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)

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.19

19. Show that [\sqrt{[x]}]=[\sqrt{x}] whenever x is a nonnegative real number.
Solution:  We have to prove the equation
[\sqrt{[x]}]=[\sqrt{x}]
Whenever x is a non-negative real number
We assume that
x=k+\epsilon
Where k is an integer and
0<=\epsilon<1
Further, we assume that
k=a^2+b
Where a is the largest integer such that
a^2<=k
Now,
a^2<=k
=a^2+b
<=x
=a^2+b+\epsilon
a^2<(a+1)^2
This implies that
[\sqrt{x}] = a
And
[\sqrt{[x]}]=[\sqrt{k}]
=a
Hence we have achieved the equality in the equation
[\sqrt{[x]}]=[\sqrt{x}]

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.

Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.17

17. Show that [x + 1/2] is the integer nearest to x (when there are two integers equidistant from x,it is the larger of the two).
Solution: We will make an assumption to prove our statement.
Let us assume that
x=[x]+r
Where 0<=r<1
Now,
x+\frac{1}{2}=[x]+r+\frac{1}{2}
We condition r on the following cases.
Case 1 :
If r<\frac{1}{2}
Then,
[x]<=x+
=[x]+r+\frac{1}{2}
<[x]+1
This implies that [x] is an integer nearest to x and
[x+\frac{1}{2}]=[x]
Case2:
If r>=\frac{1}{2}
Then,
[x]+1<=x+r+\frac{1}{2}
<[x]+2
This implies that [x]+1 is the integer nearest to x (assuming x is midway between [x] and [x+1]) and
[x+\frac{1}{2}]=[x]+1
Hence, we have proved that [x+\frac{1}{2}] is the integer nearest to x.