## NCERT Solutions for Class 12 Maths Chapter 1 – Relations and Functions Ex 1.4

#### Question 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 − b
(ii) On Z+, define * by ab
(iii) On R, define * by ab2
(iv) On Z+, define * by = |− b|
(v) On Z+, define * by a

(i) On Z+, * is defined by * 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 ab ∈ Z+, there is a unique element ab in Z+.
This means that * carries each pair (ab) to a unique element * b ab in Z+.
Therefore, * is a binary operation.
(iii) On R, * is defined by a * b = ab2.
It is seen that for each ab ∈ R, there is a unique element ab2 in R.
This means that * carries each pair (ab) to a unique element * b abin R.
Therefore, * is a binary operation.
(iv) On Z+, * is defined by * b = |a − b|.
It is seen that for each ab ∈ Z+, there is a unique element |a − b| in Z+.
This means that * carries each pair (ab) to a unique element * b = |a − b| in Z+.
Therefore, * is a binary operation.
(v) On Z+, * is defined by a * b = a.
* carries each pair (ab) to a unique element * b a in Z+.
Therefore, * is a binary operation.

#### Question 2:

For each binary operation * defined below, determine whether * is commutative or associative.
(i) On Z, define − b
(ii) On Q, define ab + 1
(iii) On Q, define
(iv) On Z+, define = 2ab
(v) On Z+, define ab
(vi) On − {−1}, define

(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 commutative.
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 * b = ab + 1.
It is known that:
ab = ba &mnForE; a, b ∈ Q
⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q
⇒ * b = * b &mnForE; 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 * b
It is known that:
ab = ba &mnForE; a, b ∈ Q
⇒ &mnForE; a, b ∈ Q
⇒ * b = * a &mnForE; a, b ∈ Q
Therefore, the operation * is commutative.
For all a, b, c ∈ Q, we have:
Therefore, the operation * is associative.
(iv) On Z+, * is defined by * b = 2ab.
It is known that:
ab = ba &mnForE; a, b ∈ Z+
⇒ 2ab = 2ba &mnForE; a, b ∈ Z+
⇒ * b = * a &mnForE; a, b ∈ Z+
Therefore, the operation * is commutative.
It can be observed that:
∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+
Therefore, the operation * is not associative.
(v) On Z+, * is defined by * b = ab.
It can be observed that:
and
∴ 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) ; where 2, 3, 4 ∈ Z+
Therefore, the operation * is not associative.
(vi) On R, * − {−1} is defined by
It can be observed that and
∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ − {−1}
Therefore, the operation * is not commutative.
It can also be observed that:
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ − {−1}
Therefore, the operation * is not associative.

#### Question 3:

Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by = min {ab}. Write the operation table of the operation∨.

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as  b = min {ab}
&mnForE; ab ∈ {1, 2, 3, 4, 5}.
Thus, the operation table for the given operation ∨ can be given as:
 ∨ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

#### Question 4:

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
(Hint: use the following table)
 * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

(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 * b = b * a. Therefore, the operation * is commutative.
(iii) (2 * 3) = 1 and (4 * 5) = 1
∴(2 * 3) * (4 * 5) = 1 * 1 = 1

#### Question 5:

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

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *′ b = H.C.F of a and b.
The operation table for the operation *′ can be given as:
 *′ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5
We observe that the operation tables for the operations * and *′ are the same.
Thus, the operation *′ is same as the operation*.

#### Question 6:

Let * be the binary operation on given by a * = L.C.M. of 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 are invertible for the operation *?

The binary operation * on N is defined as * 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 &mnForE; a, b ∈ N.
a * b = * a
Thus, the operation * is commutative.
(iii) For a, b∈ N, we have:
(* b) * c = (L.C.M of a and b) * c = LCM of ab, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of ab, and c
∴(* b) * c = a * (* 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 &mnForE; a ∈ N
⇒ a * 1 = a = 1 * a &mnForE; 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 * 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 *.

#### Question 7:

Is * defined on the set {1, 2, 3, 4, 5} by = L.C.M. of and a binary operation? Justify your 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:
 * 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 6 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5
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.

#### Question 8:

Let * be the binary operation on defined by = H.C.F. of and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

The binary operation * on N is defined as:
* 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 &mnForE; a, b ∈ N.
a * b = * a
Thus, the operation * is commutative.
For ab∈ N, we have:
(* b)* c = (H.C.F. of a and b) * c = H.C.F. of ab, and c
*(* c)= *(H.C.F. of b and c) = H.C.F. of ab, and c
∴(* b) * c = a * (* c)
Thus, the operation * is associative.
Now, an element ∈ N will be the identity for the operation * if * e = a = e* a ∈ N.
But this relation is not true for any ∈ N.
Thus, the operation * does not have any identity in N.

#### Question 9:

Let * be a binary operation on the set of rational numbers as follows:
(i) − (ii) a2 + b2
(iii) ab (iv) = (− b)2
(v) (vi) ab2
Find which of the binary operations are commutative and which are associative.

(i) On Q, the operation * is defined as * b = a − b.
It can be observed that:
and
; where
Thus, the operation * is not commutative.
It can also be observed that:
Thus, the operation * is not associative.
(ii) On Q, the operation * is defined as * b = a2 + b2.
For a, b ∈ Q, we have:
a * b = b * a
Thus, the operation * is commutative.
It can be observed that:
(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34
1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 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 * b = a + ab.
It can be observed that:
Thus, the operation * is not commutative.
It can also be observed that:
Thus, the operation * is not associative.
(iv) On Q, the operation * is defined by a * b = (a − b)2.
For ab ∈ Q, we have:
* b = (a − b)2
* a = (b − a)2 = [− (a − b)]2 = (a − b)2
∴ * b = b * a
Thus, the operation * is commutative.
It can be observed that:
Thus, the operation * is not associative.
(v) On Q, the operation * is defined as
For ab ∈ Q, we have:
∴ * b = * a
Thus, the operation * is commutative.
For a, b, c ∈ Q, we have:
∴(* b) * c = a * (* c)
Thus, the operation * is associative.
(vi) On Q, the operation * is defined as * b = ab2
It can be observed that:
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.

#### Question 10:

Find which of the operations given above has identity.

An element ∈ Q will be the identity element for the operation * if
* e = a = e * aa ∈ Q.
We are given
* b =
ab4
⇒ a*e = a⇒ae4=a⇒ e=4Similarly, it can be checked for e*a=a, we get e=4Thus, e = 4 is the identity.

#### Question 11:

Let A = × and * be the binary operation on A defined by
(ab) * (cd) = (cd)
Show that * is commutative and associative. Find the identity element for * on A, if any.

A = N × N
* 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∈ N
We have:
Therefore, the operation * is associative.
An element will be an identity element for the operation * if
, i.e., which is not true for any element in A.
Therefore, the operation * does not have any identity element.

#### Question 12:

State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N N.
(ii) If * is a commutative binary operation on N, then * (c) = (b) * a

(i) Define an operation * on N as:
* 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. = (* b) * a
= (* c) * [* is commutative]
a * (* c) [Again, as * is commutative]
= L.H.S.
∴ a * (* c) = (* b) * a
Therefore, statement (ii) is true.

#### Question 13:

Consider a binary operation * on defined as a3 + b3. 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?