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Given a real x and an integer N > 1, prove that there exists integers h and
k with 0 < k <= N such that |kx-h|<1/N. Hint: Consider the N+1 numbers tx -
[tx] for t = 0, 1, ..., N and show that some pair differs by at most 1/N.
Notation: The [tx] means greatest integer <= tx.
First, is the t = 0 a typo, because if t = 0, 0 - 0 < 1/N is trivially true.
Second, from the hint it seems t will be my k and [tx] my h. So I want to
look at:
t=1: x - [x]
t=2: 2x - [2x]
t=3: 3x - [3x]
...
t=N: Nx - [Nx]
and conclude that one of these differences is < 1/N.
I looked at some examples by fixing x, say x = 0.6 and N = 0.1, and see that
it is true:
.6 - 0 = .6
1.2 - 1 = .2
1.8 - 1 = .8
2.4 - 2 = .4
3 - 3 = 0
But I can't come up with an idea to prove it for general real x. Any hints?
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