6.8: Proving biconditionals

We also often want to prove that two statements \(P,Q\) are equivalent; i.e. that \(P \Leftrightarrow Q\text{.}\)

Fact \(\PageIndex{1}\)

The equivalence

\begin{equation*} P \leftrightarrow Q \; \Leftrightarrow \; (P \rightarrow Q) \land (Q \rightarrow P) \end{equation*}

holds; i.e. a biconditional is equivalent to the conjunction of the corresponding conditional \(P\rightarrow Q\) and its converse.

Proof

You are asked to prove this by truth table in Exercise 2.5.5.

Procedure \(\PageIndex{1}\): Proving a biconditonal

To prove \(P \Leftrightarrow Q\text{,}\) prove \(P \Rightarrow Q\) and \(Q \Rightarrow P\) separately.

As usual, this also works in the universal case since \(\forall\) distributes over \(\land\) (Proposition 4.2.2).

Example \(\PageIndex{1}\)

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. This is proved as Worked Example 6.3.1.

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 \(\PageIndex{1}\)

Attempt Exercise 6.12.10.