#### Â 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 âˆ— given below gives a binary

operation. In the event that âˆ— is not a binary operation, give justification for this.

(i) On Z

(ii) On Z

(iii) On R, define âˆ— by a âˆ— b = ab

(iv) On Z

(v) On Z

operation. In the event that âˆ— is not a binary operation, give justification for this.

(i) On Z

^{+}, define âˆ— by a âˆ— b = a â€“ b(ii) On Z

^{+}, define âˆ— by a âˆ— b = ab(iii) On R, define âˆ— by a âˆ— b = ab

^{2}(iv) On Z

^{+}, define âˆ— by a âˆ— b = |a â€“ b|(v) On Z

^{+}, define âˆ— by a âˆ— b = a**Answer**

(i) On Z+, * is defined by a * b = a âˆ’ b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 âˆ’ 2 = âˆ’1 âˆ‰ Z+

(ii) On Z+, * is defined by a * b = ab.

It is seen that for each a, b âˆˆ 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 âˆˆ 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}Â in R.
Therefore, * is a binary operation.

(iv) On Z+,* is defined by a * b = |a âˆ’ b|.

It is seen that for each a, b âˆˆ Z+, there is a unique element |a âˆ’ b| in Z+.

This means that * carries each pair (a, b) to a unique element a * b = |a âˆ’ 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 âˆˆ 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 âˆ— defined below, determine whether âˆ— is binary, commutative
or associative.

(i) On Z, define a âˆ— b = a â€“ bÂ

(ii) On Q, define a âˆ— b = ab + 1Â

(iii) On Q, define a âˆ— b = ab/2

(iv) On Z+
, define a âˆ— b = 2

^{ab}Â
(v) On Z
+
, define a âˆ— b = a

^{b}
(vi)Â On R â€“{â€“ 1}, define a âˆ— b = a/b+1

(i) On Z, * is defined by a * b = a âˆ’ b.

It can be observed that 1 * 2 = 1 âˆ’ 2 = âˆ’1 and 2 * 1 = 2 âˆ’ 1 = 1.

âˆ´1 * 2 â‰ 2 * 1, where 1, 2 âˆˆ Z

Hence, the operation * is not commut

Also, we have

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

1 * (2 * 3) = 1 * (2 âˆ’ 3) = 1 * âˆ’1 = 1 âˆ’ (âˆ’1) = 2

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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 âˆˆ Q

â‡’ ab + 1 = ba + 1 for all a, b âˆˆ Q

â‡’ a * b = a * b for all a, b âˆˆ Q

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1 Ã— 2 + 1) * 3 = 3 * 3 = 3 Ã— 3 + 1 = 10

1 * (2 * 3) = 1 * (2 Ã— 3 + 1) = 1 * 7 = 1 Ã— 7 + 1 = 8

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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 âˆˆ Q

â‡’ab/2 = ba/2

for all a, b âˆˆ Q

â‡’ a * b = b * a for all a, b âˆˆ Q

Therefore, the operation * is commutative.

For all a, b, c âˆˆ 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 âˆ’ b.

It can be observed that 1 * 2 = 1 âˆ’ 2 = âˆ’1 and 2 * 1 = 2 âˆ’ 1 = 1.

âˆ´1 * 2 â‰ 2 * 1, where 1, 2 âˆˆ Z

Hence, the operation * is not commut

Also, we have

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

1 * (2 * 3) = 1 * (2 âˆ’ 3) = 1 * âˆ’1 = 1 âˆ’ (âˆ’1) = 2

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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 âˆˆ Q

â‡’ ab + 1 = ba + 1 for all a, b âˆˆ Q

â‡’ a * b = a * b for all a, b âˆˆ Q

Therefore, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1 Ã— 2 + 1) * 3 = 3 * 3 = 3 Ã— 3 + 1 = 10

1 * (2 * 3) = 1 * (2 Ã— 3 + 1) = 1 * 7 = 1 Ã— 7 + 1 = 8

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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 âˆˆ Q

â‡’ab/2 = ba/2

for all a, b âˆˆ Q

â‡’ a * b = b * a for all a, b âˆˆ Q

Therefore, the operation * is commutative.

For all a, b, c âˆˆ Q, we have

âˆ´ (a*b)*c = a*(b*c), where a, b, c âˆˆ 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 âˆˆ Z+

â‡’ 2

^{ab}Â Â = 2^{ba}Â for all a, b âˆˆ Z+
â‡’ a * b = b * a for all a, b âˆˆ Z+

Therefore, the operation * is commutative.

It can be observed that

(1 âˆ— 2) âˆ— 3 = 2

^{1Ã—2}Â âˆ— 3 = 4 âˆ— 3 =2^{4Ã—3}Â = 2^{12}Â and
1 âˆ— (2 âˆ— 3) = 1 âˆ— 2

^{2Ã—3}Â = 1 âˆ— 2^{6}Â = 1 âˆ— 64 = 2^{1Ã—64}Â = 2^{64}
âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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}Â = 1 and 2*1 = 2^{1}Â = 2
âˆ´ 1 * 2 â‰ 2 * 1, where 1, 2 âˆˆ Z+

Therefore, the operation * is not commutative.

It can also be observed that

(2 âˆ— 3) âˆ— 4 = 2

^{3}Â âˆ— 4 = 8 âˆ— 4 = 8^{4}Â = 2^{12}Â and
2 âˆ— (3 âˆ— 4) = 2 âˆ— 3

^{4}Â = 2 âˆ— 81 = 2^{81}
âˆ´ (2 * 3) * 4 â‰ 2 * (3 * 4), where 2, 3, 4 âˆˆ Z+

Therefore, the operation * is not associative.

(vi) On R, * âˆ’{âˆ’1} is defined by a âˆ— b = a/b+1

It can be observed that

1 âˆ— 2 = 1/2+1 = 1/3 and 2 âˆ— 1 = 2/1+1 = 2/2 = 1

âˆ´1 * 2 â‰ 2*1, where 1, 2 âˆˆ R âˆ’ {âˆ’1}

Therefore, the operation * is not commutative.

It can also be observed that

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ R âˆ’{âˆ’1}

Therefore, the operation * is not associative.

3. Consider the binary operation âˆ§ on the set {1, 2, 3, 4, 5} defined by a âˆ§ b = min {a, b}. Write the operation table of the operation âˆ§.

**Answer**

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

Thus, the operation table for the given operation âˆ§ can be given as:

Page No. 25

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

(i) Compute (2 âˆ— 3) âˆ— 4 and 2 âˆ— (3 âˆ— 4)

(ii) Is âˆ— commutative?

(iii) Compute (2 âˆ— 3) âˆ— (4 âˆ— 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 âˆˆ {1, 2, 3, 4, 5}, we have a * b = b * a.

Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1

âˆ´ (2 * 3) * (4 * 5) = 1 * 1 = 1

5. Let âˆ—â€² be the binary operation on the set {1, 2, 3, 4, 5} defined by
a âˆ—â€² b = H.C.F. of a and b. Is the operation âˆ—â€² same as the operation âˆ— defined
in Exercise 4 above? Justify your answer.

**Answer**

The binary operation *â€² on the set {1, 2, 3 4, 5} is defined as a *â€² b = H.C.F of a and b.

The operation table for the operation *â€² can be given as:

We observe that the operation tables for the operations * and *â€² are the same.

Thus, the operation *â€² is same as the operation*.

6. Let âˆ— be the binary operation on N given by a âˆ— b = L.C.M. of a and b. Find

(i) 5 âˆ— 7, 20 âˆ— 16

(ii) Is âˆ— commutative?

(iii) Is âˆ— associative?

(iv) Find the identity of âˆ— in N

(v) Which elements of N are invertible for the operation âˆ—?

**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 âˆˆ N.

âˆ´ a * b = b * a

Thus, the operation * is commutative.

(iii) For a, b, c âˆˆ 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

âˆ´ (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 âˆˆ N

â‡’ a * 1 = a = 1 * a for all a âˆˆ 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 âˆ— defined on the set {1, 2, 3, 4, 5} by a âˆ— 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 âˆ‰ A,

5 * 2 = 2 * 5 = 10 âˆ‰ A,

3 * 4 = 4 * 3 = 12 âˆ‰ A,

3 * 5 = 5 * 3 = 15 âˆ‰ A,

4 * 5 = 5 * 4 = 20 âˆ‰ A

Hence, the given operation * is not a binary operation.

8. Let âˆ— be the binary operation on N defined by a âˆ— b = H.C.F. of a and b.

Is âˆ— commutative? Is âˆ— 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 âˆˆ N.

âˆ´ a * b = b * a

Thus, the operation * is commutative.

For a, b, c âˆˆ 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

âˆ´ (a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e âˆˆ N will be the identity for the operation * if a * e = a = e* a for all a âˆˆ N.

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

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

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

(i) a âˆ— b = a â€“ b

(ii) a âˆ— b = a

^{2}Â + b^{2}
(iii) a âˆ— b = a + ab

(iv) a âˆ— b = (a â€“ b)

^{2}
(v) a âˆ— b = ab/4

(vi) a âˆ— b = ab

^{2}**Answer**

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

Thus, the operation * is not associative.

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

^{2}Â + b^{2}.
For a, b âˆˆ Q, we have

a * b = a

^{2}Â + b^{2}Â = b^{2}Â + a^{2}Â = b * a
âˆ´ a * b = b * a

Thus, the operation * is commutative.

It can be observed that

(1 * 2) * 3 = (1

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

^{2}Â + 3^{2}) = 1 * (4 + 9) = 1 * 13 = 1^{2}Â + 13^{2}Â =170
âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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 + 1Ã—2 = 1 + 2 = 3

2 * 1 = 2 + 2Ã—1 = 2 + 2 = 4

âˆ´ 1 * 2 â‰ 2 * 1, where 1, 2 âˆˆ Q

Thus, the operation * is not commutative.

It can also be observed that

(1 * 2) * 3 = (1+ 1Ã—2) * 3 = (1 + 2) * 3 = 3 * 3 = 3 + 3Ã—3 = 3 + 9 = 12 and

1 * (2 * 3) = 1 * (2 + 2Ã—3 ) = 1 * (2 + 6) = 1 * 8 = 1 + 1Ã—8 =1 + 8 = 9

âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ Q

Thus, the operation * is not associative.

(iv) On Q, the operation * is defined by a * b = (a âˆ’ b)

^{2}.
For a, b âˆˆ Q, we have

a * b = (a âˆ’ b)

^{2}
b * a = (b âˆ’ a)

^{2}Â = [âˆ’ (a âˆ’ b)]^{2}Â = (a âˆ’ b)^{2}
âˆ´ a * b = b * a

Thus, the operation * is commutative.

It can be observed that

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

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

^{2}Â = 1 * (-1)^{2}Â = 1 * 1 = (1 - 1)^{2}Â = 0
âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ Q

Thus, the operation * is not associative.

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

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

âˆ´ a * b = b * a

Thus, the operation * is commutative.

For a, b, c âˆˆ Q, we have

âˆ´ (a * b) * c = a * (b * c), where a, b, c âˆˆ Q

Thus, the operation * is associative.

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

^{2}
It can be observed that

âˆ´ 1/2 * 1/3 â‰ 1/3 * 1/2, where 1/2 and 1/3 âˆˆ Q

Thus, the operation * is not commutative.

It can also be observed that

Thus, the operation * is not associative.Â

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 âˆˆ Q will be the identity element for the operation *

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

However, there is no such element e âˆˆ 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 Ã— N and âˆ— be the binary operation on A defined by

(a, b) âˆ— (c, d) = (a + c, b + d)

Show that âˆ— is commutative and associative. Find the identity element for âˆ— on
A, if any.

**Answer**

A = N Ã— 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) âˆˆ A

Then, a, b, c, d âˆˆ 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]

âˆ´ (a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) âˆˆ A

Then, a, b, c, d, e, f âˆˆ N

We have

[(a, b) âˆ— (c, d)] âˆ— (e, f) = (a + c, b + d) âˆ— (e, f) = (a + c + e, b + d + f)

and

(a, b) âˆ— [(c, d) âˆ— (e, f)] = (a, b) âˆ— (c + e, d + f) = (a + c + e, b + d + f)

âˆ´ [(a, b) âˆ— (c, d)] âˆ— (e, f) = (a, b) âˆ— [(c, d) âˆ— (e, f)]

Therefore, the operation * is associative.

Let an element e = (e

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

_{1}, a_{2}) âˆˆ A
i.e., (a

_{1}+ e_{1}, a_{2}Â + e_{2}) = (a_{1}, a_{2}) = (e_{1}Â + a_{1}, e_{2}Â + 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 âˆ— on a set N, a âˆ— a = a âˆ€ a âˆˆ N.

(ii) If âˆ— is a commutative binary operation on N, then a âˆ— (b âˆ— c) = (c âˆ— b) âˆ— a

**Answer**

(i) Define an operation * on N as a * b = a + b âˆ€ a, b âˆˆ N

Then, in particular, for b = a = 3, we have

3 * 3 = 3 + 3 = 6 â‰ 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.

âˆ´ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

13. Consider a binary operation âˆ— on N defined as a âˆ— b = a

^{3}Â + b^{3}. Choose the correct answer.
(A) Is âˆ— both associative and commutative?

(B) Is âˆ— commutative but not associative?

(C) Is âˆ— associative but not commutative?

(D) Is âˆ— neither commutative nor associative?

**Answer**

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

For, a, b, âˆˆ N, we have

a * b = a

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

It can be observed that

(1 * 2) * 3 = (1

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

^{3}+ 3^{3}) = 1* (8 + 27) = 1*35 = 1^{3}Â + 35^{3}Â = 1 + 42875 = 42876
âˆ´ (1 * 2) * 3 â‰ 1 * (2 * 3), where 1, 2, 3 âˆˆ 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