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A joke in approximating numbers raised to irrational powers

(www.andreinc.net)
104 points nomemory | 1 comments | 18 Nov 24 16:41 UTC | HN request time: 0.21s | source
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NameError ◴[18 Nov 24 16:34 UTC] No.42174036[source]▶
Reminds me of a cool proof I saw recently that there are two numbers a and b such that a and b are both irrational, but a^b is rational:

Take sqrt(2)^sqrt(2), which is either rational or not. If it's rational, we're done. If not, consider sqrt(2) ^ (sqrt(2) ^ sqrt(2)). Since (a^b)^c = a^bc, we get sqrt(2) ^ (sqrt(2))^2 = sqrt(2)^2 = 2, which is rational!

It feels like a bit of a sleight of hand, since we don't actually have to know whether sqrt(2)^sqrt(2) is rational for the proof to work.

replies(3): >>42178040 #>>42190010 #>>42191430 #
1. seanhunter ◴[20 Nov 24 07:04 UTC] No.42191430[source]▶
Well the proof I would use is let a = e and b = i(pi).

e^(i theta) = cos theta + i sin theta (Euler's identity) thus e^(i pi) = cos pi + i sin pi = -1 + i(0) = -1

We know that e and i pi are irrational (in fact i pi isn't even a real) and -1 is rational.

Therefore there exist two numbers a and b such that both a and b are irrational but a^b is rational.

In fact log of just about anything is irrational so e^(log x) works as well for just about all rational x, but Euler's identity is cool so I wanted to use that.