- #1
Silversonic
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Homework Statement
Let [itex] \theta \in [0,1] [/itex] and [itex] n \in \mathbb{Z} [/itex]. Let [itex] n\theta mod(1) [/itex] denote [itex] n\theta [/itex] minus the integer part. Show [itex] n \theta mod(1) [/itex] is a discrete subset of [0,1] if and only if theta is rational.
The Attempt at a Solution
I'm having a bit of trouble with the "only if" part of the statement. The "if" part of the statement is simple. If theta is rational, then the subset is finite (because [itex] n \theta mod(1) [/itex] will repeat itself once [itex] n [/itex] is equal to or larger than [itex] p [/itex], where [itex] \theta = q/p [/itex]). Since it is a finite subset, it is discrete.
But the only if part? Assume that it is a discrete subset, I need to show this is irrational. I'm having trouble with that. If I could show there was a surjective map from [itex] n\theta mod(1) [/itex], theta being irrational, to the subset [0,1] then that would prove it. Since no point would be isolated in that case, and the set wouldn't be discrete. I can't seem to get anywhere with that, though.
I know that [itex] n \theta mod (1) [/itex], for theta irrational, is an infinite set (not that this proves discreteness), so maybe I could go from there? It's infinite, because if it were not, then we would have that [itex] k\theta mod(1) [/itex] and [itex] n\theta mod (1) [/itex] would be mapped on to the same number, for different integers k and n.
i.e.
[itex] k\theta = p + r [/itex]
[itex] n\theta = q + r [/itex]
p,q are integers and r is the "remainder part".
equating shows
[itex] \theta = (p-q)/(k-n) [/itex]
Contradicting irrationality.
Maybe I could go somewhere with that fact that it's an infinite set? I haven't managed to though.