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Prove question

  1. Nov 22, 2008 #1
    need to prove that if [tex]0\leq X<n[/tex]

    and also 0<n so x=0.

    The attempt at a solution

    assume that x>0

    [tex]\exists n, x\ni(0,\infty) |n< x [/tex]

    contradiction !

    for X<0 theres contradiction agin .

    from here X have to be 0

    is that right ?
     
  2. jcsd
  3. Nov 22, 2008 #2
    x=1, n=2>0 satisfies [itex]0\leq x<n[/itex], yet x does not equal zero.
     
  4. Nov 22, 2008 #3
    that is right but the porve is for each X so if theres only one case its enough .
     
  5. Nov 22, 2008 #4
    What are you trying to prove? I was under the impression that it is the following:
    Whenever 0<=x<n for some positive (integer?) n, then x must be zero.
    I gave a counter example that this is not always the case. As you said, one example is enough to show that the statement above is wrong.
     
  6. Nov 22, 2008 #5
    not integer all the real num.

    so what u suggest ?
     
  7. Nov 22, 2008 #6
    My counterexample is still valid.
    I suggest, you say what you are trying to prove and what you don't like about the counter example:smile:
     
  8. Nov 22, 2008 #7
    what`s not clear i need to prove that if x a real positive num and [tex]0\leq X<n[/tex]

    for each real posiive uumber n so x=0 .
     
  9. Nov 22, 2008 #8

    HallsofIvy

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    This makes no sense. Do you mean you want to prove:
    "if [itex]0\le x< n[/itex] for all real numbers n> 0, then x= 0"?

    [/quote]The attempt at a solution

    assume that x>0

    [tex]\exists n, x\ni(0,\infty) |n< x [/tex]

    contradiction !

    for X<0 theres contradiction agin .

    from here X have to be 0

    is that right ?[/QUOTE]
     
  10. Nov 22, 2008 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I would suggest that you care enough about the problem to at least copy the problem correctly!

    The proof you give is by contradiction:
    Suppose x> 0 then there exist n> 0 such that x> n.

    But you give no proof that the last statement "there exist n> 0 such that> n" is true itself!
    You need to construct such a number. If x> 0 then what positive number is less than x?
     
  11. Nov 22, 2008 #10

    0<x<n

    x=0.01
    n=0.02

    n = all the real numbers so that n might be 0.00000000000000000001

    so x have to be 0
     
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