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1 MATHEMATICS 271 L60 SUMMER 2003 ASSIGNMENT 1 SOLUTION 1. Let P be the statement: “If a | b and a | c then a | (b + c) and a | (b − c)” (a) Is P true for all integers a, b and c? Prove your answer. (b) Write the converse of P. Is the converse of P true for all integers a, b and c? Prove your answer. (c) Write the contrapositive of P. Is the contrapositive of P true for all integers a, b and c? Explain. (d) Write the negation of P. Is the negation of P true for some integers a, b and c? Explain. Solution: (a) P is true for all integers a, b and c and here is a proof. Let a, b and c be integers so that a | b and a | c. Since a | b and a | c, there are integers m and n so that b = am and c = an, and so, b + c = am + an = a (m + n) and b − c = am − an = a (m − n). Since b + c = a (m + n) and b − c = a (m − n), where a, m + n and m − n are integers, we conclude that a | (b + c) and a | (b − c).¤ (b) The converse of P is “If a | (b + c) and a | (b − c) then a | b and a | c”. The converse of P is not true for all integers a, b and c. For example, when a = 2 and b = c = 1, we have a | (b + c) and a | (b − c) (because 2 | 2 and 2 | 0), but a - b (because 2 - 1).¤ (c) The contrapositive of P is “If a - (b + c) or a - (b − c) then a - b or a - c”. The contrapositive of P is true for all integers a, b and c. This is because the contrapositive of P is logically equivalent to P which is proven to be true in (a).¤ (d) The negation of P is “ a | b and a | c, but a - (b + c) or a - (b − c)”. The negation of P is not true for some integers a, b and c, because we have shown in (a) that P is true for all integers a, b and c.¤ 2. Prove or disprove each of the following: (a) For all real numbers x and y, if x is rational and y is irrational then x + y is irrational. (b) For all real numbers x and y, if x is rational and y is irrational then xy is irrational. (c) For all real numbers x and y, if x+y is irrational then x is irrational or y is irrational. (d) For all real numbers x and y, if xy is irrational then x is irrational or y is irrational. Solution: (a) This statement is true and here is a proof. Let x be a rational number and y be an irrational number. We show that x + y is irrational by a contradiction proof. Suppose that x + y is rational. Then, since x and x + y are rational, there are integers m, n, p, q so that x= m and x + y = pq where n 6= 0 6= q. It follows that y = (x + y) − x = pq − m = pn−mq , n n qn which implies that y is rational (for being a quotient of the integers pn − mq and qn, and note that nq 6= 0 because n 6= 0 6= q). This contradicts the assumption that y is irrational. Thus, x + y is irrational.¤ 2 √ (b) This statement is false. For example, when x = 0 and y = 2, we see that x is rational and y is irrational. However, xy = 0 is rational.¤ (c) This statement is true and here is a proof. Let x and y be real numbers so that x + y is irrational. We show that x is irrational or y is irrational by a contradiction proof. Suppose that both x and y are rational; that is, there are integers m, n, p, q so that x= m and y = pq where n 6= 0 6= q. It follows that x + y = pq + m = pn+mq , which implies n n qn that x + y is rational (for being a quotient of the integers pn + mq and qn, and note that nq 6= 0 because n 6= 0 6= q). This contradicts the assumption that x + y is irrational. Thus, x is irrational or y is irrational.¤ (c) This statement is true and here is a proof. Let x and y be real numbers so that xy is irrational. We show that x is irrational or y is irrational by a contradiction proof. Suppose that both x and y are rational; that is, there are integers m, n, p, q so that x= m and y = pq where n 6= 0 6= q. It follows that xy = pq m = pm , which implies that xy n n qn is rational (for being a quotient of the integers pm and qn, and note that nq 6= 0 because n 6= 0 6= q). This contradicts the assumption that xy is irrational. Thus, x is irrational or y is irrational.¤ 3. In this question, k, m and n are integers. Prove or disprove each of the following: (a) m is even and n is odd if and only if m + 2n is even and m + n is odd. (b) If k | m and m | n then k | n. (c) If k | mn then k | m and k | n. (d) For all natural numbers n, n2 + n + 41 is a prime. Solution: (a) This statement is true and here is a proof. (=⇒) We prove that if m is even and n is odd then m + 2n is even and m + n is odd. Suppose that m is an even integer and n is an odd integer, that is, there are integers x and y so that m = 2x and y = 2y + 1. Then, m + 2n = 2x + 2n = 2 (x + n) and m + n = 2x + 2y + 1 = 2 (x + y) + 1, which means that m + 2n is even and m + n is odd. (⇐=) We prove that if m + 2n is even and m + n is odd then m is even and n is odd. Suppose that m and n are integers so that m + 2n is even and m + n is odd, that is, m + 2n = 2s and m + n = 2t + 1 for some integers s and t. Then, m = 2 (m + n) − (m + 2n) = 2 (m + n) − 2s = 2 (m + n − s) and n = (m + 2n) − (m + n) = 2s − (2t + 1) = 2 (s − t − 1) + 1, which imply that m is even and n is odd.¤ (b) This statement is true and here is a proof. Suppose that k, m and n are integers so that k | m and m | n, which mean that m = ka and n = mb for some integers a and b. Then, n = mb = (ka) b = k (ab) which means that k | n.¤ (c) This statement is false because when k = 4 and m = n = 2, we have k | mn (because 4 | 4), but k - m (because 4 - 2).¤ (d) This statement is false because when n = 41, we have 2 n + n + 41 = 412 + 41 + 41 = 41 × (41 + 1 + 1) = 41 × 43 which is not a prime.¤ 3 4. Prove or disprove each of the following: (a) For all real numbers x, x2 − x + 1 > 0. (b) For all real numbers x and y, if x2 − x + 1 = y 2 − y + 1 then x = y. (c) For all posive real numbers x, there exists a positive number y so that 0 < y < x. (d) There exists a positive number y so that for all positive real numbers x, 0 < y < x. Solution: (a) This statement is true and here is a proof. Let x be a real number. Then ¡ ¢ ¡ ¡ ¢2 ¢2 2 x − x + 1 = x2 − 2 12 x + 14 + 34 = x − 12 + 34 ≥ 34 > 0. This is because x − 12 ≥ 0.¤ (b) This statement is false because when x = 0 and y = 1, we have x2 − x + 1 = 02 − 0 + 1 = 1 = 12 − 1 + 1 = y 2 − y + 1, but x = 0 6= 1 = y.¤ (c) This statement is true and here is a proof. Let x be a posive real number. We consider y = x2 . From 0 < 12 < 1,we get (by multiplying by the positive number x), that 0 < x2 < x, that is, 0 < y < x. (d) This statement is false because for any real number y, as shown in (c) we can find a positive real number x so that x < y.