RED. Fall 2016 Student Submitted Sample Questions

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1 RED Fall 2016 Student Submitted Sample Questions Name: Last Update: November 22, 2016 The questions are divided into three sections: True-false, Multiple Choice, and Written Answer. I will add questions to the end of each section, so the number of a question should never change you can refer to T/F question 5 no matter when you download the file (as long as it is after five questions have been put in). 1. True-False Questions 1. True or false: Let A = {1, 2, 3, 4}, let B = {6, 7} and let f : A B be given by f = {(1, 7), (2, 6), (3, 7), (4, 7)}. Then f is surjective. (Solution) True. The function is surjective because every element in B has at least one ordered pair. 2. Let A = {1, 2, 3, 4}, and B = {Russia, China, USA, Mexico} and let f : A B then A and B have the same cardinality. (Solution) True. Cardinality means that the two sets have equal number of elements, thus regardless of what the elements are they are equal in number in this example. 3. Let R be a relation on the set of people where arb implies that person a and person b share a grandparent. Then R is an equivalence relation. Proof. This statement is false. R is not transitive because I can share a grandparent with my cousin on their mom s side, and they can share a grandparent with a cousin on their dad s side, and we are not necessarily related. This statement is reflexive and symmetric, but not transitive, so not an equivalence relation. 4. There is a greatest cardinality. Proof. This statement is false. For any set, a larger cardinality can be found by taking the power set of the set. Therefore, no largest cardinality exists. 5. Let A and B be sets. If there is a surjective but not a bijective function f : A B, then A < B. Proof. False. Let the set A = {1, 2, 3} and the set B = {1, 2}. Define f : A B by f(1) = 1, f(2) = 1, and f(3) = 2. The function f is surjective, but is not bijective. However, A > B. 6. Define f : R R by f(x) = x 2 + 2x + 1. The function is injective. Proof. False. Consider f( 2) = f(1) = 3, so f is not injective. 7. Let A = {2, 3, 4, 5, 6} and R be the relation defined by (does not divide). True/False: Relation R on set A is an equivalence relationship. Proof. False. This relationship will never be reflexive regardless of a set provided, because a number will always divide itself. 8. The set Q is countably infinite. 1

2 2 Proof. True by Corollary This is derived by using theorems (The set Q + is countably infinite) and theorem (If S and T are countable sets, then S T is countable). Q = Q + Q {0} 9. Define a relation R on N as arb if and only if a 2 b 2 0. Then R is transitive, but not symmetric. Answer: True Proof. To disprove symmetry, let a = 2, b = 1 Then a 2 b 2 = 4 1 = 3 is greater than 0, so arb Now b 2 a 2 = 1 4 = 3 is less than 0, so b does not relate to a. Therefore, R is not symmetric. To show transitivity, assume arb and brc Then a 2 b 2 0 and b 2 c 2 0 Adding these inequalities yeilds a 2 c 2 0 Hence, arc and R is transitive. 10. Define a function f : N R such that f(n) = n. f is a bijection. Answer: False Proof. See that 2 R, but there is no n N such that n = 2 Thus, f is not surjective, and thus not bijective. 11. Let be the relation defined on the set S of lines in the plane by l 1 l 2 if l 1 and l 2 have slopes m 1 and m 2 with m 1 = 1/m 2. Then the relation is not symmetric, but it is reflexive and transitive. Proof. This is false. All lines have the same slope as themselves. So they aren t perpendicular to themselves. Hence the relation isn t reflexive. 12. True or false. Let A = {1, 2, 3, 4, 5} and R = {(1, 1), (3, 3), (4, 4), (5, 5)}. This relation is reflexive, symmetric, and transitive. (a) solution: False. Proof. The relation is not reflexive. A relation is reflexive if xrx for EVERY x A. 13. True or false. The function f : R R by f(n) = n is bijective. Proof. True. Assume f(n 1 ) = f(n 2 ). So (n 1 ) 3 +6 = (n 2 ) Then, (n 1 ) 3 = (n 2 ) 3. Hence, (n 1 ) = (n 2 ) so f is injective. Let b R. Then f( 3 b 6) = ( 3 b 6) 3 +6 = b. Hence, f is surjective. Since f is both injective and surjective, f is bijective. 14. Let f : A B be a function. Then f is surjective B = Im(f). Solution: Proof. Let f be a function A B. Let f : Z to Z be the function given by f(x) = 2x. So 1 Z but 1 / f(x), so f(x) isn t surjective. However, Z = 2x. Hence the statement is false. 15. For a function, f: A B, B is in all cases considered the range of f.

3 3 (a) T (b) F Proof. False, B is only the range of f when f is surjective. B is the codomain in all cases. 16. There are exactly six different functions f : {a, b, c} {0, 1}. Proof. False, there are eight, 17. Let A={1,2,3,4,5}, and B={1,2,3,4,5,6,7,8}. Define f : A B by f ={(1,6),(2,2),(3,6),(4,7),(5,1)}. Then f is injective. Proof. False; two elements from the domain map to the same element in the codomain. 18. N = Q = Z? Proof. True; all of these sets are countably infinite. 19. Let S = {1, 2, 3} and S X. Then X = 3 if and only if X S Proof. True. By the Schreder-Bernstein Theorem we know that if S X and X S then X = S 20. If f is surjective then g(f(x)) is surjective. Proof. False. Let A = {1, 2}, B = {3} and C = {4, 5} f(x) = {(1, 3), (2, 3)} g(x) = {(3, 4)} g(f(x)) = {(1, 4), (2, 4)} Since 5 is not a second component of g(f(x)) then it is not surjective. 21. Let f : A B be a function, where A = {15, 27, 38} and B = {5, 6, 7, 8}. Then f is surjective. Proof. False. Because A < B, f cannot be surjective. 22. Let A be an infinite set. If A B, then A < B. Proof. False. Let A = N and B = Z. Then A B, but A = B because there exists a bijection f : Z N given by { 2n if n 0 f(n) = 1 2n if n < [5, 5000] = R Proof. True. R = [0, 1]. There exists a bijection between [0, 1] and [5, 5000] given by f(x) = 4995x + 5. So [0, 1] = [5, 5000].

4 4 Therefore, [5, 5000] = R. 24. True/False: Let R be an equivalence relation on a set A = {1, 2, 3, 4, 5} given by R = {(1, 1), (1, 5), (2, 2), (2, 3), (3, 2), (3, 3), (4, 4), (5, 1), (5, 5)}. The set of equivalence classes of R is {{1,4,5},{2,3}}. Proof. False. The set of equivalence classes is {{1.5}, {2, 3}, {4}}.

5 5 2. Multiple Choice Questions 1. Which of the following is a bijection from: (0, 2) (c, d) a) f(x) = (d c)x + d b) f(x) = cx + 2 c) f(x) = (c d)x + c d) f(x) = ( d c 2 )x + c (Solution): D, if we take the slope of the coordinates we get d c 2 and then the x intercept is located at c when x = 0 which gives us the function of f(x) = ( d c 2 )x+c 2. Let A = {1, 2, 3, 4} and the function is defined as: f : A A f = {(1, 2), (2, 3), (3, 4), (4, 1)} Find f f a) f = {(1, 3), (2, 4), (3, 1), (4, 2)} b) f = {(1, 2), (2, 4), (3, 2), (2, 4)} c) f = {(2, 1), (4, 2), (1, 3), (4, 2)} d) f = {(1, 3), (2, 3), (3, 1), (4, 2)} (Solution): A, if we write out the set we can see f is given then follow that pattern going from one set to the next. For example with (1, 2) we get and (2, 3) we get and (3, 4) we get and (4, 1) we get and end with f = {(1, 3), (2, 4), (3, 1), (4, 2)} 3. Suppose f : R {0, 1}. What can we infer from the domain and codomain of f? (a) f is surjective. (b) f is not injective. (c) f is bijective. (d) both (a) and (b). (e) both (a) and (c). Proof. (b) is the correct answer. Since R > {0, 1}, f cannot be injective. Also, f could be a function such as f(x) = 0, so f is not necessarily surjective, ruling out the other options. 4. If a relation is a reflexive function, what must it also be? (a) bijective (b) an equivalence relation (c) transitive (d) symmetric (e) surjective (f) all of the above (g) three of the above (h) two of the above (i) none of the above Proof. The answer is (f). Let f : A A be our reflexive function. Then f contains all ordered pairs (a, a) for a A. Since is is a function, it contains only one ordered pair for each element of A, so this is all that it contains. Hence, the function is given by f(a) = a for all a A. This is bijective, hence, surjective. It is also clearly symmetric and transitive, hence an equivalence relation. 5. Which of the following relations is not a function from A = {a, b, c, d} to B = {1, 2, 3, 4, 5}?

6 6 (a) f = {(a, 1), (b, 2), (c, 3), (d, 4)} (b) f = {(a, 2), (c, 3), (c, 4), (d, 5)} (c) f = {(a, 3), (b, 4), (c, 5), (d, 1)} (d) f = {(a, 2), (b, 3), (c, 5), (d, 5)} Proof. The answer is (b) because c A is the first component in more than one element of f. Therefore, it is not a function. 6. Define f : Z Z by f(x) = x + 5. Which of the following best describes the function? (a) f if injective. (b) f is surjective. (c) f is bijective. (d) f is neither injective nor surjective. Proof. The answer is (c). Suppose f(x) = f(y). So x + 5 = y + 5. Then x = y. Therefore, f is injective. Suppose b Z Let x = b 5. Then f(b 5) = b 5+5 = b. Hence, f is surjective. Because f is injective and surjective, then f is bijective. 7. Let f be a function mapping R R show below. Choose the best description for this function. 30 f : {3x} (a) neither injective nor surjective (b) injective only (c) surjective only (d) bijective x Proof. (d) The function is bijective. 8. Let f be a function mapping R R show below. Choose the best description for this function.

7 7 10 f : { 1 2 x3 + 2x 2 } x 5 10 (a) neither injective nor surjective (b) injective only (c) surjective only (d) bijective Proof. (c) The function is surjective but not injective. 9. Let f be a function mapping R R show below. Choose the best description for this function. 10 f : {x 2 } x 5 10 (a) neither injective nor surjective (b) injective only (c) surjective only (d) bijective Proof. (a): The function is neither injective nor surjective. 10. Which of the following does NOT have the same cardinality as N? (a) {n N : 5 n} (b) Z (c) Q (d) {x R : x is a possible 5-digit zip code} (e) None of the above Answer: d - there are only 100,000 possible zip codes which is finite.

8 8 11. Define a relation R by arb if a b. Then R is... (a) Transitive Only (b) Reflexive Only (c) Symmetric Only (d) Transitive and Reflexive (e) Transitive and Symmetric (f) Symmetric and Reflexive (g) Transitive, Reflexive, and Symmetric (h) None of the above Answer: b Proof. See that a does not relate to a because a a is false. Hence, R is not reflexive. Also, let a = 0, b = 1, c = 0 Then arb and brc but a = c, so a does not relate to c Hence, R is not transitive. Now, assume arb, then a b So b a, and bra Hence, R is symmetric. 12. Let R be an equivalence relation on the set A = {1, 2, 3, 4, 5, 6}. Assume 2R4, 1R6 and 4R5. Let R be a subset of A A, what is the smallest possible set that R could be? (a) {(1, 1), (1, 6), (2, 2), (2, 4), (3, 3), (4, 4), (4, 5), (5, 5), (6, 6)} (b){(1, 1), (1, 6), (2, 2), (2, 4), (3, 3), (4, 2), (4, 4), (4, 5), (5, 4), (5, 5), (6, 1), (6, 6)} (c){(1, 1), (1, 6), (2, 2), (2, 4), (2, 5), (3, 3), (4, 2), (4, 4), (4, 5), (5, 2), (5, 4), (5, 5), (6, 1), (6, 6)} (d){(1, 1), (1, 6), (2, 2), (2, 4), (2, 5), (4, 2), (4, 4), (4, 5), (5, 2), (5, 4), (5, 5), (6, 1), (6, 6)} Proof. The correct answer is C. It provides the smallest possible equivalence relation on A. R is reflexive, symmetric, and transitive. 13. Which group of relations are all reflexive? (a),, < (b),, = (c), >, (d) =,, Proof. The correct answer is B. All of these are reflexive because x x, x x and x = x. 14. Let A = {x Z : x 3}, and let f : A Z be the function given by f(x) = 7x+2. What are the domain and range f(x)? (a) Domain {0, 1, 2, 3}; Range {2, 9, 16, 23}. (b) Domain { 3, 2, 1, 0, 1, 2, 3}; Range { 19, 12, 5, 2, 9, 16, 23}. (c) Domain { 19, 12, 5, 2, 9, 16, 23}; Range { 3, 2, 1, 0, 1, 2, 3}. (d) Domain { 3, 2, 1, 0, 1, 2, 3}; Range {Z}. Proof. The domain of f(x) is the set A = { 3, 2, 1, 0, 1, 2, 3}. When you plug in a A, you get the range of f(x). Therefore Im(f) = { 19, 12, 5, 2, 9, 16, 23}. Hence (b) is correct. 15. Which (if any) of the following functions are well defined? (a) f 1 : Let f 1 from Z 10 Z 5 be the function given by f(a) = f([a]), a Z 10, [a] Z 5.

9 9 (b) f 2 : Let f 2 from Z 4 Z 8 be the function given by f(a) = f([a]), a Z 4, [a] Z 8. (c) f 3 : Let f 3 from Z 5 Z 10 be the function given by f(a) = f([a]), a Z 5, [a] Z 10. (d) b and c (e) none of the above Proof. Let elt of Z 10 = a and elt of Z 5 = [a]. Assume a = b. Then arb, and a b (mod 1)0. So a b = 10k, k Z, and a b = 5(2)k, k Z. Hence a b (mod 5). Hence [a] = [b], and f 1 is well defined. Proof. Let elt of Z 4 = a and elt of Z 8 = [a]. Let a = 0 and b = 4. Then 0 = 0 and 4 = 0, but [0] [4]. Hence f 2 isn t well-defined. Proof. Let elements of Z 5 = a and elements of Z 10 = [a]. Let a = 0 and b = 5. Then 0 = 0 and 5 = 0, but [0] [5]. Hence f 3 isn t well-defined. Hence (a) is correct. 16. Consider the relation R = {(3, 3), (4, 4), (2, 3), (3, 2), (2, 4)} on set A = {2, 3, 4}. Is R reflexive? Symmetric? Transitive? (a) Reflexive (b) Symmetric (c) Transitive (d) both (a) and (c) (e) both (b) and (c) (f) None of the above Proof. d, none of the above. There is no (2, 2), so it can t be reflexive. There is no (4, 2), so it can t be reflexive since (2, 4) is a member of the set A. (2, 3) and (3, 2) are members of A, but there is no (2, 2) so R is not transitive either. 17. For two finite sets A = {1, 2, 3, 4, 5, 6} and B = {1, 2, 3}, how many functions f : A B can be made? (a) 58 (b) 241 (c) 729 (d) 1245 Proof. c, the number of functions that can be made will the cardinality of the codomain to the power of the cardinality of the domain: 3 6 = If there exists a bijection f : A B, then (a) A = B. (b) f is injective. (c) f is invertible. (d) All the the above.

10 10 (e) None of the above. Solution: (d). They have the same cardinality since there exists a bijection. Since there is a bijection, the function also has to also be injective. An bijective function is also invertible. 19. Let R be a relation on {1,2,3,4}. Select the most true statement. (a) R={(1,1), (2,2), (3,3)} is reflexive. (b) R={(1,1), (1,2), (2,1), (2,2), (3,3), (1,3), (3,1), (2,3), (3,2)} is symmetric. (c) R={(1,1), (1,2), (2,1), (2,2), (3,3), (1,3), (3,1), (2,3), (3,2)} is transitive. (d) R={(1,1), (1,2), (2,1), (2,2), (3,3), (1,3), (3,1), (2,3), (3,2)} is reflexive (e) Three of the above (f) Two of the above Solution: (f). In (b), whenever we have xry, it follows that yrx, so R is symmetric. In (c), (xry yrz) = xrz is true for all values in the set, so the relation is transitive. 20. Let R be an equivalence relation on the set {a, b, c} given by R = {(a, b), (b, a), (a, a), (b, b), (c, c)}. What are all the equivalence classes of R? (a) [a], [b] (b) There are no equivalence classes of R. (c) [c] (d) [a], [c] Proof. D Since [a] = [b], D contains all the equivalence classes of X. 21. Which of the following sets has the greatest cardinality? (a) N (b) Z (c) {0, 1} (d) P(( 3, 2)) Proof. D We proved that (a, b) = (0, 1) = R Therefore, using Theorem 31.5 ( 3, 2) < P(( 3, 2)). 22. Which of the functions below are injective? (a) f : R { 2} R {1}, f(x) = x 1 x+2 (b) g : R R, g(x) = x 2 (c) h : Z N, h(x) = x (d) two of the above (e) none of the above Proof. Correct Answer: (a) Let f(x) = f(y). Then x 1 x+2 = y 1 y+2, so (x 1)(y + 2) = (y 1)(x + 2), so xy y + 2x 2 = xy x + 2y 2, then 3x = 3y, so x = y, Therefore, f is injective. Proof. Let g(x) = g(y). Then let x = 2 and y = 2. Therefore g is not injective. Proof. Let h(x) = h(y). Then let x = 1 and y = 1. Therefore h is not injective. 23. Let R be a relation on A, where a, b, c A. Which of the below are necessary for an equivalence relation? (more than one bubble can be selected)

11 11 (a) arb if a = b (b) arb if a and b have the same parity. (c) arb if bra (d) arb and brc implies arc (e) none of the above Proof. a - definition of reflexive, c - definition of symmetric, d - definition of transitive. 24. Let f be a function f : N Z defined by f(x) = 3x + 2. Which of the following is true? (a) f is injective (b) f is surjective (c) f is bijective (d) All of the above are true (e) none of the above are true Proof. (a) is correct.f(x) = f(y), so 3x + 2 = 3y + 2, then 3x = 3y, so x = y. (b) is not true, since no x N can give any negative numbers in Z from the function. 25. Let n N and let R be a relation on Z given by arb if a b (mod n). Which of the following is true? (a) The relation is reflexive (b) The relation is symmetric (c) The relation is transitive (d) The relation is an equivalence relation (e) All of the Above (f) None of the Above Proof. (e) is correct. The relation a b (mod n) is an equivalence relation. Therefore it is reflexive, symmetric, and transitive.

12 12 3. Written Answer Questions 1. Prove or disprove that the following function is bijective: f(x) = 2x + 1 x f = R {0} R {0} (Solution): Show that it is injective and surjective. Proof of injective: 2x + 1 x = 2y + 1 y 2xy + y = 2xy + x y = x Thus is can be seen that because y = x it is injective. Proof of surjective: let b = 2x+1 x and solve for x, we will then get the following which can be proven to show surjectivity: f ( ) 1 = 2 1 b b 2 1 b (b 2) = 1 = b Thus for each b and x we can show that we will get a surjective function. So it can be seen that because the function is injective and surjective that it is bijective. 2. Prove or disprove: There are as many real numbers as complex numbers. Proof. This statement is true. Let C be the set of all complex numbers, and let f : C R = f(a) = c 1 r 1 c 2 r 2 c 3 r d 1 s 1 d 2 s 2 d 3 s 3..., where c n is equal to the n th digit of the whole number portion of the complex component of a C, d n is equal to the n th digit of the decimal portion of the complex component of a C, r n is equal to the n th digit of the whole number portion of the real component of a C, and s n is equal to the n th digit of the decimal portion of the real component of a C. We will show that C = R by showing that f is a bijection. First, assume f(a) = f(b). Then every digit of a and b are the same in both the complex and real parts, so a = b, and f is injective. Next, for every number in R, we can build it by using a number in C. So, f is surjective. Therefore, f is bijective, so C = R. 3. For a Z, denote the congruence class of a modulo 6 by ā, and the congruence class of a modulo 3 by [a]. Define f : Z 6 Z 3 by f(ā) = [a]. Prove that f is well defined. Proof. Assume ā = b. Then a b (mod 6). So 6 a b. Then a b = 6k = 3(2k) for some k Z. Therefore, a b (mod 3). So [a] = [b]. Hence, f(ā) = [a] is well defined. 4. Let A = {2, 3, 4, 5, 6}. Write out the relation R that expresses (does not divide) on the set A. Proof. R = {(2, 3), (2, 5), (3, 2), (3, 4), (3, 5), (4, 2), (4, 3), (4, 5), (4, 6), (5, 2), (5, 3), (5, 4), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5)}

13 13 5. Prove the cardinality of (, ) is the same as the cardinality of (0, ) Proof. First, note that (, ) = R We need to show that there exists a bijection from one set to the other. Define f : (, ) (0, ) as f(x) = e x where x (, ) Assume f(x 1 ) = f(x 2 ) Then e x1 = e x2. Taking the natural logarithm of both sides, we see that x 1 = x 2, Hence f is injective. Let b (0, ), there exists an x (, ) such that f(x) = b. Namely x = log b (, ) We see f(log(b) = e log b = b, so f is surjective. Therefore, f is a bijection, proving that the cardinality of (, ) is the same as the cardinality of (0, ). 6. Prove or disprove. Let f : N Z be given f(n) = 2 n 2. Then f is a bijection. Disproof. Since x 2 0 for all x R, we see that f(x) = 2 x 2 2. Hence, there is not value of x for which f(x) = 3, so f is not surjective. 7. Prove for Z 7, when A = 3 and B = 2, x A, y B, x + y = 5. Proof. Let a A, b B. Then a 3 (mod 7), so a 3 = 7k. Also b 2 (mod 7), so b 2 = 7l. Note k, l Z.So (a + b) 5 = 7(k + l), and a + b 5 (mod 7). Moreover a + b = 5. So x + y = 5. Hence the statement is true. 8. Prove or disprove that the set of complex numbers, C, is not countable. Proof. We know that R C. This means, by definition, that if x R then x C. Therefore, there exists an injection f : R C by the rule f(x) = x. Because this injection exists we can say that R C. Therefore, the set of complex numbers is not countable. 9. Prove or disprove: R and (0, ) have the same cardinality. If true, give a bijection as a formula. Proof. This is a true statement. We need a function f such that f : R (0, ). Consider the function f(x) = e x. The domain of f(x) is R, and the codomain (range) is (0, ), so this is indeed a function such that f : R (0, ). We assume the function is injective and let e a = e b. Then ln(e a ) = ln(e b ). And a ln(e) = b ln(e), so a = b. Therefore, our function is injective. Our function is also surjective since e ln(c) = c for all c > 0. Therefore, f is bijective, and it follows that R and (0, ) have the same cardinality. Additionally, since (0, ) R and R is infinitely uncountable, (0, ) is also infinitely uncountable. Therefore they have the same cardinality. 10. Prove the following is a bijective function.

14 14 f : R R Where f(x) = 3x + 12 Proof. Assume f(x) = f(y) Then 3x + 12 = 3y x = 3y x = y Therefore f is injective. Now, Let f( b 12 3 ) = 3( b 12 3 ) + 12 So, b = b Then f is surjective. Since f is surjective and injective, it is bijective. 11. Prove or disprove: the union of two countably infinite sets is countably infinite. Proof. Let S, T be countably infinite sets, where s S and t T Then let S T be written as follows: {s 1, t 1, s 2, t 2,...}. Starting from the left hand side, assign s 1 = 1, t 1 = 2, s 2 = 3... then each element of S T maps to exactly one element of N. Therefore the set S T is countably infinite. 12. Prove that the set Z is countably infinite. Proof. Let f be a function on Z N be defined by { 2n n > 0 f(n) = (2n 1) x 0 This function is a bijection on f : Z N. Therefore, Z = N, so Z is countably infinite.