Analysis Question: Find the supremum and infimum of S,where S is the set S = {√n − [√

Simkate

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.

Dick

It's pretty obvious that all of the elements in S are in [0,1), right? And you shouldn't have any trouble showing the infimum is 0. Just find a n where f(n)=sqrt(n)-[sqrt(n)] is 0. Showing the supremum is 1 is a little harder. You want to find a sequence of integers a_n such that f(a_n) approaches 1.

╔(σ_σ)╝

http://www.physicsforums.com/showthread.php?t=433380

JonF

I actually don’t think an ε proof will work for this, since n must be a natural number, unless you restrict ε to naturals too.

I’m not sure on this by any means but this is the approach I would take. First because S is a subset of the reals it must have a LUB. Arguing that 1 is an upper bound is easy. I would try to show that if √n − [√n]<1 you can find an √m − [√m] that’s even closer to 1 which would make 1 the smallest upper bound. These three facts together ill show that 1 is the LUB.

╔(σ_σ)╝

You are correct an epsilon argument would not work here.

Originally, I was going to use density of R. But since there are countably many irrationals in the set proposed it is obvious that I can't use it.

Since the set proposed is a subset of all irrational numbers between (0,1).


Your approach is similar to the epsilon argument and I doubt it would work.

Even the sequence approach suggested is a little hairy as it requires an epsilon argument to show convergence. And one cannot guarantee there are no "jumps". Eg. sqrt(1023) = 31.98437118..... and then sqrt(1024)=32.

JonF

Not true since transcendental numbers fall in [0,1)



But I got it, man did this take me awhile, but my idea can work. I don't wanna give it away, but here's the general idea.

For contradiction sake, assume that there is a n in N, and for all other m in N that: 1 –(√n − [√n]) < 1 – (√m − [√m]).

Simplify this, then pick a clever m in terms of n that will get rid of the radicals that cause problems.

╔(σ_σ)╝

What exactly about my statement is not true ?

I said "Since the set proposed a subset of all irrational numbers between (0,1)."

Transcendentals are irrationals right ?

I don't see where I went wrong.

Hmm...
What exactly are you getting a contradiction from ? I don't follow your argument.


Using your argument, I suceeded in showing that

(√n − [√n]) is not bounded above by any number of the form (√m − [√m]). Maybe I am missing something but that doesn't prove suprema.

Dick

Uh, pick a_n=n^2-1. That's the worst case in some sense. What is [n^2-1]? What's the limit as n->infinity of the difference? And for Simkate, please don't double post again, ok?

╔(σ_σ)╝

Either it is too late at night and my brain it shut off or I just don't understand what you mean.

Did you not say

f(n) = sqrt(n) - [sqrt(n)]

Then

f(a_n) = sqrt(n^2 -1) - [ sqrt(n^2 -1)]



Seems to me like this may not even converge.

For some large n we could find f(a_n) =0

Dick

Possibly it is too late. f(a_n) is only going to be equal to zero if sqrt(n^2-1) is an integer. What is [sqrt(n^2-1)]? There's a pretty simple answer.

╔(σ_σ)╝

I see it . n-1.

It is defintely too late for me to be thinking :(.


Your solution works. Hopefully OP can use it.
In the analysis books I have seen limits appear after suprema and the like.
Mine is certainly like that.

Inferior89

Edit: Oh lol.. I go to bed now. Missed the []
--------------------------------
sqrt(n^2 + 1) = n-1 wooot?
I think he meant that it is never an integer because for it to be an integer it need to be a quadratic number which n^2 + 1 never is.

╔(σ_σ)╝

We both need sleep

btw it was
[sqrt(n^2 -1) ] = n-1

Haha...sleeeeeeppppppppppp.:)