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Is this logical reasoning correct?

  1. Aug 13, 2008 #1

    frb

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    Suppose I want to show that two functions f and g are equal. A way to prove this could be to prove the statement:
    [tex] f(x) = n \Leftrightarrow g(x) = n[/tex]

    Is it enough to show one side of the implication?
    Prove the following statement:
    [tex] f(x) = n \Rightarrow g(x) = n[/tex]

    and reason as follows, suppose [tex] f(x) \neq n[/tex],
    [tex] \Rightarrow \exists m\neq n: f(x)=m[/tex]
    [tex]\Rightarrow g(x)=m\neq n[/tex]
    Which would mean that I have shown the converse implication, and thus I have equivalence.
     
    Last edited: Aug 13, 2008
  2. jcsd
  3. Aug 13, 2008 #2

    tiny-tim

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    Hi frb! :smile:

    Sorry, but I don't understand your proof, so I'm going to say no, it's not correct. :redface:
     
  4. Aug 13, 2008 #3

    frb

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    Suppose I have proven the statement f(x) = n implies g(x)=n.

    To obtain equivalence I have to prove f(x) != n implies g(x) != n.

    So i reason as follows, suppose f(x) != n,
    f(x) surely has another value, let this value be m, so f(x) = m, and m != n.
    f(x) = m implies g(x) = m, and that implies that g(x) != n.
     
  5. Aug 13, 2008 #4

    tiny-tim

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    ah … that seems fine …

    but if you want to prove f = g, which is the same as, for all x, f(x) = g(x),

    then if you know for all x, f(x) = n implies g(x)=n,

    then for all x, f(x) = g(x) … you don't need any more logic than that. :smile:
     
  6. Aug 13, 2008 #5

    frb

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    of course. My brain is fried due to too much studying. I tend to make things more difficult then. I thought that there should be a flaw or something. Thanks though!
     
  7. Aug 13, 2008 #6
    This proof assumes that f and g have the same domain.

    It fails if they have different domains. But then by definition the functions are not equal, so there was no need to go any further.
     
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