# 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
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
We can add $n$ to this inequality to obtain
$l+n<=x+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.