1 / 20

CMSC 203 / 0201 Fall 2002

CMSC 203 / 0201 Fall 2002. Exam #1 Review – 27 September 2002 Prof. Marie desJardins. Survey says…. Proposal. Better balance of straightforward and challenging problems on homework Grading (somewhat) less weighted towards challenging problems

pearljsmith
Download Presentation

CMSC 203 / 0201 Fall 2002

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CMSC 203 / 0201Fall 2002 Exam #1 Review – 27 September 2002 Prof. Marie desJardins

  2. Survey says…

  3. 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.

  4. Let’s make a proof • HW #2, Problem #1, exercise 1.6.20 • If f and fg are one-to-one, does it follow that g is one-to-one? Justify your answer. • General problem-solving approach to proof construction: • Restate the problem, writing the premise and conclusion in mathematical language. • Decide what type of proof to use. • Apply any relevant definitions, axioms, laws, or theorems to simplify the premise, make it look more like the conclusion, or connect (relate) multiple premises. • Carefully write down and justify each step of the proof, in a sequence of connected steps. • Write a conclusion statement. • Write “Q.E.D.”

  5. Restate the problem • If f and fg are one-to-one, does it follow that g is one-to-one? • 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: “fg 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

  6. Restate the problem • If f and fg 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)

  7. 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´)

  8. Apply relevant knowledge • Premise 1: xy ( f(x) = f(y)  x = y ) (P1) • Premise 2: ab ( f(g(a)) = f(g(b))  a = b ) (P2) • Negated conclusion: wz ( g(w) = g(z)  (w=z) ) (C´) • Suppose (3) holds. Then  w and z s.t.: • 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).

  9. 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 fg is one-to-one, it must be the case that w=z, which contradicts our earlier supposition.

  10. Write a conclusion statement • Therefore, g must be one-to-one.

  11. Write “Q.E.D.” • Q.E.D.

  12. The Proof • Theorem. If f and fg 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 fg 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.

  13. 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 fg is given by (fg)-1 = g-1f-1. • f is invertible: there exists a function f-1 such that yY, f-1(f(y)) = y. • g is invertible: there exists a function g-1 such that xX, g-1(g(x)) = x. • If fg is invertible, there must exist a function (fg)-1 s.t. fg-1(fg(x)) = x. • We wish to show that fg-1 = g-1f-1, i.e., x, fg-1(x) = g-1(f-1(x))

  14. The Second Proof • Theorem. If f, g, and fg are invertible, then (fg)-1 = g-1f-1. • Proof. Since f, g, and fg are invertible, then their inverse functions g-1, f-1, and fg-1 must exist. The inverse function (fg)-1 exhibits the property thatx, fg-1(fg(x)) = x. We show that g-1f-1 exhibits this property:x, g-1(f-1(f(g(x))) = g-1(g(x)) Since f-1 is the inverse of f = x.Therefore, g-1f-1 is the inverse of fg.Q.E.D.

  15. 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.

  16. Big-O, ,  • HW2, P4, 1.8.8(a,c) • (a) f(x) = 2x2 + x3 log x • (c) f(x) = (x4 + x2 + 1) / (x4 + 1)

  17. 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.

  18. 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.

  19. 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.

  20. Other requested topics • Sets, inclusion-exclusion • |AB| = |A| + |B| - |AB| • *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

More Related