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I finished discrete math and now I am in foundations of Computer Science class and I have 2 questions. In discrete math the teacher taught us that the set of Natural Numbers starts at 1, but in Foundations of Computer Science the teacher is saying that the set of Natural Numbers starts at 0. How can that be?

Another question is about the contradiction method of proofs.

In discrete math they taught us for $P \Rightarrow Q$

assume ${\sim} P$

therefore $Q$

therefore ${\sim}Q$

hence $Q \, \& \, {\sim}Q$ a contradiction

thus, $P$

Now in my foundations of computer science class the teacher is saying that you never start by negating $P$ and you always start by negating $Q$ and then you come up with the contradiction ${\sim}P \,\&\, P$.

So why are these 2 things different the beginning of the set of natural numbers and the way to do proof by contradiction? Thank you for your time.

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closed as too broad by GNUSupporter 8964民主女神 地下教會, Sahiba Arora, Claude Leibovici, Gautam Shenoy, B. Goddard Jan 24 '18 at 12:51

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 24 '18 at 2:52
  • $\begingroup$ See also the post: how is $\mathbb N$ actually defined ?. $\endgroup$ – Mauro ALLEGRANZA Jan 24 '18 at 11:46
  • $\begingroup$ Please, note that the proof is wrongly written. I think that you have a premise $P \to Q$ and you want to prove the conclusion $\lnot Q \to \lnot P$. Thus, you have to assume $\lnot Q$ ("therefore" is used for inference, not for assumptions) and, in order to argue by contradiction, assume $P$. Now, from premise $P \to Q$ and assumption $P$, you have (therefore: justified by modus ponens) $Q$, and thus the contradiction. $\endgroup$ – Mauro ALLEGRANZA Jan 24 '18 at 14:02
  • $\begingroup$ The first natural number can be defined to be either 0 or 1, depending on the author. Both are seen as legitimate. And there is nothing wrong with starting a proof with a negation. $\endgroup$ – Dan Christensen Jan 24 '18 at 14:04
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For the first question, it depends who you ask. Some textbooks include $0$ as a natural number, and some do not. This is why you sometimes see the notation $\mathbb{N}_0$ and $\mathbb{N}$.

For the second question, there might be a bit of confusion between proof by contradiction and proof by contrapositive. To prove $P \implies Q$, you can prove the contrapositive (just the logical negation) $\sim Q \implies \sim P$. So you assume $\sim Q$ and try to prove $\sim P$. This is equivalent to $P\implies Q$, and is called the contrapositve. It is not uncommon to see people write a proof by contradiction that is actually a proof by contrapositive.

A proof by contradiction would assume the result you want to prove is false, and then derive a contradiction to some previously established fact (like obtaining $1=0$). In this setting you would assume $P \implies Q$ is false, and then try to derive a contradiction. Assuming $P \implies Q$ is false means you assume $P$ and $\sim Q$ both hold. It's not really a proof by contradiction unless the contradiction is of a previously established fact (not something you assumed at the start of the proof).

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  • $\begingroup$ Thanks for the answer. $\endgroup$ – JRowan Jan 24 '18 at 3:05

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