroup$
    – Gerry Myerson
    Jan 8, 2014 at 1:50
  • $\begingroup$ $a/bm$ is not (necessarily) rational, since it is not a ratio of two integers. $\endgroup$
    – Igor Rivin
    Jan 8, 2014 at 1:50
  • $\begingroup$ Ahh I see. So $\frac{a}{bm}$ is likely irrational, since we've assumed m is irrational, and the product of a rational number and an irrational number is irrational. So $\frac{a}{b}$ * $\frac{1}{m}$ shows that "n is irrational" and is not a contradiction. $\endgroup$
    – James Bender
    Jan 8, 2014 at 1:53
  • $\begingroup$ $m$ is irrational and $b$, being an integer, is rational. Since of course $b \ne 0$, $bm$ is irrational (if $bm$ were rational, $m = bm/b$ would be rational). We also have $a \ne 0$ since $mn = a/b$ is non-zero, so we get that $a/(bm)$ is irrational (if it were rational, $bm = a/(a/(bm))$ would be rational). $\endgroup$
    – fkraiem
    Jan 8, 2014 at 1:54

1 Answer 1

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There's no reason to conclude that $n$ is rational: In fact, since $$n = \frac a b \cdot \frac 1 m$$ is the product of a rational number and an irrational number, we can in fact conclude that $n$ is irrational. Hence no contradiction.

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