Consider: kx + k(1-x)
Hello!
I want to prove the following:
where is the distance to the nearest integer. and k is an integer.
Here is what I did so far:
Since I know there exists a c with . Thus I can conclude that where
Now I got
I know because of that
I should now look at the cases that k is even and odd and greater or less than zero. But I donīt get to anything when I look at the last inequality to calculate The case k=0 is trivial.
Plan is to look at k greater and less 0 and k even and odd for this case and then also do the same thing for the right hand side and compare those values and it should be the same result. Then I am done. Can anyone give me a hint on how to progress from here?
Thanks.
Let k be some integer, and x be some real number. Then kx is a real number.
The max distance any real number is from the nearest integer is (1/2) .
∴ there is some integer M such that M-(1/2) ≤ kx ≤ M+(1/2).Case 1: M-(1/2) ≤ kx < M
Then
M-(1/2) ≤ kx < M → -M < -kx ≤ (1/2) - M → k - M < k - kx ≤ (1/2) + k - M
Since k - M is the closest integer to k - kx = k(1 - x), we have
For Case 1:
Case 2: kx = M (This case should be easy.)
Case 3: M < kx ≤ M+(1/2) (This case proceeds much like Case 1.