# Class 12 RD Sharma Solutions – Chapter 3 Binary Operations – Exercise 3.2

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

Solution:

We are given that a * b = L.C.M. (a, b)

â‡’ 2 * 4 = L.C.M. (2, 4) = 4

and, 3 * 5 = L.C.M. (3, 5) = 15

now, 1 * 6 = L.C.M. (1, 6) = 6

Hence, 2 * 4 = 4, 3 * 5 = 15 and 1 * 6 = 6.

### (ii) Check the commutativity and associativity of â€˜*â€™ on N.

Solution:

For Commutativity:

Let a, b âˆˆ N

a * b = L.C.M. (a, b) = L.C.M. (b, a) = b * a

Therefore, a * b = b * a âˆ€ a, b âˆˆ N

Thus * is commutative on N.

For Associativity:

Let a, b, c âˆˆ N

â‡’ a * (b * c) = a * L.C.M. (b, c) = L.C.M. (a, (b, c)) = L.C.M. (a, b, c)

And, (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, âˆ€ a, b, c âˆˆ N

Thus, * is associative on N.

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

Solution:

For commutativity:

Let a, b âˆˆ N

a * b = 1 and b * a = 1

Therefore, a * b = b * a, for all a, b âˆˆ N

Thus * is commutative on N.

For associativity:

Let a, b, c âˆˆ N

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

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

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

Thus, * is associative on N.

Hence, * is both commutative and associative on N.

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

Solution:

For Commutativity:

Let a, b âˆˆ N

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

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus * is commutative on N.

For Associativity:

Let a, b, c âˆˆ N

â‡’ 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) â‰  (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

and, (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 âˆˆ N such that a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N.

Hence * is commutative on N but not associative on N.

### Question 3. Let A be any set containing more than one element. Let â€˜*â€™ be a binary operation on A defined by a * b = b for all a, b âˆˆ A Is â€˜*â€™ commutative or associative on A?

Solution:

For Commutativity:

Let a, b âˆˆ A.

Then, a * b = b

â‡’ b * a = a

Therefore, a * b â‰  b * a

Thus, * is not commutative on A.

Now we have to check associativity:

Let a, b, c âˆˆ A

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

Therefore, a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ A

Thus, * is associative on A.

### (i) â€˜*â€™ on Z defined by a * b = a + b + a b for all a, b âˆˆ Z

Solution:

For Commutativity:

Let a, b âˆˆ Z

Then a * b = a + b + ab = b + a + ba = b * a

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

Hence, * is commutative on Z.

For Associativity:

Let a, b, c âˆˆ Z, Then,

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

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

= a + b + c + b c + a b + a c + a b c

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

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

= a + b + a b + c + a c + b c + a b c

Clearly, a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ Z

Thus, * is associative on Z.

### (ii) â€˜*â€™ on N defined by a * b = 2ab for all a, b âˆˆ N

Solution:

For Commutativity:

Let a, b âˆˆ N

a * b = 2ab = 2ba = b * a

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus, * is commutative on N

For Associativity:

Let a, b, c âˆˆ N

Then, a * (b * c) = a * (2bc) = 2a2bc

and, (a * b) * c = (2ab) * c = 2ab2c

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N.

### (iii) â€˜*â€™ on Q defined by a * b = a â€“ b for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a * b = a â€“ b

b * a = b â€“ a

Clearly, a * b â‰  b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c) = a â€“ (b â€“ c) = a â€“ b + c

and, (a * b) * c = (a â€“ b) * c = a â€“ b â€“ c

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

### (iv) â€˜âŠ™â€™ on Q defined by a âŠ™ b = a2 + b2 for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a âŠ™ b = a2 + b2 = b2 + a2 = b âŠ™ a

Clearly, a âŠ™ b = b âŠ™ a, âˆ€ a, b âˆˆ Q

Thus, âŠ™ is commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

a âŠ™ (b âŠ™ c) = a âŠ™ (b2 + c2)

= a2 + (b2 + c2)2

= a2 + b4 + c4 + 2b2c2

(a âŠ™ b) âŠ™ c = (a2 + b2) âŠ™ c

= (a2 + b2)2 + c2

= a4 + b4 + 2a2b2 + c2

Clearly, (a âŠ™ b) âŠ™ c â‰  a âŠ™ (b âŠ™ c)

Thus, âŠ™ is not associative on Q.

(v) â€˜oâ€™ on Q defined by a o b = (ab/2) for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a o b = (ab/2) = (b a/2) = b o a

Clearly, a o b = b o a, âˆ€ a, b âˆˆ Q

Thus, o is commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

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

= [a (b c/2)]/2 = (a b c)/4

and, (a o b) o c = (ab/2) o c = [(ab/2) c] /2 = (a b c)/4

Clearly, a o (b o c) = (a o b) o c, âˆ€ a, b, c âˆˆ Q

Thus, o is associative on Q.

### (vi) â€˜*â€™ on Q defined by a * b = ab2 for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a * b = ab2

b * a = ba2

Clearly, * b â‰  b * a

Thus, * is not commutative on Q.

Now we have to check associativity of *

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (bc2)

= a (bc2)2

= ab2 c4

(a * b) * c = (ab2) * c

= ab2c2

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

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

Solution:

For commutative:

Let a, b âˆˆ Q, then

a * b = a + ab

b * a = b + ba = b + ab

Clearly, a * b â‰  b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

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

= a + a (b + bc)

= a + ab + abc

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

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

= a + ab + ac + abc

Therefore, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

### (viii) â€˜*â€™ on R defined by a * b = a + b -7 for all a, b âˆˆ R

Solution:

For Commutativity:

Let a, b âˆˆ R, then

a * b = a + b â€“ 7

= b + a â€“ 7 = b * a

Clearly, a * b = b * a, for all a, b âˆˆ R

Thus, * is commutative on R.

For Associativity:

Let a, b, c âˆˆ R, then

a * (b * c) = a * (b + c â€“ 7)

= a + b + c -7 -7

= a + b + c â€“ 14

and, (a * b) * c = (a + b â€“ 7) * c

= a + b â€“ 7 + c â€“ 7

= a + b + c â€“ 14

Clearly, a * (b * c ) = (a * b) * c, for all a, b, c âˆˆ R

Thus, * is associative on R.

### (ix) â€˜*â€™ on Q defined by a * b = (a â€“ b)2 for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a * b = (a â€“ b)2

= (b â€“ a)2

= b * a

Clearly, a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c)2

= a * (b2 + c2 â€“ 2bc)

= (a â€“ b2 â€“ c2 + 2bc)2

(a * b) * c = (a â€“ b)2 * c

= (a2 + b2 â€“ 2ab) * c

= (a2 + b2 â€“ 2ab â€“ c)2

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

### (x) â€˜*â€™ on Q defined by a * b = ab + 1 for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a * b = ab + 1

= ba + 1

= b * a

Clearly, a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q.

For Associativity:

Let a, b, c âˆˆ Q, then

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

= a (bc + 1) + 1

= abc + a + 1

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

= (ab + 1) c + 1

= abc + c + 1

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Q.

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

Solution:

For Commutativity:

Let a, b âˆˆ N, then

a * b = ab

b * a = ba

Clearly, a * b â‰  b * a

Thus, * is not commutative on N.

For Associativity:

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

and, (a * b) * c = (ab) * c = (ab)c = abc

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on N.

### (xii) â€˜*â€™ on Z defined by a * b = a â€“ b for all a, b âˆˆ Z

Solution:

Let a, b âˆˆ Z, then

a * b = a â€“ b

b * a = b â€“ a

Clearly, a * b â‰  b * a

Thus, * is not commutative on Z.

For Associativity:

Let a, b, c âˆˆ Z, then

a * (b * c) = a * (b â€“ c)

= a â€“ (b â€“ c)

= a â€“ (b + c)

(a * b) * c = (a â€“ b) â€“ c

= a â€“ b â€“ c

Clearly, a * (b * c) â‰  (a * b) * c

Thus, * is not associative on Z.

### (xiii) â€˜*â€™ on Q defined by a * b = (ab/4) for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

For Associativity:

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (bc/4)

= [a (b c/4)]/4

= (a b c/16)

(a * b) * c = (ab/4) * c

= [(ab/4) c]/4

= abc/16

Clearly a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Thus, * is associative on Q.

### (xiv) â€˜*â€™ on Z defined by a * b = a + b â€“ ab for all a, b âˆˆ Z

Solution:

For Commutativity:

Let a, b âˆˆ Z, then

a * b = a + b â€“ ab

= b + a â€“ ba

= b * a

Clearly, a * b = b * a, for all a, b âˆˆ Z

Thus, * is commutative on Z.

For Associativity:

Let a, b, c âˆˆ Z

a * (b * c) = a * (b + c â€“ bc)

= a + b + c- b c â€“ ab â€“ ac + abc

(a * b) * c = (a + b â€“ ab) c

= a + b â€“ ab + c â€“ (a + b â€“ ab)

= a + b + c â€“ ab â€“ ac â€“ bc + a b c

Clearly, a * (b * c) = (a * b) * c, for all a, b, c âˆˆ Z

Thus, * is associative on Z.

### (xv) â€˜*â€™ on Q defined by a * b = gcd (a, b) for all a, b âˆˆ Q

Solution:

For Commutativity:

Let a, b âˆˆ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

Therefore, a * b = b * a, for all a, b âˆˆ N

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c âˆˆ N

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

= gcd (a, b, c)

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

= gcd (a, b, c)

Clearly, a * (b * c) = (a * b) * c, for all a, b, c âˆˆ N

Thus, * is associative on N.

### Question 5. If the binary operation o is defined by a0b = a + b â€“ ab on the set Q â€“ {-1} of all rational numbers other than 1, show that o is commutative on Q â€“ [ â€“1].

Solution:

Let a, b âˆˆ Q â€“ {-1}.

Then aob = a + b â€“ ab

= b+ a â€“ b = boa

Therefore,

aob = boa for all a, b âˆˆ Q â€“ {-1}

Thus, o is commutative on Q â€“ {-1}.

### Question 6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?

Solution:

Let a, b âˆˆ Z

a * b = 3a + 7b

and, b * a = 3b + 7a

Clearly, a * b â‰  b * a for all a, b âˆˆ Z.

Example, Let a = 1 and b = 2

1 * 2 = 3 Ã— 1 + 7 Ã— 2 = 3 + 14 = 17

2 * 1 = 3 Ã— 2 + 7 Ã— 1 = 6 + 7 = 13

Therefore, there exist a = 1, b = 2 âˆˆ Z such that a * b â‰  b * a

Thus, * is not commutative on Z.

### Question 7. On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a, b âˆˆ Z. Prove that * is not associative on Z.

Solution:

Let a, b, c âˆˆ Z

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

= a (bc + 1) + 1

= a b c + a + 1

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

= (ab + 1) c + 1

= a b c + c + 1

Clearly, a * (b * c) â‰  (a * b) * c for all a, b, c âˆˆ Z

Thus, * is not associative on Z.

### Question 8. Let S be the sum of all real numbers except âˆ’1 and let * be an operation defined by a * b = a + b + ab for all a,b âˆˆ S. Determine whether * is a binary operation on S. If yes, check its commutativity and associativity.

Solution:

Given: a * b = a + b + ab, a, b âˆˆ S = R âˆ’ {âˆ’1}

Let a, b âˆˆ S.

Thus, ab âˆˆ S and hence, a + b âˆ’ ab âˆˆ S or a * b âˆˆ S

Hence, a * b S is a binary operation.

For Commutativity:

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

Hence, * is commutative.

For Associativity:

Let a, b, c âˆˆ Z, Then,

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

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

= a + b + c + ab + ac + bc + abc      …..(a)

Now, a * (b * c) = a * (b + c + bc)

= a + b + c + bc + ac +ab +abc        …..(b)

From (a) and (b), it is clear that a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Hence, * is associative on Q.

### Question 9. On Q, the set of rational numbers, * is defined by a * b = (a – b)/2, show that * is not associative.

Solution:

Let a, b, c âˆˆ Q. Then,

(a * b) * c = * c =       …….(a)

Now, a * (b * c) = a *                       ……….(b)

From (a) and (b), it is clear that a * (b * c) â‰  (a * b) * c for all a, b, c âˆˆ Q

Hence, * is not associative on Q.

### Question 10. Let binary operation * : RÃ—Râ‡¥R is defined as a * b = 2a + b. Find (2 * 3) * 4

Solution:

Given, a * b = 2a + b

â‡’ (2 * 3) * 4 = (2 Ã— 2 + 3) * 4 = 7 * 4 = (2 Ã— 7 + 4) = 18

Hence, (2 * 3) * 4 = 18.

### Question 11. On Z, the set of integers, a binary operation * is defined as a * b = a + 3b âˆ’ 4. Prove that * is neither commutative nor associative on Z.

Solution:

For Commutativity:

a * b = a + 3b âˆ’ 4 â‰  b + 3a âˆ’ 4 = b * a

â‡’ a * b â‰  b * a

Hence * is not commutative on Z.

For Associativity:

Let a, b, c âˆˆ Z, Then,

(a * b) * c = (a + 3b âˆ’ 4) * c

= a + 3b âˆ’ 4 + 3c âˆ’ 4

= a + 3b + 3c âˆ’ 8     …….(a)

Now, a * (b * c) = a + 3(b + 3c âˆ’ 4) âˆ’ 4

= a + 3b + 9c âˆ’ 16    ……(b)

From (a) and (b),  we get a * (b * c) â‰  (a * b) * c for all a, b, c âˆˆ Q

Hence, * is not associative on Q.

### a * b = ab/5, prove that * is associative on Q.

Solution:

Let a, b, c âˆˆ Z, then,

(a * b) * c = ab/5 * c = abc/25      …..(a)

and, a * (b * c) = a * bc/5 = abc/25    ….(b)

From eq (a) and (b), we have

a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Hence, * is associative on Q.

### Question 13. The binary operation * is defined as a * b = ab/7 on the set Q of rational numbers. Prove that * is associative on Q.

Solution:

Let a, b, c âˆˆ Z, then,

(a * b) * c = ab/7 * c = abc/49      …..(a)

and, a * (b * c) = a * bc/7 = abc/49   ….(b)

From eq(a) and (b), we have

a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Hence, * is associative on Q.

### Question 14. On Q, the set of all rational numbers, a binary operation * is defined as (a + b)/2 . Show that * is not associative on Q.

Solution:

Let a, b, c âˆˆ Z, then,

(a * b) * c =  * c =       …(a)

a * (b * c) = a *      …(b)

From eq(a) and (b), we have,

a * (b * c) â‰  (a * b) * c for all a, b, c âˆˆ Q

Hence, * is not associative on Q.

### (i) * is a binary operation on S.

Solution:

Let a, b âˆˆ S

Thus, ab âˆˆ S and hence,

a + b âˆ’ ab âˆˆ S or a * b âˆˆ S

Hence, a * b S is a binary operation.

### (ii) is commutative and associative.

Solution:

For Commutativity:

a * b = a + b âˆ’ ab = b + a âˆ’ ba = b * a

Hence, * is commutative.

For Associativity:

Let a, b, c âˆˆ Z, Then,

(a * b) * c = (a + b âˆ’ ab) * c

= a + b âˆ’ ab + c + (a + b âˆ’ ab)c

= a + b + c âˆ’ ab âˆ’ ac âˆ’ bc + abc      …..(a)

Now, a * (b * c) = a * (b + c âˆ’ bc)

= a + b + c âˆ’ bc âˆ’ ac âˆ’ ab +abc        …..(b)

From eq(a) and (b), it is clear that

a * (b * c) = (a * b) * c for all a, b, c âˆˆ Q

Hence, * is associative on Q.

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