Does there exist a rational number ratio for any two irrational numbers?
Let x be an irrational number.
x / x = 1.
Also, any nonzero integer multiple of x is irrational.
Proof: If x is irrational, then by definition it cannot be expressed in the form p / q, for all integers p and q (q ~= 0). Now let us consider n*x, where n is a nonzero integer. Let us assume that n*x is rational. If this is the case, then n*x = p/q, for some integers p and q. But re-arranging this equation we get x = p/n*q. Since p/n*q is the ratio of two integers, it is a rational number. But this contradicts our initial assumption that x is irrational. Therefore, n*x must also be irrational.
So for any irrational x, n*x/m*x is a rational ratio, where n and m are integers and m ~= 0.
Your question is ambiguous. The title "Can the ratio of two irrationals be rational?" seems to ask whether there exist two irrationals whose ratio is rational. That is easy to answer: certainly. Take any irrational x, Let y= 2x. Then then y is also an irrational but the ratio of x to y is 2.
But then you ask "Does there exist a rational number ratio for any two irrational numbers?" which, as well as "do there exist any two rational numbers whose ratio is rational" (the question above), could be interpreted as: "Given any two rational numbers, is there ratio rational?" and the answer to that is, just as obviously, no. [sqrt](2) and [sqrt](3) are irrational and their ratio is not rational.
I assume a typo here on your part. Yes, I tend to make ambiguous mathematical statements. I had originally questioned my use of the word "any." I see your point in this regard. Mea culpa.
Your mathematical imagery was simpler than I thought possible for solving the problem at hand. The answer, I guess, is a standard in the field.
Hypnagogics are one of my favorite pastimes, and "help" me visualize possible physical situations.
I actually don't know how standard the solution is, or what field in particular it might apply to. The answer just made itself apparent to me, that if the ratio of two irrationals were to be rational then that must imply that one irrational is an integer multiple of the other. Proving that turned out to be easier than I thought it might be, given the simple definition of what an irrational is.
I'm quite intrigued by hypnagogics too, as you might be able to tell. Although for me they are usually spontaneous, ie I don't have any conscious control over them.
Separate names with a comma.