Exam (elaborations) TEST BANK FOR A First Course in Abstract Algebra 7th Edition By John B. Fraleigh (Solution Manual)

Exam (elaborations) TEST BANK FOR A First Course in Abstract Algebra 7th Edition By John B. Fraleigh (Solution Manual) Instructor’s Solutions Manual to accompany A First Course in Abstract Algebra Seventh Edition John B. Fraleigh University of Rhode Island CONTENTS 0. Sets and Relations 1 I. Groups and Subgroups 1. Introduction and Examples 4 2. Binary Operations 7 3. Isomorphic Binary Structures 9 4. Groups 13 5. Subgroups 17 6. Cyclic Groups 21 7. Generators and Cayley Digraphs 24 II. Permutations, Cosets, and Direct Products 8. Groups of Permutations 26 9. Orbits, Cycles, and the Alternating Groups 30 10. Cosets and the Theorem of Lagrange 34 11. Direct Products and Finitely Generated Abelian Groups 37 12. Plane Isometries 42 III. Homomorphisms and Factor Groups 13. Homomorphisms 44 14. Factor Groups 49 15. Factor-Group Computations and Simple Groups 53 16. Group Action on a Set 58 17. Applications of G-Sets to Counting 61 IV. Rings and Fields 18. Rings and Fields 63 19. Integral Domains 68 20. Fermat’s and Euler’s Theorems 72 21. The Field of Quotients of an Integral Domain 74 22. Rings of Polynomials 76 23. Factorization of Polynomials over a Field 79 24. Noncommutative Examples 85 25. Ordered Rings and Fields 87 V. Ideals and Factor Rings 26. Homomorphisms and Factor Rings 89 27. Prime and Maximal Ideals 94 28. Gr¨obner Bases for Ideals 99 iii VI. Extension Fields 29. Introduction to Extension Fields 103 30. Vector Spaces 107 31. Algebraic Extensions 111 32. Geometric Constructions 115 33. Finite Fields 116 VII. Advanced Group Theory 34. Isomorphism Theorems 117 35. Series of Groups 119 36. Sylow Theorems 122 37. Applications of the Sylow Theory 124 38. Free Abelian Groups 128 39. Free Groups 130 40. Group Presentations 133 VIII. Groups in Topology 41. Simplicial Complexes and Homology Groups 136 42. Computations of Homology Groups 138 43. More Homology Computations and Applications 140 44. Homological Algebra 144 IX. Factorization 45. Unique Factorization Domains 148 46. Euclidean Domains 151 47. Gaussian Integers and Multiplicative Norms 154 X. Automorphisms and Galois Theory 48. Automorphisms of Fields 159 49. The Isomorphism Extension Theorem 164 50. Splitting Fields 165 51. Separable Extensions 167 52. Totally Inseparable Extensions 171 53. Galois Theory 173 54. Illustrations of Galois Theory 176 55. Cyclotomic Extensions 183 56. Insolvability of the Quintic 185 APPENDIX Matrix Algebra 187 iv 0. Sets and Relations 1 0. Sets and Relations 1. { p 3,− p 3} 2. The set is empty. 3. {1,−1, 2,−2, 3,−3, 4,−4, 5,−5, 6,−6, 10,−10, 12,−12, 15,−15, 20,−20, 30,−30, 60,−60} 4. {−10,−9,−8,−7,−6,−5,−4,−3,−2,−1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} 5. It is not a well-defined set. (Some may argue that no element of Z+ is large, because every element exceeds only a finite number of other elements but is exceeded by an infinite number of other elements. Such people might claim the answer should be ?.) 6. ? 7. The set is ? because 33 = 27 and 43 = 64. 8. It is not a well-defined set. 9. Q 10. The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or 1/3. 11. {(a, 1), (a, 2), (a, c), (b, 1), (b, 2), (b, c), (c, 1), (c, 2), (c, c)} 12. a. It is a function. It is not one-to-one since there are two pairs with second member 4. It is not onto B because there is no pair with second member 2. b. (Same answer as Part(a).) c. It is not a function because there are two pairs with first member 1. d. It is a function. It is one-to-one. It is onto B because every element of B appears as second member of some pair. e. It is a function. It is not one-to-one because there are two pairs with second member 6. It is not onto B because there is no pair with second member 2. f. It is not a function because there are two pairs with first member 2. 13. Draw the line through P and x, and let y be its point of intersection with the line segment CD. 14. a.  : [0, 1] ! [0, 2] where (x) = 2x b.  : [1, 3] ! [5, 25] where (x) = 5 + 10(x − 1) c.  : [a, b] ! [c, d] where (x) = c + d−c b−a (x − a) 15. Let  : S ! R be defined by (x) = tan((x − 1 2 )). 16. a. ?; cardinality 1 b. ?, {a}; cardinality 2 c. ?, {a}, {b}, {a, b}; cardinality 4 d. ?, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}; cardinality 8 17. Conjecture: |P(A)| = 2s = 2|A|. Proof The number of subsets of a set A depends only on the cardinality of A, not on what the elements of A actually are. Suppose B = {1, 2, 3, · · · , s − 1} and A = {1, 2, 3, · · · , s}. Then A has all the elements of B plus the one additional element s. All subsets of B are also subsets of A; these are precisely the subsets of A that do not contain s, so the number of subsets of A not containing s is |P(B)|. Any other subset of A must contain s, and removal of the s would produce a subset of B. Thus the number of subsets of A containing s is also |P(B)|. Because every subset of A either contains s or does not contain s (but not both), we see that the number of subsets of A is 2|P(B)|. 2 0. Sets and Relations We have shown that if A has one more element that B, then |P(A)| = 2|P(B)|. Now |P(?)| = 1, so if |A| = s, then |P(A)| = 2s. 18. We define a one-to-one map  of BA onto P(A). Let f 2 BA, and let (f) = {x 2 A | f(x) = 1}. Suppose (f) = (g). Then f(x) = 1 if and only if g(x) = 1. Because the only possible values for f(x) and g(x) are 0 and 1, we see that f(x) = 0 if and only if g(x) = 0. Consequently f(x) = g(x) for all x 2 A so f = g and  is one to one. To show that  is onto P(A), let S  A, and let h : A ! {0, 1} be defined by h(x) = 1 if x 2 S and h(x) = 0 otherwise. Clearly (h) = S, showing that  is indeed onto P(A). 19. Picking up from the hint, let Z = {x 2 A | x /2 (x)}. We claim that for any a 2 A, (a) 6= Z. Either a 2 (a), in which case a /2 Z, or a /2 (a), in which case a 2 Z. Thus Z and (a) are certainly different subsets of A; one of them contains a and the other one does not. Based on what we just showed, we feel that the power set of A has cardinality greater than |A|. Proceeding naively, we can start with the infinite set Z, form its power set, then form the power set of that, and continue this process indefinitely. If there were only a finite number of infinite cardinal numbers, this process would have to terminate after a fixed finite number of steps. Since it doesn’t, it appears that there must be an infinite number of different infinite cardinal numbers. The set of everything is not logically acceptable, because the set of all subsets of the set of everything would be larger than the set of everything, which is a fallacy. 20. a. The set containing precisely the two elements of A and the three (different) elements of B is C = {1, 2, 3, 4, 5} which has 5 elements. i) Let A = {−2,−1, 0} and B = {1, 2, 3, · · ·} = Z+. Then |A| = 3 and |B| = @0, and A and B have no elements in common. The set C containing all elements in either A or B is C = {−2,−1, 0, 1, 2, 3, · · ·}. The map  : C ! B defined by (x) = x + 3 is one to one and onto B, so |C| = |B| = @0. Thus we consider 3 + @0 = @0. ii) Let A = {1, 2, 3, · · ·} and B = {1/2, 3/2, 5/2, · · ·}. Then |A| = |B| = @0 and A and B have no elements in common. The set C containing all elements in either A of B is C = {1/2, 1, 3/2, 2, 5/2, 3, · · ·}. The map  : C ! A defined by (x) = 2x is one to one and onto A, so |C| = |A| = @0. Thus we consider @0 + @0 = @0. b. We leave the plotting of the points in A × B to you. Figure 0.14 in the text, where there are @0 rows each having @0 entries, illustrates that we would consider that @0 · @0 = @0. 21. There are 102 = 100 numbers (.00 through .99) of the form .##, and 105 = 100, 000 numbers (.00000 through .99999) of the form .#####. Thus for .#####· · ·, we expect 10@0 sequences representing all numbers x 2 R such that 0  x  1, but a sequence trailing off in 0’s may represent the same x 2 R as a sequence trailing of in 9’s. At any rate, we should have 10@0  |[0, 1]| = |R|; see Exercise 15. On the other hand, we can represent numbers in R using any integer base n 1, and these same 10@0 sequences using digits from 0 to 9 in base n = 12 would not represent all x 2 [0, 1], so we have 10@0  |R|. Thus we consider the value of 10@0 to be |R|. We could make the same argument using any other integer base n 1, and thus consider n@0 = |R| for n 2 Z+, n 1. In particular, 12@0 = 2@0 = |R|. 22. @0, |R|, 2|R|, 2(2|R|), 2(2(2|R|)) 23. 1. There is only one partition {{a}} of a one-element set {a}. 24. There are two partitions of {a, b}, namely {{a, b}} and {{a}, {b}}. 0. Sets and Relations 3 25. There are five partitions of {a, b, c}, namely {{a, b, c}}, {{a}, {b, c}}, {{b}, {a, c}}, {{c}, {a, b}}, and {{a}, {b}, {c}}. 26. 15. The set {a, b, c, d} has 1 partition into one cell, 7 partitions into two cells (four with a 1,3 split and three with a 2,2 split), 6 partitions into three cells, and 1 partition into four cells for a total of 15 partitions. 27. 52. The set {a, b, c, d, e} has 1 partition into one cell, 15 into two cells, 25 into three cells, 10 into four cells, and 1 into five cells for a total of 52. (Do a combinatorics count for each possible case, such as a 1,2,2 split where there are 15 possible partitions.) 28. Reflexive: In order for xRx to be true, x must be in the same cell of the partition as the cell that contains x. This is certainly true. Transitive: Suppose that xRy and yRz. Then x is in the same cell as y so x = y, and y is in the same cell as z so that y = z. By the transitivity of the set equality relation on the collection of cells in the partition, we see that x = z so that x is in the same cell as z. Consequently, xRz. 29. Not an equivalence relation; 0 is not related to 0, so it is not reflexive. 30. Not an equivalence relation; 3  2 but 2  3, so it is not symmetric. 31. It is an equivalence relation; 0 = {0} and a = {a,−a} for a 2 R, a 6= 0. 32. It is not an equivalence relation; 1R3 and 3R5 but we do not have 1R5 because |1 − 5| = 4 3. 33. (See the answer in the text.) 34. It is an equivalence relation; 1 = {1, 11, 21, 31, · · ·}, 2 = {2, 12, 22, 32, · · ·}, · · · , 10 = {10, 20, 30, 40, · · ·}. 35. (See the answer in the text.) 36. a. Let h, k, and m be positive integers. We check the three criteria. Reflexive: h − h = n0 so h  h. Symmetric: If h  k so that h − k = ns for some s 2 Z, then k − h = n(−s) so k  h. Transitive: If h  k and k  m, then for some s, t 2 Z, we have h − k = ns and k − m = nt. Then h − m = (h − k) + (k − m) = ns + nt = n(s + t), so h  m. b. Let h, k 2 Z+. In the sense of this exercise, h  k if and only if h − k = nq for some q 2 Z. In the sense of Example 0.19, h  k (mod n) if and only if h and k have the same remainder when divided by n. Write h = nq1 + r1 and k = nq2 + r2 where 0  r1 n and 0  r2 n. Then h − k = n(q1 − q2) + (r1 − r2) and we see that h − k is a multiple of n if and only if r1 = r2. Thus the conditions are the same. c. a. 0 = {· · · ,−2, 0, 2, · · ·}, 1 = {· · · ,−3,−1, 1, 3, · · ·} b. 0 = {· · · ,−3, 0, 3, · · ·}, 1 = {· · · ,−5,−2, 1, 4, · · ·}, 2 = {· · · ,−1, 2, 5, · · ·} c. 0 = {· · · ,−5, 0, 5, · · ·}, 1 = {· · · ,−9,−4, 1, 6, · · ·}, 2 = {· · · ,−3, 2, 7, · · ·}, 3 = {· · · ,−7,−2, 3, 8, · · ·}, 4 = {· · · ,−1, 4, 9, · · ·} 4 1. Introduction and Examples 37. The name two-to-two function suggests that such a function f should carry every pair of distinct points into two distinct points. Such a function is one-to-one in the conventional sense. (If the domain has only one element, the function cannot fail to be two-to-two, because the only way it can fail to be two-to-two is to carry two points into one point, and the set does not have two points.) Conversely, every function that is one-to-one in the conventional sense carries each pair of distinct points into two distinct points. Thus the functions conventionally called one-to-one are precisely those that carry two points into two points, which is a much more intuitive unidirectional way of regarding them. Also, the standard way of trying to show that a function is one-to-one is precisely to show that it does not fail to be two-to-two. That is, proving that a function is one-to-one becomes more natural in the two-to-two terminology. 1. Introduction and Examples 1. i3 = i2 · i = −1 · i = −i 2. i4 = (i2)2 = (−1)2 = 1 3. i23 = (i2)11 · i = (−1)11 · i = (−1)i = −i 4. (−i)35 = (i2)17(−i) = (−1)17(−i) = (−1)(−i) = i 5. (4 − i)(5 + 3i) = 20 + 12i − 5i − 3i2 = 20 + 7i + 3 = 23 + 7i 6. (8 + 2i)(3 − i) = 24 − 8i + 6i − 2i2 = 24 − 2i − 2(−1) = 26 − 2i 7. (2 − 3i)(4 + i) + (6 − 5i) = 8 + 2i − 12i − 3i2 + 6 − 5i = 14 − 15i − 3(−1) = 17 − 15i 8. (1 + i)3 = (1 + i)2(1 + i) = (1 + 2i − 1)(1 + i) = 2i(1 + i) = 2i2 + 2i = −2 + 2i 9. (1−i)5 = 15+ 5 114(−i)+ 5·4 2·113(−i)2+ 5·4 2·112(−i)3+ 5 111(−i)4+(−i)5 = 1−5i+10i2−10i3+5i4−i5 = 1 − 5i − 10 + 10i + 5 − i = −4 + 4i 10. |3−4i| = p 32 + (−4)2 = p 9 + 16 = p 25 = 5 11. |6+4i| = p 62 + 42 = p 36 + 16 = p 52 = 2 p 13 12. |3 − 4i| = p 32 + (−4)2 = p 25 = 5 and 3 − 4i = 5(3 5 − 4 5 i) 13. | − 1 + i| = p (−1)2 + 12 = p 2 and − 1 + i = p 2(−p1 2 + p1 2 i) 14. |12 + 5i| = p 122 + 52 = p 169 and 12 + 5i = 13(12 13 + 5 13 i) 15. | − 3 + 5i| = p (−3)2 + 52 = p 34 and − 3 + 5i = p 34(−p3 34 + p5 34 i) 16. |z|4(cos 4 + i sin 4) = 1(1 + 0i) so |z| = 1 and cos 4 = 1 and sin 4 = 0. Thus 4 = 0 + n(2) so  = n 2 which yields values 0,  2 , , and 3 2 less than 2. The solutions are z1 = cos 0 + i sin 0 = 1, z2 = cos  2 + i sin  2 = i, z3 = cos  + i sin  = −1, and z4 = cos 3 2 + i sin 3 2 = −i. 17. |z|4(cos 4 + i sin 4) = 1(−1 + 0i) so |z| = 1 and cos 4 = −1 and sin 4 = 0. Thus 4 =  + n(2) so  =  4 + n 2 which yields values  4 , 3 4 , 5 4 , and 7 4 less than 2. The solutions are z1 = cos  4 + i sin  4 = 1 p 2 + 1 p 2 i, z2 = cos 3 4 + i sin 3 4 = − 1 p 2 + 1 p 2 i, z3 = cos 5 4 + i sin 5 4 = − 1 p 2 − 1 p 2 i, and z4 = cos 7 4 + i sin 7 4 = 1 p 2 − 1 p 2 i. 1. Introduction and Examples 5 18. |z|3(cos 3 + i sin 3) = 8(−1 + 0i) so |z| = 2 and cos 3 = −1 and sin 3 = 0. Thus 3 =  + n(2) so  =  3 + n2 3 which yields values  3 , , and 5 3 less than 2. The solutions are z1 = 2(cos  3 + i sin  3 ) = 2( 1 2 + p 3 2 i) = 1 + p 3i, z2 = 2(cos  + i sin ) = 2(−1 + 0i) = −2, and z3 = 2(cos 5 3 + i sin 5 3 ) = 2( 1 2 − p 3 2 i) = 1 − p 3i. 19. |z|3(cos 3 + i sin 3) = 27(0 − i) so |z| = 3 and cos 3 = 0 and sin 3 = −1. Thus 3 = 3/2 + n(2) so  =  2 + n2 3 which yields values  2 , 7 6 , and 11 6 less than 2. The solutions are z1 = 3(cos  2 + i sin  2 ) = 3(0 + i) = 3i, z2 = 3(cos 7 6 + i sin 7 6 ) = 3(− p 3 2 − 1 2i) = − 3 p 3 2 − 3 2i and z3 = 3(cos 11 6 + i sin 11 6 ) = 3( p 3 2 − 1 2i) = 3 p 3 2 − 3 2i. 20. |z|6(cos 6 + i sin 6) = 1 + 0i so |z| = 1 and cos 6 = 1 and sin 6 = 0. Thus 6 = 0 + n(2) so  = 0 + n2 6 which yields values 0,  3 , 2 3 , , 4 3 , and 5 3 less than 2. The solutions are z1 = 1(cos 0 + i sin 0) = 1 + 0i = 1, z2 = 1(cos  3 + i sin  3 ) = 1 2 + p 3 2 i, z3 = 1(cos 2 3 + i sin 2 3 ) = − 1 2 + p 3 2 i, z4 = 1(cos  + i sin ) = −1 + 0i = −1, z5 = 1(cos 4 3 + i sin 4 3 ) = − 1 2 − p 3 2 i, z6 = 1(cos 5 3 + i sin 5 3 ) = 1 2 − p 3 2 i. 21. |z|6(cos 6 + i sin 6) = 64(−1 + 0i) so |z| = 2 and cos 6 = −1 and sin 6 = 0. Thus 6 =  + n(2) so  =  6 + n2 6 which yields values  6 ,  2 , 5 6 , 7 6 , 3 2 and 11 6 less than 2. The solutions are z1 = 2(cos  6 + i sin  6 ) = 2( p 3 2 + 1 2i) = p 3 + i, z2 = 2(cos  2 + i sin  2 ) = 2(0 + i) = 2i, z3 = 2(cos 5 6 + i sin 5 6 ) = 2(− p 3 2 + 1 2i) = − p 3 + i, z4 = 2(cos 7 6 + i sin 7 6 ) = 2(− p 3 2 − 1 2i) = − p 3 − i, z5 = 2(cos 3 2 + i sin 3 2 ) = 2(0 − i) = −2i, z6 = 2(cos 11 6 + i sin 11 6 ) = 2( p 3 2 − 1 2i) = p 3 − i. 22. 10 + 16 = 26 17, so 10 +17 16 = 26 − 17 = 9. 23. 8 + 6 = 14 10, so 8 +10 6 = 14 − 10 = 4. 24. 20.5 + 19.3 = 39.8 25, so 20.5 +25 19.3 = 39.8 − 25 = 14.8. 25. 1 2 + 7 8 = 11 8 1, so 1 2 +1 7 8 = 11 8 − 1 = 3 8 . 26. 3 4 + 3 2 = 9 4 2, so 3 4 +2 3 2 = 9 4 − 2 =  4 . 6 1. Introduction and Examples 27. 2 p 2 + 3 p 2 = 5 p 2 p 32 = 4 p 2, so 2 p 2 +p 32 3 p 2 = 5 p 2 − 4 p 2 = p 2. 28. 8 is not in R6 because 8 6, and we have only defined a +6 b for a, b 2 R6. 29. We need to have x + 7 = 15 + 3, so x = 11 will work. It is easily checked that there is no other solution. 30. We need to have x + 3 2 = 2 + 3 4 = 11 4 , so x = 5 4 will work. It is easy to see there is no other solution. 31. We need to have x + x = 7 + 3 = 10, so x = 5 will work. It is easy to see that there is no other solution. 32. We need to have x + x + x = 7 + 5, so x = 4 will work. Checking the other possibilities 0, 1, 2, 3, 5, and 6, we see that this is the only solution. 33. An obvious solution is x = 1. Otherwise, we need to have x + x = 12 + 2, so x = 7 will work also. Checking the other ten elements, in Z12, we see that these are the only solutions. 34. Checking the elements 0, 1, 2, 3 2 Z4, we find that they are all solutions. For example, 3+43+43+43 = (3 +4 3) +4 (3 +4 3) = 2 +4 2 = 0. 35. 0 $ 0, 3 = 2 $ 2 +8 5 = 7, 4 = 22 $ 2 +8 2 = 4, 5 = 4 $ 4 +8 5 = 1, 6 = 33 $ 7 +8 7 = 6, 7 = 34 $ 7 +8 4 = 3 36. 0 $ 0, 2 =  $ 4 +7 4 = 1, 3 = 2 $ 1 +7 4 = 5, 4 = 22 $ 1 +7 1 = 2, 5 = 32 $ 5 +7 1 = 6, 6 = 33 $ 5 +7 5 = 3 37. If there were an isomorphism such that  $ 4, then we would have 2 $ 4+6 4 = 2 and 4 = 22 $ 2+6 2 = 4 again, contradicting the fact that an isomorphism $ must give a one-to-one correpondence. 38. By Euler’s fomula, eiaeib = ei(a+b) = cos(a + b) + i sin(a + b). Also by Euler’s formula, eiaeib = (cos a + i sin a)(cos b + i sin b) = (cos a cos b − sin a sin b) + i(sin a cos b + cos a sin b). The desired formulas follow at once. 39. (See the text answer.) 40. a. We have e3 = cos 3 + i sin 3. On the other hand, e3 = (e)3 = (cos  + i sin )3 = cos3  + 3i cos2  sin  − 3 cos  sin2  − i sin3  = (cos3  − 3 cos  sin2 ) + i(3 cos2  sin  − sin3 ). Comparing these two expressions, we see that cos 3 = cos3  − 3 cos  sin2 . b. From Part(a), we obtain cos 3 = cos3  − 3(cos )(1 − cos2 ) = 4 cos3  − 3 cos . 2. Binary Operations 7 2. Binary Operations 1. b  d = e, c  c = b, [(a  c)  e]  a = [c  e]  a = a  a = a 2. (a  b)  c = b  c = a and a  (b  c) = a  a = a, so the operation might be associative, but we can’t tell without checking all other triple products. 3. (b  d)  c = e  c = a and b  (d  c) = b  b = c, so the operation is not associative. 4. It is not commutative because b  e = c but e  b = b. 5. Now d  a = d so fill in d for a  d. Also, c  b = a so fill in a for b  c. Now b  d = c so fill in c for d  b. Finally, c  d = b so fill in b for d  c. 6. d  a = (c  b)  a = c  (b  a) = c  b = d. In a similar fashion, substituting c  b for d and using the associative property, we find that d  b = c, d  c = c, and d  d = d. 7. It is not commutative because 1−2 6= 2−1. It is not associative because 2 = 1−(2−3) 6= (1−2)−3 = −4. 8. It is commutative because ab + 1 = ba + 1 for all a, b 2 Q. It is not associative because (a  b)  c = (ab + 1)  c = abc + c + 1 but a  (b  c) = a  (bc + 1) = abc + a + 1, and we need not have a = c. 9. It is commutative because ab/2 = ba/2 for all a, b 2 Q. It is associative because a(bc) = a(bc/2) = [a(bc/2)]/2 = abc/4, and (a  b)  c = (ab/2)  c = [(ab/2)c]/2 = abc/4 also. 10. It is commutative because 2ab = 2ba for all a, b 2 Z+. It is not associative because (a b)  c = 2ab  c = 2(2ab)c, but a  (b  c) = a  2bc = 2a(2bc). 11. It is not commutative because 2  3 = 23 = 8 6= 9 = 32 = 3  2. It is not associative because a  (b  c) = a  bc = a(bc), but (a  b)  c = ab  c = (ab)c = abc, and bc 6= bc for some b, c 2 Z+. 12. If S has just one element, there is only one possible binary operation on S; the table must be filled in with that single element. If S has two elements, there are 16 possible operations, for there are four places to fill in a table, and each may be filled in two ways, and 2 · 2 · 2 · 2 = 16. There are 19,683 operations on a set S with three elements, for there are nine places to fill in a table, and 39 = 19, 683. With n elements, there are n2 places to fill in a table, each of which can be done in n ways, so there are n(n2) possible tables. 13. A commutative binary operation on a set with n elements is completely determined by the elements on or above the main diagonal in its table, which runs from the upper left corner to the lower right corner. The number of such places to fill in is n + n2 − n 2 = n2 + n 2 . Thus there are n(n2+n)/2 possible commutative binary operations on an n-element set. For n = 2, we obtain 23 = 8, and for n = 3 we obtain 36 = 729. 14. It is incorrect. Mention should be made of the underlying set for  and the universal quantifier, for all, should appear. A binary operation  on a set S is commutative if and only if a  b = b  a for all a, b 2 S. 8 2. Binary Operations 15. The definition is correct. 16. It is incorrect. Replace the final S by H. 17. It is not a binary operation. Condition 2 is violated, for 1  1 = 0 and 0 /2 Z+. 18. This does define a binary operation. 19. This does define a binary operation. 20. This does define a binary operation. 21. It is not a binary operation. Condition 1 is violated, for 2  3 might be any integer greater than 9. 22. It is not a binary operation. Condition 2 is violated, for 1  1 = 0 and 0 /2 Z+. 23. a. Yes.  a −b b a  +  c −d d c  =  a + c −(b + d) b + d a + c  . b. Yes.  a −b b a   c −d d c  =  ac − bd −(ad + bc) ad + bc ac − bd  . 24. F T F F F T T T T F 25. (See the answer in the text.) 26. We have (a b)  (c d) = (c d)  (a b) = (d c)  (a b) = [(d c) a]  b, where we used commutativity for the first two steps and associativity for the last. 27. The statement is true. Commutativity and associativity assert the equality of certain computations. For a binary operation on a set with just one element, that element is the result of every computation involving the operation, so the operation must be commutative and associative. 28.  a b a b a b a a The statement is false. Consider the operation on {a, b} defined by the table. Then (a  a)  b = b  b = a but a  (a  b) = a  a = b. 29. It is associative. Proof: [(f + g) + h](x) = (f + g)(x) + h(x) = [f(x) + g(x)] + h(x) = f(x) + [g(x) + h(x)] = f(x) + [(g + h)(x)] = [f + (g + h)](x) because addition in R is associative. 30. It is not commutative. Let f(x) = 2x and g(x) = 5x. Then (f −g)(x) = f(x)−g(x) = 2x−5x = −3x while (g − f)(x) = g(x) − f(x) = 5x − 2x = 3x. 31. It is not associative. Let f(x) = 2x, g(x) = 5x, and h(x) = 8x. Then [f − (g − h)](x) = f(x) − (g − h)(x) = f(x) − [g(x) − h(x)] = f(x) − g(x) + h(x) = 2x − 5x + 8x = 5x, but [(f − g) − h](x) = (f − g)(x) − h(x) = f(x) − g(x) − h(x) = 2x − 5x − 8x = −11x. 32. It is commutative. Proof: (f · g)(x) = f(x) · g(x) = g(x) · f(x) = (g · f)(x) because multiplication in R is commutative. 33. It is associative. Proof: [(f · g) · h](x) = (f · g)(x) · h(x) = [f(x) · g(x)] · h(x) = f(x) · [g(x) · h(x)] = [f · (g · h)](x) because multiplication in R is associative. 3. Isomorphic Binary Structures 9 34. It is not commutative. Let f(x) = x2 and g(x) = x + 1. Then (f  g)(3) = f(g(3)) = f(4) = 16 but (g  f)(3) = g(f(3)) = g(9) = 10. 35. It is not true. Let  be + and let 0 be · and let S = Z. Then 2+(3 · 5) = 17 but (2+3) · (2+5) = 35. 36. Let a, b 2 H. By definition of H, we have a  x = x  a and b  x = x  b for all