# Class 12 NCERT Solutions – Mathematics Part I – Chapter 1 Relations and Functions – Exercise 1.4 | Set 1

### (i) On Z+, define ∗ by a ∗ b = a – b

Solution:

If a, b belongs to Z+

a * b = a – b which may not belong to Z+

For eg:  1 – 3 = -2 which doesn’t belongs to Z+

Therefore, * is not a Binary Operation on Z+

### (ii) On Z+, define * by a * b = ab

Solution:

If a, b belongs to Z+

a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

### (iii) On R, define * by a * b = ab²

Solution:

If a, b belongs to R

a * b = ab which belongs to R

Therefore, * is Binary Operation on R

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

Solution:

If a, b belongs to Z+

a * b = |a – b| which belongs to Z+

Therefore, * is Binary Operation on Z+

### (v) On Z+, define * by a * b = a

Solution:

If a, b belongs to Z+

a * b = a which belongs to Z+

Therefore, * is Binary Operation on Z+

### (i) On Z, define a * b = a – b

Solution:

a) Binary:

If a, b belongs to Z

a * b = a – b which belongs to Z

Therefore, * is Binary Operation on Z

b) Commutative:

If a, b belongs to Z, a * b = b * a

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – b + c

RHS = (a – b) * c = a – b- c

Since, LHS is not equal to RHS

Therefore, * is not Associative

### (ii) On Q, define a * b = ab + 1

Solution:

a) Binary:

If a, b belongs to Q, a * b = ab + 1 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:

If a, b belongs to Q, a * b = b * a

LHS = a * b = ab + 1

RHS = b * a = ba + 1 = ab + 1

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc + 1) = abc + a + 1

RHS = (a * b) * c = abc + c + 1

Since, LHS is not equal to RHS

Therefore, * is not Associative

### (iii) On Q, define a ∗ b = ab/2

Solution :

a) Binary:

If a, b belongs to Q, a * b = ab/2 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:

If a, b belongs to Q, a * b = b * a

LHS = a * b = ab/2

RHS = b * a = ba/2

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc/2) = (abc)/2

RHS = (a * b) * c = (ab/2) * c = (abc)/2

Since, LHS is equal to RHS

Therefore, * is Associative

### (iv) On Z+, define a * b = 2ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = 2ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = 2ab

RHS = b * a = 2ba = 2ab

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)

RHS = (a * b) * c = 2ab * c = 22abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

### (v) On Z+, define a * b = ab

Solution:

a) Binary:

If a, b belongs to Z+, a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

b) Commutative:

If a, b belongs to Z+, a * b = b * a

LHS = a * b = ab

RHS = b * a = ba

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * bc = ab^c

RHS = (a * b) * c = ab * c = abc

Since, LHS is not equal to RHS

Therefore, * is not Associative

### (vi) On R – {– 1}, define a ∗ b = a / (b + 1)

Solution:

a) Binary:

If a, b belongs to R, a * b = a / (b+1) which belongs to R

Therefore, * is Binary Operation on R

b) Commutative:

If a, b belongs to R, a * b = b * a

LHS = a * b = a / (b + 1)

RHS = b * a = b / (a + 1)

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:

If a, b, c belongs to A, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1

RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)

Since, LHS is not equal to RHS

Therefore, * is not Associative

Solution:

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

(Hint: use the following table)

### (i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

Solution:

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

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

### (ii) Is ∗ commutative?

Solution:

The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.

### (iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

Solution:

(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.

Solution:

Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.

We see that the operation *’ is the same as the operation * in Exercise 4 above.

### (i) 5 ∗ 7, 20 ∗ 16

Solution:

If a, b belongs to N

a * b = LCM of a and b

5 * 7 = 35

20 * 16 = 80

### (ii) Is ∗ commutative?

Solution:

If a, b belongs to N

LCM of a * b = ab

LCM of b * a = ba = ab

a*b = b*a

Thus, * binary operation is commutative.

### (iii) Is ∗ associative?

Solution:

a * (b * c) = LCM of a, b, c

(a * b) * c = LCM of a, b, c

Since, a * (b * c) = (a * b) * c

Thus, * binary operation is associative.

### (iv) Find the identity of ∗ in N

Solution:

Let ‘e’ is an identity

a * e = e * a, for a belonging to N

LCM of a * e = a, for a belonging to N

LCM of e * a = a, for a belonging to N

e divides a

e divides 1

Thus, e = 1

Hence, 1 is an identity element

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

Solution:

a * b = b * a = identity element

LCM of a and b = 1

a = b = 1

only ‘1’ is invertible element in N.