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Section 6.8 Proving biconditionals

We also often want to prove that two statements \(P,Q \) are equivalent; that is, that \(P \lgcequiv Q \text{.}\)
As usual, this also works in the universal case since \(\forall \) distributes over \(\lgcand \) (Proposition 4.2.7).

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.

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.
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