Suppose a function f has two periods a, b with a =/= n*b or vice versa. Then is f necessarily a constant? What if a is not a rational multiple of b? Answers involving whether f is continuous or not are appreciated. Basically, here's now I started:(adsbygoogle = window.adsbygoogle || []).push({});

WLOG, suppose b > a. Then b-a < b, and either a < b-a or a > b-a. Either way, we've found a third period which does not satisfy the integral (or rational) multiplicity requirement. Since one of these periods is smaller than the other, we can subtract the smaller from the larger to get another smaller period. This process should be repeatable infinite times to get infinite discrete periods which tend to zero. But what is this enough for?

EX:

Suppose p is Pi and e is, well, e. Then p-e is a period, and e-(p-e) = 2e-p is a period. 2e-p - (p-e) = 3e-2p is a period, and so is 3e -2p + (2e-p) 5e-3p. So you can make a whole bunch of periods.... I don't know which of the periods I have listed here is the smallest (I'm too lazy to get a calculator out and do this for more than a couple runs), but I think you can see the process of shrinking the period. So you'll get periods of the form k_{i}*a - m_{i}*b, where, I believe, k_{i}/m_{i}approaches b/a.

Thoughts?

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# Function with two periods

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