$\begingroup$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.
$\endgroup$41 Answer
$\begingroup$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.
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