1. September1999 CMSC 203 / 0201 Fall 2002 Exam #1 Review – 27 September 2002 Prof. Marie desJardins

2. September1999 October 1999 Survey says…

3. 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.

4. September1999 October 1999 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: 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.”

5. September1999 October 1999 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. September1999 October 1999 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. 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´)

8. September1999 October 1999 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. 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 f  g is one-to-one, it must be the case that w=z, which contradicts our earlier supposition.

10. September1999 October 1999 Write a conclusion statement  Therefore, g must be one-to-one.

11. September1999 October 1999 Write “Q.E.D.”  Q.E.D.

12. September1999 October 1999 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. 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 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. September1999 October 1999 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. 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=1 n k 2.

16. September1999 October 1999 Big-O, ,   HW2, P4, 1.8.8(a,c)  (a) f(x) = 2x 2 + x 3 log x  (c) f(x) = (x 4 + x 2 + 1) / (x 4 + 1)

17. 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.

18. September1999 October 1999 Harder perfect numbers  2.3.16(b) Show that x=2 p-1 (2 p -1) is a perfect number when 2 p -1 is prime.  The divisors of x are 2 p-1, 2 p -1, their divisors, and the products of their divisors.  The divisors of 2 p-1 are 1, 2, 2 2, …, 2 p-1.  Since 2 p -1 is prime, its divisors are 1 and 2 p-1.

19. September1999 October 1999 Harder perfect numbers cont.  Therefore, the proper divisors of x are 1, 2, 2 2, …, 2 p-1, 2 p -1, 2(2 p -1), 2 2 (2 p -1 ), …, 2 p-2 (2 p -1).  The sum of these divisors is  i=0 p-1 2 i + (2 p -1)  i=0 p-2 2 i = 2 p -1 + (2 p -1) (2 p-1 – 1) = (2 p -1) (1 + 2 p-1 – 1) = 2 p-1 (2 p -1)  Therefore, 2 p-1 (2 p -1) is a proper number.  Q.E.D.

20. September1999 October 1999 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