CS 173: Discrete Mathematical Structures

CS 173:Discrete Mathematical Structures Cinda Heeren heeren@cs.uiuc.edu Rm 2213 Siebel Center Office Hours: M 11a-12p

CS 173 Announcements • Homework 2 returned in section this week. • Homework 3 available. Due 09/17, 8a. Cs173 - Spring 2004

New & important Like truth tables Like  Not hard, a little tedious CS 173 Set Theory - 4 Ways to prove identities • Show that A  B and that A  B. • Use a membership table. • Use previously proven identities. • Use logical equivalences to prove equivalent set definitions. Cs173 - Spring 2004

(A UB)= A  B CS 173 Set Theory - 4 Ways to prove identities Prove that • () (we did this last time) 2. () Cs173 - Spring 2004

(A UB)= A  B Haven’t we seen this before? CS 173 Set Theory - 4 Ways to prove identities Prove that using a membership table. 0 : x is not in the specified set 1 : otherwise Cs173 - Spring 2004

(A UB)= A  B (A UB)= A U B = A  B = A  B Feels fishy. CS 173 Set Theory - 4 Ways to prove identities Prove that using identities. Cs173 - Spring 2004

(A UB)= A  B (A UB)= {x : (x  A v x  B)} = A  B = {x : (x  A)  (x  B)} CS 173 Set Theory - 4 Ways to prove identities Prove that using logically equivalent set definitions. = {x : (x  A)  (x  B)} Cs173 - Spring 2004

CS 173 Set Theory - A proof for us to do together. X  (Y - Z) = (X  Y) - (X  Z). True or False? Prove your response. (X  Y) - (X  Z) = (X  Y)  (X  Z)’ = (X  Y)  (X’ U Z’) = (X  Y  X’) U (X  Y  Z’) =  U (X  Y  Z’) = (X  Y  Z’) Cs173 - Spring 2004

A U B =  A = B A  B =  A-B = B-A =  Trying to pv p --> q Assume p and not q, and find a contradiction. Our contradiction was that sets weren’t equal. CS 173 Set Theory - A proof for us to do together. Pv that if (A - B) U (B - A) = (A U B) then ______ A  B =  Suppose to the contrary, that A  B  , and that x  A  B. Then x cannot be in A-B and x cannot be in B-A. DeMorgan’s!! Then x is not in (A - B) U (B - A). Do you see the contradiction yet? But x is in A U B since (A  B)  (A U B). Thus, A  B = . Cs173 - Spring 2004

CS 173 Set Theory - Generalized Union Ex. Let U = N, and define: A1 = {2,3,4,…} A2 = {4,6,8,…} A3 = {6,9,12,…} Cs173 - Spring 2004

Primes Composites  N I have no clue. primes CS 173 Set Theory - Generalized Union Ex. Let U = N, and define: Then Cs173 - Spring 2004

CS 173 Set Theory - Generalized Intersection Ex. Let U = N, and define: A1 = {1,2,3,4,…} A2 = {2,4,6,…} A3 = {3,6,9,…} Cs173 - Spring 2004

Multiples of LCM(1,…,n) CS 173 Set Theory - Generalized Intersection Ex. Let U = N, and define: Then Cs173 - Spring 2004

B A Wrong. CS 173 Set Theory - Inclusion/Exclusion Example: How many people are wearing a watch? How many people are wearing sneakers? How many people are wearing a watch OR sneakers? What’s wrong? |A  B| = |A| + |B| - |A  B| Cs173 - Spring 2004

125 173 217 - (157 + 145 - 98) = 13 CS 173 Set Theory - Inclusion/Exclusion Example: There are 217 cs majors. 157 are taking cs125. 145 are taking cs173. 98 are taking both. How many are taking neither? Cs173 - Spring 2004

Now let’s do it for 4 sets! kidding. CS 173 Set Theory - Generalized Inclusion/Exclusion Suppose we have: B A C And I want to know |A U B U C| |A U B U C| = |A| + |B| + |C| - |A  B| - |A  C| - |B  C| + |A  B  C| Cs173 - Spring 2004

CS 173 Set Theory - Inclusion/Exclusion Example: How many people are wearing a watch? How many people are wearing sneakers? How many people are wearing a watch AND sneakers? How many people are wearing a watch OR sneakers? Cs173 - Spring 2004

f(x) = -(1/2)x - 25 domain co-domain CS 173 Functions Suppose we have: And I ask you to describe the yellow function. What’s a function? Notation: f: RR, f(x) = -(1/2)x - 25 Cs173 - Spring 2004

CS 173 Functions Definition: a function f : A  B is a subset of AxB where  a  A, ! b  B and <a,b>  f. Cs173 - Spring 2004

B A A point! A collection of points! CS 173 Functions Definition: a function f : A  B is a subset of AxB where  a  A, ! b  B and <a,b>  f. B A Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa CS 173 Functions A = {Michael, Tito, Janet, Cindy, Bobby} B = {Katherine Scruse, Carol Brady, Mother Teresa} Let f: A  B be defined as f(a) = mother(a). Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa What about the range? Some say it means codomain, others say, image. Since it’s ambiguous, we don’t use it at all. image(S) = f(S) CS 173 Functions - image & preimage For any set S  A, image(S) = {b : a  S, f(a) = b} So, image({Michael, Tito}) = {Katherine Scruse} image(A) = B - {Mother Teresa} Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa preimage(S) = f-1(S) CS 173 Functions - image & preimage For any S  B, preimage(S) = {a: b  S, f(a) = b} So, preimage({Carol Brady}) = {Cindy, Bobby} preimage(B) = A Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa  S CS 173 Functions - image & preimage What is image(preimage(S))? • S • { } • subset of S • superset of S • who knows? Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa Suppose S is {Janet, Cindy} preimage(image(S)) = A CS 173 Functions - image & preimage What is preimage(image(S))? Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa CS 173 Functions - misc. properties • f() =  • f({a}) = {f(a)} (this is a definition, actually) • f(A U B) = f(A) U f(B) • f(A  B)  f(A)  f(B) Cs173 - Spring 2004

CS 173 Functions - misc. properties f(A  B)  f(A)  f(B)? Choose an arbitrary c  f(A  B), and show that it must also be an element of f(A)  f(B). f(A  B) = {x : a  (A  B), f(a) = x} So, a (A  B) such that f(a) = c. If a  A (it is), then f(a) = c  f(A). If a  B (it is), then f(a) = c  f(B). c  f(A), and c  f(B), so c  f(A)  f(B). Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa CS 173 Functions - misc. properties • f-1() =  • f-1(A U B) = f-1(A) U f-1(B) • f-1(A  B) = f-1(A)  f-1(B) Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa Not one-to-one Every b  B has at most 1 preimage. CS 173 Functions - injection A function f: A  B is one-to-one (injective, an injection) if a,b,c, (f(a) = b  f(c) = b)  a = c Cs173 - Spring 2004

Michael Tito Janet Cindy Bobby Katherine Scruse Carol Brady Mother Teresa Not onto Every b  B has at least 1 preimage. CS 173 Functions - surjection A function f: A  B is onto (surjective, a surjection) if b  B, a  A f(a) = b Cs173 - Spring 2004

Isaak Bri Lynette Aidan Evan Isaak Bri Lynette Aidan Evan Cinda Dee Deb Katrina Dawn Cinda Dee Deb Katrina Dawn Every b  B has exactly 1 preimage. An important implication of this characteristic: The preimage (f-1) is a function! CS 173 Functions - bijection A function f: A  B is bijective if it is one-to-one and onto. Cs173 - Spring 2004

yes yes yes CS 173 Functions - examples Suppose f: R+  R+, f(x) = x2. Is f one-to-one? Is f onto? Is f bijective? Cs173 - Spring 2004

no yes no CS 173 Functions - examples Suppose f: R  R+, f(x) = x2. Is f one-to-one? Is f onto? Is f bijective? Cs173 - Spring 2004

no no no CS 173 Functions - examples Suppose f: R  R, f(x) = x2. Is f one-to-one? Is f onto? Is f bijective? Cs173 - Spring 2004

CS 173 Functions - composition Let f:AB, and g:BC be functions. Then the composition of f and g is: (g o f)(x) = g(f(x)) Cs173 - Spring 2004

CS 173 Functions - a little problem Let f:AB, and g:BC be functions. Prove that if f and g are one to one, then g o f :AC is one to one. Recall defn of one to one: f:A->B is 1to1 if f(a)=b and f(c)=b --> a=c. Suppose g(f(x)) = y and g(f(w)) = y. Show that x=w. f(x) = f(w) since g is 1 to 1. Then x = w since f is 1 to 1. Cs173 - Spring 2004

CS 173 Functions - another Let f:AB, and g:BC be functions. Prove that if f and g are onto, then g o f :AC is onto. Cs173 - Spring 2004

CS 173: Discrete Mathematical Structures