Class 12 NCERT Solutions – Mathematics Part I – Chapter 1 Relations and Functions – Exercise 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 a ∗ b = a – b
Solution:
If a, b belongs to Z+
a * b = a – b which may not belong to Z+
For eg: 1 – 3 = -2 which doesn’t belongs to Z+
Therefore, * is not a Binary Operation on Z+
(ii) On Z+, define * by a * b = ab
Solution:
If a, b belongs to Z+
a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
(iii) On R, define * by a * b = ab²
Solution:
If a, b belongs to R
a * b = ab2 which belongs to R
Therefore, * is Binary Operation on R
(iv) On Z+, define * by a * b = |a – b|
Solution:
If a, b belongs to Z+
a * b = |a – b| which belongs to Z+
Therefore, * is Binary Operation on Z+
(v) On Z+, define * by a * b = a
Solution:
If a, b belongs to Z+
a * b = a which belongs to Z+
Therefore, * is Binary Operation on Z+
Question 2: For each binary operation * defined below, determine whether * is binary, commutative or associative.
(i) On Z, define a * b = a – b
Solution:
a) Binary:
If a, b belongs to Z
a * b = a – b which belongs to Z
Therefore, * is Binary Operation on Z
b) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – b + c
RHS = (a – b) * c = a – b- c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) On Q, define a * b = ab + 1
Solution:
a) Binary:
If a, b belongs to Q, a * b = ab + 1 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab + 1
RHS = b * a = ba + 1 = ab + 1
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc + 1) = abc + a + 1
RHS = (a * b) * c = abc + c + 1
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) On Q, define a ∗ b = ab/2
Solution :
a) Binary:
If a, b belongs to Q, a * b = ab/2 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab/2
RHS = b * a = ba/2
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc/2) = (abc)/2
RHS = (a * b) * c = (ab/2) * c = (abc)/2
Since, LHS is equal to RHS
Therefore, * is Associative
(iv) On Z+, define a * b = 2ab
Solution:
a) Binary:
If a, b belongs to Z+, a * b = 2ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = 2ab
RHS = b * a = 2ba = 2ab
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * 2bc = 2a * 2^(bc)
RHS = (a * b) * c = 2ab * c = 22abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) On Z+, define a * b = ab
Solution:
a) Binary:
If a, b belongs to Z+, a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = ab
RHS = b * a = ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc = ab^c
RHS = (a * b) * c = ab * c = abc
Since, LHS is not equal to RHS
Therefore, * is not Associative
(vi) On R – {– 1}, define a ∗ b = a / (b + 1)
Solution:
a) Binary:
If a, b belongs to R, a * b = a / (b+1) which belongs to R
Therefore, * is Binary Operation on R
b) Commutative:
If a, b belongs to R, a * b = b * a
LHS = a * b = a / (b + 1)
RHS = b * a = b / (a + 1)
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to A, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1
RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)
Since, LHS is not equal to RHS
Therefore, * is not Associative
Question 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 ∧.
Solution:
^ 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.
(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) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
Solution:
Here, (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) Is ∗ commutative?
Solution:
The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
Solution:
(2 * 3) * (4 * 5) = 1 * 1 = 1
Question 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.
Solution:
Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.
*’ 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 see that the operation *’ is the same as the operation * in Exercise 4 above.
Question 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
Solution:
If a, b belongs to N
a * b = LCM of a and b
5 * 7 = 35
20 * 16 = 80
(ii) Is ∗ commutative?
Solution:
If a, b belongs to N
LCM of a * b = ab
LCM of b * a = ba = ab
a*b = b*a
Thus, * binary operation is commutative.
(iii) Is ∗ associative?
Solution:
a * (b * c) = LCM of a, b, c
(a * b) * c = LCM of a, b, c
Since, a * (b * c) = (a * b) * c
Thus, * binary operation is associative.
(iv) Find the identity of ∗ in N
Solution:
Let ‘e’ is an identity
a * e = e * a, for a belonging to N
LCM of a * e = a, for a belonging to N
LCM of e * a = a, for a belonging to N
e divides a
e divides 1
Thus, e = 1
Hence, 1 is an identity element
(v) Which elements of N are invertible for the operation ∗?
Solution:
a * b = b * a = identity element
LCM of a and b = 1
a = b = 1
only ‘1’ is invertible element in N.
Question 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 * on the set {1, 2, 3, 4, 5} is defined as
a * b = L.C.M. of a and b
Let a=3, b=5
3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set
Thus, * is not a Binary Operation.
Question 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:
If a, b belongs to N
LHS = a * b = HCF of a and b
RHS = b * a = HCF of b and a
Since LHS = RHS
Therefore, * is Commutative
Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = HCF of a, b and c
RHS = (a – b) * c = HCF of a, b and c
Since, LHS = RHS
Therefore, * is Associative
Now, 1 * a = a * 1 ≠ a
Thus, there doesn’t exist any identity element.
Question 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) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – (b – c) = a – b + c
RHS = (a – b) * c = a – b – c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a2 + b2
RHS = b * a = b2 + a2
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b2 + c2) = a2 + (b2 + c2)2
RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a + ab
RHS = b * a = b + ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)
RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iv) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = (a – b)2
RHS = b * a = (b – a)2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b – c)2 = [a – (b – c)2]2
RHS = (a * b) * c = (a – b)2 * c = [(a – b)2 – c]2
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab / 4
RHS = b * a = ba / 4
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc/4 = abc/16
RHS = (a * b) * c = ab/4 * c = abc/16
Since, LHS is equal to RHS
Therefore, * is Associative
(vi) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab2
RHS = b * a = ba2
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc)2 = a(bc2)2
RHS = (a * b) * c = (ab2) * c = ab2c2
Since, LHS is not equal to RHS
Therefore, * is not Associative
Question 10: Find which of the operations given above has identity
Solution:
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, for a ∈ Q
for (v) a * b = ab/4
Let e be an identity element
a * e = a = e * a
LHS : ae/4 = a
=> e = 4
RHS : ea/4 = a
=> e = 4
LHS = RHS
Thus, Identity element exists
Other operations doesn’t satisfy the required conditions.
Hence, other operations doesn’t have identity.
Question 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:
Given (a, b) * (c, d) = (a+c, b+d) on A
Let (a, b), (c, d), (e,f) be 3 pairs ∈ A
Commutative :
LHS = (a, b) * (c, d) = (a+c, b+d)
RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)
RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)
Since, LHS is equal to RHS
Therefore, * is Associative
Existence of Identity element:
For a, e ∈ A, a * e = a
(a, b) * (e, e) = (a, b)
(a+e, b+e) = (a, b)
a + e = a
=> e = 0
b + e = b
=> e = 0
As 0 is not a part of set of natural numbers. So, identity function does not exist.
Question 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) Let * be an operation on N, defined as:
a * b = a + b ∀ a, b ∈ N
Let us consider b = a = 6, we have:
6 * 6 = 6 + 6 = 12 ≠ 6
Therefore, this statement is false.
(ii) Since, * is commutative
LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS
Therefore, this statement is true.
Question 13: Consider a binary operation ∗ on N defined as a ∗ b = 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?
Solution:
On N, * is defined as a * b = a3 + b3
Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a3 + b3
RHS = b * a = b3 + a3
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3
Since, LHS is not equal to RHS
Therefore, * is not Associative
Thus, Option (B) is correct.