MATH 2730 Contradiction Benjamin V.C. Collins James A. Swenson
Contrapositive The contrapositive of the statement If A, then B is the statement If not B, then not A. A statement and its contrapositive are logically equivalent. Non-mathematical example: If you are a computer science major, then you are very smart, can be rephrased as If you are not very smart, then you are not a computer science major. Mathematical example: If n 2 is odd, then n is odd, can be rephrased as If n is not odd, then n 2 is not odd.
Contrapositive, Converse, and Inverse Consider the following statements: If you are a computer science major, then you are very smart. Statement If you are not very smart, then you are not a computer science major. Contrapositive If you are very smart, then you are a computer science major. Converse If you are not a computer science major, then you are not very smart. Inverse (or the Contrapositive of the Converse) The first two are logically equivalent, and the last two are logically equivalent, but the two pairs are not logically equivalent to each other.
Proof Template: To prove If A, then B Write We prove the contrapositive of the statement. Write the negation of B. Leave some space and write the negation of A. Fill in the space using definitions and known results.
Example Theorem If n 2 is odd, then n is odd. Proof. We prove the contrapositive of the statement. Assume that n is an integer that is not odd. Then n is even. That is, n = 2k for some integer k. So n 2 = 4k 2 = 2(2k 2 ). That is, n 2 = 2t for some integer t, namely t = 2k 2. So n 2 is even. Therefore, n 2 is not odd.
Non-mathematical example: If he got that score without cheating, then I ll eat my hat! Mathematical Example: If A is true and B is false, then 4 is a prime number. Since we know that 4 is not a prime number, it can t be the case that A is true and B is false. Since (A B) is logically equivalent to A B, we have proven that A implies B.
Proof Template: To prove If A, then B Write Assume A. Write Suppose, for the sake of contradiction, the negation of B. Work with both statements until you have something that is obviously false. Note: The abbreviation Sftsoc is standard for Suppose for the sake of contradiction, and can be used in all but the most formal proof settings.
Example Theorem No integer is both even and odd. If n is an integer, then n is not both even and odd. Proof. Assume that n is an integer. Suppose, for the sake of contradiction, that n is even and odd. By definition of even, there is an integer k such that n = 2k. By definition of odd, there is an integer m such that n = 2m + 1. Combining those two equations, 2k = 2m + 1. With a little algebra, we have: 2k 2m = 1, or 2(k m) = 1. Now, we have that 2(k m) is larger than 0 = 2 0, but smaller than 2 = 2 1. So k m is an integer between 0 and 1. Since no such integer exists, we have a contradiction. Therefore, n is not both even and odd, which proves the proposition.
Compare and Contrast In a proof by contrapositive, we assume not B, and show not A. In a proof by contradiction, we assume both A and not B, and show something clearly false. If your thing that is clearly false is A and not A, then your proof can be (and should be) rewritten as a proof by contrapositive.
Example Theorem Let p be a prime number. If an integer n is divisible by p, then n + 1 is not divisible by p. Proof. Assume that p is a prime, and that n is an integer that is divisible by p. Sftsoc that n + 1 is also divisible by p. Then we have n = pk for some integer k, and also n + 1 = pm for some integer m. Substituting one equation into the other, we have pk + 1 = pm. A little algebra gives us that 1 = p(m k). Since 1 is between 0 and p, we have that p(m k) is between p 0 and p 1. So m k is an integer between 0 and 1. Since no such integer exists, we have a contradiction. Therefore, p does not divide both n and n + 1.
Example Theorem There are an infinite number of primes. If P is the set of all primes, then P is infinite. Proof. Let P be the set of all primes. Sftsoc that P is finite. That is, P = n for some (finite) natural number n. So P = {p 1, p 2, p 3,..., p n }, where each p i is prime. Consider the integer n N = i=1 p i
Example Theorem There are an infinite number of primes. If P is the set of all primes, then P is infinite. Proof (continued). Clearly, N is divisible by all the primes in P. By the previous theorem, N + 1 is not divisible by any of the primes in P. So either N + 1 is prime, or N + 1 has a prime factor not in P. This contradicts the assumption that P contains all the primes. Therefore, the set P cannot be finite, which proves the proposition.
Proof Template: Proving a Set is Empty To prove a set S is empty: Suppose, for the sake of contradiction, that S. Let x S. Work to show a contradiction.
Example Theorem Let A and B be sets. If A B, then A \ B =. Proof. Suppose that A and B are sets with A B. Sftsoc that A \ B. Let x A \ B. By definition of the set difference, x A and x / B. By hypothesis, A B. By definition of subset, since x A B, we have x B. Now x B and x / B. Therefore, A \ B =.
Proof Template: Proving Uniqueness To prove that a particular object is unique: Suppose, for the sake of contradiction, that there are two such objects. Work to show a contradiction.
Example Theorem Let n be an integer. Then there is a unique integer m such that n + m = 0. Definition The integer m is called the additive inverse of n. Proof. First, note that there is at least one such integer, since n satisfies n + ( n) = 0. Sftsoc that there are two such integers. That is, there is an integer m such that n + m = 0 and an integer t m such that n + t = 0. Then we have n + m = n + t. Subtracting n from both sides, we have m = t, which contradicts our assumption that t m. Therefore, the additive inverse is unique.
Reframed as a proof by contrapositive Theorem Let n be an integer. Then there is a unique integer m such that n + m = 0. Proof. First, note that there is at least one such integer, since n satisfies n + ( n) = 0. Assume that m and t are integers such that n + m = 0 and n + t = 0. Then we have n + m = n + t. Subtracting n from both sides, we have m = t. Therefore, the additive inverse is unique.