Wollongong College Australia

Wollongong College Australia University of Wollongong Diploma in Information Technology WUCT121 Discrete Mathematics Mid-Term Test #2 Autumn Session 2008 SOLUTIONS Time Allowed: 50 minutes DIRECTIONS TO CANDIDATES Total No. of Questions: 4 All questions are to be attempted. You may answer the questions in any sequence. Questions are NOT of equal value. All Notation is as used in lectures. Working ( including all necessary reasoning ) is to be shown for all solutions. No examination aids or materials are allowed. Programmable Calculators are permitted. Electronic dictionaries are NOT permitted. This exam represents 10% of the total subject marks. ITC Education Ltd trading as Wollongong College Australia CRICOS 02723D ABN 14105312329

Question 1 (a) Define or otherwise explain the following terms in relation to the set of integers Ÿ: (i) Law of Trichotomy. "x, y Œ Ÿ (x = y, x > y or x < y) Law of Trichotomy Distributivity. "x, y, z Œ Ÿ (x (y + z) = x y + x z) Distributive Law (iii) Commutativity under ordinary multiplication. "x, y Œ Ÿ ( x y = y x ) Commutative Law for Multiplication (iv) Associativity under ordinary addition. "x, y, z Œ Ÿ ( x + ( y + z ) = ( x + y ) + z ) Associative Law for Addition (b) Explain whether the set of integers Ÿ is well-ordered. For a set S to be well-ordered, any non-empty subset of S must have a least element. In the case of the set of integers Ÿ, the set E of even integers i.e. {, -4, -2, 0, 2, 4, } is a non-empty subset of Ÿ, but E has no least member. Consequently, the set of integers Ÿ is not well-ordered. (c) Consider the set B = {-1, 0, 1}. (i) Determine, giving reasons, which of ordinary addition, ordinary subtraction, ordinary multiplication, and ordinary division is a closed operation on B. Under ordinary addition 1 + 1 = 2, but 2 œ B. Hence, B is not closed under ordinary addition. Under ordinary subtraction -1-1 = -2, but -2 œ B. Hence, B is not closed under ordinary subtraction. Under ordinary multiplication -1-1 = 1, and 1 Œ B; -1 0 = 0 and 0 Œ B; -1 1 = -1 and -1 Œ B; 0 0 = 0 and 0 Œ B; 0 1 = 0 and 0 Œ B; 1 1 = 1 and 1 Œ B. Hence, B is closed under ordinary multiplication. Under ordinary division (which excludes division by 0) -1-1 = 1, and 1 Œ B; -1 1 = -1 and -1 Œ B; 0-1 = 0 and 0 Œ B; 0 1 = 0 and 0 Œ B; 1 1 = 1 and 1 Œ B. Hence, B is closed under ordinary division. Determine, giving reasons, the Identity elements under the operations of addition and multiplication in B. "x Œ B ( 0 + x = x + 0 = x ), so 0 is the Identity for Addition in set B. "x Œ B ( 1 x = x 1 = x ), so 1 is the Identity for Multiplication in set B. WCA-WUCT121-EXMST2 1

(iii) Determine the additive inverses and the multiplicative inverses of those elements of B that have multiplicative inverse elements. "x Œ B, if $ y Œ B ( x + y = y + x = 0 ) then y is termed the additive inverse of x and vice versa. From ordinary addition 1 + 1 = 0, so 1 and 1 are additive inverses of each other. Since 0 + 0 = 0, 0 is its own additive inverse. "x Œ B, if $ y Œ B ( x y = y x = 1 ) then y is termed the multiplicative inverse of x and vice versa. From Q1(c) for ordinary multiplication -1-1 = 1 so -1 is its own multiplicative inverse and 1. Also, as 1 1 = 1, 1 is its own multiplicative inverse. 0 does not have a multiplicative inverse. Marks: (a) (i) (iv) 2 each; (b) 4; (c) (i) 2, 2, 4, 4; 2, 2; (iii) 3,3. Total marks = 34. Question 2 (a) (i) What is the smallest number m Œ Õ for which m! > 4 m? 8! = 40,320 < 4 8 = 65,536, but 9! = 362,880 > 4 9 = 262,144. So, m = 9. Use the Generalised Principle of mathematical induction to prove that n! > 4 n for all natural numbers n m. Let Claim (n) be n! > 4 n for all natural numbers n 9. Claim (9) is 9! > 4 9 Since LHS = 362880 and RHS = 262144, LHS > RHS, and so Claim (9) is true. Assume that Claim (k) is true for some k 9 i.e. k! > 4 k Try to prove that Claim (k + 1) is also true, i.e. (k+1)! > 4 (k+1) LHS = (k + 1).k! > (k+1).4 k using Claim(k) > (4).4 k since k 9 > 4 (k+1) = RHS Therefore, given Claim (k) is true, Claim (k + 1) is also true, and so by the Generalised Principle of Mathematical Induction, Claim(n) must be true for all integers n 1. End of Proof. (b) Prove by mathematical induction that for all n Œ Õ, 15 n 7 n is divisible by 8. Let Claim (n) be 15 n 7 n is divisible by 8 for all natural numbers n Œ Õ. Claim (1) is 8 15 1 7 1 Since 15 1 7 1 = 15 7 = 8 and 8 8 Claim (1) is true. Assume that Claim (k) is true for some k 8 15 k 7 k i.e. $ l Œ Ÿ such that 8l = 15 k 7 k Try to prove that Claim (k + 1) is also true, i.e. 8 15 k+1 7 k+1 WCA-WUCT121-EXMST2 2

Now 15 k+1 7 k+1 = (8 + 7) 15 k 7 7 k = 8 15 k + 7 (15 k 7 k ) = 8 15 k + 7 (8l) using Claim (k) = 8 (15 k + 7l) i.e. 8 15 k+1 7 k+1, so Claim (k + 1) is also true. Therefore, by the Principle of Mathematical Induction, Claim (n) is true for all natural numbers n. (c) Given that u n+2 = 3u n+1-2u n for all n Œ Õ, u 1 = 5 and u 2 = 3, (i) Prove or disprove the statement that u n 0 for all natural numbers. u 1 0, u 2 0, but u 3 = 3 u 2 2 u 1 = 3 3 2 5 = -1 < 0, therefore the statement that u n 0 for all natural numbers is false. Use the Strong Principle of Induction to prove that u n is odd for all n Œ Õ. Let Claim (n) be u n is odd for all n Œ Õ. Claim (1) is true as u 1 = 5 is odd. Claim (2) is true as u 2 = 3 is odd. Assume that Claim (k) and Claim (k - 1),, Claim (1) are true for some arbitrary natural number k 2, that is, u k is odd $ p Œ Õ (u k = 2p - 1) u k-1 is odd $ q Œ Õ (u k-1 = 2q - 1) Try to prove that Claim (k + 1) is also true, that is, u k+1 is odd i.e. $ r Œ Õ (u k+1 = 2r - 1) Now u k+1 = 3u k - 2u k-1 using recursive definition of u k+1 = 3(2p - 1) - 2(2q - 1) using Claim (k), Claim (k - 1) = 6p - 3-4q - 2 = 6p - 4q - 5 = 6p - 4q - 4-1 = 2(3p - 2q - 2) - 1 = 2r - 1, where r = (3p - 2q - 2) Œ Õ So u k+1 is odd if Claim (k) and Claim (k - 1) are true. Therefore, by the Strong Principle of Mathematical Induction, Claim (n) is true for all natural numbers n. Marks: (a)(i) 4; 10; (b) 10; (c)(i) 3; 14. Total marks 41. Question 3 (a) Define or otherwise explain the following logical rules: WCA-WUCT121-EXMST2 3

(i) Modus Ponens. If p and q are logical predicates then (p Ÿ (p fi q)) fi q Law of Syllogism. If p, q and r are logical predicates then ((p fi q) Ÿ (q fi r)) fi (p fi r) (b) Consider the following statement: If a product of two positive real numbers is greater than 100, then at least one of the numbers is greater than 10. Prove this statement using proof by contradiction. This statement is of the form P fi Q. To prove it by contradiction, consider the statement ~Q i.e. both of two real positive numbers are less than 10. Clearly, the product of two such numbers, each less than 10, must be less than 100. Consequently, ~Q fi ~P so that P fi Q is proven. (c) Show by a direct proof that 9y 2 + 3 12y - 1. " y Œ, 9y 2 + 3 12y - 1 9y 2-12y + 3 + 1 0 (3y 2) 2 0, but " x Œ, x 2 0 and " y Œ, 3y 2 Œ, so (3y 2) 2 0 is a true statement. Hence, the logically equivalent statement 9y 2 + 3 12y 1 is also true. (d) Consider the following statements: If 5 is greater than 8, then 29 is not a prime number. However, 5 is not greater than 8. Therefore, 29 is a prime number. Translate these statements into symbolic notation and, using a truth table, show whether or not they form a valid argument. Let p be the statement: 5 is greater than 8. Let q be the statement: 29 is a prime number. Let r be the compound statement: pfi ~q Let s be the statement: ~p The given argument then takes the form rÿ sfi q and is a valid argument iff rÿ sfi q is a tautology. This is tested in the truth table below. p q p fi ~q ~p (p fi ~q) ~p ((p fi ~q) ~p) fi q T T F F F T T F T F F T F T T T T T F F T T T F WCA-WUCT121-EXMST2 4

The final column of this table indicates that rÿ sfi q is a contingent statement and not a tautology. Consequently the argument is false. (e) (i) Prove or disprove the following statement: $x Œ, y Œ, xy = 0. This statement claims that there is a pair of real numbers that have a zero product. This can be shown to be true by providing an example of such a pair of real numbers. One example of such a pair is 0 and 0, since 0 x 0 = 0. Hence, the statement is proven to be true. Is the statement in 3(e)(i) an existential statement or a universal statement. Explain briefly. The statement in 3(e)(i) is an existential statement because it claims only that there are some real numbers that satisfy the claim. It doesn't claim that it is true for all real numbers (as would a corresponding universal statement) although this possibility is not excluded. (f) (i) Complete the tautology (P fi R) Ÿ (Q fi R) ï (..fi ) ((P fi R) Ÿ (Q fi R)) ï ((P Q) fi R) By considering all cases, prove or disprove the following statement: For any integer z Œ Ÿ, (2z) 2 is an even positive integer. By the Quotient Remainder Theorem for any z Œ Ÿ, $ q Œ Ÿ and r Œ Õ such that z = 2q + r where 0 r < 2, i.e. z = 2q or 2q +1. In words, z is either even or odd. Case 1: z is even i.e. $ q Œ Ÿ such that z =2q so (2z) 2 = (2 2q) 2 = 2(8q 2 ). Hence, (2z) 2 is an even positive integer. Case 2: z is odd i.e. $ q Œ Ÿ such that z =2q + 1 so (2z) 2 = (2 (2q + 1)) 2 = 2 (2 (2q + 1) 2 ). Hence, (2z) 2 is an even positive integer if z is an odd integer. Consequently, by the tautology in Q3(f) (i), for any integer z, (2z) 2 is an even positive integer. Note : the counter example z = 0 gives (2z) 2 =0 which is neither positive nor negative and hence the statement is disproved. Full marks have been awarded for either proof by cases or the disproof noted, as the distinction for zero being neutral has not been fully noted in lectures. Marks (a)(i) 2; 2; (b) 4; (c)6; (d) 8; (e)(i) 3; 2; (f) (i) 3; 8. Total marks = 38. Question 4 (a) Define or otherwise explain the Fundamental Theorem of Arithmetic. If a Œ Ÿ and a > 1, then a can be factorized in a unique way in the form a a = p 1 a 1 p 2 a 2 p 3 a k 3 p k where p 1, p 2,, p k are each prime numbers and a i Œ Õ for each i = 1, 2,, k. (b) (i) Write down a composite integer, m, such that m and 7 are not relatively prime. WCA-WUCT121-EXMST2 5

m = 14. The gcd(14, 7) = 7, so 14 and 7 are not relatively prime. What is the value of gcd(m, 7)? Briefly explain your answer. gcd(m, 7) = 7 because 7 has no factors other than 7 and 1, m must be a multiple of 7. (iii) What is value of lcm(m, 7)? Briefly explain your answer. The lcm(m, 7) = m since m is a multiple of 7 and m is the smallest multiple of m. (c) (i) Use the Sieve of Eratosthenes to find all the prime numbers between 1 and 80. Step 1: Write down all of the integers between 1 and 80 inclusive. Step 2: Eliminate all multiples of prime numbers less than or equal to 80. a) Eliminating multiples of 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 b) Eliminating multiples of 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 c) Eliminating multiples of 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 d) Eliminating multiples of 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 WCA-WUCT121-EXMST2 6

So the set of prime numbers between 1 and 80 is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79} Write down all the sets of twin primes that lie between 1 and 80. The sets of twin primes are: (3, 5),(5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71,73) (iii) Write down each set of twin primes that lie between 1 and 80 whose sum is a perfect square. There are two such pairs of twin primes: (17, 19), where 17 + 19 = 36 = 6 2, and (71, 73), where 71 + 73 = 144 = 12 2. (d) (i) Use the Euclidean Algorithm to find the greatest common divisor of 76 and 1657. gcd( 1657, 76) = gcd(1657, 76) 1657 = 21 76 + 61 76 = 1 61 + 15 61 = 4 15 + 1 gcd(1657, 76)= gcd(76, 61) = gcd(61, 15) = gcd(15, 1) = gcd(1, 0) = 1 15 = 15 1 + 0 Find integers m and n such that gcd( 1657, 76) = m( 1657) + n(76). From (i) 61 = 1657 21 76 15 = 76 1 61 1 = 61 4 15 Hence, 1 = 61 4 (76 1 61) 1 = 5 61 4 76 1 = 5 (1657 21 76) 4 76 1 = 5 1657 109 76 1 = ( 5) ( 1657) + ( 109) 76 so that m = 5 and n = 109. (e) A school administrator wants to assign 80 students to 13 classes so that no class has more than 12 students. Show that there must be at least 3 classes with 6 or more. If two classes each have the maximum number of students enrolled, that leaves 56 students to be allocated to 11 classes with no class having more than 12 students. Since 56 > 5 11 but < 6 11, it follows from the Generalised Pigeonhole Principle that WCA-WUCT121-EXMST2 7

there must be at one class of the 11 that has at least 5 + 1 = 6 students. Hence there must be at least 2 + 1 = 3 classes with 6 or more students. (f) (i) Define or otherwise explain the Quotient-Remainder Theorem. For any integers n and d, where d > 0, there exist integers q and r such that n = q d +r, where 0 r < d. For example, if n = 34 and d = 3 then 34 = 12 3 + 2. Using the Quotient-Remainder Theorem, or otherwise, show that for any odd natural number n Œ Õ, 8 n 2 1. [Hint: By the Quotient-Remainder Theorem, any natural number can be written as 4k, 4k +1, 4k +2 or 4k +3.] Since n is an odd natural number it must take the form of 4k + 1 or 4k + 3, as 4k and 4k +2 are even numbers. Hence there are two cases to consider. Case 1: If n = 4k + 1, then n 2 1= 16k 2 + 8k + 1 1 = 8(2k 2 + k). Case 2: If n = 4k + 3, then n 2 1= 16k 2 + 24k + 9 1 = 8(2k 2 +3 k +1). Consequently, by the tautology in Q3(g) (i), when n is an odd natural number 8 n 2 1. (g) (i) Prove or disprove the following statement: The sum of any 5 consecutive natural numbers is always divisible by 5. Let the five consecutive numbers be k, k+1, k + 2, k + 3, k +4. The sum of these numbers is 5k + 10, and 5 (5k + 10). Consequently, the statement that, The sum of any 5 consecutive natural numbers is always divisible by 5, is true. Prove or disprove the following statement: The sum of any n consecutive natural numbers is always divisible by n. Let the four consecutive numbers be k, k+1, k + 2, k + 3. The sum of these numbers is 4k + 6, but 4 F (4k + 6). Consequently, the statement that, The sum of any n consecutive natural numbers is always divisible by n, is false. Marks (a) 3; (b) (i) 2; 2; (iii) 2; (c) (i) 10; 4; (iii) 4; (d) (i) 4; 6; (e) 6; (f) (i) 2; 6; (g) (i) 3; 3. Total marks = 57. Total marks for whole paper = 34 + 41 + 38 + 57 = 170. WCA-WUCT121-EXMST2 1