Answer
Claim: For all integers m and n, if mn is even then m is even or n is even.
Contrapositive: For all integers m and n, if m is not even and n is not even, then mn is not even.
Proof: Suppose m and n are particular but arbitrarily chosen integers such that m is odd and n is odd.
By definition of odd, m = 2a + 1 and n = 2b + 1.
By substitution and algebra, mn = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(ab + a + b) + 1.
But 2(ab + a + b) + 1 is an integer because products and sums of integers are integers.
Let t = (ab + a + b), then 2(ab + a + b) + 1 = 2t + 1.
By definition of odd, it follows that mn = 2t + 1 is odd.
Therefore, mn is not even.
Work Step by Step
Method of Proof by Contraposition
1. Express the statement to be proved in the form
∀x in D, if P(x) then Q(x).
(This step may be done mentally.)
2. Rewrite this statement in the contrapositive form
∀x in D, if Q(x) is false then P(x) is false.
(This step may also be done mentally.)
3. Prove the contrapositive by a direct proof.
a. Suppose x is a (particular but arbitrarily chosen) element of D such that Q(x)
is false.
b. Show that P(x) is false.
