Answer
Statement: ∀ integers a, b, and c, if a − b is even and b − c is even, then a − c is even.
Contrapositive: ∀ integers a, b, and c, if a − c is not even, then a − b is not even or b − c is not even.
Converse: ∀ integers a, b and c, if a − c is even then a − b is even and b − c is even.
Inverse: ∀ integers a, b, and c, if a − b is not even or b − c is not even, then a − c is not even.
The statement is true (and hence its contrapositive is also true because the contrapositive is logically equivalent to the statement), but its converse and inverse are false.
As a counterexample, let a = 3, b = 2, and c = 1. Then a − c = 2, which is even, but a − b = 1 and b − c = 1, so it is not the case that both a − b and b − c are even.
The converse and inverse and logically equivalent to each other so proving that one is false proves them both to be false.
Work Step by Step
A statement of the form: $\forall x \in D$, if P(x) then Q(x),
has as its contrapositive statement: $\forall x \in D$, if ~Q(x) then ~P(x),
as its converse statement: $\forall x \in D$, if Q(x) then P(x),
and as its inverse statement: $\forall x \in D$, if ~P(x) then ~Q(x).
