Open in App
Not now

# Class 12 NCERT Solutions – Mathematics Part I – Chapter 1 Relations and Functions – Exercise 1.4 | Set 2

• Last Updated : 05 Apr, 2021

### Question 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.

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 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?

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 â‰  a

Thus, there doesn’t exist any identity element.

### Question 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 = a2 + b2

(iii) a âˆ— b = a + ab

(iv) a âˆ— b = (a â€“ b)2

(v) a âˆ— b = ab / 4

(vi) a âˆ— 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 âˆˆ Q will be the identity element for the operation * if

a * e = a = e * a, for a âˆˆ 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.

### Show that âˆ— is commutative and associative. Find the identity element for âˆ— 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 âˆˆ 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 âˆˆ 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 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

Solution:

(i) Let * be an operation on N, defined as:

a * b =  a + b âˆ€ a, b âˆˆ N

Let us consider b = a = 6, we have:

6 * 6 = 6 + 6 = 12 â‰  6

Therefore, this statement is false.

(ii) Since, * is commutative

LHS = a âˆ— (b âˆ— c) = a * (c * b) = (c * b) * a = RHS

Therefore, this statement is true.

### Question 13: Consider a binary operation âˆ— on N defined as a âˆ— b = a3+ b3. 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?

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.

My Personal Notes arrow_drop_up
Related Articles