Chapter V - Algebraic Structures - Exercises [PDF]

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Algebraic Structures - Exercises NGUYEN CANH Nam1 1 Faculty of Applied Mathematics Department of Applied Mathematics and Informatics Hanoi University of Technologies [email protected]



HUT - 2010



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 1



In each of following cases, for the binary operator ∗ on IR, consider its properties : commutative, associative, identity element, inverse element. a) x ∗ y = xy + 1 1 b) x ∗ y = xy 2 c) x ∗ y = |x|y



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution



a) We have x ∗ y = xy + 1 = yx + 1 = y ∗ x for all x, y ∈ IR, so ∗ is commutative. (1 ∗ 2) ∗ 3 = (1 × 2 + 1) ∗ 3 = 3 × 3 + 1 = 10 6= 1 ∗ (2 ∗ 3) = 1 ∗ (2 × 3 + 1) = 1 × 7 + 1 = 8 So ∗ isn’t associative. Assume that ∗ has the identity element e. Then 1 2 ∗ e = 2e + 1 = 2 ⇔ e = . Moreover 2 2 3 ∗ e = 3e + 1 = 3 ⇔ e = . 3 Hence, there doesn’t exist the identity element. Therefore there doesn’t exist the inverse element of an element x neither.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution



b) We have 1 1 x ∗ y = xy = yx = y ∗ x for all x, y ∈ IR, so ∗ is 2 2 commutative. 1 1 1 (x ∗ y ) ∗ z = ( xy ) ∗ z = xyz = x ∗ ( yz) = x ∗ (y ∗ z), so 2 4 2 ∗ is associative. Denote e the identity element. Then 1 x ∗ e = xe = x ⇔ e = 2. 2 Consider an element x, denote x −1 the inverse element. 2 We then have x ∗ x −1 = e. So if x 6= 0 then x −1 = . There x doesn’t exist the inverse element of 0.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution



c) We have 1 ∗ 2 = |1|2 6= |2|1 = 2 ∗ 1, so ∗ isn’t commutative. (2 ∗ 3) ∗ 4 = (23 ) ∗ 4 = 8 ∗ 4 = 84 2 ∗ (3 ∗ 4) = 2 ∗ (34 ) = 2 ∗ 81 = 281 6= 84 So ∗ isn’t associative. Denote e the identity element. Then −1 ∗ e = −1 ⇔ 1e = −1 (impossible). Hence, there doesn’t exist the identity element. Therefore there doesn’t exist the inverse element of an element x neither.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 2



Suppose F is set of open intervals of IR and contains the empty set ∅ (∅ is considered as an open interval). Show that a) F is closed under operator intersection (∩) on P(IR) b) F isn’t closed under operator union (∪) on P(IR)



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution



a) It is known that the interval (a, a), for all a ∈ IR, is the empty set ∅. Consider now two any intervals (a, b), (c, d) ∈ F , we have If a ≤ b ≤ c If a ≤ c < b If c ≤ a < d If c ≤ a ≤ b If a ≤ c ≤ d



≤d ≤d ≤b ≤d ≤b



or c ≤ d ≤ a ≤ b then (a, b) ∩ (c, d) = ∅. then (a, b) ∩ (c, d) = (c, b). then (a, b) ∩ (c, d) = (a, d). then (a, b) ∩ (c, d) = (a, b). then (a, b) ∩ (c, d) = (c, d).



Hence, in all cases, (a, b) ∩ (c, d) ∈ F , i.e., F is closed under operator intersection (∩) on P(IR). b) We have (1; 2), (2; 3) ∈ F , but (1; 2) ∪ (2; 3) 6∈ F . So F isn’t closed under operator union (∪).



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 3



Suppose X , Y are sets and ∗ : Y × Y → Y is a binary operator. Show that ∗ deduce a binary operator on F(X , Y ) (the collection of mapping from X to Y .)



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



We define a binary operator ∆ on F(X , Y ) as follows : For all f , g ∈ F(X , Y ), (f ∆g)(x) = f (x) ∗ g(x) for all x ∈ X Obviously (f ∆g) is a mapping from X to Y .



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 4



Let X be a set and ∗ be an operator on X defined by x ∗ y = x for all x, y ∈ X . Prove (X , ∗) is a semigroup. Is it commutative? Does it have the identity element?



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution



It is obvious that ∗ is closed. Moreover, for any x, y ∈ X , we have ( (x ∗ y ) ∗ z = x ∗ z = x ⇒ (x ∗ y ) ∗ z = x ∗ (y ∗ z). x ∗ (y ∗ z) = x ∗ y = x So ∗ is associative. Then (X , ∗) is a semigroup. If X has two distinct elements then (X , ∗) isn’t commutative and it doesn’t have the identity element. Indeed, we have x ∗ y = x 6= y = y ∗ x and e ∗ x = e 6= x for all x ∈ X , x 6= e.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 5



Let X be a semigroup with its multiplication (·). a) Show that if a · b = b · a for all a, b ∈ X then (a · b)n = an · bn , ∀a, b ∈ X , ∀n ∈ IN, n ≥ 1, where x n = x · x · · · x. b) Suppose (a · b)2 = a2 · b2 . Could we deduce a · b = b · a?



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 6



√ √ Show that the set IR[ 3] = {a = b 3 | a, b ∈ IR, a2 + b2 6= 0} with ordinary multiplication is a group.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution. First, is a binary operator on √ we prove that the multiplication √ √ IR[ 3].Indeed, for any a + b 3 and c + d 3 we have √ √ √ √ (a + b 3)(c + d 3) = (ac + 3bd) + (ad + bc) 3 ∈ IR[ 3]. It is not so difficult (!) to prove √ that the multiplication is √ associative and 1 = 3 is the identity element of IR[ 3]. √1 + 0 √ Now, for any a + b 3 ∈ IR[ 3], we have √ √ b a − x= 2 3 ∈ IR[ 3] and a − 3b2 a2 − 3b2   √ √ a b (a + b 3) − 3 a2 − 3b2 a2 − 3b2   √ √ a b = − 2 3 (a + b 3) = 1. 2 2 2 a − 3b a − 3b √ b a − 2 3 is the inverse element of That means 2 2 2 a − 3b√ a − 3b √ a + b 3.Hence (IR[ 3], .) is a group. NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 7



Let X , Y be sets, and ∗Y × Y → Y be a binary operator which is commutative, associative and invertible, and f : X → Y be a bijective mapping. We equip X with a binary operator ◦ as follows x1 ◦ x2 = f −1 (f (x1 ) ∗ f (x2 )) for all x1 , x2 ∈ X . Prove that ◦ is commutative, associative and invertible.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 8



Prove that if a ring has an identity then it has only one identity.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



Suppose that there are two identity e and f . We have e =e·f =f



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Mathematics I - Chapter 4 - Exercises



Exercise 9



Define a new addition ◦ and multiplication ? on the integers ZZ by a ◦ b = a + b − 1 and a ? b = a + b − ab, where the operations on the right-hand side of the equal sign are ordinary addition, subtraction and multiplication. Prove that (ZZ, ◦, ?) is a commutative ring with identity.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution. First we verify that ◦ and ? are binary operations: If a, b ∈ ZZ, then a ◦ b = a + b − 1 ∈ Z Z. Thus the set ZZ is closed under addition. If a, b ∈ ZZ, then a ? b = a + b − ab ∈ Z Z. Thus ZZ is closed under multiplication. We note for later purposes that a ? b = b ? a from the definition of ?. Next we need to show that (Z Z, ◦, ?) is a ring. We need to verify all the defining properties of a ring. Throughout the following, let a, b, c ∈ Z Z. Associativity of addition: We conclude from a◦(b ◦c) = a◦(b +c −1) = a+(b +c −1)−1 = a+b +c −2 and (a◦b)◦c = (a+b −1)◦c = (a+b −1)+c −1 = a+b +c −2, that a ◦ (b ◦ c) = (a ◦ b) ◦ c for all a, b, c ∈ ZZ. NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution. Commutativity of addition: a ◦ b = a + b − 1 = b + a − 1 = b ◦ a. We identify an element 0Z ∈ (Z Z, ◦, ?)such that a ◦ 0Z = a for all a ∈ (ZZ, ◦, ?) : a + 0Z = a ⇔ a + 0Z − 1 = a ⇔ 0Z = 1. Thus 0Z = 1 is the zero element in (Z Z, ◦, ?). The equation a ◦ x = 0Z has a solution in (ZZ, ◦, ?) because it is equivalent to the equation a + x − 1 = 1 which has a solution x in ZZ. Associativity of multiplication: We conclude from a?(b?c) = a?(b+c−bc) = a+(b+c−bc)−a(b+c−bc) = a+b+c−ab−bc−ca+abc



and (a?b)?c = (a+b−ab)?c = (a+b−ab)+c−(a+b−ab)c = a+b+c−ab−bc−ca+abc,



that a ? (b ? c) = (a ? b) ? c for all a, b, c ∈ ZZ. NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution. Distributivity: Since a?(b◦c) = a?(b+c −1) = a+(b+c −1)−a(b+c −1) = 2a+b+c −ab−ac −1



and (a ? b) ◦ (a ? c) = (a + b − ab) ◦ (a + c − ac) = (a + b − ab) + (a + c − ac) − 1 = 2a + b + c − ab − ac − 1. we have a ? (b ◦ c) = (a ? b) ◦ (a ? c) for all a, b, c ∈ ZZ. From this we conclude, using the commutativity of the operation ?, that (b ◦ c) ? a = (b ? a) ◦ (c ? a). This shows the distributive law.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution. To this point, we have shown that (Z Z, ◦, ?) is a ring. Next we verify the additional properties which promote the ring (ZZ, ◦, ?) to a commutative ring with identity. Commutativity of multiplication: a ? b = a + b − ab = b + a − ba = b ? a. Existence of identity: we need to identify an identity 1Z . We conclude 1Z = 0 from a?1Z = a ⇔ a+1Z −a1Z = a ⇔ 1Z −a1Z = 0 ⇔ (1−a)1Z = 0 for any a 6= 0. The commutativity of the ring (ZZ, ◦, ?) ensures that a = 1Z ? a is also satisfied. So, 1Z is the identity of (ZZ, ◦, ?).



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 10



Let R be a ring. Then the following hold If a + b = a + c, (a, b, c ∈ R) then b = c.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



If a + b = a + c then b = 0 + b = ((−a) + a) + b = (−a) + (a + b) = (−a) + (a + c) = ((−a) + a) + c = 0 + c = c



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 11



√ √ Prove that the set ZZ[ 2] = {a + b 2 | a, b ∈ Z Z} with the addition and the multiplication of numbers is a commutative and unitary ring. Is it a field?



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 12



Let a and b be integers and let p be a prime. Let p divides ab. Then either p divides a or p divides b.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



For the sake of argument suppose p does not divide b. Then since the gcd(p, b) divides p then gcd(p, b) can only be 1. Hence there exist x, y ∈ Z Z such that px + by = 1. But then apx + aby = a. Clearly p divides apx and p divides ab (by hypothesis) and so p divides a.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 13



Let a, b and n be integers. If n divides a − b prove that n divides a2 − b2 and a3 − b3 . If n divides a + b prove that n divides a3 + b3 but does not necessarily divide a2 + b2 .



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



a2 − b2 = (a − b)(a + b), a3 − b3 = (a − b)(a2 + ab + b2 ). a3 + b3 = (a + b)(a2 − ab + b2 ). 3 divides 1 + 2 but 3 does not divide 12 + 22 .



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 14



If a | (b + c) and gcd(b, c) = 1, prove that gcd(a, b) = 1 = gcd(a, c).



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Exercise 15



If c | ab and gcd(c, a) = d, prove that c | db.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises



Solution.



We have d = cu + av for some u, v ∈ Z Z. Hence db = cbu + abv . Moreover ab = cw for some w ∈ Z Z. Therefore db = cbu + cw = c(bu + w), equivalently c | db.



NGUYEN CANH Nam



Mathematics I - Chapter 4 - Exercises