0
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ $n=a/(bm)$ proves nothing about $n$, since $m$ is assumed to be irrational. $\endgroup$ – Gerry Myerson Jan 8 '14 at 1:50
  • $\begingroup$ $a/bm$ is not (necessarily) rational, since it is not a ratio of two integers. $\endgroup$ – Igor Rivin Jan 8 '14 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 '14 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 '14 at 1:54
3
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.