RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations

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

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

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

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

(iv) ‘×6‘ on S = {1, 2, 3, 4, 5} defined by a × 6 b = Remainder when a b is divided by 6.

(v) ‘+6’ on S = {0, 1, 2, 3, 4, 5} defined by a +6 b

(vi) ‘⊙’ on N defined by a ⊙ b= ab + ba for all a, b ∈ N

(vii) ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ Q

Solution:

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

Let a, b ∈ N. Then,

a^b ∈ N      [∵ ab≠0 and a, b is positive integer]

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

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

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

⇒ a * b = 3^-1

=  ∉ Z

Thus, * is not a binary operation on Z.

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

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

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

So, * is not a binary operation on N.

(iv) Given ‘×₆‘ on S = {1, 2, 3, 4, 5} defined by a ×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.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ S

Thus, ×₆ is not a binary operation on S.

(v) Given ‘+₆’ 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.

⇒ For a = 2 and b = 3,

a ×₆ b = 2 ×₆ 3 = remainder when 6 divided by 6 = 0 ≠ Thus, ×₆ is not a binary operation on S.

(vi) Given ‘⊙’ on N defined by a ⊙ b= a^b + b^a for all a, b ∈ N

Let a, b ∈ N. Then,

ab, ba ∈ N

⇒ a^b + b^a ∈ N      [∵Addition is binary operation on N]

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

(vii) Given ‘*’ on Q defined by a * b = (a – 1)/ (b + 1) for all a, b ∈ 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 – 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 − 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 – b

If a = 1 and b = 2 in Z⁺, then

a * b = a – b

= 1 – 2

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

For a = 1 and b = 2,

a * b ∉ Z⁺

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

(ii) Given Z⁺, define * by a*b = a b

Let a, b ∈ Z⁺

⇒ a, b ∈ Z⁺

⇒ a * b ∈ Z⁺

Thus, * is a binary operation on R.

(iii) Given on R, define by a*b = ab²

Let a, b ∈ R

⇒ a, b² ∈ R

⇒ ab² ∈ R

⇒ a * b ∈ R

Thus, * is a binary operation on R.

(iv) Given on Z⁺ define * by a * b = |a − b|

Let a, b ∈ Z⁺

⇒ | a – b | ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore,

a * b ∈ Z⁺, ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

(v) Given on Z⁺ define * by a * b = a

Let a, b ∈ Z⁺

⇒ a ∈ Z⁺

⇒ a * b ∈ Z⁺

Therefore, a * b ∈ Z⁺ ∀ a, b ∈ Z⁺

Thus, * is a binary operation on Z⁺.

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

Let a, b ∈ R

⇒ a, 4b² ∈ R

⇒ a + 4b² ∈ R

⇒ a * b ∈ R

Therefore, a *b ∈ R, ∀ a, b ∈ 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 − 3. Find the value of 3 * 4.

Solution:

Given:

a * b = 2a + b – 3

3 * 4 = 2 (3) + 4 – 3

= 6 + 4 – 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 ∉ {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 ∈ M

Let A =\ and B = 

Now, AB = 

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

⇒ ac ∈ R and bd ∈ R

⇒ 

⇒ A * B ∈ 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 ∈ 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 ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

⇒ ab = 

Therefore, a * b ∉ S

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

Question 9. The binary operation & : R × R → R is defined as a*b = 2a + b

Solution:

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

Now,

(2*3) = 2 × 2 + 3

         = 4 + 3

(2*3)*4 = 7*4 = 2 × 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 ∈ N. Find 5*7.

Solution:

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

Now,

5*7 = LCM (5, 7)

       = 35

(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.

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 = 2^ab for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N

a * b = 2^ab = 2^ba = 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 * (2^bc) = 2a^2bc

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

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 = a^b for all a, b ∈ N

Solution:

For Commutativity:

Let a, b ∈ N, then

a * b = a^b

b * a = b^a

Clearly, a * b ≠ b * a

Thus, * is not commutative on N.

For Associativity:

a * (b * c) = a * (bc) = 

and, (a * b) * c = (ab) * c = (a^b)^c = a^bc

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.

Question 12. On the set Q of all rational numbers if a binary operation * is defined as 

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.

Question 15. 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. Prove that:

(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.

Question 1. Find the identity element in the set I+ of all positive integers defined by a * b = a + b for all a, b ∈ I⁺.

Solution:

Let e be the identity element in I⁺ with respect to * such that

a * e = a = e * a, ∀ a ∈ I⁺

a * e = a and e * a = a, ∀ a ∈ I⁺

a + e = a and e + a = a, ∀ a ∈ I⁺

e = 0, ∀ a ∈ I⁺

Hence, 0 is the identity element in I⁺ with respect to *.

Question 2. Find the identity element in the set of all rational numbers except – 1 with respect to * defined by a * b = a + b + ab

Solution:

Let e be the identity element in I+ with respect to * such that

a * e = a = e * a, ∀ a ∈ Q – {-1}

a * e = a and e * a = a, ∀ a ∈ Q – {-1}

a + e + ae = a and e + a + ea = a, ∀ a ∈ Q – {-1}

e + ae = 0 and e + ea = 0, ∀ a ∈ Q – {-1}

e (1 + a) = 0 and e (1 + a) = 0, ∀ a ∈ Q – {-1}

e = 0, ∀ a ∈ Q – {-1} [because a not equal to -1]

Hence, 0 is the identity element in Q – {-1} with respect to *.

Question 3. If the binary operation * on the set Z is defined by a*b = a + b – 5, then find the identity element with respect to *.

Solution: 

We are given the binary operator * defined on Z as

a*b = a + b – 5 for all a, b ∈ Q

Let e be the identity elements with respect to *

Then, a*e = e*a = a  [By identity property]

⇒ a + e – 5 = a

⇒ e = 5

Therefore, the required identity element with respect to * is 5.

Question 4. On the set Z integers, if the binary operation * is defined by a*b = a + b + 2, then find the identity elements.

Solution: 

The binary operator * is defined on Z, and is given by

a*b = a + b +2 for all a, b ∈ Z.

Let a ∈ Z and e ∈ Z be the identity element with respect to *, then

a*e = e*a = a    [By identity property]

⇒ a + e + 2 = a

⇒ e = -2 ∈ Z

Therefore, the identity element with respect to * is -2.

Question 1. Let * be a binary operation on Z defined by a * b = a + b – 4 for all a, b ∈ Z.

(i) Show that * is both commutative and associative.

(ii) Find the identity element in Z

(iii) Find the invertible element in Z. 

Solution:

(i) First we will prove commutativity of * 
Let a, b ∈ Z. 
a * b = a + b – 4 
= b + a – 4 
= b * a 

⇒ a * b = b * a, ∀ a, b ∈ Z 
So we can say that, * is commutative on Z. 

Now we will prove associativity of Z. 
Let a, b, c ∈ Z. 
a * (b * c) = a * (b + c – 4) 
= a + b + c -4 – 4 
= a + b + c – 8 
⇒ (a * b) * c = (a + b – 4) * c 
= a + b – 4 + c – 4 
= a + b + c – 8 

⇒ a * (b * c) = (a * b) * c, for all a, b, c ∈ Z 
So we can say that, * is associative on Z.

(ii) We have to find identity element in Z. 
Let x be the identity element in Z with respect to * such that 
a * x = a = x * a ∀ a ∈ Z 
a * x = a and x * a = a, ∀ a ∈ Z 
a + x – 4 = a and x + a – 4 = a, ∀ a ∈ Z 
x = 4, ∀ a ∈ Z 
So we can say that, 4 is the identity element in Z with respect to *.

(iii) We have to find the invertible element in Z. 
Let a ∈ Z and b ∈ Z be the inverse of a. So, 
a * b = x = b * a 
a * b = x and b * a = x 
a + b – 4 = 4 and b + a – 4 = 4 
b = 8 – a ∈ Z 
So we can say that, 8 – a is the inverse of a ∈ Z

Question 2. Let * be a binary operation on Q₀ (set of non-zero rational numbers) defined by a * b= (3ab/5) for all a, b ∈ Q₀. Show that * is commutative as well as associative. Also, find its identity element, if it exists.

Solution:

Firstly we will prove commutativity of * 
Let a, b ∈ Q₀ 
a * b = (3ab/5) 
= (3ba/5) 
= b * a 
⇒ a * b = b * a, for all a, b ∈ Q_0. 

Now we will prove associativity of * 
Let a, b, c ∈ Q₀ 
a * (b * c) = a * (3bc/5) 
= [a (3 bc/5)] /5 
= 3 abc/25 
(a * b) * c = (3 ab/5) * c 
= [(3 ab/5) c]/ 5 
= 3 abc /25 
⇒ a * (b * c) = (a * b) * c, for all a, b, c ∈ Q0 
So we can say that * is associative on Q₀. 

Now we will find the identity element. 
Let x be the identity element in Z with respect to * such that 
a * x = a = x * a ∀ a ∈ Q₀ 
a * x = a and x * a = a, ∀ a ∈ Q₀ 
3ax/5 = a and 3xa/5 = a, ∀ a ∈ Q₀ 
x = 5/3 ∀ a ∈ Q₀ [a ≠ 0] 
So we can say that, 5/3 is the identity element in Q₀ with respect to *.

Question 3. Let * be a binary operation on Q – {-1} defined by a * b = a + b + ab for all a, b ∈ Q – {-1}. Then,

(i) Show that * is both commutative and associative on Q – {-1}

(ii) Find the identity element in Q – {-1}

(iii) Show that every element of Q – {-1} is invertible. Also, find inverse of an arbitrary element.

Solution:

(i) First we will check commutativity of * 
Let us assume that a, b ∈ Q – {-1} 
a * b = a + b + ab 
= b + a + ba 
= b * a 
⇒ 
a * b = b * a, ∀ a, b ∈ Q – {-1} 

Now we will prove associativity of * 
Let us assume that a, b, c ∈ Q – {-1}, 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 
= (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 
⇒ a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1} 
So we can say that, * is associative on Q – {-1} 

(ii) Let us assume that x be the identity element in I+ with respect to * such that 
a * x = a = x * a, ∀ a ∈ Q – {-1} 
a * x = a and x * a = a, ∀ a ∈ Q – {-1} 
a + x + ax = a and x + a + xa = a, ∀ a ∈ Q – {-1} 
x + ax = 0 and x + xa = 0, ∀ a ∈ Q – {-1} 
x (1 + a) = 0 and x (1 + a) = 0, ∀ a ∈ Q – {-1} 
x = 0, ∀ a ∈ Q – {-1} [a ≠ -1] 
so we can say that , 0 is the identity element in Q – {-1} with respect to *. 

(iii) Let us assume that a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then, 
a * b = e = b * a 
a * b = e and b * a = e 
a + b + ab = 0 and b + a + ba = 0 
b (1 + a) = – a Q – {-1} 
b = -a/1 + a Q – {-1} [a ≠ -1] 
So we can say that, -a/1 + a is the inverse of a ∈ Q – {-1}.

Question 4. Let A = R₀ × R, where R₀ denote the set of all non-zero real numbers. A binary operation ‘O’ is defined on A as follows: (a, b) O (c, d) = (ac, bc + d) for all (a, b), (c, d) ∈ R₀ × R.

(i) Show that ‘O’ is commutative and associative on A

(ii) Find the identity element in A

(iii) Find the invertible element in A

Solution:

(i) Let us assume that X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R₀ and b, d ∈ R 
X O Y = (ac, bc + d) 
Y O X = (ca, da + b) 
⇒ X O Y = Y O X, ∀ X, Y ∈ A 
⇒ O commutative on A. 

Now we have to check associativity of O 
Let X = (a, b), Y = (c, d) and Z = (e, f), ∀ a, c, e ∈ R0 and b, d, f ∈ R 
⇒X O (Y O Z) = (a, b) O (ce, de + f) 
= (ace, bce + de + f) 
⇒ (X O Y) O Z = (ac, bc + d) O (e, f) 
= (ace, (bc + d) e + f) 
= (ace, bce + de + f) 
⇒ X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A 

(ii) Let us assume that E = (x, y) be the identity element in A with respect to O, ∀ x ∈ R₀ and y ∈ R 
X O E = X = E O X, ∀ X ∈ A 
X O E = X and EOX = X 
⇒(ax, bx +y) = (a, b) and (xa, ya + b) = (a, b) 
We know that , (ax, bx + y) = (a, b) 
ax = a 
x = 1 
bx + y = b 
y = 0 [x = 1] 
we know that, (xa, ya + b) = (a, b) 
xa = a 
x = 1 
ya + b = b 
y = 0 [since x = 1] 
So we can say that (1, 0) is the identity element in A with respect to O. 

(iii) Let us assume that F = (m, n) be the inverse in A ∀ m ∈ R₀ and n ∈ R 
X O F = E and F O X = E 
(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0) 
As we know that (am, bm + n) = (1, 0) 
am = 1 
m = 1/a 
bm + n = 0 
n = -b/a [m = 1/a] 
We know that (ma, na + b) = (1, 0) 
ma = 1 
m = 1/a 
na + b = 0 
n = -b/a 
So we can say that, the inverse of (a, b) ∈ A with respect to O is (1/a, -1/a).

Question 5. Let ‘*’ be a binary operation on the set of Q₀ of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q₀

(i) show that ‘*’ is both commutative and associative.

(ii) Find the identity element in Q₀ .

(iii) Find the invertible element of Q₀.

Solution:

(i) We have to show, ‘*’ is commutative. 
Let a, b ∈ Q₀. 
a o b = ab/2 = ba/2 
⇒ b o a 
⇒ a o b = b o a, ∀ a, b ∈ Q₀. 
So, o is commutative on Q_0. 

Now, we will show, ‘*’ is Associative. 
Let a, b, c ∈ Q₀ 
a o (b 0 c) = a o (bc/2) 
= (a(bc/2))/2 
= abc /4 
⇒ (a o b) o c = (ab/2) o c 
= abc/4 
⇒ a o (b o c) = (a o b) o c ∀ a, b, c ∈ Q_0. 
So, we can say that o is associative on Q_0. 

(ii) Let x be the identify element in Q₀ with respect to * such that 
a o x = a x o a ,∀ a ∈ Q₀ 
⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q₀ 
x = 2 ∈ Q_0,∀ a ∈ Q₀ 
So, we can say that, 2 is the identity element in Q₀ with respect to o. 

(iii) Let us assume that a ∈ Q₀ and b ∈ Q₀ be the inverse of a. 
⇒ a o b = e = b o a = e 
⇒ ab/2 = 2 and ba/2 = 2 
⇒ b = 4/a ∈ Q₀ 
So, we can say that, 4/a is the inverse of a∈ Q₀.

Question 6. On R -{1}, a binary operation * is defined by a*b = a+b-ab . Prove that * is commutative and associative. Find the identity element for * on R-{1}. Also, prove that every element of R-{1} is invertible.

Solution:

Firstly we will find commutative. 
Let us assume that a, b ∈ R -{1} 
a * b = a + b – ab 
= b + a -ba 
= b*a 
⇒ a * b = b + a ,∀ a , b ∈ R – {1} 
So , we can say that * is commutative on R-{1} 

Now , we will find Associative. 
Let assume that a , b , c ∈ R – {1} 
a * (b * c ) = a * (b + c – bc) 
=a + b + c – bc -a(b + c – bc) 
=a + b + c – bc – ab – ac + abc 
(a * b) * c = (a + b – ab ) * c 
= a + b – ab + c – (a + b – ab)c 
= a + b + c – ab – ac – bc + abc 
⇒ a * (b * c) = (a * c )* c , ∀ a , b , c ∈ R – {1} 
So we can say that , * is associative on R-{1} 

Now we will find identity element. 
Let assume that x be the identity element in R-{1} with respect to * 
a * x = a = x * a , ∀ a ∈ R-{1} 
a * x = a and x * a = a, ∀ a ∈ R-{1} 
⇒ a + x – ax = a and x + a – xa = a , ∀ a ∈ R-{1} 
x(1 – a) = 0 , ∀ a ∈ R-{1} 
⇒ x = 0 [ a ≠ 1 ⇒ 1 – a ≠ 0 ] 
So we can say that , x = 0 will be the identity element with respect to * . 

Now lets find inverse element. 
Let’s assume that b ∈ R-{1} be the inverse element of a ∈ R-{1} 
a * b = b * a = x 
⇒ a + b -ab = 0 [e=0] 
⇒b(1 – a) = -a 
⇒ b = -a /(1 – a) ≠ 1 [ if -a/(1-a) = 1 ⇒ -a = 1 – a ⇒ 1≠ 0] 
So we can say that , b = -a/(1 – a) is the inverse of a ∈ R-{1} with respect to *.

Question 7.Let R₀ denote the set of all non zero real number and let A = R₀ x R₀ . If  ‘*’ is a binary operation on A defined by ( a, b) * (c ,d) = (ac , bd) for all (a , b)(c , d)  ∈  A.

(i) Show that ‘*’ is both commutative and associative on A.

(ii) Find the identity element in A.

(iii) Find the invertible element in A.

Solution:

In the question we have given (a, b) * (c ,d) = (ac , bd) for all (a,b)(c,d) ∈ A. 
(i) Let us assume that , (a,b)(c,d) ∈ A. So, 
(a, b) * (c ,d) = (ac , bd) 
=(ca , bd) [ ac = ca and bd = db ] 
=(c , d)*(a , b) 
⇒ (a, b) * (c,d) = (ac,bd) 
So we can say that , ‘*’ is commutative on A. 

⇒ Now we will find associativity on A. 
Let us assume that , (a,b),(c,d),(e,f) ∈ A. 
⇒ ((a,b)*(c,d))*(e,f) = (ac , bd)*(e,f) 
=(ace , bdf) –(i) 
Now (a,b)*((c,d)*(e,f)) =(a,b)*(ce,df) 
=(ace , bdf) –(ii) 
From equation (i) and (ii). 
((a,b)*(c,d))*(e,f) = (a,b)*((c,d)*(e,f)) 
So we can say that , ‘*’ is associative on A. 

(ii) Let find identity element in A. 
Let assume that (x,y) ∈ A be the identity element with respect to *. 
(a,b) * (x,y) = (x,y)*(a,b) = (a,b) for all (a,b) ∈ A. 
⇒ (ax , by) = (a,b) 
⇒ ax = a & by = b 
⇒ x = 1 & y = 1 
So we can say that (1,1) will be identity element. 

(iii) Now we will find invertible element in A. 
Let assume that (c,d) ∈ A be the inverse of (a,b) ∈ A 
(a,b)*(c,d) = (c,d)*(a,b) = x 
(ac , bd) = (1,1) [e = (1,1) ] 
ac = 1 & bd = 1 
c = 1/a & d = 1/b 
So we can say that (1/a ,1/b) will be the inverse of (a,b) with respect to *.

Question 8. Let * be the binary operation on N defined by a*b = H.C.F of a and b. 

Is  * commutative? Is * associative? Does there exist identity for this binary operation on N?

Solution:

The binary operation * on N can be defined as: 
a*b = H.C.F of a and b 
And we also know that , HCF(a,b) = HCF(b,a) . a,b ∈ N. 
So we can say that , a * b = b * a 
So , the operation * is commutative. 

For a,b,c ∈ N. So we have. 
(a * b) * c = (HCF(a,b))*c = HCF(a,b,c) 
a * (b * c) = a * (HCF(a,b)) = HCF(a,b,c) 
So it can be said that (a * b) * c = a * (b * c) 
So we can say that , the operation * is associative. 

Now , an element e ∈ N will be the identity for the operation. 
* if a * e = a = e * a ,∀ a ∈ N. 
But we can say that , this relation is not true for any a ∈ N. 
So we can say that , the operation * does not have any identity in N. 

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