Converse and Equivalence

What you'll get from this section. First, the converse of an implication — what it is, and the crucial fact that it need not be true. Then what happens when an implication and its converse both hold: the two statements become equivalent, written $P \Leftrightarrow Q$, "if and only if". This also puts a formal name to the "is the same as" language we've used since Section 2.

The converse

The converse of $P \Rightarrow Q$ is the implication with the arrow reversed:

You simply swap the condition and the conclusion. (We met it briefly in Section 4, as the contrapositive's deceptive look-alike; here it gets its due.)

The single most important fact about the converse is this: a true implication need not have a true converse. $P \Rightarrow Q$ and $Q \Rightarrow P$ are different claims, and one can hold while the other fails.

Some examples make this vivid:

• "If it is a dog, then it is a mammal" is true. Its converse, "if it is a mammal, then it is a dog", is false — a cat is a mammal but no dog.
• "If $n$ is a multiple of $4$, then $n$ is even" is true. Its converse, "if $n$ is even, then $n$ is a multiple of $4$", is false — $6$ is even but not a multiple of $4$.
• "If $x = 3$, then $x^2 = 9$" is true. Its converse, "if $x^2 = 9$, then $x = 3$", is false — $x$ could be $-3$. (Exactly the caution from Section 1.)

Don't confuse the converse with the contrapositive. The contrapositive $\neg Q \Rightarrow \neg P$ (flip and negate) is always equivalent to $P \Rightarrow Q$. The converse $Q \Rightarrow P$ (flip only) is a separate question, whose answer might be yes or might be no.

So whether the converse holds is genuinely new information. And when it does hold — when both an implication and its converse are true — we have something worth a name.

Equivalence

When $P \Rightarrow Q$ and its converse $Q \Rightarrow P$ are both true, the statements $P$ and $Q$ are equivalent, written

We read $P \Leftrightarrow Q$ as "$P$ if and only if $Q$", often shortened to "$P$ iff $Q$". It says $P$ and $Q$ are true in exactly the same situations: they stand or fall together.

Why "if and only if"?

The phrase spells out the two halves exactly, using Section 2's readings:

• "$P$ only if $Q$" means $P \Rightarrow Q$ — the original implication.
• "$P$ if $Q$" means $Q \Rightarrow P$ — the converse.

So "$P$ if and only if $Q$" is precisely the implication together with its converse: $P \Leftrightarrow Q$.

A necessary and sufficient condition

Again from Section 2: from $P \Rightarrow Q$, $P$ is sufficient for $Q$; from the converse $Q \Rightarrow P$, $P$ is necessary for $Q$. So when both hold,

$P \Leftrightarrow Q$ says that $P$ is a necessary and sufficient condition for $Q$

(and, symmetrically, $Q$ is necessary and sufficient for $P$). "Necessary and sufficient" and "if and only if" are two names for the same thing.

Examples

A genuine equivalence. "$n$ is even $\Leftrightarrow n^2$ is even." Both directions hold:

• forward ($n$ even $\Rightarrow n^2$ even): if $n = 2k$ then $n^2 = 4k^2 = 2(2k^2)$, even;
• converse ($n^2$ even $\Rightarrow n$ even): this is exactly the lemma from Section 5.

Since the implication and its converse both hold, the two statements are equivalent.

Repairing an earlier converse. We just saw that "$x^2 = 9 \Rightarrow x = 3$" has a false converse. Widen the conclusion, though, and the converse is mended:

a true equivalence — each side forces the other.

An everyday one. "A whole number is divisible by $10$ if and only if its last digit is $0$." True both ways.

A non-equivalence. "$n$ is a multiple of $4 \Rightarrow n$ is even" is true, but its converse fails ($6$ is even, not a multiple of $4$) — so it is not an equivalence. Equivalence is a genuinely stronger claim than implication: it demands the converse as well.

Proving an equivalence

To establish $P \Leftrightarrow Q$ you must prove both the implication and its converse — $P \Rightarrow Q$ and $Q \Rightarrow P$. They are separate jobs and often need quite different arguments (the even/$n^2$ example used a direct proof one way and a contrapositive the other). Proving a single direction never establishes the biconditional.

You can sometimes run a chain of equivalences, $P \Leftrightarrow R \Leftrightarrow S \Leftrightarrow Q$, to get from one end to the other — but only if every link is genuinely two-way. A single one-directional step breaks the whole chain.

Looking back

We can now be precise about the "sameness" used throughout this course. Each time we said one statement "is the same as" another — the contrapositive $(P \Rightarrow Q) \Leftrightarrow (\neg Q \Rightarrow \neg P)$, De Morgan's laws, the quantifier-negation rules — we were asserting an equivalence. They read as "becomes" only because we hadn't yet named the relationship.

And the warning sharpens to a single line: the contrapositive is equivalent to the original; the converse is not. The two coincide precisely when $P \Leftrightarrow Q$ happens to hold.

Common mistakes

1. Assuming the converse comes free. $P \Rightarrow Q$ tells you nothing about $Q \Rightarrow P$ — the converse must be checked on its own.

2. Confusing converse with contrapositive. Flip-only is the converse (maybe true, maybe not); flip-and-negate is the contrapositive (always equivalent).

3. Proving only one direction of an equivalence. $P \Leftrightarrow Q$ needs both $P \Rightarrow Q$ and $Q \Rightarrow P$.

4. Treating $\Rightarrow$ as $\Leftrightarrow$. An implication runs one way; a biconditional runs both.

Summary

• The converse of $P \Rightarrow Q$ is $Q \Rightarrow P$; a true implication need not have a true converse.
• When an implication and its converse both hold, $P$ and $Q$ are equivalent: $P \Leftrightarrow Q$, "$P$ if and only if $Q$".
• "Only if" is the implication, "if" is the converse; together they make the equivalence — equivalently, $P$ is necessary and sufficient for $Q$.
• To prove $P \Leftrightarrow Q$, prove both directions.
• Contrapositive: always equivalent. Converse: not automatic.

Next: Section 7 — Miscellaneous.