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]