Solving a Little Problem on Irrational Values & Functions

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The discussion revolves around understanding the function f defined by f(kx) = kx - [kx], where k is an integer in the interval [0, N]. It is clarified that f has N+1 values that fall within the intervals [i/N, (i+1)/N] for integers i in [0, N-1]. The key point is that for each integer k, there exists an integer i such that f(kx) is included in one of these intervals, confirming that the values of f cover the entire range from 0 to 1. Additionally, it is emphasized that there are two values k1 and k2 for which the difference in their function values is bounded by 1/N. This understanding is crucial for solving the exercise effectively.
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Hi everyone,

I’m studying the correction of an exercise and I there are some points I can’t see:
In this exercise, we introduce a irrational x and function f so that f(kx)=kx-[kx] with k an integer included in the interval [0,N].

It is said that f has N+1 values (Ok) that are included in the intervals [i/N,(i+1)/N] for i integers included in [0,N-1]… I can’t see this last point. If anyone could help me understand this point…

Thanks in advance!
 
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y-[y]\in [0,1]
\cup\{[i/N,i+1/N]:i\in \mathbb{Z}\cap [0,N-1]\}=[0,1].
Is it saying anything more than that?
 
Nothing else, they just say that for each k integer included in [0,N], x an irrational, there exists an i integer included in [0,N-1] so that f(kx) is included in the interval [i/N,(i+1)/N] (and then, for the rest of the exercise, since there are N+1 values and N intervals, there are two values included in the same [i/N,(i+1)/N]).

But I don't know why f(kx) is included in [i/N,(i+1)/N]...
 
It's not saying f(kx) is included in any particular [i/N,(i+1)/N], only that it's somewhere in [0,1] and since the sets [i/N,(i+1)/N] cover the whole of [0,1] it's got to be in one of them.
 
Ok Martin I got it. (In fact, the important in this exercise is that there exists two values k1 and k2 so that abs(f(k1x)-f(k2x))<=1/N.)

Thanks very much!
 
Yes I guessed that. Glad to be of help.
 
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