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Proving a property of real numbers

  1. Jul 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Given x<y for some real numbers x and y. Prove that there is at least
    one real z satisfying x<z<y


    2. Relevant equations

    This is an exercise from Apostol's Calculus Vol. 1. The usual laws of
    algebra, inequalities, a brief discussion on supremum, infimum and
    the Archimedean property preceeded this exercise.


    3. The attempt at a solution

    I'm trying to solve as many problems from the first chapter, which I had
    just glossed over during a first read. Proving theorems in a rigorous
    fashion is not exactly my forte. It becomes all the more difficult when
    I'm asked to prove something "intuitively obvious."

    I think I have the solution in fragments.

    Let S be the set of all numbers greater than x. x is a lower bound for
    this set, and therefore S has a greatest lower bound - inf S.

    [tex]inf S \geq x [/tex]

    [The following argument assumes that x = inf S]

    Consider some [tex]y \in S[/tex]. I can always find at least one [tex]z \in S[/tex] such that z < y, because if there are no z's satisfying that inequality then y would be inf S.

    The only problem here seems to be to prove that inf S = x. The first choice is a proof by contradiction.

    Assume inf S > x...:uhh:

    Am I on the right track?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 3, 2007 #2

    Dick

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    How about proving some stuff about (x+y)/2?
     
  4. Jul 3, 2007 #3
    :redface: Yes, thank you...I really don't know why I was thinking about lower bounds and stuff.

    Was my earlier "proof" correct? It will be complete when I prove x = inf S, right?
     
  5. Jul 3, 2007 #4

    Dick

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    The thing is that I think you might find this property useful for proving x=inf S.
     
  6. Jul 3, 2007 #5
    :confused:

    I thought you were suggesting this...

    x<(x+y)/2
    y>(x+y)/2

    Therefore x<(x+y)/2<y - this being a much shorter solution than what I had written down earlier. I'm not sure how (x+y)/2 relates to inf S.
     
  7. Jul 3, 2007 #6

    Dick

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    Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S. Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.
     
  8. Jul 3, 2007 #7
    Ah...
    Assume inf S > x. This implies that inf S belongs to S.

    But then x<(inf S + x)/2 < inf S and it also belongs to S, which would imply inf S is not the infimum. A contradiction!

    But trying to prove that x = inf S this way seems redundant since it relies on the fact that for all real x and y, x<(x+y)/2<y, which was what I set to prove in the first place.

    And that's what I think you tried to convey in your last post. :biggrin:
     
  9. Jul 3, 2007 #8

    Dick

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    Yes. If there were a gap in the real numbers between x and y, then inf S would be y. So you want to show there are no gaps before you prove inf S=x.
     
  10. Jul 3, 2007 #9
    Got it now. Thank you.
     
  11. Jul 4, 2007 #10

    HallsofIvy

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    ?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

    In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.
     
    Last edited: Jul 4, 2007
  12. Jul 4, 2007 #11

    Dick

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    I didn't say x in is S. S is defined to be the set of all y such that y>x. I find it difficult to accept that z is not in S if x<z<y.

    I really don't think you read the thread.
     
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