For any we denote the greatest integer less than or equal to by . Prove the following properties of the function :
- for any integer .
- or .
- .
- .
- Proof. Let for some integer . Then,
But, we defined . Thus, for any
- Proof. If , then for some . Hence, and
On the other hand, if , then let . This gives us,
- Proof. Let and , then we have
So, adding, we obtain,
Thus,
- Proof. By part (c) we have,
If , then let ,
Thus, .
On the other hand, if , then let , andThus,
- 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,
Awesome proofs good. Can i have any clue for proving box(x/n)=box(box(x)/n)
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!
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.
Thanks in advance
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
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.
Thanks! Fixed now.