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How do you formally show subsequent element in a series is larger?

  1. May 10, 2010 #1
    As an example, for the sequence:
    [tex]s_{1} =2[/tex]
    We see:
    When [tex]s_{n} >1 [/tex] then,
    But how do I formally show that this last inequality is true? This example is fine or just one similar to it. Thanks!

    EDIT: Many of those superscripts are suppose to be subscripts. I don't seem to be able to edit them so there correct though. Sorry, hopefully you can still figure it out. If, it doesn't have any subscript at all, then whatever is the superscript is suppose to be the subscript. However, if there is a subscript then the superscript is correct.
    Last edited by a moderator: May 13, 2010
  2. jcsd
  3. May 10, 2010 #2
    If s_n > 1, is (s_n-1)^2>0?
  4. May 10, 2010 #3
    Yes, but I'm not quite sure how to use that to formally explain that the inequality is true. Also, again, I know that what you asked is true, but how do you formally show this?
  5. May 10, 2010 #4


    Staff: Mentor

    You might try showing this by mathematical induction. Clearly, S2 > S1, so that establishes your base case.

    Assume that Sn > Sn - 1, and see if you can use this assumption and your formula for Sn to show that Sn + 1 > Sn.
  6. May 10, 2010 #5
    You're not sure how to show that the square of a non-zero number is greater than zero?
  7. May 11, 2010 #6
    Not formally, no. And that's what this thread was about to begin with. Is it ok just to state that the square of a non-zero, real number is positive?

    Thanks, I think I've got it mostly figured out.
  8. May 12, 2010 #7


    User Avatar
    Science Advisor

    Surely, for any real x, [itex]x^2\ge 0[/itex] which means "either [itex]x^2= 0[/itex] or [itex]x^2> 0[/itex]. Further, if [itex]x^2= 0[/itex] then, taking the square root of both sides, x= 0. Therefore, if [itex]x\ne 0[/itex], [itex]x^2> 0[/itex].
    That's a "formal" proof.

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