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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

Formulas and rules

Contrapositive: to prove P → Q, prove ¬Q → ¬P
Contradiction: assume ¬S, derive R ∧ ¬R, conclude S
Contradiction for P → Q: assume P ∧ ¬Q
P ↔ Q: prove P → Q and Q → P
Euclid's lemma: p prime and p | ab ⇒ p | a or p | b
Euclid's number: N = p₁p₂⋯pₖ + 1

Common mistakes

Before the test, make sure you can

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.

Question 1 (level 1)

A proof by contrapositive of P → Q proceeds by:

  1. Assuming ¬Q and proving ¬P
  2. Assuming ¬P and proving ¬Q
  3. Assuming P and proving Q
  4. Assuming Q and proving P
Show the answer and steps

Answer: A) Assuming ¬Q and proving ¬P

  1. P → Q is logically equivalent to ¬Q → ¬P.
  2. So you may instead give a direct proof of ¬Q → ¬P.
  3. Assuming ¬P and proving ¬Q proves the inverse, which is not equivalent.
Question 2 (level 1)

A proof by contradiction of a statement S begins by:

  1. Assuming S and deriving a contradiction
  2. Assuming ¬S and deriving a contradiction
  3. Assuming S and then deriving S once again
  4. Proving the converse of S instead of S
Show the answer and steps

Answer: B) Assuming ¬S and deriving a contradiction

  1. If ¬S leads to something impossible, ¬S cannot be true.
  2. So S must be true.
  3. Assuming S and reaching a contradiction would instead show S is false.
Question 3 (level 3)

With the primes 2, 3, 5, 7, 11, 13, Euclid's number is N = 30031 = 59 × 509. What does this show?

  1. 59 and 509 must be among the six listed primes
  2. There are only finitely many primes after all
  3. Euclid's proof contains an error
  4. 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

  1. Compute 2·3·5·7·11·13 + 1 = 30030 + 1 = 30031 = 59 × 509.
  2. The proof only needs some prime factor of N that is not on the list.
  3. 59 and 509 are primes not among 2, ..., 13, exactly as the argument predicts.