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SUP question

  1. Nov 25, 2008 #1
    S={3/2,5/3,7/4,9/5,11/6 ...}
    the formula for this series is S(n)=2n+1/n+1
    the limit for it as n->infinity gives me 2

    i show that 2 is upper bound

    2n+1/n+1<2 => 2>1 (always true)

    now i need to show that 2 is the "least upper bound".
    if 2 is not the least upper bound then there is a certain "x" for which
    2n+1/n+1<2-x (2-x is the least upper bound)

    now they are doing some thing really odd
    2-x<2n+1/n+1

    why???
    this move is illegal
    2-x cannot be smaller
    its supposed to be the "least upper bound"

    ??
     
  2. jcsd
  3. Nov 25, 2008 #2
    ok i found out why they do this move

    2-x<2n+1/n+1
    its to prove that 2-x is not the least upper bound.
    they develop this innequality to this point
    n>1/x -1

    and here they conclude that 2-x is not the least upper bound.
    why they conclude that 2-x is not least upper bound?
     
  4. Nov 25, 2008 #3

    Mark44

    Staff: Mentor

    You're obviously in a fairly high mathematics class, so why would you write 2n + 1/n + 1, when you almost certainly mean (2n + 1)/(n + 1).

    It would be legitimate to interpret what you've written as:
    [tex]2n + \frac{1}{n} + 1[/tex]. If you write rational expressions on one line, put parentheses around the terms in the numerator and around those in the denominator.

    You're not a new member of this forum, so it might behoove you to learn some LaTeX to format what you post something like this:
    [tex]S(n) = \frac{2n + 1}{n + 1}[/tex]
    Your conclusion above is certainly true, but it's not at all clear that (2n + 1)/(n + 1) < 2 implies this conclusion. You need to show that the first inequality is true.
     
  5. Nov 25, 2008 #4

    Mark44

    Staff: Mentor

    They are doing a proof by contradiction. One way to prove that p ==> q, (where p and q are statements) is to assume that p is true and q is false. If you arrive at a contradiction, that means your assumption that q was false must actually have been false. IOW, q must be true.
     
  6. Nov 25, 2008 #5
    i know that its a proof by contradiction
    i cant understand the last step of it:

    i got
    n>1/(x -1)
    they say
    "since n exists satisfying the above inequality our claim is proves
    2 is Least Upper Bound"

    i cant see the logic of this line
    can you explain it in simpler words?
     
  7. Nov 25, 2008 #6
    there may be a "n" that satisfy our contradiction
    but there maybe another "n" whose not

    ??
     
  8. Nov 25, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    There exist positive x so that is true. And you want to prove that statement is NOT true.

    Since you haven't show exactly what "they" are saying, I can only guess that this is what they want to prove.

    IF there exist such an x it can't. But they want to prove that there is NO such x.

    What is "supposed to be the "least upper bound" "? Not x certainly! It is "2" that they are trying to prove is the least upper bound.

    (2n+1)/(n+1)= 2- 1/n. Given any positive number x, 1/x is a positive real number so, by the Archimedian property, there exist an integer N such N> 1/x. Then 1/N< x, -1/N> -1/x and 2- 1/N> 2- 1/x. Thus, 2- 1/x cannot be an upper bound on (2n+1)/(n+1) for any positive x.
     
  9. Nov 25, 2008 #8
    i know that the i thought as "odd" was for prooving that x-2 is not LUB
    if this contradiction expression is proved then 2-x is not a LUB(then 2 is LUB)
    we came to the last part

    n>1/(x -1)


    i know that n can be :0 1 2 3 5 ... infinity
    i dont know what the properties of X(in the article its epsilon)
    https://www.math.purdue.edu/academic/files/courses/2007fall/MA301/MA301Ch6.pdf (on page 4)

    so without that last piece about "what numbers could be taken by x"
    i cant say that this expression is true

    n>1/(x -1)
    ??
     
    Last edited: Nov 26, 2008
  10. Nov 25, 2008 #9

    Vid

    User Avatar

    What are you even asking? Both the solution Halls gave and the solution in that link explain everything rather clearly.
     
  11. Nov 26, 2008 #10
    i cant understand the lst part of the proof

    why does this expression is valid?
    n>1/(x -1)
     
  12. Nov 26, 2008 #11
    i cannot understand this words of hallsofivy
    "(2n+1)/(n+1)= 2- 1/n. Given any positive number x, 1/x is a positive Real Number so, by the Archimedian property, there exist an integer N such N> 1/x. Then 1/N< x, -1/N> -1/x and 2- 1/N> 2- 1/x. Thus, 2- 1/x cannot be an upper bound on (2n+1)/(n+1) for any positive x.
    "

    is there simpler expanation?
     
  13. Nov 26, 2008 #12
    and in the article "Since there exist n ∈ N satisfying the above inequality.."

    why its satisfying the inequality ??
     
  14. Nov 26, 2008 #13
  15. Nov 26, 2008 #14

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    By basic inequality arithmetic?

    [tex]n > \frac{1}{ \epsilon } -1[/tex] add 1 to both sides

    [tex] n+1 > \frac{1}{ \epsilon } [/tex] raise both sides to the (-1) power (which flips the inequality)

    [tex] \frac{1}{n+1} < \epsilon [/tex] add 1 to both sides and subtract epsilon, 1/(n+1) from both sides

    [tex]1 - \epsilon < 1 - \frac{1}{n+1}[/tex]

    then notice

    [tex]1-\frac{1}{n+1} = \frac{n+1}{n+1} - \frac{1}{n+1} = \frac{n}{n+1}[/tex]

    This is pretty easy stuff, and it sounds like you need to review your basic inequality arithmetic
     
  16. Nov 26, 2008 #15
    ok what next

    1-e<n/(n+1)

    so what makes this expression true?
     
  17. Nov 26, 2008 #16

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You already have that
    [tex]1 - \epsilon < 1 - \frac{1}{n+1}[/tex]
    and then
    [tex]1-\frac{1}{n+1}= \frac{n}{n+1}[/tex]
    Therefore
    [tex]1-\epsilon< \frac{n}{n+1}[/tex]
     
  18. Nov 26, 2008 #17
    my proffesor said that there is no general proof ,he said that
    when i develop this expression
    n>1/(e -1)
    i say
    "for every epsilon that could be inputed
    i can put an "n" which sutisfies this expression"

    is that correct?
     
  19. Nov 26, 2008 #18
    i think i got it

    if n>1/(e -1)

    then 1-e is not the least upper bound
    and if there is one case where its not the least upper bound ("n>1/(e -1)") then its not
    the least upper bound at all
    am i correct?
     
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