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Mtk3013 chapter 2-3

• 1. Homework Solution The statements  (P  Q) and (  P)  (  Q) are logically equivalent , so we write  (P  Q)  (  P)  (  Q). P Q  (P  Q) (  P)  (  Q)  (P  Q)  (  P)  (  Q) true true false false true true false false false true false true false false true false false true true true
• 2. SET THEORY MTK3013 DISCRETE MATHMATICS
• 3.
• 4.
• 5.
• 6.
• 7. Examples for Sets We are now able to define the set of rational numbers Q: Q = {a/b | a  Z  b  Z + } or Q = {a/b | a  Z  b  Z  b  0} And how about the set of real numbers R? R = {r | r is a real number} That is the best we can do.
• 8. Subsets A  B “A is a subset of B” A  B if and only if every element of A is also an element of B. We can completely formalize this: A  B   x (x  A  x  B) Examples: A = {3, 9}, B = {5, 9, 1, 3}, A  B ? true A = {3, 3, 3, 9}, B = {5, 9, 1, 3}, A  B ? false true A = {1, 2, 3}, B = {2, 3, 4}, A  B ?
• 9.
• 10.
• 11. Cardinality of Sets If a set S contains n distinct elements, n  N , we call S a finite set with cardinality n . Examples: A = {Mercedes, BMW, Porsche}, |A| = 3 B = {1, {2, 3}, {4, 5}, 6} |B| = 4 C =  |C| = 0 D = { x N | x  7000 } |D| = 7001 E = { x N | x  7000 } E is infinite!
• 12. The Power Set 2 A or P(A) “power set of A” 2 A = {B | B  A} (contains all subsets of A) Examples: A = {x, y, z} 2 A = {  , {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}} A =  2 A = {  } Note: |A| = 0, |2 A | = 1
• 13.
• 14. Cartesian Product The ordered n-tuple (a 1 , a 2 , a 3 , …, a n ) is an ordered collection of objects. Two ordered n-tuples (a 1 , a 2 , a 3 , …, a n ) and (b 1 , b 2 , b 3 , …, b n ) are equal if and only if they contain exactly the same elements in the same order , i.e. a i = b i for 1  i  n. The Cartesian product of two sets is defined as: A  B = {(a, b) | a  A  b  B} Example: A = {x, y}, B = {a, b, c} A  B = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
• 15.
• 16. Set Operations Union: A  B = {x | x  A  x  B} Example: A = {a, b}, B = {b, c, d} A  B = {a, b, c, d} Intersection: A  B = {x | x  A  x  B} Example: A = {a, b}, B = {b, c, d} A  B = {b}
• 17. Set Operations Two sets are called disjoint if their intersection is empty, that is, they share no elements: A  B =  The difference between two sets A and B contains exactly those elements of A that are not in B: A-B = {x | x  A  x  B} Example: A = {a, b}, B = {b, c, d}, A-B = {a}
• 18. Set Operations The complement of a set A contains exactly those elements under consideration that are not in A: -A = U-A Example: U = N , B = {250, 251, 252, …} -B = {0, 1, 2, …, 248, 249}
• 19.
• 20. Set Operations Method II: Membership table 1 means “x is an element of this set” 0 means “x is not an element of this set” A B C B  C A  (B  C) A  B A  C (A  B)  (A  C) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1
• 21. Set Operations Take-home message: Every logical expression can be transformed into an equivalent expression in set theory and vice versa.
• 22. Exercises Question 1: Given a set A = {x, y, z} and a set B = {1, 2, 3, 4}, what is the value of | 2 A  2 B | ? Question 2: Is it true for all sets A and B that (A  B)  (B  A) =  ? Or do A and B have to meet certain conditions? Question 3: For any two sets A and B, if A – B =  and B – A =  , can we conclude that A = B? Why or why not?
• 24. Functions A function f from a set A to a set B is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. If f is a function from A to B, we write f: A  B (note: Here, “  “ has nothing to do with if… then)
• 25. Functions If f:A  B, we say that A is the domain of f and B is the codomain of f. If f(a) = b, we say that b is the image of a and a is the pre-image of b. The range of f:A  B is the set of all images of elements of A. We say that f:A  B maps A to B.
• 26. Functions Let us take a look at the function f:P  C with P = {Linda, Max, Kathy, Peter} C = {Boston, New York, Hong Kong, Moscow} f(Linda) = Moscow f(Max) = Boston f(Kathy) = Hong Kong f(Peter) = New York Here, the range of f is C.
• 27. Functions Let us re-specify f as follows: f(Linda) = Moscow f(Max) = Boston f(Kathy) = Hong Kong f(Peter) = Boston Is f still a function? {Moscow, Boston, Hong Kong} What is its range?
• 28. Functions Other ways to represent f: x f(x) Linda Moscow Max Boston Kathy Hong Kong Peter Boston Linda Max Kathy Peter Boston New York Hong Kong Moscow
• 29. Functions If the domain of our function f is large, it is convenient to specify f with a formula , e.g.: f: R  R f(x) = 2x This leads to: f(1) = 2 f(3) = 6 f(-3) = -6 …
• 30. Functions Let f 1 and f 2 be functions from A to R . Then the sum and the product of f 1 and f 2 are also functions from A to R defined by: (f 1 + f 2 )(x) = f 1 (x) + f 2 (x) (f 1 f 2 )(x) = f 1 (x) f 2 (x) Example: f 1 (x) = 3x, f 2 (x) = x + 5 (f 1 + f 2 )(x) = f 1 (x) + f 2 (x) = 3x + x + 5 = 4x + 5 (f 1 f 2 )(x) = f 1 (x) f 2 (x) = 3x (x + 5) = 3x 2 + 15x
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