### Chapter 1 Relations and Functions – Exercise 1.4 | Set 1

**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 = 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}**

**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 = a

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

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

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

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

^{2}RHS = b * a = ba

^{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 * (bc)

^{2}= a(bc^{2})^{2}RHS = (a * b) * c = (ab

^{2}) * c = ab^{2}c^{2}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 commutativeLHS = 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 = a**^{3}+ b^{3}. Choose the correct answer.

^{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:**

On N, * is defined as a * b = a

^{3}+ b^{3}

Commutative:If a, b belongs to Z, a * b = b * a

LHS = a * b = a

^{3}+ b^{3}RHS = b * a = b

^{3}+ a^{3}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 * (b

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

^{3}+ b^{3}) * c = (a^{3}+ b^{3})^{3}+ c^{3}Since, LHS is not equal to RHS

Therefore, * is not Associative

Thus, Option (B) is correct.

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