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CSE 215: Foundations of Computer Science Unit 10: Proofs by Contradiction and Contraposition Kevin McDonnell Stony Brook University – CSE 215 1 • In a direct proof you start with the hypothesis of a statement and make one deduction after another until you reach the conclusion. • Indirect proofs are more roundabout. One kind of indirect proof, argument by contradiction, is based on the fact that either a statement is true or it is false but not both. • So if you can show that the assumption that a given statement is not true leads logically to a contradiction, impossibility, or absurdity, then that assumption must be wrong: and, hence, the given statement must be true. • This method of proof relies on reducing a given assumption to an impossibility or absurdity. Kevin McDonnell Stony Brook University – CSE 215 2 1. Suppose the statement to be proved is false. That is suppose that the negation of the statement is true. 2. Show that this supposition leads logically to a contradiction. 3. Conclude that the statement to be proved is true. Kevin McDonnell Stony Brook University – CSE 215 3 • Use proof by contradiction to show that there is no greatest integer. • Proof: Suppose not. That is, suppose there is a greatest integer N. • Then N n for every integer n. • Let M = N + 1. Now M is an integer since it is a sum of integers. • Also M > N since M = N + 1. Thus M is an integer that is greater than N. • So N is the greatest integer and N is not the greatest integer, which is a contradiction. Kevin McDonnell Stony Brook University – CSE 215 4 • Use proof by contradiction to show that there is no integer that is both even and odd. • Proof: Suppose not. That is, suppose that there is at least one integer n that is both even and odd • By definition of even, n = 2a for some integer a, and by definition of odd, 2b + 1 for some integer b. • Consequently, 2a = 2b + 1 and so 2a – 2b = 1 2(a – b) = 1 a – b = 1/2 • Since a and b are integers, the difference a – b must also be an integer. But a – b = 1/2 and 1/2 is not an integer, which is a contradiction. Kevin McDonnell Stony Brook University – CSE 215 5 • Prove that the sum of any rational number and any irrational number is irrational. • Proof: Suppose not. That is, suppose that there is a rational number r and an irrational number s such that r + s is rational. • By definition of rational, r = a / b and r + s = c / d for some integers a, b, c and d with b ≠ 0 and d ≠ 0. • By substitution and so Kevin McDonnell Stony Brook University – CSE 215 6 • Now bc – ad and bd are both integers and bd ≠ 0 by the zero product property. • Hence, s is a quotient of the two integers bc – ad and bd. • Therefore, s is rational, which contradicts the supposition that s is irrational. Kevin McDonnell Stony Brook University – CSE 215 7 • A second form of indirect argument, argument by contraposition, is based on the logical equivalence between a statement and its contrapositive. • To prove a statement by contraposition, you take the contrapositive of the statement, prove the contrapositive by a direct proof, and conclude that the original statement is true. • The underlying reasoning is that since a conditional statement is logically equivalent to its contrapositive, if the contrapositive is true then the statement must also be true. Kevin McDonnell Stony Brook University – CSE 215 8 1. Express the statement to be proved in the form ∀x in D, if P(x) then Q(x) 2. Rewrite this statement in the contrapositive form ∀x in D, if Q(x) is false then P(x) is false 3. Prove the contrapositive by a direct proof. a. Suppose x is an element of D such that Q(x) is false. b. Show that P(x) is false. Kevin McDonnell Stony Brook University – CSE 215 9 • Use proof by contraposition to show that for all integers n, if n2 is even then n is even. • Proof: Contrapositive: For all integers n, if n is not even then n2 is not even. • Suppose n is any odd integer. By definition of odd, n = 2k + 1 for some integer k. • By substitution and algebra: n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 • Now, 2k2 + 2k is an integer because products and sums of integers are integers. • So n2 = 2 (an integer) + 1, and thus, by definition of odd, n2 is odd. Kevin McDonnell Stony Brook University – CSE 215 10 • Observe that any proof by contraposition can be recast in the language of proof by contradiction. • In a proof by contraposition, the statement ∀x in D, if P(x) then Q(x) is proved by giving a direct proof of the equivalent statement ∀x in D, if ~Q(x) then ~P(x) • Exactly the same sequence of steps can be used as the heart of a proof by contradiction for the given statement. • The only thing that changes is the context in which the steps are written down. • To rewrite the proof as a proof by contradiction, you suppose there is an x in D such that P(x) and ~Q(x). Kevin McDonnell Stony Brook University – CSE 215 11 • You then follow the steps of the proof by contraposition to deduce the statement ~P(x). • But ~P(x) is a contradiction to the supposition that P(x) and ~Q(x). (Because to contradict a conjunction of two statements, it is only necessary to contradict one of them.) • When you use proof by contraposition, you know exactly what conclusion you need to show, namely the negation of the hypothesis. In proof by contradiction it may be difficult to know what contradiction to aim for. • The disadvantage of contraposition as compared with contradiction is that you can use contraposition only for a specific class of statements – those that are universal and conditional. Kevin McDonnell Stony Brook University – CSE 215 12 • Use proof by contradiction to show that for all integers n, if n2 is even then n is even. • Proof: Suppose not. That is, suppose that there is an integer n such that n2 is even and n is not even. • Therefore, n is odd and can be written n = 2k + 1 for some integer k. • By substitution and algebra, n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. • But 2k2 + 2k is an integer (because products and sums of integers are integer). So n2 = 2 (an integer) + 1, and thus, by definition of odd, n2 is odd. This is a contradiction that n2 is even. • Compare this proof with the one based on contraposition. Kevin McDonnell Stony Brook University – CSE 215 13 • Prove that is irrational. Because this is not a universal conditional statement, we cannot use proof by contraposition. • Let’s try proof by contradiction. Kevin McDonnell Stony Brook University – CSE 215 14 Kevin McDonnell Stony Brook University – CSE 215 15 p ap a • Prove that for any integer a and any prime number p, if p | a then p ∤ (a + 1). Kevin McDonnell Stony Brook University – CSE 215 16 • Prove that the set of prime numbers is infinite. Kevin McDonnell Stony Brook University – CSE 215 17 Kevin McDonnell Stony Brook University – CSE 215 18 • Prove that if a product of two positive real numbers is greater than 100, then at least one of the numbers is greater than 10. Kevin McDonnell Stony Brook University – CSE 215 19 • Prove that Kevin McDonnell + is irrational. Stony Brook University – CSE 215 20 Kevin McDonnell Stony Brook University – CSE 215 21