## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.16

16. Show that $-[-x]$ is the least integer greater than or equal to $x$ when $x$ is a real number.
Solution:  We define the greatest integer, denoted by $[x]$ , as the largest integer less than or equal to $x$ satisfying the inequation
$[x]<=x<[x]+1$ …..(1)
Where $x$ is a real number
We assume that
$y=-x$
Then, from (1) we get,
$[y]<=y<[y]+1$
We take negative sign which will reverse the inequality, so
$-[y]>=-y>-[y]-1$
Therefore, we get,
$-[-x]>=-(-x)>-[-x]-1$
$-[-x]>=x>-([-x]+1)$
This implies $-[-x]$ that is the least integer greater than or equal to $x$ whenever $x$ is a real number.

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.15

15. Show that if x and y are positive real numbers,then $[xy] >= [x] [y]$. What is the situation when both x and y are negative? When one of x and y is negative and the other positive?
Solution: We have to prove the inequality
$[xy]>=[x][y]$
Where $x$ and $y$ are positive real numbers.
We prove the above inequality by making assumptions.
Let us assume that
$x=a+r$
$y=b+s$
Where $a,b$ are integers and $r,s$ are real numbers such that $0<=r,s<1$
Now, using the values of $x$ and $y$ , we get,
$[xy]=[(a+r)(b+s)]$
$[xy]=[ab+as+br+rs]$
$[xy]=ab+[as+br+rs]$
Whereas
$[x][y]=[a][b]$
$[x][y]=ab$
This is because $a,b$ are integers and
$[r]=0$
$[s]=0$
Where $0<=r,s<1$
Thus we have
$[xy]<=[x][y]$
If one of $x$ and $y$ is positive and the other negative, then depending on what the numbers are, the order sign can go in either direction.

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.14

14. Show that $[2x] + [2y] >= [x] + [y] + [x + y]$ whenever x and y are real numbers.
Solution: We have to show that
$[2x]+[2y]>=[x]+[y]+[x+y]$
Where $x,y$ are real numbers
We write $x,y$ as
$x=n+\epsilon$
$y=m+\delta$
Where $n,m$ are integers and $\epsilon,\delta$ are non-negative real numbers less than 1.
There are two possibilities for the right hand side.
$n+m+(n+m)=2n+2m$
Or
$n+m+(n+m+1)=2n+2m+1$
If
$\epsilon+\delta>=1$
There are two possibilities for the left hand side.
$\lceil2x\rceil=\begin{cases} 2n & \epsilon<1/2\\ 2n+1 & otherwise \end{cases}$
And
$\lceil2y\rceil=\begin{cases} 2m &\delta<1/2\\ 2m+1 & otherwise \end{cases}$
We are trying to prove that the right hand side is always less than or equal to the left hand side. We construct a proof by contradiction and assume that right hand side is greater than left hand side. So the right hand side must be $2n+2m+1$ and the left hand side must be $2n+2m$
But then,
$\epsilon+\delta>=1$
So at least one of them must be equal to or greater than $1/2$
So the left hand side cannot equal $2n+2m$
Hence we arrive at a contradiction and end the proof.

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.13

13. Show that $[x + y] >= [x] + [y]$ for all real numbers x and y.

Solution: We consider the following inequation

$[x+y]>=[x]+[y]$
We know by the definition of greatest integer function that
$[x]<=x$ ……(1)
$[y]<=y$ ……(1)
Adding the inequations in (1) gives us
$[x]+[y]<=x+y$
Since greatest integer function preserves the order, therefore,
$[x+y]>=[[x]+[y]]$
$[x+y]>=[x]+[y]$
Hence, we have proved that
$[x+y]>=[x]+[y]$