**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.