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)

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