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

• Last Updated : 30 Apr, 2021

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

Solution:

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

⇒ 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 ‘*’ on N.

Solution:

For Commutativity:

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.

For Associativity:

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)

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, ∀ a, b, c ∈ N

Thus, * is associative on N.

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

Solution:

For commutativity:

Let a, b ∈ N

a * b = 1 and b * a = 1

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

Thus * is commutative on N.

For associativity:

Let a, b, c ∈ 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 ∈ 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 ∈ Q

Solution:

For Commutativity:

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.

For Associativity:

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

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 ∈ N such that a * (b * c) ≠ (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 ‘*’ be a binary operation on A defined by a * b = b for all a, b ∈ A Is ‘*’ commutative or associative on A?

Solution:

For Commutativity:

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.

(i) ‘*’ on Z defined by a * b = a + b + a b for all a, b ∈ Z

Solution:

For Commutativity:

Let a, b ∈ Z

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

Therefore, a * b = b * a, ∀ a, b ∈ Z

Hence, * is commutative on Z.

For Associativity:

Let a, b, c ∈ 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, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) ‘*’ on N defined by a * b = 2ab for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N

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

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

For Associativity:

Let a, b, c ∈ N

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

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

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

(iii) ‘*’ on Q defined by a * b = a – b for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

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

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

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(iv) ‘⊙’ on Q defined by a ⊙ b = a2 + b2 for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a ⊙ b = a2 + b2 = b2 + a2 = b ⊙ a

Clearly, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a ⊙ (b ⊙ c) = a ⊙ (b2 + c2)

= a2 + (b2 + c2)2

= a2 + b4 + c4 + 2b2c2

(a ⊙ b) ⊙ c = (a2 + b2) ⊙ c

= (a2 + b2)2 + c2

= a4 + b4 + 2a2b2 + c2

Clearly, (a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) ‘o’ on Q defined by a o b = (ab/2) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

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

Clearly, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q.

For Associativity:

Let a, b, c ∈ 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, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi) ‘*’ on Q defined by a * b = ab2 for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = ab2

b * a = ba2

Clearly, * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to check associativity of *

Let a, b, c ∈ Q, then

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

= a (bc2)2

= ab2 c4

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

= ab2c2

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(vii) ‘*’ on Q defined by a * b = a + ab for all a, b ∈ Q

Solution:

For commutative:

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba = b + ab

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.

For Associativity:

Let a, b, c ∈ 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) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii) ‘*’ on R defined by a * b = a + b -7 for all a, b ∈ R

Solution:

For Commutativity:

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7 = b * a

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

Thus, * is commutative on R.

For Associativity:

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

and, (a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

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

Thus, * is associative on R.

(ix) ‘*’ on Q defined by a * b = (a – b)2 for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = (a – b)2

= (b – a)2

= b * a

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

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)2

= a * (b2 + c2 – 2bc)

= (a – b2 – c2 + 2bc)2

(a * b) * c = (a – b)2 * c

= (a2 + b2 – 2ab) * c

= (a2 + b2 – 2ab – c)2

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x) ‘*’ on Q defined by a * b = ab + 1 for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

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

Thus, * is commutative on Q.

For Associativity:

Let a, b, c ∈ 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) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi) ‘*’ on N defined by a * b = ab for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N, then

a * b = ab

b * a = ba

Clearly, a * b ≠ 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) ≠ (a * b) * c

Thus, * is not associative on N.

(xii) ‘*’ on Z defined by a * b = a – b for all a, b ∈ Z

Solution:

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

(xiii) ‘*’ on Q defined by a * b = (ab/4) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

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

Thus, * is commutative on Q

For Associativity:

Let a, b, c ∈ 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 ∈ Q

Thus, * is associative on Q.

(xiv) ‘*’ on Z defined by a * b = a + b – ab for all a, b ∈ Z

Solution:

For Commutativity:

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

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

Thus, * is commutative on Z.

For Associativity:

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – bc)

= a + b + c- b c – ab – ac + abc

(a * b) * c = (a + b – ab) c

= a + b – ab + c – (a + b – ab)

= a + b + c – ab – ac – bc + a b c

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

Thus, * is associative on Z.

(xv) ‘*’ on Q defined by a * b = gcd (a, b) for all a, b ∈ Q

Solution:

For Commutativity:

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

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

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ 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 ∈ N

Thus, * is associative on N.

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

Solution:

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – b = boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}.

Question 6. 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

and, b * a = 3b + 7a

Clearly, a * b ≠ b * a for all a, b ∈ Z.

Example, 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

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 ∈ Z. Prove that * is not associative on Z.

Solution:

Let a, b, c ∈ 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) ≠ (a * b) * c for all a, b, c ∈ Z

Thus, * is not associative on Z.

Question 8. Let S be the sum of all real numbers except −1 and let * be an operation defined by a * b = a + b + ab for all a,b ∈ 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 ∈ S = R − {−1}

Let a, b ∈ S.

Thus, ab ∈ S and hence, a + b − ab ∈ S or a * b ∈ 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 ∈ 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 ∈ 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 ∈ Q. Then,

(a * b) * c = * c =       …….(a)

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

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

Hence, * is not associative on Q.

Question 10. Let binary operation * : R×R⇥R is defined as a * b = 2a + b. Find (2 * 3) * 4

Solution:

Given, a * b = 2a + b

⇒ (2 * 3) * 4 = (2 × 2 + 3) * 4 = 7 * 4 = (2 × 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 − 4. Prove that * is neither commutative nor associative on Z.

Solution:

For Commutativity:

a * b = a + 3b − 4 ≠ b + 3a − 4 = b * a

⇒ a * b ≠ b * a

Hence * is not commutative on Z.

For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + 3b − 4) * c

= a + 3b − 4 + 3c − 4

= a + 3b + 3c − 8     …….(a)

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

= a + 3b + 9c − 16    ……(b)

From (a) and (b),  we get a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

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

Solution:

Let a, b, c ∈ 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 ∈ 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 ∈ 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 ∈ 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 ∈ Z, then,

(a * b) * c =  * c =       …(a)

a * (b * c) = a *      …(b)

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

a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

(i) * is a binary operation on S.

Solution:

Let a, b ∈ S

Thus, ab ∈ S and hence,

a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

(ii) is commutative and associative.

Solution:

For Commutativity:

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

Hence, * is commutative.

For Associativity:

Let a, b, c ∈ 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 eq(a) and (b), it is clear that

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

Hence, * is associative on Q.

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