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Hello ladies and gentlemen! A friend of mine and I have been thinking about this particular issue: under what circumstances is the product of two irrational numbers rational?

For example, multiplying $\sqrt{2}$ by any nonzero rational multiple of $\sqrt{2}$ or its inverse we obtain a rational number. Moreover, whenever we multiply an irrational number by a rational multiple of its inverse we obviously obtain a rational number.

I've also thought of the following case: if $q= \frac{m}{n}$ is in lowest terms where m and n are not k-th powers of integers, $\large q^{\frac{k-l}{k}} \cdot q^{\frac{l}{k}}$ is also rational.

In that direction, what is the best we know? i.e., is it known exactly when the product of two irrational numbers is rational? Or even, is it known exactly when the product of two transcendental numbers is algebraic?

I am a freshman and lessons started just one week ago so please be tolerant :) References are welcome and appreciated. Thanks in advance.

Steve Pap
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    If we knew this exactly, the irrationality of $e\pi $ and the like would not be open problems. – ocg Dec 28 '13 at 16:54
  • We know very little about the numbers you wrote. – mathlove Dec 28 '13 at 16:55
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    You've already given the complete answer: "whenever we multiply an irrational number by a rational multiple of its inverse we obviously obtain a rational number." – Hagen von Eitzen Dec 28 '13 at 16:59
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    You say nonzero rational multiple of $\sqrt2$ but zero is a perfectly good rational number. – Dan Rust Dec 28 '13 at 17:00
  • @DanielRust the zero multiple of $\sqrt{2}$ is not irrational! – Steve Pap Dec 28 '13 at 17:51
  • @HagenvonEitzen Right, I don't know how I failed to notice that it was actually a perfect answer xD – Steve Pap Dec 28 '13 at 17:59
  • @StevePap I think it's better to consider, for some chosen irrational $p$, all real numbers (not just irrational) $q$ such that $qp$ is rational. The reason this is better is because, if we call the above set $R_p$, then $R_p$ forms an abelian group under addition (with identity being zero). Of course, it's rather easy to see that $R_p$ has only one rational element which is zero (if we extend to all real $p$, then every element of $R_p$ is rational). – Dan Rust Dec 28 '13 at 18:08

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If $a$ is any rational number, and $b$ is any irrational number, then $c=a/b$ is irrational (it's pretty easy to prove that; I can give specifics if necessary) so the product of the two irrational numbers $b$ and $c$ is rational. And every case of a product of two irrational numbers being rational is an instance of exactly that situation, as you'll see if you think it through.

Michael Hardy
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  • 'product of the two irrational numbers is rational' really? How about $(\sqrt 3-\sqrt 2)\cdot \sqrt 2$? – mathlove Dec 28 '13 at 17:18
  • @mathlove : You're really not paying attention very well at all. I didn't say that you get a rational number in EVERY case in which you multiply two irrational numbers. Rather, I wrote about what happens in cases in which you do get a rational number. – Michael Hardy Dec 28 '13 at 17:30
  • I understand what you mean. hmm, maybe typo? $b$ and $c$ instead of $a$ and $c$? – mathlove Dec 28 '13 at 17:36
  • You must mean the product of the irrationals $b$ and $c$ is rational. And I assume the intention was $a \neq 0$. The whole argument seems a tautology though. – Macavity Dec 28 '13 at 17:38
  • It's really a tautology as Macavity pointed out. I was asking for something more concrete, which as HagenvonEitzen pointed out in the comments, I had already mentioned in the question xD – Steve Pap Dec 28 '13 at 18:00
  • Typo fixed: The product of the two irrational numbers $b$ and $c$ is rational. – Michael Hardy Dec 28 '13 at 20:26
  • That it's a "tautology" was the point of my sentence beginning with "And every[...]". – Michael Hardy Dec 28 '13 at 20:26