6.6: Proving the contrapositive
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Recall. Modus tollens: P \rightarrow Q \Leftrightarrow \neg Q \rightarrow \neg P\text{.}
To prove P \Rightarrow Q\text{,} you can instead prove \neg Q \Rightarrow \neg P\text{.}
In Worked Example 6.3.1, we proved that the square of an even number is also even. Therefore, this also constitutes a proof of the contrapositive statement: if the square of a number is odd, then that number is also odd.
Prove that every prime number larger than 2 is odd.
Solution
We want to prove the following universally quantified conditional (“for all p” omitted, domain is positive integers).
conditional | if (p is prime and p>2) then p is odd. |
contrapositive | if p is not odd, then not (p is prime and p>2) |
DeMorgan Subsitution | if p is not odd, then (p is not prime or p \le 2) |
These are all equivalent.
Let's prove the last statement: as in the procedure for proving conditionals with a disjunction, start by assuming that p is not odd and p \gt 2\text{.} We must then show that p is not prime. Since p is not odd, it is divisible by 2\text{.} But since p \gt 2\text{,} p is divisible by a number other than 1 and p itself. Therefore, p is not prime.
Check your understanding. Attempt Exercise 6.12.8.