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Are there any basic rules and crucial things to be known about employing the method of contradiction to prove anything?

user516517
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  • If you want to prove that something is true, you assume it isn't and then reach a contradiction. Note that you should only assume that the statement in question is false; if you make other assumptions then you cannot determine which assumption led to the contradiction. – Dave Dec 26 '17 at 19:15

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In addition to Dave's comment: when you reach a contradiction, make sure that what you conclude is the negation of that very assumption, and not anything that was inferred from that assumption on your way to the contradiction.

So, for example, suppose you assumed statement $P$, from which you inferred statement $Q$, which in turn lead to both $R$ and $\neg R$, i.e. a contradiction .... then the only thing you can conclude is $\neg P$; you cannot conclude $\neg Q$

Bram28
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  • So here's a problem which with I have trouble making an assumption: prove that either xf(x)>0 for all x>p or xf(x) < 0 – user516517 Dec 27 '17 at 04:30
  • Sorry that was incomplete, so I had t – user516517 Dec 27 '17 at 04:31
  • So here's a problem which with I have trouble making an assumption: prove that either xf(x)>0 for all x>p or xf(x)<0 for all x>p . ( given:f(p)=0). So then what should I assume so as to contradict it and imply that the claim is true? – user516517 Dec 27 '17 at 04:37
  • @user516517 In formal logic notation, the claim you want to prove is $\forall x>p: xf(x)>0 \lor \forall x>p: xf(x)<0$. So, negating that: $\neg (\forall x>p:xf(x)>0 \lor \forall x >p: xf(x)<0)=\neg \forall x>p:xf(x)>0 \land \neg \forall x>p: xf(x)<0 = \exists x>p:\neg xf(x)>0 \land \exists x>p:\neg xf(x)<0=\exists x>p:xf(x) \le 0 \land \exists x>p:xf(x) \ge0$ – Bram28 Dec 27 '17 at 17:00