# Find Infimum & Supremum of S: Justify Your Claims

• Simkate
In summary: N.This is the idea I was going for. But if n is large enough, there may be an a_n that is closer to 1 than any other number in N, but still doesn't make 1 the smallest upper bound. This is where the contradiction comes in. If 1 is the smallest upper bound, then a_n cannot get closer to 1 than 1. So there must be another smallest upper bound.
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.

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.

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.

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JonF said:
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 tihe smallest upper bound. These three facts together ill show that 1 is the LUB.

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.

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╔(σ_σ)╝ said:
Since the set proposed is a subset of all irrational numbers between (0,1).
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 want to 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.

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

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.

But I got it, man did this take me awhile, but my idea can work. I don't want to 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.

Hmm...

(√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.

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?

Dick said:
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?

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

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╔(σ_σ)╝ said:
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

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.

Dick said:
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.

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.

╔(σ_σ)╝ said:
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.

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.

Inferior89 said:
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.

We both need sleep

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

Haha...sleeeeeeppppppppppp.:)

## 1. What is the definition of infimum and supremum?

The infimum of a set S is the greatest lower bound of S, meaning that it is the largest number that is less than or equal to every element in S. The supremum of a set S is the least upper bound of S, meaning that it is the smallest number that is greater than or equal to every element in S.

## 2. How do you find the infimum and supremum of a set?

To find the infimum and supremum of a set, you must first order the elements in the set from least to greatest. Then, the infimum will be the first element in the set, and the supremum will be the last element in the set.

## 3. Can a set have multiple infimum and supremum values?

Yes, a set can have multiple infimum and supremum values if there are duplicate elements in the set. In this case, all equal elements will be the infimum and supremum of the set.

## 4. How do you justify your claims of the infimum and supremum values?

To justify your claims, you must show that the infimum and supremum values satisfy the definitions. This means that the infimum must be less than or equal to all elements in the set, and the supremum must be greater than or equal to all elements in the set.

## 5. Can the infimum and supremum values of a set change?

Yes, the infimum and supremum values of a set can change if the elements in the set change. For example, adding a larger element to the set can change the supremum value, and removing a smaller element can change the infimum value.

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