Math 2534 Solution to Test 3A Spring 2010 Problem 1: (10pts) Prove that R is a transitive relation on Z when given that mrpiff m pmod d (ie. d ( m p) ) Solution: The relation R is transitive, if arb and brc, then arc where a,b,c are integers. Since we are given that arb and brc, we have that d a b and d b c. By definition of divisible we also have that for integers q and k, dq = a b and dk = b c. If we add dq and dk we have that dq + dk = a b + b c = a c. So we have dm = a c, where the integer m = q + k. By definition of divisible we have that d divides a c which gives us that arc and that R is transitive. Problem 2: (12pts) a) Prove the following using complete sentences: If the relations R and S are both transitive then R S is also transitive. Proof: We need to use the definition of transitive which states that IF (x,y) and (y,z), then (x,z). Suppose that (x,y) and (y,z) are elements of R S. Then by definition of intersection we have that (x,y) and (y,z) are elements of R and (x,y) and (y,z) are elements of S. Since R and S are given to be transitive we know that (x,z) is an element of R and (x, z) is an element of S. Therefore by definition of intersection we have that (x,z) must also be an element of R S. Therefore R S is transitive. b) If the relations R and S are both transitive then explain why R S is not guaranteed to also be transitive. ( you may want to look at a counter example) Example: Let S = { (1,3), (3,5), (1,5) and let R = { (5, 2), (2, 1), (5,1)} R S ={ (1,3), (3,5), (1,5), (5, 2), (2, 1), (5,1)} Notice this union is not transitive, for instance there is no (1,1), (3,1), etc. Problem 3: (5pts) 1 1 1 Given that f : A B, and g: B C, If ( g f) (6) = ( f f)(3), and g (6) = 5, find f(3). Explain your reasoning. 1 1 1 1 1 ( g f ) (6) = ( f ( g (6)) = f (5) = ( f f )(3) = Id(3) = 3 1 so f (5) = 3 and by definition of inverse we have f (3) = 5
Problem 4: (10pts) Prove that Cardinality between sets is a symmetric relation using the formal definition that two sets A and B have the same cardinality if and only if there is a bijection f so that f maps A to B. Proof: In order to verify that cardinality is symmetric relation, we need to use the definition of what it means for two sets to relate to each other. ( ie two sets A and B have the same cardinality if and only if there is a bijection f so that f maps A to B.) We will show that IF ARB then BRA. We will assume that ARB which means that there is a bijection F that maps A to B. Since F is a bijection we know that the inverse function F -1 that maps B to A does exist and is also a bijection. Therefore BRA and the cardinality relation is symmetric. x Problem 5: (8pts) Let f(x) = f( x) = defined on the R-{3/2} 2x 3 a) Is f(x) onto? (Determine if the definition can be satisfied?) b) If f -1 (x) exist then give the function and its domain. If f -1 (x) does not exist then do the necessary modifications in order for f -1 (x) to be defined. Proof: To determine if f(x) is onto we must satisfy the definition which states: For all elements y in the co-domain, there exist an element x in the domain so that f(x) = y. a Consider a to be in the domain and b to be in co-domain. We will try to confirm that b = 2a 3 3b By solving for a. We get that a = where a is not defined for b = ½ 2b 1 Therefore f(x) is not onto. However we may make modifications to create the following inverse function. Let f -1 3x = defined on R -{1/ 2} 2x 1 Problem 6: (10pts) A) Given the following relation R draw a Hasse Diagram that illustrates R. R = {(a,a),(e,e),(b,b),(c,c),(d,d),(a,e),(a,c),(a,d),(b,c),(b,d),(e,c)} See drawing from class. B) Given the set A = {1,2,3,4} that is partitioned by an equivalence relation R so that A={1} {2,3} {4}, find R in terms of ordered pairs. R = { (1,1), (2,2), (3,3), (2,3), (3,2), (4,4)
Problem 7: (9pts) Let R be a relation on a set A = {a, b, c, d} defined by R = {(a, b), (c, c), (c, d),(d, c)} a) Is R transitive? Justify your answer. No, R needs (d,d) to be transitive. b) Make the necessary minimal deletions or additions to make R anti symmetric? Delete (d,c) c) Make the necessary minimal deletions or additions to make R symmetric? Add (b,a) Problem 8: (12pts) Verify that R is a partial order on all sets if for sets A and B, that ARB if and only if A B. Proof: To show that R is a partial order we need to verify that R is reflexive, anti-symmetric, and transitive. To show that R is reflexive, we need to show that ARA, ie. We will verify that A Awhich is true since every set is a subset of itself. R has been verified as reflexive. To show that R is anti-symmetric we need to show that IF ARB and BRA, then A = B. We are assuming that A B and B A Therefore we know that A = B since this is the precise definition for equal sets and R is therefore anti-symmetric To show that R is transitive we need to show that IF ARB and BRC, then ARC. We are assuming that A B and B C, Therefore we know that x A x B since A is a subset of B and we also know that x A x B x C since B is a subset of C. Therefore by definition of subsets we know that A C. Therefore R is transitive. Since we have shown that R is reflexive, anti-symmetric and transitive, we know that R is a partial order by definition.
Problem 9: (5pts) An equivalence relation R is defined on the set of integers so that arb if and only if a bmod 3. Since R partitions the integers, verify that if x is in the equivalence class [y] and y is in the equivalence [z], then [x] = [z] Solution: Since x is in the equivalence class of y that means that x and y have the same remainder when divided by 3. But we also know that y is the same equivalence as z, which means they have the same remainder when divided by 3. We also know that Zmod 3 partitions the integers into 3 classes according to the remainder value. Therefore x, y and z must be in the same equivalence class and [x] = [y] = [z]. Problem 10: (8pts) The functions f( n) = 1+ 5n and gn ( ) = 4 n are defined on the natural numbers. Verify that f(x) is Big O g(x) by demonstrating that it fits the definition. You may assume the inequality is true. Solution: The definition of Big O is as follows: The function f(x) is Big) g(x) if they are both defined for the same domain and there exists positive constants C and K so that f( x) Cgx ( ) for all x > K. Using PMI we can show that for f( n) = 1+ 5n and gn ( ) = 4 n, 1+ 5n 4 n for n >1 Therefore choose C = 1 and choose K = 1. (See me at the final exam to change grade if k = 1 is marked wrong. Sorry!) Problem 11:(5pts) If A and B are finite sets and n(a) > n(b), by the Pigeon Hole Principle f : A B is never one to one. But f : A B will always be onto. Confirm or deny this statement and justify your conclusion. The Pigeon Hole Principle guarantees that f cannot be one to one. However there is no guarantee that f is onto. You can come up with a counter example. Let A = { a,b,c,d,e} and B = { 1, 3} Define a mapping f from A to B. Let f(x) = 3 for all x in A.
Problem 12; Let C and D be disjoint subsets of set A and f : C B and g: D B. Define a function h(x) as follows: f( x) if x C hx ( ) = gx ( ) if x D Determine if the following is true or false and justify you conclusions. a) If f and g are both one to one, then h is one to one. It possible that f and g are each one to one but that f and g have the exact same range. Therefore h(x) would have more than one element mapping to a single element in the range. False b) If h is onto, then f and g are also onto. False See example below where h(x) is an onto mapping on the entire real number line but f and g are not. Let h(x) = 1/ x x 1 2x+ 1 x> 1 Pledge: I have neither given nor received help on this test. Signature