From: UtarEmpire
x is an irrational number if and only if the set { nx mod 1 | n is a natural number } is dense in [0,1].
Let x be real and A = { nx mod 1 | n natural }.
Suppose x is rational. Then x = a/b for some integers a, b. Thus bx mod 1 = 0, so |A| <= b, so A is finite and hence can't be dense in [0, 1]. Contrapositively, if A is dense in [0, 1], x is irrational.
Suppose x is irrational and suppose for contradiction that A is not dense in [0, 1]. Then there exist y in [0, 1] and e > 0 such that the open e-ball about y is a subset of [0, 1] that's disjoint from A. Pick the least natural n such that nx > 1 and consider the sequence
y, (y - nx) mod 1, (y - 2nx) mod 1, (y - 3nx) mod 1, ...
(I define "p mod 1" for real p < 0 to mean "the least real z >= 0 such that a - z is an integer".) Clearly, the distance between each pair of adjacent elements (call it d) is the same. Then given a sufficiently long prefix of this sequence (say, of the first ceiling(1/min(e, d)) elements—don't make me calculate an exact length, just agree that it exists), the e-balls about the elements cover [0, 1]. Thus one of these balls includes x. But this contradicts the choice of e and y, since it implies that there exists m such that (mx mod 1) is in the open e-ball about y. Thus, A is dense in [0, 1]. QED.
So what do I win?
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He lives in cyberspace and wants to destroy the network system.