## 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
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.

## 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 $x$.