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

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by habsfan31 View Post
    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 n\in\mathbb{R}^+\,,\,\,[n]=n-\{n\} , with \{n\}= the fractional part of n, thus

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

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

    Tonio
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    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?
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    Quote Originally Posted by habsfan31 View Post
    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 \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
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    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.
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    Quote Originally Posted by habsfan31 View Post
    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 n\in\mathbb{N}\,,\,\,\sqrt{n}-[\sqrt{n}]\geq 0 , so if 0=\sqrt{n}-[\sqrt{n}] for some natural n then automatically 0 is the MINIMUM of the set S

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

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

    Tonio
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    Thanks, i got it!
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    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?
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    Quote Originally Posted by habsfan31 View Post
    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
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    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?
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    Quote Originally Posted by habsfan31 View Post
    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 \sqrt{n^2-1}-[\sqrt{n^2-1}]=\{\sqrt{n^2-1}\}\xrightarrow [n\to\infty]{}1 , for \epsilon:=1-b choose

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

    Tonio
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  12. #12
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    Quote Originally Posted by steveausten View Post
    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.
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