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