excersie 1.4

Question 1:

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z⁺, define * by a * b = a − b

(ii) On Z⁺, define * by a * b = ab

(iii) On R, define * by a * b = ab²

(iv) On Z⁺, define * by a * b = |a − b|

(v) On Z⁺, define * by a * b = a

solutions =

 

(i) On Z⁺, * is defined by a * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2 
= −1 ∉ Z⁺.

(ii) On Z⁺, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z⁺, there is a unique element ab in Z⁺.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z⁺.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R.

This means that * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

(iv) On Z⁺, * is defined by a * b = |a − b|.

It is seen that for each a, b ∈ Z⁺, there is a unique element |a − b| in Z⁺.

This means that * carries each pair (a, b) to a unique element a * b = 
|a − b| in Z⁺.

Therefore, * is a binary operation.

(v) On Z⁺, * is defined by a * b = a.

* carries each pair (a, b) to a unique element a * b = a in Z⁺.

Therefore, * is a binary operation.

 

Question 2:

For each binary operation * defined below, determine whether * is commutative or associative.

(i) On Z, define a * b = a − b

(ii) On Q, define a * b = ab + 1

(iii) On Q, define a * b 

(iv) On Z⁺, define a * b = 2^ab

(v) On Z⁺, define a * b = a^b

(vi) On R − {−1}, define                                                                                                  solutions

(i) On Z, * is defined by a * b = a − b.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1.

It is known that:

ab = ba " a, b ∈ Q

⇒ ab + 1 = ba + 1 " a, b ∈ Q

⇒ a * b = a * b " a, b ∈ Q

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by a * b 

It is known that:

ab = ba " a, b ∈ Q

⇒ " a, b ∈ Q

⇒ a * b = b * a " a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

∴

Therefore, the operation * is associative.

(iv) On Z⁺, * is defined by a * b = 2^ab.

It is known that:

ab = ba " a, b ∈ Z⁺

⇒ 2^ab = 2^ba " a, b ∈ Z⁺

⇒ a * b = b * a " a, b ∈ Z⁺

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z⁺

Therefore, the operation * is not associative.

(v) On Z⁺, * is defined by a * b = a^b.

It can be observed that:

 and 

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z⁺

Therefore, the operation * is not commutative.

It can also be observed that:

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z⁺

Therefore, the operation * is not associative.

(vi) On R, * − {−1} is defined by

It can be observed that and 

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ R − {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ R − {−1}

Therefore, the operation * is not associative.

 

Question 3:

Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by a ∨b = min {a, b}. Write the operation table of the operation∨.

solutions =

 

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as a ∨ b = min {a, b}

" a, b ∈ {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation ∨ can be given as:

∨	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

 

Question 4:

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(ii) Is * commutative?

(iii) Compute (2 * 3) * (4 * 5).

(Hint: use the following table)

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

solutions =

(i) (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1

∴(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5:

Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer

solutions=

 

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b.

The operation table for the operation *′ can be given as:

*′	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

We observe that the operation tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

Question 6:

Let * be the binary operation on N given by a * b = L.C.M. of a and b. Find

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

(iii) Is * associative? (iv) Find the identity of * in N

(v) Which elements of N are invertible for the operation *?

solutions=

 

The binary operation * on N is defined as a * b = L.C.M. of a and b.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:

L.C.M of a and b = L.C.M of b and a " a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

(iii) For a, b, c ∈ N, we have:

(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c

a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a " a ∈ N

⇒ a * 1 = a = 1 * a " a ∈ N

Thus, 1 is the identity of * in N.

(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a.

Here, e = 1

This means that:

L.C.M of a and b = 1 = L.C.M of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

 

Question 7:

Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.

solutions=

 

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b.

Then, the operation table for the given operation * can be given as:

*	1	2	3	4	5
1	1	2	3	4	5
2	2	2	6	4	10
3	3	6	3	12	15
4	4	4	12	4	20
5	5	10	15	20	5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

 

Question 8:

Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

solutions=

The binary operation * on N is defined as:

a * b = H.C.F. of a and b

It is known that:

H.C.F. of a and b = H.C.F. of b and a " a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ N, we have:

(a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c

a *(b * c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a a ∈ N.

But this relation is not true for any a ∈ N.

Thus, the operation * does not have any identity in N.

 

Question 9:

Let * be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a − b (ii) a * b = a² + b²

(iii) a * b = a + ab (iv) a * b = (a − b)²

(v) (vi) a * b = ab²

Find which of the binary operations are commutative and which are associative.

solutions=

 

(i) On Q, the operation * is defined as a * b = a − b.

It can be observed that:

and 

∴ ; where

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(ii) On Q, the operation * is defined as a * b = a² + b².

For a, b ∈ Q, we have:

∴a * b = b * a

Thus, the operation * is commutative.

It can be observed that:

Thus, ,the operation * is not associative.

(iii) On Q, the operation * is defined as a * b = a + ab.

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(iv) On Q, the operation * is defined by a * b = (a − b)².

For a, b ∈ Q, we have:

a * b = (a − b)²

b * a = (b − a)² = [− (a − b)]² = (a − b)²

∴ a * b = b * a

Thus, the operation * is commutative.

It can be observed that:

Thus, the operation * is not associative.

(v) On Q, the operation * is defined as 

For a, b ∈ Q, we have:

∴ a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have:

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

(vi) On Q, the operation * is defined as a * b = ab²

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

 

Question 10:

Find which of the operations given above has identity.

solutions=

An element e ∈ Q will be the identity element for the operation * if

a * e = a = e * a, a ∈ Q.

However, there is no such element e ∈ Q with respect to each of the six operations satisfying the above condition.

Thus, none of the six operations has identity.

Question 11:

Let A = N × N and * be the binary operation on A defined by

(a, b) * (c, d) = (a + c, b + d)

Show that * is commutative and associative. Find the identity element for * on A, if any.

solutions=

 

A = N × N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

Therefore, the operation * is associative.

An element will be an identity element for the operation * if

, i.e., which is not true for any element in A.

Therefore, the operation * does not have any identity element.

 

Question 12:

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N, a * a = a  a * N.

(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a

solutions=

 

(i) Define an operation * on N as:

a * b = a + b a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

 

Question 13:

Consider a binary operation * on N defined as a * b = a³ + b³. Choose the correct answer.

(A) Is * both associative and commutative?

(B) Is * commutative but not associative?

(C) Is * associative but not commutative?

(D) Is * neither commutative nor associative?

solutions=

 

On N, the operation * is defined as a * b = a³ + b³.

For, a, b, ∈ N, we have:

a * b = a³ + b³ = b³ + a³ = b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.