# Class 12-science RD SHARMA Solutions Maths Chapter 3 - Binary Operations

## Binary Operations Exercise Ex. 3.1

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

We have,

A ⊙ b = a^{b} + b^{a} for all a, b, ϵ N

Let a ϵ N and b ϵ N

⇒ a^{b} ϵ N and b^{a} ϵ N

⇒ a^{b} + b^{a} ϵ N

⇒ a ⊙ b ϵ N

Thus, the operation ‘⊙’ defines a binary relation on N

### Solution 1(vii)

### Solution 2

### Solution 3

It is given that, a*b = 2a + b - 3

now,

3*4 = 2 × 3 + 4 - 3

= 10 - 3

= 7

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## Binary Operations Exercise Ex. 3.2

### Solution 1

### Solution 2

### Solution 3

### Solution 4(i)

### Solution 4(ii)

### Solution 4(iii)

### Solution 4(iv)

'⊙'
on Q defined by a⊙b = a^{2}
+ b^{2} for all a, b ϵ Q

Commutativity:

For a, b ϵ Q

a⊙b
= a^{2} + b^{2} = b^{2} + a^{2} = b⊙a

So, '⊙' is commutative on Q.

Associativity:

For a, b, c ϵ Q

(a⊙b) ⊙c
= (a^{2} + b^{2}) ⊙c = (a^{2}
+ b^{2})^{2} + c^{2 }

a⊙(b ⊙c)
=a ⊙(
b^{2} +c^{2})= a^{2} +(b^{2} + c^{2 })^{2}

(a⊙b) ⊙c ≠ a⊙(b ⊙c)

So, '⊙' is not associative on Q.

### Solution 4(v)

### Solution 4(vi)

### Solution 4(vii)

### Solution 4(viii)

### Solution 4(ix)

### Solution 4(x)

### Solution 4(xi)

### Solution 4(xii)

### Solution 4(xiii)

### Solution 4(xiv)

'*' on Q defined by a*b = a + b - ab for all a, b ϵ Z

Commutativity:

For a, b ϵ Z

a*b = a + b - ab = b + a - ba = b*a

So, '*' is commutative on Z.

Associativity:

For a, b, c ϵ Z

(a*b) *c = (a + b - ab) *c

= a + b - ab + c + ac + bc - abc

a*(b*c )= a*( b + c - bc)

= a + b +c - bc + ab + ac + - abc

(a*b) *c ≠ a*(b*c )

So, '*' is not associative on Z.

### Solution 4(xv)

'*' on Q defined by a*b = gcd (a, b) for all a, b ϵ N

Commutativity:

For a, b ϵ Q

a*b = gcd (a, b) = gcd (b, a) = b*a

So, '*' is commutative on N.

Associativity:

For a, b, c ϵ N

(a*b) *c = (gcd (a, b)) *c

= gcd (a, b, c)

=a*( gcd (b, c))

=a*(b*c)

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

So, '*' is associative on N.

### Solution 5

### Solution 6

### Solution 7

### Solution 8

Now consider (a * b) * c.

Thus, we have, (a * b) * c = (a + b + ab) * c

= a + b + ab + c +(a + b + ab)c

= a + b + ab + c + ac + bc + abc

= a + b + c + ab + ac + bc + abc ---(i)

### Solution 9

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

## Binary Operations Exercise Ex. 3.3

### Solution 1

### Solution 2

### Solution 3

### Solution 4

## Binary Operations Exercise Ex. 3.4

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

## Binary Operations Exercise Ex. 3.5

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## Binary Operations Exercise MCQ

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

## Binary Operations Exercise Ex.3VSAQ

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20