# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

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

Solution:

(i) On Z+, define ∗ by a ∗ b = a – b

On Z+ = {1, 2,3 , 4, 5,…….}
Let a = 1 and b = 2
Therefore, a ∗ b = a – b = 1 – 2 = -1 ∉ Z+
operation * is not a binary operation on Z+.

(ii) On Z+, define ∗ by a ∗ b = ab
On Z+ = {1, 2,3 , 4, 5,…….}
Let a = 2 and b = 3
Therefore, a ∗ b = a b = 2 * 3 = 6 ∈ Z+
operation * is a binary operation on Z+

(iii) On R, define ∗ by a ∗ b = ab2
R = { – ∞, ……, -1, 0, 1, 2,……, ∞}
Let a = 1.2 and b = 2

Therefore, a ∗ b = ab2 = (1.2) x 22 = 4.8 ∈ R
Operation * is a binary operation on R.

(iv) On Z+, define ∗ by a ∗ b = | a – b |
On Z+ = {1, 2,3 , 4, 5,…….}
Let a = 2 and b = 3
Therefore, a ∗ b = a b = 2 * 3 = 6 ∈ Z+
operation * is a binary operation on Z+

(v) On Z+, define ∗ by a ∗ b = a
On Z+ = {1, 2, 3, 4, 5,…….}
Let a = 2 and b = 1
Therefore, a ∗ b = 2 ∈ Z+
Operation * 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 = ab/2
(iv) On Z+, define a ∗ b = 2ab
(v) On Z+, define a ∗ b = ab
(vi) On R – {– 1}, define a ∗ b = a/(b+1).

Solution:
(i) On Z, define a ∗ b = a – b
Step 1: Check for commutative
Consider ∗ is commutative, then
a ∗ b = b * a
Which means, a – b = b – a (not true)
Therefore, ∗ is not commutative.
Step 2: Check for Associative.
Consider ∗ is associative, then
(a ∗ b)* c = a * (b * c)
LHS = (a ∗ b)* c = (a – b) * c
= a – b – c
RHS = a * (b * c) = a – (b- c)
= a – (b – c)
= a – b + c
This implies LHS ≠ RHS
Therefore, ∗ is not associative.

(ii) On Q, define a ∗ b = ab + 1
Step 1: Check for commutative
Consider ∗ is commutative, then
a ∗ b = b * a
Which means, ab + 1 = ba + 1
or ab + 1 = ab + 1 (which is true)

a ∗ b = b * a for all a, b ∈ Q
Therefore, ∗ is commutative.
Step 2: Check for Associative.
Consider ∗ is associative, then
(a ∗ b)* c = a * (b * c)
LHS = (a ∗ b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
RHS = a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
This implies LHS ≠ RHS
Therefore, ∗ is not associative.

( 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 ∗?

Solution:

(i) 5 ∗ 7 = LCM of 5 and 7 = 35
20 ∗ 16 = LCM of 20 and 16 = 80
(ii) Is ∗ commutative?
a ∗ b = L.C.M. of a and b
b ∗ a = L.C.M. of b and a
a ∗ b = b ∗ a
Therefore ∗ is commutative.
(iii) Is ∗ associative?
For a,b, c ∈ N
(a ∗ b) * c = (L.C.M. of a and b) * c = L.C.M. of a, b and c
a ∗ (b * c) = a * (L.C.M. of b and c) = L.C.M. of a, b and c
(a ∗ b) * c = a ∗ (b * c)
Therefore, operation ∗ associative.
(iv) Find the identity of ∗ in N
Identity of * in N = 1
because a * 1 = L.C.M. of a and 1 = a
(v) Which elements of N are invertible for the operation ∗?
Only the element 1 in N is invertible for the operation * because 1 * 1/1 = 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?

Solution:
The operation ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b

Suppose, a = 2 and b = 3
2 * 3 = L.C.M. of 2 and 3 = 6
But 6 does not belongs to the set A.
Therefore, given operation * is not a binary operation.

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?
Solution:
The operation ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b
a * b = H.C.F. of a and b = H.C.F. of b and a = b * a
Therefore, operation * is commutative.
Again, (a *b)*c = (HCF of a and b) * c = HCF of (HCF of a and b) and c = a * (b *c)
(a *b)*c = a * (b *c)
Therefore, the operation is associative.
Now, 1 * a = a * 1 ≠ a
Therefore, there does not exist any 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 = a2 + b2
(iii) a ∗ b = a + ab
(iv) a ∗ b = (a – b)2
(v) a ∗ b = ab/4
(vi) a ∗ b = ab2
Find which of the binary operations are commutative and which are associative.

Solution:
(i) a ∗ b = a – b
a ∗ b = a – b = – (b – a) = -b * c ≠ b * a (Not commutative)
(a * b) * c = (a – b) * c = (a – (b – c) = a – b + c ≠ a * (b *c) (Not associative)

(ii) a ∗ b = a2 + b2
a ∗ b = a2 + b2 = b2 + a2 = b * a (operation is commutative)
Check for associative:
(a * b) * c = (a2 + b2) * c2 = (a2 + b2) + c2a * (b *c)

= a * (b2 + c2) = a2* (b2 + c2)2
(a * b) * c ≠ a * (b *c) (Not associative)
(iii) a ∗ b = a + ab
a ∗ b = a + ab = a(1 + b)
b * a = b + ba = b (1+a)
a ∗ b ≠ b * a
The operation * is not commutative
Check for associative:
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c
a * (b *c) = a * (b + bc ) = a + a(b + bc)
(a * b) * c ≠ a * (b *c)
The operation * is not associative

(iv) a ∗ b = (a – b)2
a ∗ b = (a – b)2
b * a = (b – a)2
a ∗ b = b * a
The operation * is commutative.

Check for associative:
(a * b) * c = (a – b)2
* c = ((a – b)2 – c)2
a * (b *c) = a * (b – c )2
= (a – (b – c)2)2
(a * b) * c ≠ a * (b *c)
The operation * is not associative

(v) a ∗ b = ab/4
b * a = ba/2 = ab/2
a ∗ b = b * a
The operation * is commutative.
Check for associative:
(a * b) * c = ab/4 * c = abc/16
a * (b *c) = a * (bc/4) = abc/16
(a * b) * c = a * (b *c)
The operation * is associative.

(vi) a ∗ b = ab2
b ∗ a = ba2
a ∗ b ≠ b ∗ a
The operation * is not commutative.
Check for associative:
(a * b) * c = (ab2) * c = ab2 c2
a * (b *c) = a * (b c2) = ab2 c4
(a * b) * c ≠ a * (b *c)
The operation * is not associative.

10. Find which of the operations given above has identity.
Solution: Let I be the identity.
(i) a * I = a – I ≠ a
(ii) a * I = a2 – I2 ≠ a
(iii) a * I = a + a I ≠ a
(iv) a * I = (a – I) 2 ≠ a
(v) a * I = aI/4 ≠ a
Which is only possible at I = 4 i.e. a * I = aI/4 = a(4)/4 = a
(vi) a * I = a I2 ≠ a
Above identities does not have identity element except (V) at b = 4.

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.
Solution: A = N x N and * is a binary operation defined on A.
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
The operation * is commutative
Again, ((a, b) * (c, d)) * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)
(a, b) * ((c, d)) * (e, f)) = (a, b) * (c+e, e+f) = (a+c+e, b+d+f)
=> ((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d)) * (e, f))
The operation * is associative.
Let (e, f) be the identity function, then
(a, b) * (e, f) = (a + e, b + f)

For identity function, a = a + e => e = 0 and b = b + f => f = 0
As zero is not a part of set of natural numbers. So identity function does not exist.
As 0 ∉ N, therefore, identity-element does not exist.

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
Solution:
(i) Given: * being a binary operation on N, is defined as a * a = a ∀ a ∈ N
Here operation * is not defined, therefore, the given statement is not true.
(ii) Operation * being a binary operation on N.
c * b = b * c
(c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * c) = (c * b) * a, therefore the given statement is true

13. Consider a binary operation ∗ on N defined as a ∗ b = a3 + b3
. Choose the correct
(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?
Solution:
A binary operation ∗ on N defined as a ∗ b = a3 + b3
,
Also, a ∗ b = a3 + b3 = b3 + a3 = b * a
The operation * is commutative.
Again, (a ∗ b)*c = (a3 + b3) * c

= ( a3 + b3)3+ c3

a * (b * c) = a * (b3 + c3)

= a3 + (b3 + c3)3
(a ∗ b)*c ≠ a * (b * c)
The operation * is not associative.
Therefore, option (B) is correct.