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Introduction to Analysis by Bilodeau. Problem 1.3.3

  1. Feb 21, 2017 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    I couldn't find more informative title!

    I find difficulties with proofs. So my solution might be weird

    The problem says
    "Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤ε<x. Show that x=0."

    My attempt was:
    Assume 0≤x and x <ε
    Case 1:
    0 <x
    For evey ε, 0<x and x <ε is false

    Case 2:
    0=x
    For evey ε, 0=x and x <ε is true


    My professor was not happy with making cases to proof the statement, and he said this is not a real proof. He suggested starting with the assumption 0 <x and from that I proof 0=x is the wanted fixed non negative real number that satisfies the statement.

    I am confused. How am I supposed to proof it?
     
  2. jcsd
  3. Feb 21, 2017 #2

    BvU

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    Can you check the problem statement ?
    1 is a positive real number, but 0≤ε<0 is definitely not right for ε = 1
     
  4. Feb 21, 2017 #3

    PeroK

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    You've got this the wrong way round. That would force ##x## to be infinite. Instead, you should have:

    ##\forall \epsilon > 0, \ x < \epsilon##

    For the solution, did you consider a proof by contradiction?
     
  5. Feb 21, 2017 #4
    Oh sorry, you are right.

    The question should be the following:
    "Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."

    Sorry, I wrote the wrong way.

    The question should be the following:
    "Suppose that x is a fixed non-negative real number such that, for all positive real numbers ε, 0≤x<ε. Show that x=0."
     
  6. Feb 21, 2017 #5

    PeroK

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    Any ideas about a proof? First, perhaps, can you see and explain why it's true?
     
  7. Feb 21, 2017 #6
    Discussion:
    x fixed non-negative real number
    ε is a positive real number

    There exist no fixed non-negative real number x such that it's less than every positive number ε, except 0.

    My Proof:

    Assume 0≤x and x <ε
    Case 1:
    0 <x
    For every ε, 0<x and x <ε is false (Since there is always a positive number that is less than x "i.e x >ε" when 0<x. Another way to state it would be that since x and ε here are both positive real numbers, there will always be x >ε an x <ε which is a contradiction with one of the axioms)

    Case 2:
    0=x
    For every ε, 0=x and x <ε is true (Since x <ε, ε≠0 "Axiom". Thus 0<ε is always true since by definition a positive real number is ε>0 )
     
  8. Feb 21, 2017 #7

    PeroK

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    I think you've got the idea, but this proof is not really right. You need to think about a specific ##\epsilon## if ##x > 0##. Any ideas?
     
  9. Feb 21, 2017 #8
    Well. This is confusing for me! I am supposed to find a fixed x, why should I search for a fixed ε?

    Since x>0 is assumed, 0<x<ε. There is no such fixed ε that satisfy 0<x<ε for all x.
     
  10. Feb 21, 2017 #9

    PeroK

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    Okay, I claim that ##10^{-6}## is a number that is less than every positive number. How would you disprove that?
     
  11. Feb 21, 2017 #10
    I would say that 10-7 is less than 10-6, thus 10-6 is not less than every positive number
     
  12. Feb 21, 2017 #11

    PeroK

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    Okay, so what about ##10^{-8}##?
     
  13. Feb 21, 2017 #12
    A number that is less than 10-7, but yet not less than 10-9, which mean it is not less than every positive number
     
  14. Feb 21, 2017 #13

    PeroK

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    So, using what you've done for these numbers, what about any number ##x > 0##? Why isn't that less than every positive number?
     
  15. Feb 21, 2017 #14
    Because there is always a positive real number that is less than x. It also can be shown by number line: Since x>0, x/2>0 and thus x/2<x. So there is always a positive number that is less than another positive number.
     
  16. Feb 21, 2017 #15

    PeroK

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    Yes. Note you also have ##x \not < x##. In other words, any positive ##x## is not less than itself.

    Can you do the proof now?
     
  17. Feb 21, 2017 #16
    Proof:
    let x>0
    there exit ε for any x, 0<x<ε | this is false since there exit x>ε, which contradicts with an Axiom.
    Thus x=0 is the only non-negative real number x that satisfies 0≤x<ε (0=0<ε).
     
  18. Feb 21, 2017 #17

    PeroK

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    What happened to ##x/2##?
     
  19. Feb 21, 2017 #18
    for x>0, x/2>0 which is still not less than all positive numbers ε
    for x=0, x/2=0 which is less than all positive numbers
     
  20. Feb 21, 2017 #19

    PeroK

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    No, you haven't understood the idea at all. It's difficult now to give you any more hints. Let me give you a written answer and you can try to translate that into maths:

    Suppose ##x## is greater than zero, then ##x## is not less than ##x/2##, which is itself a positive number, so ##x## is not less than all positive numbers ...
     
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