How Do I Negate 'If n Is Even, Then n² Is Even' — What Is the New Hypothesis?

You stare at the statement "If n is even, then n^2 is even." A textbook asks for its negation. You write down: "If n is odd, then n^2 is odd."

Wrong — though the wrong answer is a very close cousin of the right one. What you wrote is the contrapositive, not the negation. These are fundamentally different objects, and the confusion comes from the same three words ("even", "odd", "implies") appearing in both.

This page disentangles the two.

The negation of an implication is not another implication

Here is the key fact. The negation of "P \Rightarrow Q" is not another if-then statement. It is:

\lnot(P \Rightarrow Q) \;\equiv\; P \land \lnot Q.

In words: the only way an implication fails is if P is true and Q is false. So the negation says "P holds and Q does not hold" — a conjunction, not an implication.

Why: an implication P \Rightarrow Q is the claim "whenever P holds, Q also holds." It is false only in the situation where P holds but Q fails. Negating it means asserting that this failure situation actually occurs — so the negation is "P is true, and Q is false." No if-then; just two facts joined by "and."

For our statement "If n is even, then n^2 is even," the negation is:

\text{"There exists an integer } n \text{ such that } n \text{ is even and } n^2 \text{ is odd."}

Not an if-then. A witness claim: there is a specific n that is even with an odd square. (This happens to be false — no such n exists. The original statement is true, so its negation is false.)

What the new "hypothesis" is — there isn't one

The negation of an implication has no hypothesis. It has two conditions, both required to hold simultaneously:

n is even (the original P).
n^2 is odd (the original \lnot Q).

You do not "assume" anything first. You claim both things at once, about the same n. If you are trying to prove the negation is true, your job is to produce such an n. If you are trying to show the negation is false (i.e., the original implication is true), your job is to show no such n can exist.

Why students confuse negation with the contrapositive

Both involve negating P and Q, and both produce something that contains "odd" where the original had "even." That shared surface makes them feel similar. But watch the structure:

Original: If n is even, then n^2 is even. (An implication.)
Contrapositive: If n^2 is odd, then n is odd. (Still an implication — but P and Q swapped, both negated.)
Inverse: If n is odd, then n^2 is odd. (An implication — both P and Q negated, not swapped.)
Negation: There exists an integer n such that n is even and n^2 is odd. (A conjunction with an existential witness — not an implication at all.)

The negation is the only one of the four that is not an implication. That alone distinguishes it from the others.

Negating "for all" adds an existential quantifier

The original statement is not just "if n is even, then n^2 is even" for some n — it is a claim about every integer n. Written out fully:

\forall n \in \mathbb{Z}: \; (n \text{ even}) \Rightarrow (n^2 \text{ even}).

Now negate:

\lnot \forall n: \; (n \text{ even}) \Rightarrow (n^2 \text{ even}) \quad\equiv\quad \exists n: \; \lnot\big( (n \text{ even}) \Rightarrow (n^2 \text{ even}) \big)

\equiv\quad \exists n: \; (n \text{ even}) \land (n^2 \text{ odd}).

The universal quantifier flips to an existential quantifier, and the inner implication becomes a conjunction. (See Negation Lab — ∀ Becomes ∃¬, and ∃ Becomes ∀¬ for the quantifier-flip rule in isolation.)

So the proper negation reads: "There exists an integer n such that n is even and n^2 is odd."

This is the "witness" form. To prove the negation true, produce such an n. To prove the negation false — which is to prove the original true — show no such n can exist.

When would you ever want the negation?

Two scenarios:

1. Disproving a claim. If someone claims "if n is even, then n^2 is even," and you disagree, you would produce a negation: find an even n whose square is odd. (You cannot — the claim is true.) But for other statements, counter-examples are exactly what you need. "Every prime is odd"? Negation: "there exists an even prime." Witness: 2. Claim demolished.

2. Proof by contradiction. In a contradiction proof, you assume the negation of the goal and derive an impossibility. For an if-then goal, the negation is "P and \lnot Q" — so you assume P holds and Q fails, and derive a contradiction. This is structurally different from proof by contrapositive (where you assume \lnot Q and derive \lnot P, no contradiction required). See Proof by Contradiction for the technique.

When would you want the contrapositive?

The contrapositive is what you use when you want to prove the same statement via a different route. The negation is what you use when you want to disprove the statement entirely, or set up a contradiction proof.

So the two questions — "what is the negation?" and "what is the contrapositive?" — have different purposes:

"Negate this" → produce the witness claim "P and \lnot Q hold for some case."
"Contrapositive of this" → produce the equivalent implication "\lnot Q \Rightarrow \lnot P."

Mixing them up usually shows up as students calling the contrapositive "the negation" or vice versa. The structural test is quick: if your answer contains "if... then...", it is not the negation of an implication.

The specific answer for the n² example

Original: "If n is even, then n^2 is even."
Contrapositive: "If n^2 is odd, then n is odd."
Negation: "There exists an integer n such that n is even and n^2 is odd."

The contrapositive has a new hypothesis (n^2 is odd) — which is what a proof by contrapositive would start from. The negation has no hypothesis at all — it is a single existential claim conjoining "n is even" and "n^2 is odd."

If your proof begins with "Assume n is odd" — that is the inverse, not the negation or the contrapositive. (It is also not equivalent to the original.) The inverse negates both parts but does not swap them; it is the hidden fourth sibling and belongs to the converse family.

Summary in one line

The negation of "if P then Q" is not another implication — it is the single conjoined claim "P is true and Q is false." There is no new hypothesis to assume. If you want a new hypothesis to start reasoning from, you are not negating — you are building the contrapositive, and that has its own article above.