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Math Help - Proof by contradiction.

  1. #16
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    Re: Proof by contradiction.

    i could be wrong (i'm sort of skimming here) but the problem is not with the latter part of your statement, but with the beginning bit.

    the original statement starts out: "if three numbers x,y,z are chosen between 0 and 1 with 0 < x < y < z < 1....."

    this is equivalent to: for all (x,y,z) such that 0 < x < y < z < 1....

    the negation of which is: there exists some (x,y,z) with 0 < x < y < z < 1 such that (negation of rest of statement).

    x,y and z can be chosen arbitrarily in the original sentence. the negation of an arbitrary choice is a specific choice, which is an assertion of existence.
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  2. #17
    Member anonimnystefy's Avatar
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    Re: Proof by contradiction.

    hi Deveno

    yes,that should be right because (\lnot(\forall\ P)\ P \Rightarrow Q )\Leftrightarrow ((\exists\ \lnot P) \lnot(P \Rightarrow Q))\Leftrightarrow ((\exists\ \lnot P)\ P \wedge \lnot Q)
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  3. #18
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    Re: Proof by contradiction.

    Quote Originally Posted by anonimnystefy View Post
    hi Deveno
    yes,that should be right because (\lnot(\forall\ P)\ P \Rightarrow Q )\Leftrightarrow ((\exists\ \lnot P) \lnot(P \Rightarrow Q))\Leftrightarrow ((\exists\ \lnot P)\ P \wedge \lnot Q)
    Did you read reply #15?
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  4. #19
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    Re: Proof by contradiction.

    hi Plato

    yes i have but i was replying to post #16.
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  5. #20
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    Re: Proof by contradiction.

    Plato your reply was awesome. a bit terse, though.
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  6. #21
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    Re: Proof by contradiction.

    So, we agree on the answer to the original question.

    Quote Originally Posted by anonimnystefy View Post
    (\lnot(\forall\ P)\ P \Rightarrow Q )\Leftrightarrow ((\exists\ \lnot P) \lnot(P \Rightarrow Q))\Leftrightarrow ((\exists\ \lnot P)\ P \wedge \lnot Q)
    This should be written as follows:

    \neg((\forall x)\,P\Rightarrow Q)\Leftrightarrow(\exists x)\,\neg(P\Rightarrow Q)\Leftrightarrow(\exists x)\,P\land\neg Q
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