CSE 215: Foundations of Computer Science Recitation Exercises Set #5 Stony Brook University. Name: ID#: Section #: Score: / 4

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1 CSE 215: Foundations of Computer Science Recitation Exercises Set #5 Stony Brook University Name: ID#: Section #: Score: / 4 Unit 10: Proofs by Contradiction and Contraposition 1. Prove the following statement by contradiction: The product of any nonzero rational number and any irrational number is irrational. Suppose not. That is, suppose a nonzero rational number x and an irrational number y such that xy is rational. By definition of rational, x = a/b and xy = c/d for some integers a, b, c and d with b 0 and d 0. Also, a 0 because x is nonzero. By substitution, xy = (a/b)y = c/d. Solving for y gives y = (bc)/(ad). Now, bc and ad are integers because they are products of integers, and ad 0 by the zero product property. Thus, by definition of rational, y is rational, which contradicts the supposition that y is irrational. 2. Prove by contraposition: For all positive integers n, r and s, if rs n, then r n or s n. Suppose not. That is, suppose r, s and n are integers and r > n and s > n. Note that r and s are positive because n cannot be negative. Multiply both sides of r > n by s, and both sides of s > n by n: rs > ns and ns > n n = n. By transitivity of the less-than operator, rs > n, which is a contradiction of the assumption that rs n. CSE 215 R01/R02/R06 Page 1 of 5

2 3. Prove the following statement first by contraposition and then by contradiction: For all integers a, b and c, if a bc then a b. (Note that means does not divide. ) a. Proof by contraposition. Suppose a, b and c are any integers such that a b. We must show that a bc. By definition of divides, b = ak for some integer k. Then bc = (ak)c = a(kc). But kc is an integer because it is a product of the integers k and c. Hence a bc by definition of divisibility b. Proof by contradiction. Suppose not. That is, suppose integers a, b and c such that a bc and a b. Since a b, there exists an integer k such that b = ak by definition of divides. Then bc = (ak)c = a(kc) by the associative law of algebra. But kc is an integer because it is a product of integers, and so a bc by definition of divides. Thus a bc and a bc, which is a contradiction. CSE 215 R01/R02/R06 Page 2 of 5

3 4. Prove the following statement if it s true, or disprove it if it s false: The positive square root of a positive irrational number is irrational. The statement is true. The statement can be written formally as: positive real numbers r, if r is irrational, then r is irrational. Proof by contraposition. Suppose r is any positive real number such that r is rational. We must show that r is rational. By definition of rational, r = a for some integers a and b with b 0. Then r = ( r) 2 = ( ) a 2 b b = a 2. b 2 But both a 2 and b 2 are integers because they are products of integers, and b 2 0 by the zero product. Thus, r is rational. 5. Suppose a is an integer and p is a prime number such that p a and p (a + 3). What can you deduce about p? Why? We can deduce that p = 3. Proof: Let a be any integer, and let p be any prime number such that p a and p (a + 3). By definition of divisibility, there are integers r and s so that a = pr and a + 3 = ps. Then 3 = (a + 3) a = ps pr = p(s r). So 3 = p(s r). But s r is an integer (because it is a difference of integers), and so by definition of divisibility p 3. But since 3 is a prime number, its only positive divisors are 1 and itself. So p = 1 or p = 3. However, since p is a prime number, p 1. Hence p = 3. CSE 215 R01/R02/R06 Page 3 of 5

4 6. Use proof by contradiction to show that for any integer n, it is impossible for n to equal both 3q 1 + r 1 and 3q 2 + r 2, where q 1, q 2, r 1 and r 2, are integers, 0 r 1 < 3, 0 r 2 < 3, and r 1 r 2. Suppose not. That is, suppose there is an integer n such that n = 3q 1 + r 1 = 3q 2 + r 2, where q 1, q 2, r 1 and r 2 are integers, 0 r 1 < 3, 0 r 2 < 3 and r 1 r 2. By interchanging the labels for r 1 and r 2 if necessary, we may assume that r 2 > r 1. Then 3(q 1 q 2 ) = r 2 r 1. Since r 2 > r 1, r 2 r 1 > 0. Now, both r 1 and r 2 are less than 3. This means r 2 r 1 = 1 or r 2 r 1 = 2. This means 3(q 1 q 2 ) = 1 or 3(q 1 q 2 ) = 2. The first case implies that 3 1, which means 3 1. The second case implies that 3 2, which means 3 2. These results contradict the fact that 3 is greater than both 1 and 2. Thus in either case we have reached a contradiction, so the original statement is true. CSE 215 R01/R02/R06 Page 4 of 5

5 7. Use proof by contradiction, the quotient-remainder theorem, division into cases and the result of the previous question to prove that for all integers n, if n 2 is divisible by 3 then n is divisible by 3. Suppose not. That is, suppose there is an integer n such that n 2 is divisible by 3 and n is not divisible by 3. By definition of divisible, n 2 = 3q for some integer q, and by the quotient-remainder theorem and the previous question, n = 3k + 1 or n = 3k + 2 for some integer k. Case 1 (n = 3k + 1 for some integer k): In this case, n 2 = (3k + 1) 2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1 Let s = 3k 2 + 2k. Then n 2 = 3s + 1 and s is an integer because it is a sum and products of integers. So n 2 = 3q = 3s + 1 for some integers q and s, which contradicts the result of the previous question. Case 2 (n = 3k + 2 for some integer k): In this case, n 2 = (3k + 2) 2 = 9k k + 4 = 3(3k 2 + 6k + 1) + 1. Then n 2 = 3t + 1 and t is an integer because it is a sum of products of integers. So n 2 = 3q = 3t + 1 for some integers q and t, which contradicts the result of the previous question. Thus in either case, a contradiction is reached, which shows that the original statement is true. CSE 215 R01/R02/R06 Page 5 of 5

MATH 271 Summer 2016 Practice problem solutions Week 1

Part I MATH 271 Summer 2016 Practice problem solutions Week 1 For each of the following statements, determine whether the statement is true or false. Prove the true statements. For the false statement,