6.8: Proving biconditionals

Elementary Foundations: An Introduction to Topics in Discrete Mathematics · LibreTexts

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

    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.

    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.

    This page titled 6.8: Proving biconditionals was last modified on Tue, 01 Sep 2026 03:35:22 GMT and is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Jeremy Sylvestre via source content that was edited to the style and standards of the LibreTexts platform.