# 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 = ab ^{2}**

R = { – ∞, ……, -1, 0, 1, 2,……, ∞}

Let a = 1.2 and b = 2

Therefore, a ∗ b = ab^{2} = (1.2) x 2^{2} = 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 = 2^{ab}**

**(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?**

**Justify your answer.**

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

a ∗ b = a^{2} + b^{2} = b^{2} + a^{2} = b * a (operation is commutative)

Check for associative:

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

= a * (b^{2} + c^{2}) = a2* (b^{2} + c^{2})^{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 = ab ^{2}**

b ∗ a = ba

^{2}

a ∗ b ≠ b ∗ a

The operation * is not commutative.

Check for associative:

(a * b) * c = (ab

^{2}) * c = ab

^{2}c

^{2}

a * (b *c) = a * (b c

^{2}) = ab

^{2}c

^{4}

(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 = a^{2} – I^{2} ≠ 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 I^{2} ≠ 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 = 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?

Solution:

A binary operation ∗ on N defined as a ∗ b = a^{3} + b^{3}

,

Also, a ∗ b = a^{3} + b^{3} = b^{3} + a^{3} = b * a

The operation * is commutative.

Again, (a ∗ b)*c = (a^{3} + b^{3}) * c

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

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

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

(a ∗ b)*c ≠ a * (b * c)

The operation * is not associative.

Therefore, option (B) is **correct**.