**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)