**18. Show that if m and n are integers, then [(x + n)/m]=[([x]+n)/m] whenever x is a real number.
**Solution: We have to show that if m,n are integers, then

[(x+n)/m]=[([x]+n)/m

Whenever x is a real number

If x is a real number, then (x+n) will also be a real number.

We prove the property which we shall use subsequently

[x/m]=[[x]/m]

Where x is a real number and m is an integer

We assume that

[x]=l

By division algorithm, we have integers q,r such that

l=mq+r

Where

0<=r<m-1

So we have

q=[[x]/m]

Because

[x]<=x<[x]+1

It follows that

x=[x]+\epsilon

Where

0<= \epsilon<1

We see that

[x/m]=[([x]+\epsilon)/m]

=[(l+\epsilon)/m]

=[((mq+r)+\epsilon)/m]

=[q+(r+\epsilon)/m]

Because

0<=\epsilon<1

We have

0<=r+\epsilon

<(m-1)+1

=m

It follows that

[x/m]=[q]

This ends the proof of the property.

Now using this property, we get

[(x+n)/m]=[([x+n])/m]….(1)

Now we prove another important property

[x+n]=[x]+n….(2)

Where x is a real number and m is an integer

We assume that

[x]=l

So that l is an integer

This implies that

l<=x<l+1

We can add n to this inequality to obtain

l+n<=x+n<l+1+n

This shows that

l+n=[x]+n

Is the greatest integer less than or equal to x+n

Hence,

[x+n]=[x]+n

Using (2) in (1) we get

[(x+n)/m]=[([x]+n)/m]

This is our desired result.