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

• Sep 29th 2010, 03:25 PM
habsfan31
Find the supremum and infimum of S, where S is the set S = {√n − [√n]}
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)

I found my infimum to be 0 and my supremum to be 1, but how do i go about proving them? Help please.
• Sep 29th 2010, 04:05 PM
tonio
Quote:

Originally Posted by habsfan31
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)

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

Remember that for $\displaystyle n\in\mathbb{R}^+\,,\,\,[n]=n-\{n\}$ , with $\displaystyle \{n\}=$ the fractional part of n, thus

$\displaystyle \sqrt{n}-[\sqrt{n}]=\{\sqrt{n}\}$

Since $\displaystyle \sqrt{100}=10\,,\,then\,\,0\in S\Longrightarrow 0=min(S)$ (why?), and since $\displaystyle \{\sqrt{n^2-1}\}\xrightarrow [n\to\infty]{}1$ then $\displaystyle 1=sup(S)$ (not maximum!...why?)

Tonio
• Sep 29th 2010, 04:20 PM
habsfan31
Thanks for the quick reply, but i got that part, its the "why's" that i dont get. How to prove the inf and sup, and why do you say that 1 isnt the supremum?
• Sep 29th 2010, 05:26 PM
tonio
Quote:

Originally Posted by habsfan31
Thanks for the quick reply, but i got that part, its the "why's" that i dont get. How to prove the inf and sup, and why do you say that 1 isnt the supremum?

I didn't say that: 1 IS the supremum, but it is NOT the maximum since it cannot be $\displaystyle \sqrt{n}-[\sqrt{n}]=1$ for no natural n.

Then you have to check the zero thing (it's easy), and for the other one you need to calculate an easy limit...

Tonio
• Sep 29th 2010, 05:33 PM
habsfan31
I feel like an idiot cause you say checking for the zero thing is easy, and to check the other one all you need to calculate is an easy limit, yet i am lost. I cant find anything in my textbook to help me with this problem.
• Sep 30th 2010, 04:47 AM
tonio
Quote:

Originally Posted by habsfan31
I feel like an idiot cause you say checking for the zero thing is easy, and to check the other one all you need to calculate is an easy limit, yet i am lost. I cant find anything in my textbook to help me with this problem.

It's not in your textbook that you need to look for...

1) It's easy to check that for $\displaystyle n\in\mathbb{N}\,,\,\,\sqrt{n}-[\sqrt{n}]\geq 0$ , so if $\displaystyle 0=\sqrt{n}-[\sqrt{n}]$ for some natural n then automatically 0 is the MINIMUM of the set S

2) As already noted, $\displaystyle \sqrt{n}-[\sqrt{n}]=\{\sqrt{n}\}\Longrightarrow 0\leq \sqrt{n}-[\sqrt{n}]\leq 1\Longrightarrow sup(S)\leq 1$.

Now, check that $\displaystyle \lim\limits_{n\to\infty}\sqrt{n^2-1}-[\sqrt{n^2-1}]=\lim\limits_{n\to\infty}\{\sqrt{n^2-1}\}=1$ .

Tonio
• Sep 30th 2010, 09:13 AM
habsfan31
Thanks, i got it!
• Sep 30th 2010, 11:28 AM
habsfan31
I took a different approach, not sure if its right:

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?
• Sep 30th 2010, 02:56 PM
tonio
Quote:

Originally Posted by habsfan31
I took a different approach, not sure if its right:

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?

Well, yes it is right, but only because you just wrote the definitions of supremum and infimum...

Tonio
• Sep 30th 2010, 03:00 PM
habsfan31
What about if i prove the infimum as follows:

By definition of [x], x≥[x] so the terms of the sequence are always positive thus 0 is indeed a lower bound.

Now 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. So inf S = 0.

and for the supremum, i do sort of the same:

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) or this where are use the limit you displayed above?
• Sep 30th 2010, 03:17 PM
tonio
Quote:

Originally Posted by habsfan31
What about if i prove the infimum as follows:

By definition of [x], x≥[x] so the terms of the sequence are always positive thus 0 is indeed a lower bound.

Now 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. So inf S = 0.

and for the supremum, i do sort of the same:

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) or this where are use the limit you displayed above?

As $\displaystyle \sqrt{n^2-1}-[\sqrt{n^2-1}]=\{\sqrt{n^2-1}\}\xrightarrow [n\to\infty]{}1$ , for $\displaystyle \epsilon:=1-b$ choose

$\displaystyle n\in\mathbb{N}\,\,s.t.\,\,|\{\sqrt{n^2-1}\}-1|<\epsilon\Longrightarrow \{\sqrt{n^2-1}\}>1-\epsilon=b$ , contradiction.

Tonio
• Sep 30th 2010, 03:40 PM
habsfan31
Quote:

Originally Posted by steveausten
I don't imagine I have seen every one of the angles of this problem the way in which you have pointed them out. Youre a accurate star, a rock star guy. You have got a great deal to say and know so much about the subject that i think you tough to just teach a class about it.

Thanks
Steve

Thanks for your sarcastic remark, thank the lord for an ignore list.