Section 6.8 Proving biconditionals
We also often want to prove that two statements \(P,Q\) are equivalent; i.e. that \(P \lgcequiv Q\text{.}\)
Fact 6.8.1.
The equivalence
\begin{equation*}
P \lgcbicond Q \; \lgcequiv \; (P \lgccond Q) \lgcand (Q \lgccond P)
\end{equation*}
holds; i.e. a biconditional is equivalent to the conjunction of the corresponding conditional \(P\lgccond Q\) and its converse.
Proof.
Procedure 6.8.2. Proving a biconditional.
To prove \(P \lgcequiv Q\text{,}\) prove \(P \lgcimplies Q\) and \(Q \lgcimplies P\) separately.
As usual, this also works in the universal case since
\(\forall\) distributes over
\(\lgcand\) (
Proposition 4.2.6).
Worked Example 6.8.3.
Prove: A number is even if and only if its square is even.
Solution.
We want to prove that the following quantified biconditional (“for all \(n\)” omitted, domain is nonnegative, whole numbers).
- biconditional
\(n\) is even if and only if \(n^2\) is even.
- conditional and converse
(if \(n\) is even then \(n^2\) is even) and (if \(n^2\) is even then \(n\) is even)
- contrapositive and converse
(if \(n^2\) is odd then \(n\) is odd) and (if \(n^2\) is even then \(n\) is even)
- conditional and inverse
(if \(n\) is even then \(n^2\) is even) and (if \(n\) is odd then \(n^2\) is odd)
These are all equivalent, so we could prove any one pair.
Original conditional.
Converse.
If \(n^2\) is even, then there exists an integer \(m\) such that \(n^2 = 2m\text{,}\) so that \(n = \sqrt{2m}\) … ? We seem to be stuck.
Inverse.
If \(n\) is odd, then there exists an integer \(m\) such that \(n = 2m+1\text{.}\) Then, \(n^2 = 4m^2 + 4m + 1\) is odd.
Checkpoint 6.8.4.
