RD Sharma Solutions for Class 12 Maths Chapter 3 – Binary Operations (Free PDF Download)

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Q. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.
(i) On Z⁺, defined * by a * b = a – b

(ii) On Z⁺, define * by a*b = ab

(iii) On R, define * by a*b = ab²

(iv) On Z⁺ define * by a * b = |a − b|

(v) On Z⁺ define * by a * b = a

(vi) On R, define * by a * b = a + 4b²

Here, Z⁺ denotes the set of all non-negative integers.

Solution:

(i) Given On Z⁺, defined * by a * b = a – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

= -1 ∉ Z⁺ [because Z⁺ is the set of non-negative integers]

For a = 1 and b = 2,

a * b ∉ Z⁺

Thus, * is not a binary operation on Z⁺.

(ii) Given Z⁺, define * by a*b = a b

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ | a – b | ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

So, * is a binary operation on Z⁺.

(v) Given on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(vi) Given On R, define * by a * b = a + 4b²

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ R

So, * is a binary operation on R.

Q. Let S = {a, b, c}. Find the total number of binary operations on S.

Solution:

Number of binary operations on a set with n elements is

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is 3³²

Q. Let * be a binary operation on the set I of integers, defined by a * b = 2a + b − 3. Find the value of 3 * 4.

Solution: Given a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 3

= 7

Q. Let ‘*’ be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b ∈ N
 

(i) Find 2 * 4, 3 * 5, 1 * 6.

(ii) Check the commutativity and associativity of ‘*’ on N.

Solution:

(i) Given a * b = 1.c.m. (a, b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5)

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) We have to prove commutativity of *

Let a, b ∈ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

Therefore

a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c ) = a * l.c.m. (b, c)

= l.c.m. (a, (b, c))

= l.c.m (a, b, c)

(a * b) * c = l.c.m. (a, b) * c

= l.c.m. ((a, b), c)

= l.c.m. (a, b, c)

Therefore

(a * (b * c) = (a * b) * c, ∀ a, b , c ∈ N

Thus, * is associative on N.

Q. Let A be any set containing more than one element. Let ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?

Solution:

Let a, b ∈ A

Then, a * b = b

b * a = a

Therefore a * b ≠ b * a

Thus, * is not commutative on A

Now we have to check associativity:

Let a, b, c ∈ A

a * (b * c) = a * c

= c

Therefore

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ A

Thus, * is associative on A

Q. Determine which of the following binary operation is associative and which is commutative:

(i) * on N defined by a * b = 1 for all a, b ∈ N

(ii) * on Q defined by a * b = (a + b)/2 for all a, b ∈ Q

Solution:

(i) We have to prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

Then a * (b * c) = a * (1)

= 1

(a * b) *c = (1) * c

= 1

Therefore a * (b * c) = (a * b) *c for all a, b, c ∈ N

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

Let a, b ∈ N

a * b = (a + b)/2

= (b + a)/2

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

a * (b * c) = a * (b + c)/2

= [a + (b + c)]/2

= (2a + b + c)/4

Now, (a * b) * c = (a + b)/2 * c

= [(a + b)/2 + c] /2

= (a + b + 2c)/4

Thus, a * (b * c) ≠ (a * b) * c

If a = 1, b= 2, c = 3

1 * (2 * 3) = 1 * (2 + 3)/2

= 1 * (5/2)

= [1 + (5/2)]/2

= 7/4

(1 * 2) * 3 = (1 + 2)/2 * 3

= 3/2 * 3

= [(3/2) + 3]/2

= 4/9

Therefore, there exist a = 1, b = 2, c = 3 ∈ N such that a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

Q. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?

Solution:

Let a, b ∈ Z

a * b = 3a + 7b

b * a = 3b + 7a

Thus, a * b ≠ b * a

Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2

= 3 + 14

= 17

2 * 1 = 3 × 2 + 7 × 1

= 6 + 7

= 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

So, * is not commutative on Z.