Final Review

1. Final Review

2. List of Logical Equivalences pT  p; pF  p Identity Laws pT  T; pF  F Domination Laws pp  p; pp  p Idempotent Laws (p)  p Double Negation Law pq  qp; pq  qp Commutative Laws (pq) r  p (qr); (pq)  r  p  (qr) Associative Laws

3. List of Equivalences p(qr)  (pq)(pr) Distribution Laws p(qr)  (pq)(pr) (pq)(p  q) De Morgan’s Laws (pq)(p  q) Miscellaneous p  p  T Or Tautology p p  F And Contradiction (pq)  (p  q) Implication Equivalence pq(pq)  (qp) Biconditional Equivalence

4. The Proof Process Assumptions -Definitions -Already-proved equivalences -Statements (e.g., arithmetic or algebraic) Logical Steps Conclusion (That which was to be proved)

5. Prove: (pq)  q  pq (pq)  q Left-Hand Statement  q  (pq) Commutative  (qp)  (q q) Distributive  (qp)  T Or Tautology  qp Identity  pq Commutative Begin with exactly the left-hand side statement End with exactly what is on the right Justify EVERY step with a logical equivalence

6. Predicate Calculus: Quantifiers Universe of Discourse, U: The domain of a variable in a propositional function. Universal Quantification of P(x) is the proposition:“P(x) is true for all values of x in U.” Existential Quantification of P(x) is the proposition: “There exists an element, x, in U such that P(x) is true.”

7. Universal Quantification of P(x) xP(x) “for all x P(x)” “for every x P(x)” Defined as: P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U Example: Let P(x) denote x2  x If U is x such that 0 < x < 1 then xP(x) is false. If U is x such that 1 < x then xP(x) is true.

8. Existential Quantification of P(x) xP(x) “there is an x such that P(x)” “there is at least one x such that P(x)” “there exists at least one x such that P(x)” Defined as: P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U Example: Let P(x) denote x2  x If U is x such that 0 < x  1 then xP(x) is true. If U is x such that x < 1 then xP(x) is true.

9. Quantifiers • xP(x) • True when P(x) is true for every x. • False if there is an x for which P(x) is false. • xP(x) • True if there exists an x for which P(x) is true. • False if P(x) is false for every x.

10. Negation (it is not the case) • xP(x) equivalent to xP(x) • True when P(x) is false for every x • False if there is an x for which P(x) is true. •  xP(x) is equivalent to xP(x) • True if there exists an x for which P(x) is false. • False if P(x) is true for every x.

11. Examples 2a Let T(a,b) denote the propositional function “a trusts b.” Let U be the set of all people in the world. Everybody trusts Bob. xT(x,Bob) Could also say: xU T(x,Bob)  denotes membership Bob trusts somebody. xT(Bob,x)

12. Examples 2b Alice trusts herself. T(Alice, Alice) Alice trusts nobody. x T(Alice,x) Carol trusts everyone trusted by David. x(T(David,x)  T(Carol,x)) Everyone trusts somebody. x y T(x,y)

13. Quantification of Two Variables(read left to right) • xyP(x,y) or yxP(x,y) • True when P(x,y) is true for every pair x,y. • False if there is a pair x,y for which P(x,y) is false. • xyP(x,y) or yxP(x,y) • True if there is a pair x,y for which P(x,y) is true. • False if P(x,y) is false for every pair x,y.

14. Quantification of Two Variables • xyP(x,y) • True when for every x there is a y for which P(x,y) is true. • (in this case y can depend on x) • False if there is an x such that P(x,y) is false for every y. • yxP(x,y) • True if there is a y for which P(x,y) is true for every x. • (i.e., true for a particular y regardless (or independent) of x) • False if for every y there is an x for which P(x,y) is false. • Note that order matters here • In particular, if yxP(x,y) is true, then xyP(x,y) is true. • However, if xyP(x,y) is true, it is not necessary that yxP(x,y) is true.

15. Examples 3a Let L(x,y) be the statement “x loves y” where U for both x and y is the set of all people in the world. Everybody loves Jerry. xL(x,Jerry) Everybody loves somebody. x yL(x,y) There is somebody whom everybody loves. yxL(x,y)

16. Examples 3b1 There is somebody whom Lydia does not love. xL(Lydia,x) Nobody loves everybody. (For each person there is at least one person they do not love.) xyL(x,y) There is somebody (one or more) whom nobody loves y x L(x,y)

17. Basic Number Theory Definitionsfrom Chapter 2 • Z = Set of all Integers • Z+ = Set of all Positive Integers • N = Set of Natural Numbers (Z+ and Zero) • R = Set of Real Numbers • Addition and multiplication on integers produce integers. (a,b  Z)  [(a+b)  Z]  [(ab)  Z]

18. Number Theory Defs (cont.)  = “such that” • n is even is defined as k  Z  n = 2k • n is odd is defined as k  Z  n = 2k+1 • x is rational is defined as a,b  Z  x = a/b, b0 • x is irrational is defined as a,b  Z  x = a/b, b0 or a,b  Z, x  a/b, b0 • p  Z+ is prime means that the only positive factors of p are p and 1. If p is not prime we say it is composite.

19. Methods of Proof p q (Example: if n is even, then n2 is even) • Direct proof: Assume p is true and use a series of previously proven statements to show that q is true. • Indirect proof: Show q p is true (contrapositive), using any proof technique (usually direct proof). • Proof by contradiction: Assume negation of what you are trying to prove (pq). Show that this leads to a contradiction.

20. Example of an Indirect Proof Prove: If n3 is even, then n is even. Proof: The contrapositive of “If n3 is even, then n is even” is “If n is odd, then n3 is odd.” If the contrapositive is true then the original statement must be true. Assume n is odd.Then kZ  n = 2k+1. It follows that n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1. (4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an even integer. Therefore n3 is odd. Assumption,Definition, Arithmetic, Conclusion

21. Example: Proof by Contradiction Prove: The sum of an irrational number and a rational number is irrational. Proof:Let q be an irrational number and r be a rational number.Assume that their sum is rational,i.e., q+r=s where s is a rational number. Then q = s-r. But by our previous proof the sum of two rational numbers must be rational, so we have an irrational number on the left equal to a rational number on the right. This is a contradiction. Therefore q+r can’t be rational and must be irrational.

22. Structure of Proof by Contradiction • Basic idea is to assume that the opposite of what you are trying to prove is true and show that it results in a violation of one of your initial assumptions. • In the previous proof we showed that assuming that the sum of a rational number and an irrational number is rational and showed that it resulted in the impossible conclusion that a number could be rational and irrational at the same time. (It can be put in a form that implies n  n is true, which is a contradiction.)

23. Approaches to Set Proofs • Membership tables (similar to truth tables) • Convert to a problem in propositional logic, prove, then convert back • Use set identities for a tabular proof (similar to what we did for the propositional logic examples but using set identities) • Do a logical argument (similar to what we did for the number theory examples)

24. Prove (AB)  (AB) = B (AB)  (AB) = {x | x(AB)(AB)} Set builder notation = {x | x(AB)  x(AB)} Def of  = {x | (xA  xB)  (xA  xB)} Def of  x2 and Def of complement = {x | (xB  xA )  (xB  xA )} Commutative x2 = {x | (xB  (xA  xA )} Distributive = {x | (xB  T } Or tautology = {x | (xB } Identity = B Set Builder notation

25. Prove (AB)  (AB) = B (Using Set Identities) (AB)  (AB) = (BA)  (BA) Commutative Law x2 =B  (A  A) Distributive Law =B  U Definition of U =B Identity Law

26. Prove (AB)  (AB) = B Proof: We must show that (AB)  (AB)  B and that B  (AB)  (AB) . First we will show that (AB)  (AB)  B. Let e be an arbitrary element of (AB)  (AB). Then either e (AB) or e (AB). If e (AB), then eB and eA. If e (AB), then eB and eA. In either case e B.

27. Prove (AB)  (AB) = B Now we will show that B  (AB)  (AB). Let e be an arbitrary element of B. Then either e AB or e AB. Since e is in one or the other, then e  (AB)  (AB).

28. f a1 f a2 a3 A f f a1 b1 f a2 a3 One-to-one? b2 b3 A B f b4 b1 b2 One-to-one? b3 B Functions: One-to-one function A function f is said to be one-to-one, or injective, if and only if f(x) = f(y) implies that x=y for all x and y in the domain of f. a0,a1  A f(a0) = f(a1)  a0 = a1 ORa0  a1  f(a0)  f(a1)

29. f a1 f a2 a3 A f f a1 b1 f b2 a2 a3 A B f b1 b2 b3 B Onto Function A function f from A to B is called onto, or surjective, if and only if for every element bB there is an element aA with f(a) = b. bB  aA such that f(a) = b

30. f a1 f a2 a3 A f f a1 b1 f b2 a2 a3 Bijection? A B f Bijection? b1 b2 b3 B One-to-one Correspondence The function f is a one-to-one correspondence or a bijection, if it is both one-to-one and onto.

31. Correspondence Diagrams: One-to-One or Onto? 1 2 3 4 a b c d a b c d 1 2 3 4 a b c 1 2 3 One-to-one, not onto One-to-one, and onto Onto, not one-to-one 1 2 3 4 a b c d 1 2 3 4 a b c Neither one-to-one nor onto Not a function!

32. f b a f-1 Inverse Function, f-1 Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that if f(a) = b, then f-1(b) = a. Example: f(a) = 3(a-1) f-1(b) = (b/3)+1

33. Examples Is each of the following (on the real numbers): a function? one-to-one? Onto? Invertible? f(x) = 1/x not a function f(0) undefined f(x) = x not a function since not defined for x<0 f(x) = x2 is a function, not 1-to-1 (-2,2 both go to 4), not onto since no way to get to the negative numbers, not invertible

34. Sequence • A sequence is a discrete structure used to represent an ordered list. • A sequence is a function from a subset of the set of integers (usually either the set {0,1,2,. . .} or {1,2, 3,. . .}to a set S. • We use the notation an to denote the image of the integer n. We call an a term of the sequence. • Notation to represent sequence is {an}

35. Examples • {1, 1/2, 1/3, 1/4, . . .} or the sequence {an} where an = 1/n, nZ+ . • {1,2,4,8,16, . . .} = {an} where an = 2n, nN. • {12,22,32,42,. . .} = {an} where an = n2, nZ+

36. Summations • Notation for describing the sum of the terms am, am+1, . . ., an from the sequence, {an} n am+am+1+ . . . + an =  aj j=m • j is the index of summation (dummy variable) • The index of summation runs through all integers from its lower limit, m, to its upper limit, n.

37. Summations follow all the rules of multiplication and addition! c(1+2+…+n) = c + 2c +…+ nc

38. Telescoping Sums Example

39. Closed Form Solutions A simple formula that can be used to calculate a sum without doing all the additions. Example: Proof: First we note that k2 - (k-1)2 = k2 - (k2-2k+1) = 2k-1. Since k2-(k-1)2 = 2k-1, then we can sum each side from k=1 to k=n

40. Proof (cont.)

41. Big-O Notation • Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers. We say that f(x) is O(g(x)) if there are constants CN and kR such that |f(x)|  C|g(x)| whenever x > k. • We say “f(x) is big-oh of g(x)”. • The intuitive meaning is that as x gets large, the values of f(x) are no larger than a constant time the values of g(x), or f(x) is growing no faster than g(x). • The supposition is that x gets large, it will approach a simplified limit.

42. Show that 3x3+2x2+7x+9 is O(x3) Proof: We must show that  constants CN and kR such that |3x3+2x2+7x+9|  C|x3| whenever x > k. Choose k = 1 then 3x3+2x2+7x+9  3x3+2x3+7x3+9x3 = 21x3 So let C = 21. Then 3x3+2x2+7x+9  21 x3when x  1.

43. Show that n! is O(nn) Proof: We must show that  constants CN and kR such that |n!|  C|nn| whenever n > k. n! = n(n-1)(n-2)(n-3)…(3)(2)(1)  n(n)(n)(n)…(n)(n)(n) n times =nn So choose k = 0 and C = 1

44. General Rules • Multiplication by a constant does not change the rate of growth. If f(n) = kg(n) where k is a constant, then f is O(g) and g is O(f). • The above means that there are an infinite number of pairs C,k that satisfy the Big-O definition. • Addition of smaller terms does not change the rate of growth. If f(n) = g(n) + smaller order terms, then f is O(g) and g is O(f). Ex.: f(n) = 4n6 + 3n5 + 100n2 + 2 is O(n6).

45. General Rules (cont.) • If f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then f1(x)f2(x) is O(g1(x)g2(x)). • Examples: 10xlog2x is O(xlog2x) n!6n3 is O(n!n3) =O(nn+3)

46. Example: Big-Oh Not Symmetric • Order matters in big-oh. Sometimes f is O(g) and g is O(f), but in general big-oh is not symmetric. Consider f(n) = 4n and g(n) = n2. f is O(g). • Can we prove that g is O(f)? Formally,  constants CN and kR such that |n2|  C|4n| whenever n > k? • No. To show this, we must prove that negation is true for all C and k. CN, kR, n>k such that n2 > C|4n|.

47. CN, kR, n>k such that n2 > 4nC. • To prove that negation is true, start with arbitrary C and k. Must show/construct an n>k such that n2 > 4nC • Easy to satisfy n > k, then • To satisfy n2>4nC, divide both sides by n to get n>4C. Pick n = max(4C+1,k+1), which proves the negation.