The converse is the statement, “for all integers \(a\) and \(b\text{,}\) if \(a\) is odd or \(b\) is odd, then \(a + b\) is odd.” This is false! How do we prove it is false? We need to prove the negation of the converse. Let's look at the symbols. The converse is

\begin{equation*} \forall a \forall b ((O(a) \vee O(b)) \imp O(a+b))\text{.} \end{equation*}

We want to prove the negation:

\begin{equation*} \neg \forall a \forall b ((O(a) \vee O(b)) \imp O(a+b))\text{.} \end{equation*}

Simplify using the rules from the previous sections:

\begin{equation*} \exists a \exists b ((O(a) \vee O(b)) \wedge \neg O(a+b))\text{.} \end{equation*}

As the negation passed by the quantifiers, they changed from \(\forall\) to \(\exists\text{.}\) We then needed to take the negation of an implication, which is equivalent to asserting the if part and not the then part.

Now we know what to do. To prove that the converse is false we need to find two integers \(a\) and \(b\) so that \(a\) is odd or \(b\) is odd, but \(a+b\) is not odd (so even). That's easy: 1 and 3. (remember, “or” means one or the other or both). Both of these are odd, but \(1+3 = 4\) is not odd.