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I have a question regarding proof by contradiction (more like the way of writing it).

Let try to prove an easy proposition : For all integer $n$, if $n$ is even, then $n+1$ is odd.

  1. Negate the whole proposition : There exists an integer $n$ such that $n$ is even and $n+1$ is even. Since $n+1 = 2k$ for some integer $k$, $n = 2k-1$, a contradiction.

  2. Let $n$ be an integer. Assume that n is even. We want to show that $n+1$ is odd. Suppose to the contrary that $n+1$ is even. Then $n+1=2k$ which gives $n = 2k-1$, a contradiction.

So the first way (1) negate the whole thing and get a contradiction. Along the way, there are steps setting up quantifier "there exists".

The second way, first set up the direct proof (there is a step doing for all quantifier). Then proceed to suppose a contradiction later.

The question is : Are they both a valid proof ? Are they both called a contradiction ? Or the first one is the contradiction ?(In this case, what is the second approach called ? Or it is NOT a valid proof ?)

user117375
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  • They are both valid proofs by contradictions. The difference is subtle but they are equivalent. In the first you assume an exception and get a contradiction. In the second you show that $n$ is not that exception but that because $n$ is arbitrarty there can be no exception. – fleablood Jul 14 '20 at 01:19
  • This isn't part of the question but I wouldn't say $n = 2k-1$ is a contradiction that $n$ is even unless you prove $2k -1$ is always odd. – fleablood Jul 14 '20 at 01:20

3 Answers3

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Formally proof by contradiction is following: $$[(\neg P \Rightarrow Q) \land (\neg P \Rightarrow \neg Q)]\Rightarrow P$$ You first case directly use this formula from scratch. In second you use it from middle. Whenever you use it, imho, you can say, that you use proof by contradiction.

zkutch
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Let $P(n)$ be that $n$ is even. So you want to prove the desired result of: $$\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))$$

  1. Negate the whole proposition : There exists an integer $n$ such that $n$ is even and $n+1$ is even. Since $n+1 = 2k$ for some integer $k$, $n = 2k-1$, a contradiction.

Because a contradiction is derived under the assumption that there exists some integer $n$ where $P(n)\wedge P(n+1)$ holds, therefore the desired result is proven.  A few extra steps may make this clearer.

$$\begin{split}\because\quad& \Big(\exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\Big)\to\bot\\&\neg \exists n\in\Bbb Z~.\big(P(n)\wedge P(n+1)\big)\\&\forall n\in\Bbb Z~.\neg\big(P(n)\wedge P(n+1)\big)\\\hline\therefore\quad &\forall n\in\Bbb N~.\big(P(n)\to\neg P(n+1)\big)\end{split}$$

  1. Let $n$ be an integer. Assume that n is even. We want to show that $n+1$ is odd. Suppose to the contrary that $n+1$ is even. Then $n+1=2k$ which gives $n = 2k-1$, a contradiction.

Because, for any integer $n$, we derive a contradiction under the assumption that $P(n+1)$ holds while also under the assumption that $P(n)$ holds; therefore the desired result is proven.

$$\begin{split}\because\quad&\forall n\in\Bbb Z~.\Big(P(n)\to\big(P(n+1)\to\bot\big)\Big)\\\hline\therefore\quad&\forall n\in\Bbb Z~.(P(n)\to\neg P(n+1))\end{split}$$

Graham Kemp
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This is a useful question. Allow me to streamline the contrast between the two setups:

  1. Assume (to the contrary) that there exists an integer $n$ such that $n$ is even and $n+1$ is even.
  2. Let $n$ be an integer, supppose that $n$ is even, and assume (to the contrary) that $n+1$ is even.

When the two assumptions are negated, statements (1) and (2) become equivalent.

ryang
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