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

• Last Updated : 19 Jan, 2021

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

(i) â€˜*â€™ on N defined by a * b = ab for all a, b âˆˆ N.

(ii) â€˜Oâ€™ on Z defined by a O b = ab for all a, b âˆˆ Z.

(iii) â€˜*â€™ on N defined by a * b = a + b â€“ 2 for all a, b âˆˆ N

(iv) â€˜Ã—6â€˜ on S = {1, 2, 3, 4, 5} defined by a Ã— 6 b = Remainder when a b is divided by 6.

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

(vi) â€˜âŠ™â€™ on N defined by a âŠ™ b= ab + ba 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 â€˜*â€™ on N defined by a * b = ab for all a, b âˆˆ N.

Let a, b âˆˆ N. Then,

ab âˆˆ N      [âˆµ abâ‰ 0 and a, b is positive integer]

â‡’ a * b âˆˆ N

Therefore,

a * b âˆˆ N, âˆ€ a, b âˆˆ N

Thus, * is a binary operation on N.

(ii) Given â€˜Oâ€™ on Z defined by a O b = ab for all a, b âˆˆ Z.

Both a = 3 and b = -1 belong to Z.

â‡’ a * b = 3-1

âˆ‰ Z

Thus, * is not a binary operation on Z.

(iii)  Given â€˜*â€™ on N defined by a * b = a + b â€“ 2 for all a, b âˆˆ N

If a = 1 and b = 1,

a * b = a + b â€“ 2

= 1 + 1 â€“ 2

= 0 âˆ‰ N

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

So, * is not a binary operation on N.

(iv) Given â€˜Ã—6â€˜ on S = {1, 2, 3, 4, 5} defined by a Ã—6 b = Remainder when a b is divided by 6.

Consider the composition table,

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

â‡’ For a = 2 and b = 3,

a Ã—6 b = 2 Ã—6 3 = remainder when 6 divided by 6 = 0 â‰  S

Thus, Ã—6 is not a binary operation on S.

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

Consider the composition table,

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

â‡’ For a = 2 and b = 3,

a Ã—6 b = 2 Ã—6 3 = remainder when 6 divided by 6 = 0 â‰  Thus, Ã—6 is not a binary operation on S.

(vi) Given â€˜âŠ™â€™ on N defined by a âŠ™ b= ab + ba for all a, b âˆˆ N

Let a, b âˆˆ N. Then,

ab, ba âˆˆ N

â‡’ ab + ba âˆˆ N      [âˆµAddition is binary operation on N]

â‡’ a âŠ™ b âˆˆ N

Thus, âŠ™ is a binary operation on 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 =

[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.

### Question 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) On Z+, defined * by a * b = a â€“ 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 âˆ’ 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 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 * by a*b = 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 on R, define by a*b = 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 on Z+ define * by a * b = |a âˆ’ 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.

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

Solution:

Given:

a * b = 2a + b â€“ 3

3 * 4 = 2 (3) + 4 â€“ 3

= 6 + 4 â€“ 3

= 7

### Question 4. Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.

Solution:

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

If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 âˆ‰ {1, 2, 3, 4, 5}.

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

### Question 5. 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

### Question 6. Find the total number of binary operations on {a, b}.

Solution:

We have,

S = {a, b}

The total number of binary operation on S = {a, b} in

### Question 7. Prove that the operation * on the set

M= defined by A + B = AB is a binary operation.

Solution:

We have,

and

A + B = AB for all A, B âˆˆ M

Let A =\ and B =

Now, AB =

Therefore, a âˆˆ R, b âˆˆ R, c âˆˆ R and d âˆˆ R

â‡’ ac âˆˆ R and bd âˆˆ R

â‡’

â‡’ A * B âˆˆ M

Hence, the operator * defines a binary operation on M

### Question 8. Let S be the set of all rational numbers of the form  where m âˆˆ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation

Solution:

S = set of rational numbers of the form where m âˆˆ Z and n = 1, 2, 3

Also, a * b = ab

Let a âˆˆ S and b âˆˆ S

â‡’ ab =

Therefore, a * b âˆ‰ S

Hence, the operator * does not defines a binary operation on S

### Question 9. The binary operation & : R Ã— R â†’ R is defined as a*b = 2a + b

Solution:

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

Now,

(2*3) = 2 Ã— 2 + 3

= 4 + 3

(2*3)*4 = 7*4 = 2 Ã— 7 + 4

= 14 + 4

= 18

### Question 10. Let * be a binary operation on N given by a*b = LCM(a, b) for all a, b âˆˆ N. Find 5*7.

Solution:

It is given that a*b = LCM (a, b)

Now,

5*7 = LCM (5, 7)

= 35

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