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Exercise 1.4

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}

1.i Determine whether or not * given by {a * b = a - b} gives a binary operation.

Consider two positive integers {a, b} ∈ Z+

Now, when {a \gt b}, then {a * b = a - b \gt 0}

⇒ a - b ∈ Z+

However, when {a \lt b}, then {a * b = a - b \lt 0}

{⇒ a - b} ∉ Z+ as {a - b \lt 0}

∴ * is not a binary operation on Z+

1.ii Determine whether or not * given by {a * b = ab} gives a binary operation.

Consider two positive integers {a, b} ∈ Z+

As we know, the product of two positive numbers is always positive.

⇒ As a ∈ Z+ and b ∈ Z+, the product ab also belongs to Z+. i.e., ab ∈ Z+ (∵ the product of two positive integers is also a positive integer).

∴ * is a binary operation on Z+

1.iii Determine whether or not * given by {a * b = ab^2} gives a binary operation.

Consider two positive integers {a, b} ∈ Z+

As b ∈ R, b^2 ∈ R (∵ the square of a real number is also a real number)

As it is given that a ∈ R and also as we’ve seen above that b^2 ∈ R,

{⇒ ab^2} ∈ R

∴ * is a binary operation on R

1.iv Determine whether or not * given by {a * b = |a - b|} gives a binary operation.

Consider two positive integers {a, b} ∈ Z+

Now when {a - b \le 0} then {|a - b| \gt 0}

And also when {a - b \geq 0} then {|a - b| \gt 0}

As it is given that a ∈ Z+ and b ∈ Z+, |a - b| ∈ Z+ (∵ The modulus of any number, whether it is positive or negative, is always positive)

∴ * is a binary operation on Z+

1.v Determine whether or not * given by {a * b = a} gives a binary operation.

Consider two positive integers {a, b} ∈ Z+

As a ∈ Z+, the result a * b = a ∈ Z+

∴ * is a binary operation on Z+

2. For each operation * defined below, determine whether * is binary, 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 = \dfrac{ab}{2}}

(iv)

On Z+, define {a * b = 2^{ab}}

(v)

On Z+, define {a * b = a^b}

(vi)

On R – {-1}, define {a * b = \dfrac{a}{b + 1}}

2.i For the operation * defined on Z as {a * b = a – b} determine whether * is binary, commutative or associative.

To check whether * on Z is binary.

Consider two integers {a, b} ∈ Z

As we know, the difference of two integers is always an integer.

So, as a ∈ Z and b ∈ Z, {a - b} ∈ Z

∴ * is a binary operation on Z

To check whether * on Z is commutative.

Consider two integers a, b ∈ Z

Now, {a * b = a - b} and {b * a = b - a}

As we know, a - b ≠ b - a

⇒ a * b ≠ b * a

∴ The binary operation * is not commutative on the set Z.

To check whether * on Z is associative.

Consider three integers {a, b} and c ∈ Z

We have {(a * b) * c = (a - b) * c}

{= (a - b) - c}

{= a - b - c}

Now, {a * (b * c) = a * (b - c)}

{= a - (b - c)}

{= a - b + c}

Clearly, a - b - c ≠ a - b + c

⇒ (a * b) * c ≠ a * (b * c)

∴ The binary operation * is associative on the set Z.

2.ii For the operation * defined on Q as {a * b = ab + 1} determine whether * is binary, commutative or associative.

To check whether * on Q is binary.

Consider two rational numbers {a, b} ∈ Q

As we know, the product of two rational numbers is also a rational number.

Thus, when a ∈ Q and b ∈ Q, ab ∈ Q

{⇒ ab + 1} ∈ Q

∴ * is binary operation on Q

To check whether * on Q is commutative.

Consider two rational numbers {a, b} ∈ Q

Now, {a * b = ab + 1}

And {b * a = ba + 1}

We know that the product of two rational numbers is commutative.

{⇒ ab = ba}

{⇒ ab + 1 = ba + 1}

{⇒ a * b = b * a}

∴ The binary operation * is commutative on the set Q

To check whether * on Q is associative.

consider three rational numbers a, b, c ∈ Q

Now {(a * b) * c = (ab + 1) * c}

{= (ab + 1) × c + 1}

{= abc + c + 1}

Now, {a * (bc + 1)}

{= a × (bc + 1) + 1}

{= abc + a + 1}

Clearly {abc + c + 1} ≠ {abc + a + 1}

{⇒ (a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q

2.iii For the operation * defined on Q as {a * b = \dfrac{ab}{2}} determine whether * is binary, commutative or associative.

To check whether * on Q is binary.

Consider two rational numbers {a, b} ∈ Q

As we know, the product of two rational numbers is also a rational number.

Thus, when a ∈ Q and b ∈ Q, ab ∈ Q

{⇒ \dfrac{ab}{2}} ∈ Q

∴ * on Q is binary.

To check whether * on Q is commutative.

Consider two elements {a, b} ∈ Q

Now {a * b = \dfrac{ab}{2}}

And {b * a = \dfrac{ba}{2}}

As we know, the product of two rational numbers is commutative.

{⇒ ab = ba}

{⇒ \dfrac{ab}{2} = \dfrac{ba}{2}}

{⇒ a * b = b * a}

∴ The binary operation * is commutative on the set Q.

To check whether * on Q is associative

Consider three rational numbers a, b, c ∈ Q

Now {(a * b) * c = \dfrac{ab}{2} * c}

{= \dfrac{\dfrac{ab}{2}\times{c}}{2}}

{= \dfrac{abc}{4}}

{a * (b * c) = a * \dfrac{bc}{2}}

{= \dfrac{a\times\dfrac{bc}{2}}{2}}

{= \dfrac{abc}{4}}

{⇒ (a * b) * c = a * (b * c)}

∴ The binary operation * is associative on the set Q.

2.iv For the operation * defined on Z+ as {a * b = 2^{ab}} determine whether * is binary, commutative or associative.

To check whether * on Z+ is binary.

Consider two positive integers {a, b} ∈ Z+

As we know, when a ∈ Z+ and b ∈ Z+, ab ∈ Z+

⇒ 2^{ab} ∈ Z+

∴ * on Z+ is binary.

To check whether * on Z+ is commutative.

Consider any two positive integers a, b ∈ Z+

As multiplication on Z+ is commutative,

We have, {ab = ba}

{⇒ 2^{ab} = 2^{ba}}

{⇒ a * b = b * a}

∴ The binary operation * is commutative on the set Z+.

To check whether * on Z+ is associative

Consider any three positive integers a, b, c ∈ Z+

Now {(a * b) * c = 2^{ab} * c}

{= 2^{2^{ab} × c}}

Also, {a * (b * c) = a * 2^{bc}}

{= 2^{a × 2^{bc}}}

So, {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Z+.

2.v For the operation * defined on Z+ as {a * b = a^b} determine whether * is binary, commutative or associative.

To check whether * on Z+ is binary.

Consider two positive integers a, b ∈ Z+

As we know, when a, b ∈ Z+, a^b ∈ Z+

∴ * on Z+ is binary.

To check whether * on Z+ is commutative.

Consider any two positive integers {a, b} ∈ Z+.

We have {a * b = a^b}

and {b * a = b^a}

Clearly {a * b} ≠ {b * a}

∴ The binary operation * is not commutative on the set Z+.

To check whether * on Z+ is associative.

Consider any three positive integers, {a, b, c} ∈ Z+

We have {(a * b) * c = a^b * c}

{= (a^b)^c}

{= a^{bc}}

Now, {a * (b * c) = a * b^c}

{= a^{b^c}}

So, {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Z+.

2.vi For the operation * defined on R – {-1} as {a * b =\dfrac{a}{b + 1}}, determine whether * is binary, commutative or associative.

To check whether * on R – {-1} is binary.

Consider any two real numbers {a, b} ∈ R – {-1}.

We have

{a * b = \dfrac{a}{b + 1}}

The above equation is valid for all {a, b} ∈ R – {-1}.

∴ * on R – {-1} is binary.

To check whether * on R – {-1} is commutative.

Consider two real numbers {a, b} ∈ R – {-1}

Now {a * b = \dfrac{a}{b + 1}}

{b * a = \dfrac{b}{a + 1}}

So, a * b ≠ b * a

∴ The binary operation * is not commutative on the set R – {-1}

To check whether * on R – {-1} is associative.

Consider any three real numbers, a, b, c ∈ R – {-1}

We have {(a * b) * c = \dfrac{a}{b + 1} * c}

{= \dfrac{\dfrac{a}{b + 1}}{c + 1}}

{= \dfrac{a}{(b + 1)\times(c + 1)}}

Now {a * (b * c) = a * \dfrac{b}{c + 1}}

{= \dfrac{a}{\dfrac{b}{c + 1} + 1}}

{= \dfrac{a}{\dfrac{b + c + 1}{c + 1}}}

{= \dfrac{{a}\times{(c + 1)}}{b + c + 1}}

{⇒ (a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set R – {-1}.

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 ∧

The set is given as {1, 2, 3, 4, 5} and the binary operation ∧ is given as a ∧ b = min {a, b}

The following is the operation table of the operation ∧ on the set {1, 2, 3, 4, 5}

∧

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

4. Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).

(i)

Compute {(2 * 3) * 4} and {2 * (3 * 4)}

(ii)

Is * commutative?

(iii)

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

(Hint: use the following table)

Table 1.2

*

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

4.(i) To compute {(2 * 3) * 4} and {2 * (3 * 4)}

{(2 * 3) * 4}

{= 1 * 4} {(∵ 2 * 3 = 1} from the table)

{= 1~(∵ 1 * 4 = 1} from the table)

And {2 * (3 * 4)}

{= 2 * 1} {(∵ 3 * 4 = 1} from the table)

{= 1~(∵ 2 * 1 = 1} from the table)

4.(ii) To check whether * is commutative

From the given table it can be noticed that n^{th} row = n^{th} column

{⇒ a * b = b * a}

∴ The binary operation * is commutative on the set {1, 2, 3, 4, 5} {∀ a, b} ∈ {1, 2, 3, 4, 5}

4.(iii) To compute {(2 * 3) * (4 * 5)}

{(2 * 3) * (4 * 5)}

{= 1 * 1} {(∵ 2 * 3 = 1} and {4 * 5 = 1} from the given table)

{= 1~(∵ 1 * 1 = 1} from the given table)

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.

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

To check whether *′ same as * defined in the exercise 4 above, let’s first build the table for *′.

Operation table for *′

*′

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

It can be noted that the operation tables for both * and *′ are same.

∴ The operations * and *′ are the same.

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 *?

i. To find {5 * 7}, {20 * 16}

We’ve {5 * 7} = L.C.M. of 5 and 7 = 35

2

20,

16

2

10,

8

5,

4

Now {20 * 16} = L.C.M. of 20 and 16 = 80

ii. To check whether * is commutative on N:

We have {a * b} = L.C.M of a and b

And {b * a} = L.C.M of b and a

As we know, L.C.M. of a and b = L.C.M. of b and a

{⇒ a * b = b * a}

∴ The binary operation * is commutative on the set N

iii. To check whether * is associative on N:

We have {(a * b) * c} = (L.C.M. of a and b) * c

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

= L.C.M. of a, b and c

Similarly {a * (b * c)} = a * (L.C.M. of b and c)

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

= L.C.M. of a, b and c

Thus we see that {(a * b) * c = a * (b * c)}

∴ The binary operation * is associative on the set N

iv. To find the identity of * in N:

An element e will be the identity of * if it satisfies the condition

{a * e = e * a = a}

⇒ L.C.M. of a and e = L.C.M of e and a = a

As we know, in the set of natural numbers N, the L.C.M. of any element a and 1 is a.

Thus, we have L.C.M. of a and 1 = L.C.M. of 1 and a {= a}

{⇒ a * 1} {= 1 * a} {= a~∀~a} ∈ N

∴ 1 is the identity element of the binary operation * on the set N.

To find which elements are invertible for the operation *

Any element a ∈ N is invertible with respect to the binary operation *, if there exists an element b ∈ N such that

{a * b} {= e} {= b * a}

Where e is the identity element under the binary operation * and as we have seen in the previous case, {e = 1}

As {a * b} {= b * a}

⇒ L.C.M of {a} and b = L.C.M. of b and a

This condition is satisfied only when both a and b are equal to 1.

∴ The only invertible element in the binary operation * on the set N is 1.

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.

The binary operation * is defined as

{a * b} = L.C.M of a and b where {a, b} ∈ {1, 2, 3, 4, 5}

The corresponding operation table will be as follows:

L.C.M

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

Considering {a = 3} and {b = 4}, We have

{a * b} = L.C.M. of a and b

= L.C.M. of 3 and 4

= 12

As 12 ∉ {1, 2, 3, 4, 5}, the operation * is not a binary operation on the set {1, 2, 3, 4, 5}

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?

To check whether * is commutative:

For the operation * to be commutative, we should have

{a * b = b * a}

Now, {a * b} = H.C.F of a and b

and {b * a} = H.C.F. of b and a

We know that H.C.F of a and b = H.C.F of b and a

{∴ a * b = b * a}

∴ The binary operation * is commutative.

To check whether * is associative.

Consider there elements a, b and c ∈ N

We have {a * (b * c)} {= a * (}H.C.F. of b and c) = H.C.F. of {a, b} and c.

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

This can be well understood by considering the following example.

Consider three numbers 3, 6 and 11 ∈ N. The calculation of their H.C.F’s will be as follows:

Factors of 3 = 1, 3

Factors of 6 = 1, 2, 3, 6

Factors of 11 = 1, 11

Now {(3 * 6) * 11}

= (H.C.F of 3 and 6) * 11

= 3 * 11

= H.C.F. of 3 and 11

= 1

and {3 * (6 * 11)}

= 3 * (H.C.F. of 6 and 11)

= 3 * 1

= H.C.F. of 3 and 1

= 1

Thus we see that {(3 * 6) * 11 = 3 * (6 * 11)}

∴ The binary operation * is associative on the set N.

Identity element of the binary operation *.

For the binary operation *, the identitiy element will be an element e ∈ N such that ∀ a ∈ N, {a * e = e * a = a}

⇒ H.C.F of a and e = H.C.F. of e and a {= a}

However, the set N does not have an element that satisfies the above condition.

∴ The binary operation * does not have an identity element.

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

(i)

{a * b = a – b}

(ii)

{a * b = a^2 + b^2}

(iii)

{a * b = a + ab}

(iv)

{a * b = (a – b)^2}

(v)

{a * b = \dfrac{ab}{4}}

(vi)

{a * b = ab^2}

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

(i).a To check whether {a * b = a - b} is commutative:

Consider two rational numbers a, b ∈ Q.

We have, {a * b = a - b}

and {b * a = b - a}

As {a - b} ≠ {b - a}, it is clear that {a * b} ≠ {b * a}.

∴ The binary operation * is not commutative on the set Q.

(i).b To check whether {a * b = a - b} is associative.:

Consider three rational numbers a, b, c ∈ Q.

We have, {(a * b) * c} {= (a - b) * c} {= a - b - c}

and {a * (b * c))} {= (a * (b - c)} {= a - (b - c)} {= a - b + c}

Clearly {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q.

(ii).a. To check whether {a * b = a^2 + b^2} is commutative.

Consider two rational numbers a, b ∈ Q.

We have {a * b = a^2 + b^2}

and {b * a = b^2 + a^2}

So, {a * b} {= a^2 + b^2} {= b^2 + a^2} {= b * a}

∴ The binary operation * is commutative on the set Q.

(ii).b To check whetehr {a * b = a^2 + b^2} is associative.

Consider three rational numbers {a, b, c} ∈ Q.

We have {(a * b) * c} {= (a^2 + b^2) * c} {= (a^2 + b^2)^2 + c^2}

and {a * (b * c)} {= a * (b^2 + c^2)} {= a^2 + (b^2 + c^2)^2}

So, it is clear that {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q.

(iii).a To check whetehr a * b = a + ab is commutative.

Consider two rational numbers a, b ∈ Q

We have {a * b = a + ab}

and {b * a = b + ba}

So, it is clear that {a * b} ≠ {b * a}.

∴ The binary operation * is not commutative on the set Q.

(iii).b To check whether {a * b = a + ab} is associative on the set Q.

Consider three rational numbers a, b, c ∈ Q

We have {(a * b) * c} {= (a + ab) * c} {= a + ab + (a + ab)c} {= a + ab + ac + abc}

and {a * (b * c)} {= a * (b + bc)} {= a + a(b + bc)} {= a + ab + abc}

So, clearly {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q.

(iv).a To check whether {a * b = (a - b)^2} is commutative.

Consider two rational numbers a, b ∈ Q

We have {a * b} {= (a - b)^2} {= a^2 - 2ab + b^2}

and {b * a} {= (b - a)^2} {= b^2 - 2ba + a^2} {= a^2 - 2ab + b^2}

So, it is clear that {a * b = b * a}

∴ The binary operation * is commutative on the set Q.

(iv).b To check whether {a * b = (a - b)^2} is associative.

Consider three rational numbers a, b and c ∈ Q

We have {(a * b) * c} {= (a - b)^2 * c} {= ((a - b)^2 - c)^2} {= (a^2 - 2ab + b^2 - c)^2}

and {a * (b * c)} {= a * (b - c)^2} {= (a - (b - c)^2)^2} {= (a - (b^2 - 2bc + c^2)^2} {= (a - b^2 + 2bc + c^2)^2}

So, clearly {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q.

(v).a To check whether {a * b = \dfrac{ab}{4}} is commutative.

Consider two rational numbers a, b ∈ Q

We have {a * b = \dfrac{ab}{4}}

and {b * a = \dfrac{ba}{4} = \dfrac{ab}{4}}

So, clearly {a * b = b * a}

∴ The binary operation * is commutative on the set Q

(v).b To check whether {a * b = \dfrac{ab}{4}} is associative.

Consider three rational numbers a, b and c ∈ Q

We have {(a * b) * c} {= \dfrac{ab}{4} * c} {= \dfrac{\dfrac{ab}{4} × c}{4}} {= \dfrac{abc}{16}}

and {a * (b * c)} {= a * \dfrac{bc}{4}} {= \dfrac{a × \dfrac{bc}{4}}{4}} {= \dfrac{abc}{16}}

So, it is clear that {(a * b) * c = a * (b * c)}

∴ The binary operation * is associative on the set Q.

(vi).a To check whether {a * b = ab^2} is commutative.

Consider two rational numbers {a, b} ∈ Q

We have {a * b = ab^2}

and {b * a = ba^2}

Clearly, {a * b} ≠ {b * a}

∴ The binary operation * is not commutative on the set Q

(vi).b To check whether {a * b = ab^2} is associative.

Consider three rational numbers {a, b} and c ∈ Q

We have {(a * b) * c} {= ab^2 * c} {= ab^2c^2}

and {a * (b * c)} {= a * bc^2} {= a × (bc^2)^2} {= ab^2c^4}

So, it is clear that {(a * b) * c} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set Q.

A check for whether these binary operations are commutative and/or associative is summarised in the following table.

*

{a * b}

Commutative?

Associative?

(i)

{a - b}

❌

❌

(ii)

{a^2 + b^2}

✔

❌

(iii)

{a + ab}

❌

❌

(iv)

{(a - b)^2}

✔

❌

(v)

{\dfrac{ab}{4}}

✔

✔

(vi)

{ab^2}

❌

❌

10. Find which of the operations given above has identity.

Note: If the birnary operation * is not commutative, then we can directly conclude that there is no identity element for the binary operation * on the set Q. However, if the binary operation * is binary, it does not necessarily mean that there is an identity. We have to ensure that it satisfy the condition {a * e = e * a = a}

10.i. Identity element for the binary operation * defined by {a * b = a - b}

An element e ∈ Q will be an identity element of the binary operation * on the set Q, if for any arbitrary element a ∈ Q, we have

{a * e = e * a = a}

Now, {a * e} {= a - e}

and {e * a} {= e - a}

Obviously there is no element e ∈ Q such that {a - e = e - a = a}

∴ The binary operation * does not have an identity element on the set Q.

10.ii. Identity element for the binary operation * defined by {a * b = a^2 + b^2}

An element e ∈ Q will be an identity element of the binary operation * on the set Q, if for any arbitrary element a ∈ Q, we have

{a * e = e * a = a}

Now, {a * e} {= a^2 + e^2}

and {e * a} {= e^2 + a^2}

Obviously there is no element e ∈ Q such that {a^2 + e^2 = e^2 + a^2 = a}

∴ The binary operation * does not have an identity element on the set Q.

10.iii. Identity element for the binary operation * defined by {a * b = a + ab}

An element e ∈ Q will be an identity element of the binary operation * on the set Q, if for any arbitrary element a ∈ Q, we have

{a * e = e * a = a}

Now, {a * e} {= a + ae}

and {e * a} {= e + ea}

Obviously there is no element e ∈ Q such that {a + ae = e + ea = a}

∴ The binary operation * does not have an identity element on the set Q.

10.iv. Identity element for the binary operation * defined by {a * b = (a - b)^2}

{a * e = e * a = a}

Now, {a * e} {= (a - e)^2}

and {e * a} {= (e - a)^2}

Obviously there is no element e ∈ Q such that {(a - e)^2 = (e - a)^2 = a}

∴ The binary operation * does not have an identity element on the set Q.

10.v. Identity element for the binary operation * defined by {a * b = \dfrac{ab}{4}}

{a * e = e * a = a}

Now, {a * e} {= \dfrac{ae}{4}}

and {e * a} {= \dfrac{ea}{4}}

Obviously the element 4 ∈ Q exists such that {\dfrac{a × 4}{4} = \dfrac{4 × a}{4} = a}

∴ 4 is the identity element for the binary operation * on the set Q.

10.vi. Identity element for the binary operation * defined by {a * b = ab^2}

{a * e = e * a = a}

Now, {a * e} {= ae^2}

and {e * a} {= ea^2}

Obviously there is no element e ∈ Q such that {ae^2 = ea^2 = a}

∴ The binary operation * does not have an identity element on the set Q.

Whether the binary operation * has the identity element or not is summarized in the following table.

S. No.

{a * b}

Has Identity?

(i)

{a - b}

❌

(ii)

{a^2 + b^2}

❌

(iii)

{a + ab}

❌

(iv)

{(a - b)^2}

❌

(v)

{\dfrac{ab}{4}}

Yes – 4

(vi)

{ab^2}

❌

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.

To check whether the binary operation * defined by {(a, b) * (c, d) = (a + c, b + d)} is commutative.

Consider four elements a, b, c, d ∈ N, such that {(a, b), (c, d)} ∈ A where A = N × N

Now, {(a, b) * (c, d)} {= (a + c, b + d)}

and {(c, d) * (a, b)} {= (c + a, d + b)} {= (a + c, b + d)} {(∵ c + a = a + c} and {d + b = b + d)}

So, clearly {(a, b) * (c, d)} {= (c, d) * (a, b)} {∀~a, b, c, d} ∈ N

∴ The binary operation * is commutative on the set A.

To check whether teh binary operation * defined by {(a, b) * (c, d) = (a + c, b + d)} is associative.

Consider six elements {a, b, c, d, e, f} ∈ N, such that {(a, b), (c, d), (e, f)} ∈ A

Now {((a, b) * (c, d)) * (e, f)} {= (a + c, b + d) * (e, f)} {= (a + c + e, b + d + f)}

and {(a, b) * ((c, d) * (e, f))} {= (a, b) * (c + e, d + f)} {= (a + c + e, b + d + f)}

So, obviously {((a, b) * (c, d)) * (e, f)} {= (a, b) * ((c, d) * (e, f))}

∴ The binary operation * is associative on the set A

To find the identity element of the binary operation defined by {(a, b) * (c, d) = (a + c, b + d)}

The identity element of the binary operation * should be an element {(e, e)} such that {(a, b) * (e, e) = (e, e) * (a, b) = (a, b)} where {a, b, e} ∈ N such that {(a, b), (e, e)} ∈ N

We have {(a, b) * (e, e)} {= (a + e, b + e)}

We’ll have {(a + e, b + e) = (a, b)} only when {e = 0}.

Similary, {(e, e) * (a, b)} {= (e + a, e + b)}

We’ll have {(a + e, b + e) = (a, b)} only when {e = 0}

However, as 0 ∉ N, the identity element {(e, e)} for the binary operation * does not existing on the set A = N × N.

∴ The binary operation * does not have an identity element {(e, e)} on the set A = N × N

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}

To check the truth value of the statement (i)

Let’s assume that the arbitrary binary operation * is defined as {a * a = a × a} for an arbitrary number a ∈ N

Now {a * a = a × a = a^2}

As we know, a^2 ≠ a.

∴ The arbitrary binary operation defined as {a * a = a~∀~a} ∈ N is not true.

∴ The statement (i) is False.

To check the truth value of the statement (ii)

It is given in the statement that the binary operation * is commutative.

This implies that for any two elements a, b ∈ N, we have {a * b = b * a}

Now, consider any three elements a, b and c ∈ N.

We have

{a * (b * c)}

{= (b * c) * a} {(∵ *} is commutative)

{= (c * b) * a} {(∵ *} is commutative, we’ve {b * c = c * b)}

{⇒ a * (b * c) = (c * b) * a}

∴ The given statement (ii) is True.

13. Consider a binary operation * on N defined as {a * b = a^3 + b^3}. 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?

To check whether the binary operation * defined as {a * b = a^3 + b^3} is commutative.

Consider any natural numbers a, b ∈ N

We have {a * b = a^3 + b^3}

and {b * a} {= b^3 + a^3} {= a^3 + b^3}

∴ The binary operation * is commutative on the set N

To check whether the binary operation * defined as {a * b = a^3 + b^3} is associative.

Consider any three natural numbers {a, b, c} ∈ N

we have {(a * b) * c} {= (a^3 + b^3) * c} {= (a^3 + b^3)^3 + c^3}

and {a * (b * c)} {= a * (b^3 + c^3)} {= a^3 + (b^3 + c^3)^3}

So, clearly {(a * b) * c)} ≠ {a * (b * c)}

∴ The binary operation * is not associative on the set N.

To summarise, the binary operation * is commutative but not associative on the set N.

So, the option B is the correct answer.