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In article <wzxTg.336$zf3.306@fed1read03>,
"vsgdp" <hello@xxxxxxxx> wrote:
> 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,
N = .1? N is an integer.
> 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?
The basic idea here is this: think of N+1 points in [0,1]
arranged in order. Could the differences of successive pairs all
be > 1/N?
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