# Find the supremum and infimum of S, where S is the set S = {√n − [√n]}

1. Sep 29, 2010

### cooljosh2k2

1. The problem statement, all variables and given/known data
Find the supremum and infimum of S, where S is the set

S = {√n − [√n] : n belongs to N} .

Justify your claims. (Recall that if x belongs to R, then [x] := n where n is the largest integer less than or equal to x. For example, [7.6] = 7 and [8] = 8)

3. The attempt at a solution

I found my infimum to be 0 and my supremum to be 1, but how do i go about proving them? Help please.

2. Sep 29, 2010

### ╔(σ_σ)╝

Well, first show 0 is a lower bound of the set. This is not difficult based on the definition of the floor function. After that, show that it is the greatest lower bound by using an epsilon argument.

Then you do the same for the upper bound which is slightly more complicated.

3. Sep 30, 2010

### cooljosh2k2

Thanks, i got it!

4. Sep 30, 2010

### cooljosh2k2

Is it sufficient enough to say for the supremum:

If 1 is an upper bound of S that satisfies the stated condition and if v < 1, then ε = 1-v. Since ε > 0, there exists Sε ∈ S, such that v = 1- ε < Sε, therefore v is not an upper bound of S, and we conclude that 1 = sup S

and for the infimum:

If 0 is a lower bound of S and if t > 0, then ε = t-0. Since ε > 0, there exists Sε ∈ S, such that t = ε - 0 > Sε. Therefore, t is not a lower bound of S, and we conclude that 0 = inf S

Is this right?

5. Sep 30, 2010

### ╔(σ_σ)╝

No what you have there is not a proof. You just stated there exist without proof. With your " proof" I could show that 2 is the supremum:-).

Here is how your proof should look like...
By definition of [x], x>=[x] so the terms of the sequence are always positive thus 0 is indeed a lower bound.

Suppose a was a lower bound and a>0. Taking n=4 we see that a cannot be a lower bound since a>0 . Thus our assumption that a is a lower bound is false.

6. Sep 30, 2010

### ╔(σ_σ)╝

Proofs require a bit more work. Your thinking is good but your steps are not sufficient; try to use what I gave you.

Last edited: Sep 30, 2010
7. Sep 30, 2010

### cooljosh2k2

So would i be able to use the same reasoning for the supremum:

By definition of {x}, {x} < 1 since x - [x] = {x}, therefore {x} never reaches 1. So 1 is an upper bound.

Let b be an upper bound where b < 1. If you take ?, (what value would i be able to use here to prove this, since 1 isnt part of the set)

8. Sep 30, 2010

### cooljosh2k2

Can i use a limit to prove sup S = 1, since 1 isnt contained in the set?

9. Sep 30, 2010

### ╔(σ_σ)╝

First of all, you can't just copy my proof and expect everything to work out. I told you that proving the supremum was a bit more complicated.

I gave you the above proof to show you the general structure of your arguments. I expected you to gain some more incite into the problem.

Secondly,I don't understand your notation what is {x} and how exactly is it equal to x-[x] ? Usually when you put curly brackets it denotes a set.

I will give you an outline of what do you.

1) Show 1 is an upperbound of the set
2) Show that it is the least upper bound that is if a<1 then a is not an upper bound. Use the same sort of epsilon arguments you had originally.

In my proof the reason I picked a number is because it provided me with a straightforward contradiction. In the supremum case you can't use this method directly but you can use a modified version since as you already noted 1 is not in the set.

This is where assuming that a<1 and a is an upperbound will be helpful. As you previously did, a=1- $$\alpha$$. Now try deriving a contradiction.