chapter v - algebraic structures - exercises

Algebraic Structures - Exercises

NGUYEN CANH Nam1

1Faculty of Applied MathematicsDepartment of Applied Mathematics and Informatics

Hanoi University of [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, identityelement, inverse element.

a) x ∗ y = xy + 1

b) x ∗ y =12

xy

c) x ∗ y = |x |y


Solution

a) We havex ∗ y = xy + 1 = yx + 1 = y ∗ x for all x , y ∈ IR, so ∗ iscommutative.(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

2 ∗ e = 2e + 1 = 2⇔ e =12

. Moreover

3 ∗ e = 3e + 1 = 3⇔ e =23

.Hence, there doesn’t exist the identity element. Thereforethere doesn’t exist the inverse element of an element xneither.


Solution

b) We have

x ∗ y =12

xy =12

yx = y ∗ x for all x , y ∈ IR, so ∗ iscommutative.(x ∗ y) ∗ z = (

12

xy) ∗ z =14

xyz = x ∗ (12

yz) = x ∗ (y ∗ z), so∗ is associative.Denote e the identity element. Then

x ∗ e =12

xe = x ⇔ e = 2.

Consider an element x , denote x−1 the inverse element.

We then have x ∗ x−1 = e. So if x 6= 0 then x−1 =2x

. Theredoesn’t exist the inverse element of 0.


Solution

c) We have1 ∗ 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. Thereforethere doesn’t exist the inverse element of an element xneither.


Exercise 2

Suppose F is set of open intervals of IR and contains the emptyset ∅ (∅ 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)


Solution

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

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

Hence, in all cases, (a,b) ∩ (c,d) ∈ F , i.e., F is closedunder operator intersection (∩) on P(IR).

b) We have (1; 2), (2; 3) ∈ F , but (1; 2) ∪ (2; 3) 6∈ F . So F isn’tclosed under operator union (∪).


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 ) (thecollection of mapping from X to Y .)


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 .


Exercise 4

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


Solution

It is obvious that ∗ is closed. Moreover, for any x , y ∈ X , wehave{

(x ∗ y) ∗ z = x ∗ z = xx ∗ (y ∗ z) = x ∗ y = x

⇒ (x ∗ y) ∗ z = x ∗ (y ∗ z).

So ∗ is associative. Then (X , ∗) is a semigroup.If X has two distinct elements then (X , ∗) isn’t commutative andit doesn’t have the identity element.Indeed, we have x ∗ y = x 6= y = y ∗ x and e ∗ x = e 6= x for allx ∈ X , x 6= e.


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, wherexn = x · x · · · x .

b) Suppose (a · b)2 = a2 · b2. Could we deduce a · b = b · a?


Exercise 6

Show that the set IR[√

3] = {a = b√

3 | a,b ∈ IR,a2 + b2 6= 0}with ordinary multiplication is a group.


Solution.

First, we prove that the multiplication is a binary operator onIR[√

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 isassociative and 1 = 1 + 0

√3 is the identity element of IR[

√3].

Now, for any a + b√

3 ∈ IR[√

3], we have

x =a

a2 − 3b2 −b

a2 − 3b2

√3 ∈ IR[

√3] and

(a + b√

3)

(a

a2 − 3b2 −b

a2 − 3b2

√3)

=

(a

a2 − 3b2 −b

a2 − 3b2

√3)

(a + b√

3) = 1.

That meansa

a2 − 3b2 −b

a2 − 3b2

√3 is the inverse element of

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 whichis commutative, associative and invertible, and f : X → Y be abijective mapping. We equip X with a binary operator ◦ asfollows

x1 ◦ x2 = f−1(f (x1) ∗ f (x2)) for all x1, x2 ∈ X .

Prove that ◦ is commutative, associative and invertible.


Exercise 8

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


Solution.

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

e = e · f = f


Exercise 9

Define a new addition ◦ and multiplication ? on the integers ZZby

a ◦ b = a + b − 1 and a ? b = a + b − ab,

where the operations on the right-hand side of the equal signare ordinary addition, subtraction and multiplication. Prove that(ZZ, ◦, ?) is a commutative ring with identity.


Solution.

First we verify that ◦ and ? are binary operations:If a,b ∈ ZZ, then a ◦ b = a + b − 1 ∈ ZZ. Thus the set ZZ isclosed under addition.If a,b ∈ ZZ, then a ? b = a + b − ab ∈ ZZ. Thus ZZ is closedunder multiplication.

We note for later purposes that a ? b = b ? a from the definitionof ?. Next we need to show that (ZZ, ◦, ?) is a ring. We need toverify all the defining properties of a ring. Throughout thefollowing, let a,b, c ∈ ZZ.

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 ∈ (ZZ, ◦, ?)such that a ◦ 0Z = afor all a ∈ (ZZ, ◦, ?) :

a + 0Z = a⇔ a + 0Z − 1 = a⇔ 0Z = 1.

Thus 0Z = 1 is the zero element in (ZZ, ◦, ?).The equation a ◦ x = 0Z has a solution in (ZZ, ◦, ?) becauseit is equivalent to the equation a + x − 1 = 1 which has asolution x in ZZ.Associativity of multiplication: We conclude froma?(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.


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 theoperation ?, that (b ◦ c) ? a = (b ? a) ◦ (c ? a). This showsthe distributive law.


Solution.

To this point, we have shown that (ZZ, ◦, ?) is a ring. Next weverify 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 theidentity of (ZZ, ◦, ?).


Exercise 10

Let R be a ring. Then the following hold

If a + b = a + c, (a,b, c ∈ R) then b = c.


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


Exercise 11

Prove that the set ZZ[√

2] = {a + b√

2 | a,b ∈ ZZ} with theaddition and the multiplication of numbers is a commutative andunitary ring. Is it a field?


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.


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 ∈ ZZ 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.


Exercise 13

Let a,b and n be integers. If n divides a− b prove that n dividesa2 − b2 and a3 − b3. If n divides a + b prove that n dividesa3 + b3 but does not necessarily divide a2 + b2.


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.


Exercise 14

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


Exercise 15

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


Solution.

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


chapter v - algebraic structures - exercises

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