Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.17

17. Show that [x + 1/2] is the integer nearest to x (when there are two integers equidistant from x,it is the larger of the two).
Solution: We will make an assumption to prove our statement.
Let us assume that
x=[x]+r
Where 0<=r<1
Now,
x+\frac{1}{2}=[x]+r+\frac{1}{2}
We condition r on the following cases.
Case 1 :
If r<\frac{1}{2}
Then,
[x]<=x+
=[x]+r+\frac{1}{2}
<[x]+1
This implies that [x] is an integer nearest to x and
[x+\frac{1}{2}]=[x]
Case2:
If r>=\frac{1}{2}
Then,
[x]+1<=x+r+\frac{1}{2}
<[x]+2
This implies that [x]+1 is the integer nearest to x (assuming x is midway between [x] and [x+1]) and
[x+\frac{1}{2}]=[x]+1
Hence, we have proved that [x+\frac{1}{2}] is the integer nearest to x.

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