Rational or not
If x and y are irrational, prove that x^y can be rational.
You only need to find one example to make the proof.
You only need to find one example to make the proof.
Labels: mathschallenge
posted by Chris at
6:29 AM
5 Comments:
Hi Cam. Are you sure that log(9)/log(2) is irrational? I expect that it probably is.
Whatever, that certainly like a reasonable response.
I've got to go out for an hour or so. I'll post the argument I saw then. It's very similar to your response.
My post was a response to your first post. I don;t think I'd be stiking my neck too far out if I accepted, that at least one pair of primes (a,b) have logarithms that are independently irrational.
The version I saw didn't use logarithms. It only used sqrt(2).
Hi Cam. I've got to do some work now (intermittently). But thanks for that log stuff. I don't recall having seen that before - very nice.
Hate working, love this stuff.
The thing that caused me to post was, let y = rt2 = Sqrt(2) and x = rt2^rt2. If x is rational, we're finished, but if it is irrational then x^y = (rt2^rt2)^rt2 = rt2^(rt2^2) = rt2^2 = 2 and we're finished. Either way we've go at least one case that satisfies the problem. I rather like that argument. Of course, we haven't determined if rt2^rt2 is irrational, we've only said either it is or it isn't and hope that the law of [the] excluded middle holds.
I think your answer is better, as it is immediately obvious that there are at least countably infinite cases that do it.
... your answer is even better than I acknowledged. Mine only demonstrated existence.
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