Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.18

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.

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