# NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.4

#### NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.4 Math

Page No: 24

**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

(ii) On Z

(iii) On R, define ∗ by a ∗ b = ab

(iv) On Z

(v) On Z

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 = ab

^{2}(iv) On Z

^{+}, define ∗ by a ∗ b = |a – b|(v) On Z

^{+}, define ∗ by a ∗ b = a**Answer**

(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 = ab

^{2}.
It is seen that for each a, b ∈ R, there is a unique element ab

^{2}in R.
This means that * carries each pair (a, b) to a unique element a * b = ab

^{2}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.

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

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

Therefore, * is a binary operation.

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 = a

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

(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 commut

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 for all a, b ∈ Q

⇒ ab + 1 = ba + 1 for all a, b ∈ Q

⇒ a * b = a * b for all 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 =ab/2

It is known that: ab = ba for all a, b ∈ Q

⇒ab/2 = ba/2

for all a, b ∈ Q

⇒ a * b = b * a for all a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have

Exercise 1.1

Exercise 1.2

Exercise 1.3

Miscellaneous Exercise

Class 12 Math NCERT Solutions

**Answer**(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 commut

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 for all a, b ∈ Q

⇒ ab + 1 = ba + 1 for all a, b ∈ Q

⇒ a * b = a * b for all 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 =ab/2

It is known that: ab = ba for all a, b ∈ Q

⇒ab/2 = ba/2

for all a, b ∈ Q

⇒ a * b = b * a for all a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have

∴ (a*b)*c = a*(b*c), where a, b, c ∈ Q

Therefore, the operation * is associative.

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

^{ab}.
It is known that: ab = ba for all a, b ∈ Z+

⇒ 2

^{ab}= 2^{ba}for all a, b ∈ Z+
⇒ a * b = b * a for all a, b ∈ Z+

Therefore, the operation * is commutative.

It can be observed that

(1 ∗ 2) ∗ 3 = 2

^{1×2}∗ 3 = 4 ∗ 3 =2^{4×3}= 2^{12}and
1 ∗ (2 ∗ 3) = 1 ∗ 2

^{2×3}= 1 ∗ 2^{6}= 1 ∗ 64 = 2^{1×64}= 2^{64}
∴ (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

1*2 = 1

^{2}= 1 and 2*1 = 2^{1}= 2
∴ 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 = 8 ∗ 4 = 8^{4}= 2^{12}and
2 ∗ (3 ∗ 4) = 2 ∗ 3

^{4}= 2 ∗ 81 = 2^{81}
∴ (2 * 3) * 4 ≠ 2 * (3 * 4), where 2, 3, 4 ∈ Z+

Therefore, the operation * is not associative.

(vi) On R, * −{−1} is defined by a ∗ b = a/b+1

It can be observed that

1 ∗ 2 = 1/2+1 = 1/3 and 2 ∗ 1 = 2/1+1 = 2/2 = 1

∴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.

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 ∧.

**Answer**

The binary operation ∧ on the set {1, 2, 3, 4, 5} is defined as a ∧ b = min {a, b} for all a, b ∈ {1, 2, 3, 4, 5}.

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

Page No. 25

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)

**Answer**

(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

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.

**Answer**

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:

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

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

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 ∗?

**Answer**

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 for all 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 for all a ∈ N

⇒ a * 1 = a = 1 * a for all 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 *.

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.

**Answer**

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:

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.

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?

**Answer**

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 for all 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 for all a ∈ N.

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

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

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 = ab/4

(vi) a ∗ b = ab

^{2}**Answer**

(i) On Q, the operation * is defined as a * b = a − b. It can be observed that:

Thus, the operation * is not associative.

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

^{2}+ b^{2}.
For a, b ∈ Q, we have

a * b = a

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

Thus, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1

^{2}+ 2^{2}) * 3 = (1 + 4) * 3 = 5 * 3 = 5^{2}+ 3^{2}= 3^{4}and
1 * (2 * 3) = 1 * (2

^{2}+ 3^{2}) = 1 * (4 + 9) = 1 * 13 = 1^{2}+ 13^{2}=170
∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

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

It can be observed that

1 * 2 = 1 + 1×2 = 1 + 2 = 3

2 * 1 = 2 + 2×1 = 2 + 2 = 4

∴ 1 * 2 ≠ 2 * 1, where 1, 2 ∈ Q

Thus, the operation * is not commutative.

It can also be observed that

(1 * 2) * 3 = (1+ 1×2) * 3 = (1 + 2) * 3 = 3 * 3 = 3 + 3×3 = 3 + 9 = 12 and

1 * (2 * 3) = 1 * (2 + 2×3 ) = 1 * (2 + 6) = 1 * 8 = 1 + 1×8 =1 + 8 = 9

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

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

(1 * 2) * 3 = (1 - 2)

^{2}* 3 = (-1)^{2}* 3 = 1 * 3 = (1 - 3)^{2}= (-2) = 4 and
1 * (2 * 3) = 1 * (2 - 3)

^{2}= 1 * (-1)^{2}= 1 * 1 = (1 - 1)^{2}= 0
∴ (1 * 2) * 3 ≠ 1 * (2 * 3), where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(v) On Q, the operation * is defined as a*b = ab/4.

For a, b ∈ Q, we have a*b = ab/4 = ba/4 = b*a

∴ a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have

∴ (a * b) * c = a * (b * c), where a, b, c ∈ Q

Thus, the operation * is associative.

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

^{2}
It can be observed that

∴ 1/2 * 1/3 ≠ 1/3 * 1/2, where 1/2 and 1/3 ∈ Q

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.

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

**Answer**

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

if a * e = a = e * a, for all 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.

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.

**Answer**

A = N × N and * 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

[(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)

∴ [(a, b) ∗ (c, d)] ∗ (e, f) = (a, b) ∗ [(c, d) ∗ (e, f)]

Therefore, the operation * is associative.

Let an element e = (e

_{1}, e_{2}) ∈ A will be an identity element for the operation *
if a * e = a = e * a for all a = (a

_{1}, a_{2}) ∈ A
i.e., (a

_{1}+ e_{1}, a_{2}+ e_{2}) = (a_{1}, a_{2}) = (e_{1}+ a_{1}, e_{2}+ a_{2})
Which is not true for any element in A.

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

Page No. 26

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

**Answer**

(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.

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?

**Answer**

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

For, a, b, ∈ N, we have

a * b = a

^{3}+ b^{3}= b^{3}+ a^{3}= b * a [Addition is commutative in N]
Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1

^{3}+ 2^{3}) * 3 = (1 + 8) * 3 = 9 * 3 = 9^{3}+ 3^{3}= 729 + 27 = 756
and 1* (2 * 3) = 1 * (2

^{3}+ 3^{3}) = 1* (8 + 27) = 1*35 = 1^{3}+ 35^{3}= 1 + 42875 = 42876
∴ (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.

Exercise 1.2

Exercise 1.3

Miscellaneous Exercise

Class 12 Math NCERT Solutions