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Intro to Proofs Unit 4: Contrapositive and Contradiction
Indirect proofs handle statements that resist a direct attack. Contrapositive proof flips a conditional into an easier one, and contradiction shows the opposite of a claim is impossible.
Covers: Proof by contrapositive, Choosing a proof method, Proof by contradiction, Irrationality proofs, Infinitely many primes, If-and-only-if proofs, Counterexamples and non-constructive proofs
Study guide
Because P → Q is equivalent to ¬Q → ¬P, you can prove a conditional by assuming the conclusion is false and showing the hypothesis must be false. This helps when the negated conclusion is easier to work with. For 'if n² is even, then n is even', a direct proof starts with n² = 2k and gets stuck, since there is no simple formula for n. The contrapositive, 'if n is odd, then n² is odd', starts from n = 2k + 1 and finishes with one line of algebra. Always state clearly at the start that you are proving the contrapositive, and write it out explicitly.
Proof by contradiction proves a statement S by assuming ¬S and deriving something impossible, a statement R together with ¬R. For a conditional P → Q, assume P and ¬Q together. The classic example is the irrationality of √2: suppose √2 = a/b in lowest terms; then a² = 2b², so a is even; writing a = 2c gives b² = 2c², so b is even too, contradicting lowest terms. Contradiction is the natural tool for 'there is no...' and 'is irrational' statements, because their negations hand you a concrete object to work with.
Euclid's proof that there are infinitely many primes is another model contradiction. Assume the primes are exactly p₁, ..., pₖ and let N = p₁p₂⋯pₖ + 1. Each pᵢ leaves remainder 1 when dividing N, so N has a prime factor missing from the list, a contradiction. Note that N need not itself be prime: 2·3·5·7·11·13 + 1 = 30031 = 59 × 509. A key lemma in many of these proofs is Euclid's lemma: if a prime p divides ab, then p divides a or p divides b, and in particular p | a² implies p | a.
A biconditional P ↔ Q needs two proofs, one for each direction, and either direction may be done directly, by contrapositive, or by contradiction. Be careful not to prove one direction twice by pairing a statement with its own contrapositive. To disprove a universal claim, give a counterexample, such as 2¹¹ − 1 = 2047 = 23 × 89 against 'p prime implies 2ᵖ − 1 prime'. Some existence proofs are non-constructive: they show an object exists, often by cases, without identifying it, as in the argument that some irrational a and b have aᵇ rational.
Key ideas
- P → Q and ¬Q → ¬P are equivalent, so proving either proves both.
- Use the contrapositive when the negated conclusion gives a better starting point.
- In a contradiction proof of P → Q, assume P and ¬Q together.
- A contradiction is any statement of the form R ∧ ¬R.
- Irrationality and 'no such object' claims are natural for contradiction.
- An if-and-only-if proof needs both directions.
- Non-constructive proofs show existence without producing the object.
Formulas and rules
Common mistakes
- Assuming the conclusion is false but forgetting to keep the hypothesis in a contradiction proof.
- Proving the inverse ¬P → ¬Q instead of the contrapositive ¬Q → ¬P.
- Claiming Euclid's number N must itself be prime.
- Proving one direction of an iff twice by using its contrapositive.
- Forgetting to assume lowest terms in an irrationality proof, so no contradiction appears.
Before the test, make sure you can
- Write the contrapositive of a conditional and prove it directly.
- Explain when contrapositive or contradiction is the better method.
- Prove that √2 (or √3) is irrational by contradiction.
- Reproduce Euclid's proof that there are infinitely many primes.
- Prove both directions of an if-and-only-if statement.
- Disprove false claims with explicit, verified counterexamples.
- Distinguish constructive from non-constructive existence proofs.
Words to know
- Proof by contrapositive
- Proving P → Q by proving ¬Q → ¬P.
- Proof by contradiction
- Proving S by showing ¬S leads to a contradiction.
- Contradiction
- A statement R ∧ ¬R, which can never be true.
- Irrational number
- A real number that cannot be written as a/b with integers a, b and b ≠ 0.
- Lowest terms
- A fraction a/b where a and b share no common factor greater than 1.
- Euclid's lemma
- If a prime divides a product, it divides at least one factor.
- Euclid's theorem
- There are infinitely many primes.
- Biconditional proof
- A proof of P ↔ Q consisting of proofs of P → Q and Q → P.
- Counterexample
- A specific case showing a universal statement is false.
- Constructive proof
- An existence proof that explicitly produces the object.
- Non-constructive proof
- An existence proof that shows an object exists without identifying it.
- Reductio ad absurdum
- Latin name for proof by contradiction: reduce the assumption to an absurdity.
- Mersenne number
- A number 2ⁿ − 1; prime n does not guarantee it is prime.
- Fundamental Theorem of Arithmetic
- Every integer greater than 1 factors into primes uniquely, up to order.
Sample practice questions
30 more questions like these, from level 1 to 5, are in the practice set.
A proof by contrapositive of P → Q proceeds by:
- Assuming ¬Q and proving ¬P
- Assuming ¬P and proving ¬Q
- Assuming P and proving Q
- Assuming Q and proving P
Show the answer and steps
Answer: A) Assuming ¬Q and proving ¬P
- P → Q is logically equivalent to ¬Q → ¬P.
- So you may instead give a direct proof of ¬Q → ¬P.
- Assuming ¬P and proving ¬Q proves the inverse, which is not equivalent.
A proof by contradiction of a statement S begins by:
- Assuming S and deriving a contradiction
- Assuming ¬S and deriving a contradiction
- Assuming S and then deriving S once again
- Proving the converse of S instead of S
Show the answer and steps
Answer: B) Assuming ¬S and deriving a contradiction
- If ¬S leads to something impossible, ¬S cannot be true.
- So S must be true.
- Assuming S and reaching a contradiction would instead show S is false.
With the primes 2, 3, 5, 7, 11, 13, Euclid's number is N = 30031 = 59 × 509. What does this show?
- 59 and 509 must be among the six listed primes
- There are only finitely many primes after all
- Euclid's proof contains an error
- N need not be prime; its prime factors are new primes
Show the answer and steps
Answer: D) N need not be prime; its prime factors are new primes
- Compute 2·3·5·7·11·13 + 1 = 30030 + 1 = 30031 = 59 × 509.
- The proof only needs some prime factor of N that is not on the list.
- 59 and 509 are primes not among 2, ..., 13, exactly as the argument predicts.