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CMSC 203 / 0201 Fall 2002 Exam #1 Review – 27 September 2002 Prof. Marie desJardins September1999 Survey says… 1(a) 1(b) 1( c) Moving too fast 3(a) 3(b) 3( c) 3(d) Don't understand lecture 4 -2 (m. worse) -- 2 (much better) Worried about grade Considering dropping TRUE FALSE 13 12 F 23 3T 1 24 F 6 25 6 1 Lecture OK but not HW Not enough time on HW No trouble AVG: 20 1 19 24 -0.57692 T T F T T F F F T T F -1 0 September1999 October 1999 Proposal Better balance of straightforward and challenging problems on homework Grading (somewhat) less weighted towards challenging problems More time in class on developing and writing well structured proofs More time in class for students to try to solve problems that we then work through as a class In return: Students agree to review chapter readings before class so we can spend less time on the basics without losing everybody. September1999 October 1999 Let’s make a proof HW #2, Problem #1, exercise 1.6.20 If f and fg are one-to-one, does it follow that g is one-to-one? Justify your answer. General problem-solving approach to proof construction: 1. Restate the problem, writing the premise and conclusion in mathematical language. 2. Decide what type of proof to use. 3. Apply any relevant definitions, axioms, laws, or theorems to simplify the premise, make it look more like the conclusion, or connect (relate) multiple premises. 4. Carefully write down and justify each step of the proof, in a sequence of connected steps. 5. Write a conclusion statement. 6. Write “Q.E.D.” September1999 October 1999 Restate the problem If f and fg are one-to-one, does it follow that g is one-toone? PREMISE 1: “f is one-to-one” iff f(x) = f(y) x = y for all x,y in the domain of f. Used: Definition of one-to-one PREMISE 2: “fg is one-to-one” iff f(g(a)) = f(g(b)) a = b for all x, y in the domain of g. Used: Definition of one-to-one and composition CONCLUSION: “g is one-to-one” iff g(w) = g(z) w = z for all w,z in the domain of g. Used: Definition of one-to-one September1999 October 1999 Restate the problem If f and fg are one-to-one, does it follow that g is one-to-one? Show that if xy ( f(x) = f(y) x = y ) (P1) and ab ( f(g(a)) = f(g(b)) a = b ), (P2) then wz ( g(w) = g(z) w = z ). (C) September1999 October 1999 Select a proof type Direct proof Work from premises to conclusions Indirect proof Negate the conclusion and derive a contradiction In this case, the negated conclusion is wz ( g(w) = g(z) w = z ) or wz ( g(w) = g(z) w = z ) or wz ( g(w) = g(z) (w=z) ) (C´) September1999 October 1999 Apply relevant knowledge Premise 1: xy ( f(x) = f(y) x = y ) Premise 2: ab ( f(g(a)) = f(g(b)) a = b ) Negated conclusion: wz ( g(w) = g(z) (w=z) ) Suppose (3) holds. Then w and z s.t.: (P1) (P2) (C´) g(w) = g(z) (1) w <> z (2) Since g(w) = g(z), it must be the cas that f(g(w)) = f(g(z)), therefore w=z by (P2), which contradicts (2). September1999 October 1999 Construct a sequence of steps Suppose that g is not one-to-one. Then (by the definition of “one-to-one”) there must exist some values w and z in the domain of g such that g(w) = g(z) and w < > z. But since g(w) = g(z), it must be the case that f(g(w)) = f(g(z)). Since fg is one-to-one, it must be the case that w=z, which contradicts our earlier supposition. September1999 October 1999 Write a conclusion statement Therefore, g must be one-to-one. September1999 October 1999 Write “Q.E.D.” Q.E.D. September1999 October 1999 The Proof Theorem. If f and fg are one-to-one, then g is one-to-one. Proof. Suppose that g is not one-to-one. Then (by the definition of “one-to-one”) there must exist some values w and z in the domain of g such that g(w) = g(z) and w < > z. But since g(w) = g(z), it must be the case that f(g(w)) = f(g(z)). Since fg is one-to-one, it must be the case that w=z, which contradicts our earlier supposition. Therefore, g must be one-to-one. Q.E.D. September1999 October 1999 Another proof problem HW2, P2, exercise 1.6.56 Suppose that f is an invertible function from Y to Z and g is an invertible function from X to Y. Show that the inverse of the composition fg is given by (fg)-1 = g-1f-1. f is invertible: there exists a function f-1 such that yY, f-1(f(y)) = y. g is invertible: there exists a function g-1 such that xX, g-1(g(x)) = x. If fg is invertible, there must exist a function (fg)-1 s.t. fg-1(fg(x)) = x. We wish to show that fg-1 = g-1f-1, i.e., x, fg-1(x) = g-1(f-1(x)) September1999 October 1999 The Second Proof Theorem. If f, g, and fg are invertible, then (fg)-1 = g-1f-1. Proof. Since f, g, and fg are invertible, then their inverse functions g-1, f-1, and fg-1 must exist. The inverse function (fg)-1 exhibits the property that x, fg-1(fg(x)) = x. We show that g-1f-1 exhibits this property: x, g-1(f-1(f(g(x))) = g-1(g(x)) = x. Since f-1 is the inverse of f Therefore, g-1f-1 is the inverse of fg. Q.E.D. September1999 October 1999 The really hard one… HW2, P3 part 2, *1.7.22 Use the technique given in Exercise 19, together with the result of Exercise 13b, to find a formula for k=1n k2. September1999 October 1999 Big-O, , HW2, P4, 1.8.8(a,c) (a) f(x) = 2x2 + x3 log x (c) f(x) = (x4 + x2 + 1) / (x4 + 1) September1999 October 1999 Perfect numbers 2.3.16(a) Show that 6 and 28 are perfect. The divisors of 6 are 1, 2, and 3. 1+2+3=6; therefore, 6 is a perfect number. The divisors of 28 are 1, 2, 4, 7, and 14. 1+2+4+7+14 = 28; therefore, 28 is a perfect number. September1999 October 1999 Harder perfect numbers 2.3.16(b) Show that x=2p-1(2p-1) is a perfect number when 2p-1 is prime. The divisors of x are 2p-1, 2p-1, their divisors, and the products of their divisors. The divisors of 2p-1 are 1, 2, 22, …, 2p-1. Since 2p-1 is prime, its divisors are 1 and 2p-1. September1999 October 1999 Harder perfect numbers cont. Therefore, the proper divisors of x are 1, 2, 22, …, 2p-1, 2p-1, 2(2p-1), 22(2p-1 ), …, 2p-2(2p-1). The sum of these divisors is i=0p-1 2i + (2p-1) i=0p-2 2i = 2p-1 + (2p-1) (2p-1 – 1) = (2p-1) (1 + 2p-1 – 1) = 2p-1 (2p-1) Therefore, 2p-1 (2p-1) is a proper number. Q.E.D. September1999 October 1999 Other requested topics Sets, inclusion-exclusion |AB| = |A| + |B| - |AB| *Algorithms, complexity *Quantifiers – 1.3.13 – S(x), F(x), A(x,y) (b) Every student has asked Prof. G. a question. (c) Every faculty member has either asked Prof. M a question or been asked a question by Prof. M. There are exactly two students who have asked Prof. dJ a question. Functions Integers and division September1999 October 1999