Proof by contradiction.

**Deveno** · October 6th 2011, 09:13 AM

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.

**anonimnystefy** · October 6th 2011, 09:34 AM

hi Deveno

yes,that should be right because $(\lnot(\forall\ P)\ P \Rightarrow Q )\Leftrightarrow ((\exists\ \lnot P) \lnot(P \Rightarrow Q))\Leftrightarrow ((\exists\ \lnot P)\ P \wedge \lnot Q)$

**Plato** · October 6th 2011, 09:41 AM

Originally Posted by anonimnystefy

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?

**anonimnystefy** · October 6th 2011, 10:04 AM

hi Plato

yes i have but i was replying to post #16.

**Deveno** · October 6th 2011, 10:06 AM

Plato your reply was awesome. a bit terse, though.

**emakarov** · October 6th 2011, 01:11 PM

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

Originally Posted by anonimnystefy

$(\lnot(\forall\ P)\ P \Rightarrow Q )\Leftrightarrow ((\exists\ \lnot P) \lnot(P \Rightarrow Q))\Leftrightarrow ((\exists\ \lnot P)\ P \wedge \lnot Q)$

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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