# RD Sharma Solutions For Class 12 Maths Exercise 3.1 Chapter 3 Binary Operations

RD Sharma Solutions Class 12 Maths Exercise 3.1 Chapter 3 Binary Operations is provided here. Types of binary operations, commutativity and associativity of given functions with illustrative examples are the main topics which are explained under exercise 3.1. The exercise-wise solutions are prepared with explanations for each step to make it easier for students while solving problems. It helps the students self analyse their knowledge about the concepts, which are discussed under each chapter. In order to obtain a better academic score in the board exam, students can use RD Sharma Solutions Class 12 Maths Chapter 3 Binary Operations Exercise 3.1 free PDF, from the links which are provided below.

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### Access RD Sharma Solutions For Class 12 Maths Chapter 3 – Function Exercise 3.1

1. Determine whether the following operation define a binary operation on the given set or not:

(i) ‘*’ on N defined by a * b = abÂ for all a, b âˆˆ N.

(ii) ‘O’ on Z defined by a O b = abÂ for all a, b âˆˆ Z.

(iii) Â ‘*’ on N defined by a * b = a + b – 2 for all a, b âˆˆ N

(iv) ‘Ã—6‘Â onÂ S = {1,Â 2,Â 3,Â 4,Â 5}Â definedÂ by a Ã—6Â b = RemainderÂ whenÂ a bÂ isÂ dividedÂ byÂ 6.

(v) â€˜+6â€™ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b

(vi) ‘âŠ™’ on N defined by a âŠ™ b= abÂ + baÂ for all a, b âˆˆ N

(vii) â€˜*â€™ on Q defined by a * b = (a â€“ 1)/ (b + 1) for all a, b âˆˆ Q

Solution:

(i) Given ‘*’ on N defined by a * b = abÂ for all a, b âˆˆ N.

LetÂ a,Â bÂ âˆˆÂ N.Â Then,

abÂ âˆˆÂ NÂ Â Â Â Â Â [âˆµÂ abâ‰ 0Â andÂ a, bÂ isÂ positiveÂ integer]

â‡’Â aÂ *Â bÂ âˆˆÂ N

Therefore,

aÂ *Â bÂ âˆˆÂ N,Â âˆ€Â a,Â bÂ âˆˆÂ N

Thus, * is a binary operation onÂ N.

(ii) Given ‘O’ on Z defined by a O b = abÂ for all a, b âˆˆ Z.

BothÂ a = 3Â andÂ b = -1Â belongÂ toÂ Z.

â‡’ aÂ *Â b = 3-1

= 1/3 âˆ‰ Z

Thus, * is not a binary operation onÂ Z.

(iii) Â Given ‘*’ on N defined by a * b = a + b – 2 for all a, b âˆˆ N

IfÂ aÂ = 1 andÂ b = 1,

a * b = a + b –Â 2

= 1 + 1 –Â 2

= 0Â âˆ‰ N

Thus, there exist a = 1 and b = 1 such that a * bÂ âˆ‰ N

So, * is not a binary operation onÂ N.

(iv) Given ‘Ã—6‘Â onÂ S = {1,Â 2,Â 3,Â 4,Â 5}Â definedÂ by a Ã—6Â b = RemainderÂ whenÂ a bÂ isÂ dividedÂ byÂ 6.

Consider the composition table,

 X6 1 2 3 4 5 1 1 2 3 4 5 2 2 4 0 2 4 3 3 0 3 0 3 4 4 2 0 4 2 5 5 4 3 2 1

Here all the elements of the table are not in S.

â‡’ ForÂ a = 2Â andÂ b = 3,

a Ã—6Â b = 2 Ã—6Â 3 = remainderÂ whenÂ 6Â dividedÂ byÂ 6 = 0 â‰  S

Thus,Â Ã—6Â isÂ notÂ aÂ binaryÂ operationÂ onÂ S.

(v) Given â€˜+6â€™ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b

Consider the composition table,

 +6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

Here all the elements of the table are not in S.

â‡’ ForÂ a = 2Â andÂ b = 3,

a Ã—6Â b = 2 Ã—6Â 3 = remainderÂ whenÂ 6Â dividedÂ byÂ 6 = 0 â‰  Thus,Â Ã—6Â isÂ notÂ aÂ binaryÂ operationÂ onÂ S.

(vi) Given ‘âŠ™’ on N defined by a âŠ™ b= abÂ + baÂ for all a, b âˆˆ N

LetÂ a,Â b âˆˆ N.Â Then,

ab,Â baÂ âˆˆ N

â‡’ abÂ + baÂ âˆˆ NÂ Â Â Â Â Â [âˆµAddition is binary operation on N]

â‡’ a âŠ™ b âˆˆ N

Thus,Â âŠ™Â isÂ aÂ binaryÂ operationÂ onÂ N.

(vii) Given â€˜*â€™ on Q defined by a * b = (a â€“ 1)/ (b + 1) for all a, b âˆˆ Q

If a = 2 and b = -1 in Q,

a * b = (a â€“ 1)/ (b + 1)

= (2 â€“ 1)/ (- 1 + 1)

= 1/0 [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

2. 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) OnÂ Z+, defined * byÂ aÂ *Â bÂ =Â aÂ â€“Â b

(ii) On Z+, define * byÂ a*bÂ =Â ab

(iii) OnÂ R, define * byÂ a*bÂ =Â ab2

(iv) OnÂ Z+Â define * byÂ aÂ *Â bÂ = |aÂ âˆ’Â b|

(v) On Z+ define * by a * b = a

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

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

Solution:

(i) Given OnÂ Z+, defined * byÂ aÂ *Â bÂ =Â aÂ â€“Â b

If a = 1 and b = 2 in Z+, then

a * b = a â€“ b

= 1 â€“ 2

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

For a = 1 and b = 2,

a * b âˆ‰ Z+

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

(ii) Given Z+, define * byÂ a*bÂ =Â a b

Let a, b âˆˆ Z+

â‡’ a, b âˆˆ Z+

â‡’ a * b âˆˆ Z+

Thus, * is a binary operation on R.

(iii) Given onÂ R, define byÂ a*bÂ =Â ab2

Let a, b âˆˆ R

â‡’ a, b2 âˆˆ R

â‡’ ab2 âˆˆ R

â‡’ a * b âˆˆ R

Thus, * is a binary operation on R.

(iv) Given onÂ Z+Â define * byÂ aÂ *Â bÂ = |aÂ âˆ’Â b|

Let a, b âˆˆ Z+

â‡’ | a â€“ b | âˆˆ Z+

â‡’ a * b âˆˆ Z+

Therefore,

a * b âˆˆ Z+, âˆ€ a, b âˆˆ Z+

Thus, * is a binary operation on Z+.

(v) Given on Z+ define * by a * b = a

Let a, b âˆˆ Z+

â‡’ a âˆˆ Z+

â‡’ a * b âˆˆ Z+

Therefore, a * b âˆˆ Z+ âˆ€ a, b âˆˆ Z+

Thus, * is a binary operation on Z+.

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

Let a, b âˆˆ R

â‡’ a, 4b2 âˆˆ R

â‡’ a + 4b2 âˆˆ R

â‡’ a * b âˆˆ R

Therefore, a *b âˆˆ R, âˆ€ a, b âˆˆ R

Thus, * is a binary operation on R.

3. Let * be a binary operation on the set I of integers, defined byÂ aÂ *Â bÂ = 2aÂ +Â bÂ âˆ’ 3. Find the value of 3 * 4.

Solution:

Given aÂ *Â bÂ = 2aÂ +Â bÂ â€“ 3

3 * 4 = 2 (3) + 4 â€“ 3

= 6 + 4 â€“ 3

= 7

4. Is * defined on the set {1, 2, 3, 4, 5} byÂ aÂ *Â bÂ = LCM ofÂ aÂ andÂ bÂ a binary operation? Justify your answer.

Solution:

 LCM 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 5 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5

In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.

If we considerÂ aÂ = 2 andÂ bÂ = 3,Â a * b =Â LCM ofÂ aÂ andÂ bÂ = 6 âˆ‰Â {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

5. LetÂ SÂ = {a,Â b,Â c}. Find the total number of binary operations onÂ S.

Solution:

Number of binary operations on a set withÂ nÂ elements is

$$\begin{array}{l}n^{n^{2}}\end{array}$$

Here,Â SÂ = {a,Â b,Â c}

Number of elements inÂ SÂ = 3

Number of binary operations on a set with 3 elements is

$$\begin{array}{l}3^{3^{2}}\end{array}$$