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Question on set notation

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove or disprove the following
    (i) ##\forall a\in\mathbb{R}[(\forall \epsilon>0,a<\epsilon)\Leftrightarrow a\leq 0]##




    2. The attempt at a solution
    Can't we disprove the above statement by telling ##a\leq 0 \nRightarrow (\forall \epsilon>0,a<\epsilon)## through a counter example like ##a\leq 0 \Rightarrow (\epsilon=0,a\leq \epsilon)## or something?
     
  2. jcsd
  3. Mar 12, 2015 #2

    PeroK

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    Can you write out in words what the proposition is saying?

    What would you have to show to prove it? Hint: there would be two parts to proving it.

    Can you write down what property a countereaxmple would have? Hint: it could have one of two properties.

    Does it hold for a = 0? Does it hold for a = -1? Does it hold for a = 1?
     
  4. Mar 12, 2015 #3
    The proposition in words:
    for all a belongs to real number set , for all ##\epsilon## >0 and a<##\epsilon## if and only if ##a\leq 0## .

    I have to show both ##\forall \epsilon >0,a<\epsilon\Rightarrow a\leq 0## and ##a\leq 0\Rightarrow \forall \epsilon >0,a<0## to prove the proposition.

    A counter example should disprove the proposition.
     
  5. Mar 12, 2015 #4

    PeroK

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    That's good.

    What about a = 0, 1, -1? Does the proposition hold for these values of a?
     
  6. Mar 12, 2015 #5
    The proposition holds for a=-1,0. But it doesn't hold for a=1.
     
  7. Mar 12, 2015 #6

    PeroK

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    Why does it fail for a = 1?
     
  8. Mar 12, 2015 #7
    Because a=1 is not less than for all ##\epsilon>0##
     
  9. Mar 12, 2015 #8

    PeroK

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    Take a step back. We have a proposition:

    ##\forall a \ \ A \Leftrightarrow B##

    That's means that (if the proposition holds) then for each a we have either: A(a) true and B(a) true; or A(a) false and B(a) false.

    For a = 1, what can you say about A(1) and B(1)?
     
  10. Mar 12, 2015 #9
    I see. The proposition holds for a=1 too, because A(1) false and B(1) false. Thanks...
    Thank you for showing me the better way to look at the question.:smile:
     
  11. Mar 12, 2015 #10
    Then the proposition is true for all a, so that we can't disprove it. We have to prove it. Thanks again @PeroK
     
  12. Mar 12, 2015 #11

    PeroK

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    Also, when I first asked you to describe the proposition in words, you could have said:

    The proposition states that:

    "Any real number is less than or equal to 0 iff it is less than every positive number".

    Put like that, it's clear that the proposition holds.
     
  13. Mar 12, 2015 #12
    Yeah, it's really clearer...
     
  14. Mar 12, 2015 #13

    PeroK

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    Here's a tip. This is something I do when dealing with propositions and logic:

    I use "true" and "false" to relate to individual statements. E.g. ##a > 0## can be true or false.

    And, I say a proposition "holds" or "fails". E.g. the proposition holds for a = 1.
     
  15. Mar 12, 2015 #14
    Just now I felt what is called as "Enlightenment...."
     
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