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Section 6.6 Proving the contrapositive
Recall.
Modus tollens :
\(P \lgccond Q \lgcequiv \lgcnot Q \lgccond \lgcnot P\text{.}\)
Procedure 6.6.1 . Proof by proving the contrapositive.
To prove \(P \lgcimplies Q\text{,}\) you can instead prove \(\lgcnot Q \lgcimplies \lgcnot P\text{.}\)
Example 6.6.2 .
In
Worked Example 6.3.2 , 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.
Worked Example 6.6.3 .
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 substitution
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.
