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psol11

# psol11 - Math 55 Homework 11 Solutions 7.4 2 Z Z 4 Just add...

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Unformatted text preview: Math 55 - Homework 11 Solutions 7.4 2. Z Z. 4. Just add arrows from each to dot to itself, if none exists. 6. a b c d 10. same as 2. 12. The matrix representing {(a, a) | a A} is In , and we know that MRS = MR MS . Since the reflexive closure of R is R = R {(a, a) | a A}, we have MR = MR In . 16. (a) yes (b) no (c) yes (d) no (e) yes (f) no 26. (a) add {(a, a), (b, b), (c, c), (d, d), (e, b)} (b) add {(c, b), (b, b), (c, c), (e, e)} (c) {a, b, c, d, e}2 (d) {a, b, c, d, e}2 7.5 2. (a) yes (b) yes (c) no (d) no (e) no 6. Let B be the set of R-equivalence classes of A, and let f : A B be given by f(a) = [a]R. This clearly does what we want. 1 10. Showing R is reflexive and symmetric is trivial from its definition. For transitivity, suppose (a, b)R(c, d) and (c, d)R(e, f). By definition ad = bc and cf = de or, since all of these integers are positive, c e a = = b d f Therefore, af = be and (a, b)R(e, f) as desired. 12. (a) Certainly f (n) (x) = f (n) (x), f (n) (x) = g (n) (x) g (n) (x) = f (n) (x), and f (n) (x) = g (n) (x) g (n) (x) = h(n) (x) f (n) (x) = h(n) (x). So the relation is reflexive, symmetric, and transitive. d (b) dx4 (x4 ) = 4!, and for polynomials f, f (4) = 4! iff f = x4 + Ax3 + Bx2 + Cx + D. 4 16. yes, it is. 18. (a) no (b) yes (c) yes 22. (a) Every class is of the form {f : Z Z | f(a) = C} for some C Z. (b) Every class is of the form {g : Z Z | C Z x Z g(x) = f(x) + C} for some f : Z Z. 26. (a) {n Z | 2 | n} (b) {n Z | 3 | (n + 1)} (c) {n Z | 6 | (n + 2)} d) {n Z | 8|(n + 4)} 7.6 2. (a) yes (b) yes (c) no 2 34. (a) If a and b are both greatest, then a b and b a. By the antisymmetry of , it must be that a = b. 8 7 t s t s t p s i t p s i r q v u t s t s t p s i t p s i r q v u t s t s t p s i t p s i r q v u t s t s t p s i t p s i r q v u F E @ 9 T S R Q P I 4 3 4 3 4 3 ( ' 4 3 ( ' 4 3 2 ( 1 ' 0 " 3 ) ! 2 ( 1 ' 0 " ) ! 4 3 4 3 4 3 ( ' 4 3 ( ' 4 3 2 ( 1 ' 0 " 3 ) ! 2 ( 1 ' 0 " ) ! 4 3 4 3 4 3 ( ' 4 3 ( ' 4 3 2 ( 1 ' 0 " 3 ) ! 2 ( 1 ' 0 " ) ! 4 3 4 3 4 3 ( ' 4 3 ( ' 4 3 2 ( 1 ' 0 " 3 ) ! 2 ( 1 ' 0 " ) ! f e d c y x w y x w y x w y x w h g 30. (a) (N, ) (b) (N, ) (c) (Z, ). 26. (a) m, l (b) a, b, c (c) no (d) no (e) k, l, m (f) k (g) none (h) none 18. (b) Same argument: If a and b are both least, then b a and a b. 3 r q v u v u v u r q v u v u v u r q v u v u v u r q v u v u v u B A D C 6 5 6 5 6 5 5 6 5 6 5 6 5 5 6 5 6 5 6 5 5 6 5 6 5 6 5 5 H G V U X W 2 ( 1 ' 0 " ) ! & % 0 " ) ! & % 0 " ) ! & \$ % # & \$ % # & \$ % # % # 2 ( 1 ' 0 " ) ! & % 0 " ) ! & % 0 " ) ! & \$ % # & \$ % # & \$ % # % # 2 ( 1 ' 0 " ) ! & % 0 " ) ! & % 0 " ) ! & \$ % # & \$ % # & \$ % # % # 2 ( 1 ' 0 " ) ! & % 0 " ) ! & % 0 " ) ! & \$ % # & \$ % # & \$ % # % # b a ` Y y x w x w x w y x w x w x w y x w x w x w y x w x w x w ...
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