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Sagot :
The greatest integer function returns the largest integer smaller than the number you provide it. That is, if n ≤ x < n + 1, where n is an integer, then the "greatest integer of x" is [x] = n.
• Let n be even. Then we can write n = 2k for some integer k ≥ 0. Now,
[n/2] = [k] = k
while
[-n/2] = [-k] = -k
so that
[n/2] - [-n/2] = 2k = n
• Let n be odd. Then n = 2k + 1 for some integer k ≥ 0. Every odd integer occurs between two even integers, so that
n - 1 < n < n + 1
or equivalently,
2k < n < 2k + 2
so that
k < n/2 < k + 1
It follows that [n/2] = k.
Similarly, if we negative the previous inequality, we have
-k > -n/2 > -(k + 1), or -k - 1 < -n/2 < -k
which means [-n/2] = -k - 1.
So we make the same conclusion,
[n/2] - [-n/2] = k - (-k - 1) = 2k + 1 = n
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