Proof by Contraposition vs. Proof by Contradiction

These two get conflated more than any other pair in an intro proofs course. They start differently, they end differently, and a question that asks for one will not accept the other.

The short answer

Contraposition is a direct proof in disguise

There is nothing indirect about the reasoning. You are proving a different statement — the contrapositive ¬q → ¬p — by an ordinary direct argument, and you are entitled to do that because the contrapositive and the original are logically equivalent. They are false in exactly the same case and true in exactly the same cases.

So the shape is:

The announcement in line one matters. Say you are proving the contrapositive, and say what it is. A reader who does not know why you began by assuming the negation of the conclusion will read the proof as backwards.

Contradiction assumes the whole thing is false

Here you do not switch statements. You suppose the claim itself fails and show that the world breaks.

To deny p → q you need p ∧ ¬q, since that is precisely when a conditional is false. That is why you get both facts to work with — and it is the practical advantage of the method. Then you reason until something impossible appears: an integer that is both even and odd, a fraction in lowest terms whose numerator and denominator share a factor, 0 = 1, a set that is a member of itself, anything.

Name the contradiction explicitly. "This contradicts the assumption that a/b was in lowest terms" is a finished proof; "contradiction!" with nothing named is where credit disappears.

The same theorem, both ways

Claim. For every integer n, if n² is even then n is even.

Here p is "n² is even" and q is "n is even".

By contraposition. We prove: if n is odd, then n² is odd.
Assume n is odd, so n = 2m + 1 for some integer m. Then

n² = (2m + 1)² = 4m² + 4m + 1 = 2(2m² + 2m) + 1,

which is odd. Therefore if n² is even, n is even. ∎

By contradiction. Suppose n² is even but n is odd.
Since n is odd, n = 2m + 1, so n² = 2(2m² + 2m) + 1 is odd. But we assumed n² is even, and no integer is both even and odd. Contradiction. Therefore n is even. ∎

The algebra is identical. The difference is entirely structural: the first proof never mentions that n² is even and simply arrives at "n² is odd" as its conclusion. The second carries "n² is even" along the whole way and uses it at the end to collide with what it derived.

Notice which is cleaner. When the contradiction you reach is just "¬p contradicts p", contraposition was the better tool — the extra assumption bought nothing. That is the usual case for statements of this shape.

When contradiction is the right choice

Two situations.

First: there is no conditional to contrapose. "√2 is irrational" is not an if-then statement, so contraposition has nothing to grip. Contradiction supplies the hypothesis you lack — by supposing √2 is rational, you get an equation to manipulate where before you had none:

Suppose √2 = a/b with a, b integers, b ≠ 0, and the fraction in lowest terms. Then 2b² = a², so a² is even, so a is even (by the theorem above). Write a = 2c: then 2b² = 4c², so b² = 2c², so b is even too. But a and b were in lowest terms and cannot both be even. Contradiction. ∎

The "in lowest terms" clause is not scenery — it is the tripwire the whole proof is built to hit. A contradiction proof usually plants one like this early.

Second: you genuinely need both p and ¬q in hand. If the derivation requires facts from the hypothesis and from the denied conclusion at the same time, contraposition cannot give you that; it only ever hands you ¬q.

How to spot a mislabeled proof

The most common muddle looks like this:

"Suppose for contradiction that n² is even and n is odd. Then n² is odd. This contradicts our assumption. ∎"

That is contraposition wearing a contradiction costume. The proof assumed p, never used it, derived ¬p, and then pointed at the collision. Everything except the first clause and the last is a contraposition. It is not wrong — it proves the theorem — but if the question asked which technique was used, the answer that scores is determined by what the proof actually did with its assumptions.

Exam questions in this area usually ask for the opening line rather than the whole proof, precisely because the opening line is where the two methods separate. For "if p then q", the contraposition opener is "Assume ¬q" and the contradiction opener is "Assume p and ¬q". Those are different sentences, and only one of them answers each question.

Where these go wrong

• Assuming p in a contraposition. You get ¬q and nothing else. Bringing p along makes it a contradiction proof, whatever the heading says.
• Deriving something other than ¬p in a contraposition. The target is fixed. Reaching a generic absurdity means you drifted into contradiction.
• Assuming only ¬q in a contradiction. The denial of p → q is p ∧ ¬q. Dropping p throws away half of what you are entitled to.
• Not naming the contradiction. "This is a contradiction" is not a finished sentence. Say which two things collide.
• Negating the conclusion incorrectly. If q is "n is even and n > 2", then ¬q is "n is odd or n ≤ 2" — De Morgan applies, and the resulting proof needs both cases. Getting this wrong makes the rest unsalvageable.
• Trying to contrapose a non-conditional. "√2 is irrational" has no hypothesis. Contraposition is only defined for p → q.
• Confusing either with a converse proof. Proving q → p is neither method — it is proving a different theorem. Contraposition proves ¬q → ¬p, which is not the same statement as q → p.

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