CMSC 203 / 0201 Fall 2002

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
Lets 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) ? ?(wz) ) (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) ? ?(wz)
  ) (C)
• Suppose (3) holds. Then ? w and z s.t.
• g(w) g(z) (1)
• w ltgt z (2)
• Since g(w) g(z), it must be the cas that
  f(g(w)) f(g(z)), therefore wz 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 lt gt 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
  wz, 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 lt gt 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 wz,
  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 ?k1n 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. 1236
  therefore, 6 is a perfect number.
• The divisors of 28 are 1, 2, 4, 7, and 14.
  124714 28 therefore, 28 is a perfect
  number.

18
Harder perfect numbers

• 2.3.16(b) Show that x2p-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 ?i0p-1 2i
  (2p-1) ?i0p-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