# Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.20

20. Show that if m is a positive integer,then $[mx] = [x] + [x + (1/m) ] + [x + ( 2/m) ] +$ · · ·$+ [x + (m - 1) /m]$ whenever $x$ is a real number.
Solution:

We have to show that m if is a positive integer, then
$[mx]=[x]+[x+(1/m)]+[x+(2/m)]+...+[x+(m-1)/m]$….(1)
Whenever $x$ is a real number
We consider
$x=[x]+{x}$
Where $[x]$ is the greatest integer part of $x$ and ${x}$ is the fractional part of $x$
We also know that
$[integer+(...)]=integer+[(...)]$
We now, consider the right hand side of (1)
$RHS=[[x]+{x}]+[[x]+{x}+1/m]+[[x]+{x}+2/m]+...$
$+[[x]+{x}+(m-1)/m]$
$=m[x]+[{x}]+[{x}+1/m]+[{x}+2/m]+...+[{x}+(m-1)/m]$
$=m[x]+[{x}]+[{x}+1/m]+[{x}+2/m]+...+[{x}+p/m]+...$
$+[{x}+(m-1)/m]$
Where $p$ is the minimum value such that
$[{x}+p/m]=1$
Proceeding further, we get,
$RHS=m[x]+0+0+0+...p$ times$+[{x}+p/m]+...$
$+[{x}+(m-1)/m]$
$=m[x]+1+1+1+...(m-p)$ times
$=m[x]+m-p$
Now, since,
$[{x}+p/m]=1$
Implies that
$[{x}+(p/m)-1]=0$
$[{x}-((m-p)/m)]0$
Which in turn implies that
${x}=(m-p)/m+(<1/m)$ for $p$ to be minimum
$m{x}=m-p+(<1)$
$[m{x}]=m-p$ …. (2)

Therefore, by (2) we get

$RHS=m[x]+[m{x}]$
$=[m[x]]+[m{x}]$
$=[m[x]+m{x}]$
$=[mx]$
Hence, we have prove (1)