1. (16 points) Circle T if the corresponding statement is True or F if it is False.
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1 Name Solution Key Show All Work!!! Page 1 1. (16 points) Circle T if the corresponding statement is True or F if it is False. T F The sequence {1, 1, 1, 1, 1, 1...} is an example of an Alternating sequence. T F The Weak and Strong Principles of Mathematical Induction are logically equivalent. T F In general, iterative algorithms use memory more efficiently than their equivalent recursive representations. T F n = 10 (n + 1) 1. T F ,000 = 1,001,000. T F Algorithms whose order is O(n ) are less efficient than those of order O( n ). T F If a, b, and p are positive Integers and (a MOD p) = (b MOD p), then a = b. T F If p and q are positive Integers, then LCM( p,q)gcd( p,q) = pq.. (8 points) Let {a n } and {b n } be the sequences defined, for n > 0, by: a n = (n 1), and b n = (n + 1). Find c 1, c, c 3, and c 4 when c n = (a n )(b n ). c n = (a n )(b n ) = (n 1)(n + 1) = n 1, hence c 1 = 1 1 = 1 1 = 0; c = 1 = 4 1 = 3; c 3 = 3 1 = 9 1 = 8; c 4 = 4 1 = 16 1 = (8 points) Write out the Euclidean Algorithm and trace its steps to calculate GCD(19,4). INPUT (A,B) (where A > B) STEP 1 3 SET X = A MOD B X IF TEST (X > 0) THEN TEST T T F SET A = B A SET B = X B ELSE OUTPUT(B) OUTPUT hence GCD(19,4) = 1.
2 Name Solution Key Show All Work!!! Page 4. (15 points) (a) Give a Recursive Definition for the odd integers. (Basis) Assume 1 is in the set. (Inductive) If p is in the set, then (p + ) and (p ) are in the set. (b) Find the Big-Oh of the algorithm with complexity: (n + 1)(4n 3 + 5) + [n 4 + n(log n)](3n ). (n + 1)(4n 3 + 5) + [n 4 + n(log n)](3n ) < (n + n )(4n 3 + 5n 3 ) + [n 4 + n(n)](3n ) < (n )(9n 3 ) + (n 4 + n 4 )(3n ) = 18n 5 + 9n 6 < 7n 6 hence (n + 1)(4n 3 + 5) + [n 4 + n(log n)](3n ) is O(n 6 ). 5. (8 points) (a) Given a = and b = , then GCD(a,b) = and LCM(a,b) = (b) List out the search intervals in applying the Binary Search algorithm to find 8 in the list: Midpoint (0+0)div (10+0)div (10+5)div (10+7)div = 10 = 5 = 7 = 8 Test Interval [0,10] [5,10] [7,10] Done!
3 Name Solution Key Show All Work!!! Page 3 6. (15 points) Prove one of the two Theorems below using Mathematical Induction. Theorem 1: For all integers n > 0, 3 i = n 3 n Theorem : If a 0 = 0, a 1 = 10, and a = 0, then a n = a n 1 + a n + a n 3 is divisible by 10, for all n >. Theorem 1 Proof: (Weak Induction) Basis Step: Show true for n = 0. Now, when n = 0, we have LHS = 3 i = 3 0 = 1 and RHS = = (3 1)/ = / = 1. Thus, 1 = LHS = RHS, so the formula holds for n = 0. k Inductive Step: Assume 3 i = , for some integer k > 0. Show 3 i. = k Now (k + 1) = ( k ) + 3 (k + 1) = [3 (k + 1) 1]/ + 3 = [3 (k + 1) 1 + ()3 (k + 1) ] / = [(3)3 (k + 1) 1] / = [3 (k + 1)+1 1] /. k ( k + 1) 1 (k + 1) + 1 Therefore, for all integers n > 0, n 3 i = 3 n QED Theorem Proof: (Strong Induction) Basis Step: Show true for n = 3. Now, a 3 = a 0 + a 1 + a = = 30 = 3(10). Since 3 is an Integer, we see that a 3 is divisible by 10. Inductive Step: Assume a i is divisible by 10 for all < i < k. Show a k is divisible by 10. Now, a k = a k 1 + a k + a k 3, but the Induction Hypothesis allows us to assume a k 1, a k, a k 3 are each divisible by 10. Thus, there are Integers p, q, and r with a k 1 = 10p, a k = 10q, and a k 3 = 10r. Applying this, we see that a k = a k 1 + a k + a k 3 = 10p + 10q + 10r = 10(p + q + r). Moreover, since p, q, and r are Integers, we see that (p + q + r) is an Integer. Thus a k is divisible by 10. Therefore, a n = a n 1 + a n + a n 3 is divisible by 10, for all n >. QED
4 Name Solution Key Show All Work!!! Page 4 7. (15 points) Prove one of the two Theorems below: Theorem 1: If a, b, p, and r are Integers with a = bq + r, then GCD(a,b) = GCD(b,r). Theorem : The product of odd integers is odd. Theorem 1 Proof: Let a, b, p, and r are Integers with a = bq + r. We will show GDC(a,b) = GCD(b,r) by showing (Case 1) GDC(a,b) < GCD(b,r), and (Case ) GDC(a,b) > GCD(b,r). (Case 1) Let X = GDC(a,b). Thus X is a common divisor of a and b, so there are Integers w and x such that a = wx and b = xx. Now, a = bq + r implies r = a bq = wx xxq = (w xq)x. Since w, x, q are Integers, we have that (w xq) is an Integer, so X divides r. Since X also divides b, we see that X is a common divisor of b and r. Thus, X = GCD(a,b) < GCD(b,r). (Case ) Let Y = GDC(b,r). Thus Y is a common divisor of b and r, so there are Integers y and z such that b = yy and r = zy. Now, a = bq + r implies a = yyq + zy = (yq + z)y. Since y, z, q are Integers, we have that (yq + z) is an Integer, so Y divides a. Since Y also divides b, we see that Y is a common divisor of a and b. Thus, GCD(a,b) > GCD(b,r) = Y. Putting Case 1 and Case together, we conclude, therefore, that GCD(a,b) = GCD(b,r). QED Theorem Proof: Assume A and B are odd Integers; that is, there exist integers p and q, such that A = p + 1 and B = q + 1. Show AB is and odd integer. Now, AB = (p + 1)(q + 1) = 4pq + 4p + 4q + 1 = (pq + p + q) + 1. Since p and q are Integers, it follows that (pq + p + q) is an Integer. Thus AB = (integer) + 1. Therefore AB is odd. QED
5 Name Solution Key Show All Work!!! Page 5 8. (15 points) Prove one of the two Theorems below by Contradiction or Contraposition. Theorem 1: is irrational. Theorem : The Prime numbers form an infinite set. Theorem 1 Proof: (Contradiction) Assume is Rational. That is, = p/q where p and q are Integers, q is non-zero, and GCD(p,q) = 1. Now, = p/q implies = p / q, so p = q, implying divides p. Hence, we can assert that divides p, so there is an Integer m with p = m, so p = (m) = 4m = q. However, 4m = q yields m = q, so q is divisible by, implying q is divisible by. But, this is a contradiction, since p divisible by and q divisible by implies the assumption that GCD(p,q) = 1 is false. Therefore is irrational. QED Theorem Proof: (Contradiction) Assume the set of Primes is finite; that is, there is a largest Prime number, say P. Now, given the set of Primes, {, 3, 5, 7,..., P}, we construct the Integer M = (3)(5)(7)...(P). Clearly M is divisible by every Prime, by construction. Since M is an Integer, it follows that (M + 1) is an Integer. By the Fundamental Theorem of Arithmetic, we assert that (M + 1) is divisible by a Prime, but since M is divisible by every Prime, it follows that the Prime that divides (M + 1) must also divide M, hence it must also divide 1. This is a contradiction, since no Prime divides 1. Therefore the set of Primes is infinite. QED
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