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# Homework solutions Math 311

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This 6 page Bundle was uploaded by Akilah Fuller on Thursday September 15, 2016. The Bundle belongs to Math 311 at Jackson State University taught by Dr. Wright in Fall 2016. Since its upload, it has received 7 views. For similar materials see Abstract Algebra in Mathematics at Jackson State University.

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Date Created: 09/15/16

F16-311 Selected Homework Solutions C. Wright Homework 2: Chapter 2 (A6) a ▯ b = ja ▯ bj on the set fn : n ▯ 0g Solution: By denition, the absolute avlue function is uniquely dened. Also by de- nition, ja ▯ bj ▯ 0. Therefore this is an operation on this set. (B4) x ▯ y = jx ▯ yj 1. Commutative: y ▯ x = jy ▯ xj = j ▯ (x ▯ y)j = jx ▯ yj = x ▯ y 2. Identity: No identity since if x < 0, there is no y such that jx + yj = x. (B7) x ▯ y = xy (on the set of positive real numbers) x + y + 1 1. Commutative: Commutativity follows since the multiplication and addition op- erations on the real numbers are commutative. xy yx x ▯ y = = = y ▯ x x + y + 1 y + x + 1 2. Associative: ▯ xy ▯ xy xyz (x ▯ y) ▯ z = ▯ z = xyx+y+1 = x + y + 1 x+y+1 + z + 1 xy + (z + 1)(x + y + 1) ▯ ▯ yz x ▯ (y ▯ z) = x ▯ yz = y+z+1 = xyz y + z + 1 yz + z + 1 yz + (x + 1)(y + z + 1) y+z+1 With some more algebraic manipulation, these are both equal to xyz xy + xz + yz + x + y + z + 1 3. No identity: We will rst do some preliminary scratch work to get a sense of what we are dealing with here. The identi,ty if it exists, must be the same identity for all real numbers under this operation. Let’s try some examples where we let x = ▯1;0;3: ▯e ▯1 ▯ e = = ▯1 e Note that this means that everything acts like the identity for ▯1. 0 0 ▯ e = = 0 e + 1 1 Again, everything acts like the identity for 0. 3 ▯ e = 3 3e = 3 3 + e + 1 e = 1 4 + e But this is impossible since e 6= 4 + e. So there is no identity for 3. An alternative approach is a purely algebraic approach. Suppose there exists an identity e 2 R such that x ▯ e = e. As usual, when we set up the equation, we can nd an appropriate e if we can solve for e: x ▯ e = x xe = x x + e + 1 xe = x + xe + x 2 0 = x + x 0 = x(x + 1) which implies x = 0;▯1. In other words the only way our initial condition is satised is if x = 0 or x = ▯1, and there are no restrictions on what e could be in those two cases (i.e. any e would work.) Since there is not a single identity that works for all real numbers, the identity property is not satised. 2 MATH311 Intro. to Abstract Algebra - Homework 1 Instructor: Dr. C. Wright Due: Mond,ay August 29, 2016 Instructions: Show all your work. 1. Write the negation of each statement. (a) If a and b are relatively prime then a and b are prime. Solution : a and b are relatively prime and a is not prime or b is not prime. (b) Every linear system of equations has no solution, or it has a unique solution. Solution : There is a linear system of equations that has a solution and it is not unique. (c) oFr all integers x and y, there exists an integer t such that x ▯ y = t. Solution : There exist integers x and y such that for all integers t, x ▯ y 6= t. 2. Write the contrapositive of each statement. (a) If f is bijective and a homomorphism, then f is an isomorphism. Solution : If f is not an isomorphism, then f is not bijective or f is not a homo- morphism. (b) If G is an abelian group, then G is the center and for all g;h 2 G, gh = hg. Solution : If G is not the center or there exists g;h 2 G such that gh 6= hg, then G is not an abelian group. (c) If N is a normal subgroup of G, then for any elements a;b in G, a 2 bN implies aN = bN. Solution : If there exists elements a;b in G such that a 2 bN but aN 6= bN, then N is not a normal subgroup. 3. Describe how to prove a biconditional statement P i Q. Then outline how you would prove: A = B , A ▯ B = B ▯ A. Solution : oT prove P i Q we must prove P ! Q and its converse Q ! P. oFr this example, we must prove (a) A = B =) A ▯ B = B ▯ A Outline: Assume A = B. Show that A▯B ▯ B ▯A, then show B ▯A ▯ A▯B. 1 (b) A ▯ B = B ▯ A =) A = B Outline: Assume A ▯ B = B ▯ A. Prove A ▯ B and B ▯ A. 4. List the elements in each set. (a) fy 2 R j y = 1g = f▯1;1g (b) fy 2 C j y = 1g = f1;▯1;i;▯ig ( Xk ) (c) 3i ▯ 2 j k = 1;2;3 = f1;5;12g i=1 ( ) Y▯1 (d) (q ▯ q ) j n = 1;2;3;4 = fq ▯ 1;(q ▯ q)(q ▯ 1);(q ▯ q )(q ▯ q)(q ▯ 3 3 k=0 4 3 4 2 4 4 1);(q ▯ q )(q ▯ q )(q ▯ q)(q ▯ 1)g 5. List at least 3 elements in each set, if possible. (a) fx 3 : ▯ 8 ▯ x ▯ 8; x 2 Rg Solution: 4, -64, 1.45 Note: If ▯8 ▯ x ▯ 8, then ▯8 ▯ x ▯ 8 . So the entire set is actually the real interavl [▯512;512]. (b) fx 3 : jxj ▯ 8; x 2 Zg = f▯512;:::;▯64;▯27;▯8;▯1;0;1;8;27;64;:::;512g (c) fx 2 R : jx j ▯ 8g = [▯2;2] 3 (d) fx 2 Z : jx j ▯ 8g = f▯2;▯1;0;1;2g 2 6. List at least 3 elements in each set, if possible. (a) 4 + 3Z = f4 + 3k : k 2 Zg, ▯11;4;49 (b) fx 2 Z j x ▯ 2 is a multiple of 5g, ▯78; 7; 52 2 2 (c) f(a;b;a + b ) j a;b 2 Ng, (3;4;25); (1;1;2); (5;10;125) (d) f2 5 m j n;m 2 Zg, 2; 2 ▯2 5 ; 20 (e) f(x;y) 2 R 2 : y ▯ x = 0g, (▯2;4); (▯;▯ ); (8;64) 0 2 2 2 (f) nf 2 F(R) j f (x) = 2x + 5g, x + 5xo x + 5x ▯ 4; x + 5x + 22 a 4 7 5 (g) 2 Q : b = 6 1; 9n 2 Z s:t: b = na , 8 ; 21 ; 125 b 7. The congruence mod n relation is dened as a ▯ b mod n i a ▯ b is a multiple of n, i.e. 9k 2 Z such that a ▯ b = kn. Prove that this is an equiavlence relation. ▯ Reexive: Let a 2 Z. Certain,ly a ▯ a = 0 = 0 ▯ n is a multiple of n, so a ▯ a mod n. ▯ Symmetric: Let a;b 2 Z such that a ▯ b mod n. Then there exists an integer k such that a ▯ b = nk. Then b ▯ a = ▯(a ▯ b) = (▯k) ▯ n is an integer multiple of n, so by denition b ▯ a mod n. ▯ rTansitive: Let a;b;c 2 Z such that a ▯ b mod n and b ▯ c mod n. Then there exist integers k;‘ such that a▯b = kn and b▯c = ‘n. Adding the two equations, we obtain a ▯ c = (a ▯ b) + (b ▯ c) = kn + ‘n = (k + ‘)n, an integer multiple of n. Hence, a ▯ c mod n. 8. Dene a relation ▯ on R as follows: (a;b) ▯ (c;d) i a + b = c + d . 2 2 (a) Prove that ▯ is an equiavlence relation. Solution : ▯ Reexive: Let (a;b) 2 R . The relation is reexive if (a;b) ▯ (a;b). Certainly 2 2 2 2 a + b = a + b , so ▯ is reexive. 2 2 2 2 2 ▯ Symmetric: Let (a;b);(c;d) 2 R such that (a;b) ▯ (c;d), i.e. a +b = c +d . The relation is symmetric if (c;d) ▯ (a;b). Certainly the previous equation implies that c + d = a + b , so 2 (c;d) ▯ (e;f), therefore ▯ is symmetric. ▯ rTansitive: Let (a;b);(c;d);(e;f) 2 R 2 such that (a;b) ▯ (c;d) and (c;d) ▯ (e;f). That is, a + b = c + d and 2 2 c + d = e + f . The relation is 2 2 2 2 transitive if (a;b) ▯ (e;f). By the transitivity of equali,ty a + b = e + f , so (a;b) ▯ (e;f), and ▯ is transitive. 3 (b) Dene and describe the equiavlence classes. Solution : oFr any element (a;b) 2 R , its equiavlence class is dened as [(a;b)] = f(x;y) 2 R j (x;y) ▯ (a;b)g = f(x;y) 2 R j x + y = a + b g 2 2 2 Remark: The way this is written should remind us of the equation of a circle. The class can be rewritten as p [(a;b)] = f(x;y) 2 R j x + y = r ; where 2 2 r = a + b g:2 So the equiavlence classes are points on the same circle of a xed radius. 9. Explain why the graph of an equiavlence relation always contains the line y = x. Solution : Since it is equiavlence relation, it is reexive, meaning for all x 2 R, (x;x) is in the relation. But these are exactly all the points on the line y = x. 10. Let A;B be sets. Prove: If A ▯ B then A \ B = B. Solution: This statement is false, and thus cannot be proved. In fact, the conclusion should be A \ B = A. 11. Find the generalized union of the following family of sets: ▯▯ ▯ ▯ 1 0;3 + : n 2 Z >0 n Solution: Note that 3+ 1 is largest when n = 1, then the avlues decrease. So the union n of all these sets is ▯ ▯ [ 1 0;3 + = [0;4]: n2Z + n 12. In the mapping ▯ from A into B, where A = f1;3;5g, B = fx;y;zg and ▯ is dened as follows: ▯(1) = x; ▯(3) = x; ▯(5) = x, state, using set notation (a) the domain = f1;3;5g (b) the codomain = fx;y;zg (c) the image ▯(A) = fxg 4

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