Proofs:Biconditionals

A biconditional of two propositions P and Q takes the form " $P$ if and only if $Q$ ". This can also be written as $P\iff Q$, which is equivalent to $P\implies Q\land Q\implies P$. When proving a biconditional statement, we need to prove that $P\implies Q$ and $Q\implies P$ are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " $P\implies Q$ and $Q\implies P$ " is logically equivalent to " $P\implies Q$ and $\neg P\implies \neg Q$ ", " $Q\implies P$ and $\neg Q\implies \neg P$ ", or " $\neg P\implies \neg Q$ and $\neg Q\implies \neg P$ ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.

Example of a biconditional proof: "For $x\in \mathbb {Z} ,3x+3$ is odd if and only if $x$ is even". Let $P$ = "3x + 3 is odd" and $Q$ = "x is even".

If x is even, then 3x + 3 = 3(2k) + 3 for some integer k. 3(2k) + 3 = 3(2k + 1). 3 is odd and 2k+1 is odd, so their product is also odd. Thus, if x is even, 3x + 3 is odd. Therefore, the statement $Q\implies P$ is true.

If x is odd, then 3x + 3 = 3(2k + 1) + 3 for some integer k. 3(2k + 1) + 3 = 3(2k + 1 + 1) = 3(2k + 2) = 3(2)(k+1). Thus 3x + 3 is a multiple of 2 and is even if x is odd. So, the statement $\neg Q\implies \neg P$, which is logically equivalent to $P\implies Q$, is true.

Therefore, $P\iff Q$.

Resources

• Biconditional Statements, Mathematics LibreTexts
• Logical Biconditional, Wikipedia