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# Prove some properties of the greatest integer function

For any we denote the greatest integer less than or equal to by . Prove the following properties of the function :

1. for any integer .
2. or .
3. .
4. .

1. Proof. Let for some integer . Then,

But, we defined . Thus, for any

2. Proof. If , then for some . Hence, and

On the other hand, if , then let . This gives us,

3. Proof. Let and , then we have

Thus,

4. Proof. By part (c) we have,

If , then let ,

Thus, .
On the other hand, if , then let , and

Thus,

5. Proof. By part (c) we have

And,

So, putting these together we have,

If , then, let , so

Thus, .
Next, if , then let , giving us,

Thus,

Finally, if , then let , and we have

So,

1. Anonymous says:

Alternatively, you can use the idea of a partition. It is easy to show that for each “jump” the value of the function always increases by one. Therefore, you only have to prove that the function values are the same for a certain initial value, and that the partition is equivalent.

2. ekaveera says:

Awesome proofs good. Can i have any clue for proving box(x/n)=box(box(x)/n)

3. Anonymous says:

Hello,

I think proof (b) is wrong, when you multiplied by (-1) you did not multiply n. Here is they way i proved it:

[x]<x multiplying by -1

-[x]>-x>-[x]-1 Now we proved that -[x]-1 is less than -x.

What remains to be proved is that (-[x] – 1) is both an integer and the maximum integer < -x

The first one is quite obvious and the second one can be proved by absurd

BTW, thks for your blog, is helping a lot with my calculus!

4. Sebastian says:

On the b proof, on the line that says -n-1<-x<n shouldn't that n be negative since im assuming you multiplied all by -1?

And can i ask why you pick -n-1 and equate it with the [-x] i dont get that step.

• Sebastian says:

Also on c how did you get the m+n+1 from the m+n+2, can you just take that like that?
I dont think you can because thats a segment on a line and you cant know that x+y is not on the m+n+1 to m+n+2 segment.

Maybe thats a dumb question but i thought i should ask anyway XD

@Sebastian: Your query can be dealt within two cases:
CASE:1 When x+y is in between m+n and m+n+1,then [x+y]=m+n=[x]+[y].
CASE:2 When x+y is in between m+n+1 and m+n+2,then [x+y]=m+n+1=[x]+[y]+1

5. Anonymous says:

Just to point out a typo. In section \emph{e}, right after the line “If …” where you define the first inequality, the right side of the inequality says instead of , as well as the next line after that one. And after that everything else is fine.

• RoRi says:

Thanks! Fixed now.

• Anonymous says:

In part E, on the second case you handle ([x] +[x] +[x]+1), your first inequality has 3x between 3n+1 and 3n+1, so I figured it was a typo.