# RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations

Here we provide RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 1.1 Solutions Chapter 3 Binary Operations book pdf download. Now you will get step-by-step solutions to each question.

### 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 = ab for all a, b ∈ N.

Let a, b ∈ N. Then,

ab ∈ 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 = ab 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= ab + ba for all a, b ∈ N

Let a, b ∈ N. Then,

ab, ba ∈ N

⇒ ab + ba ∈ 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 = 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)= 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.

### 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 ∈ Q0.

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 *.

### Question3. 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 Q.

(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 Q0.

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 ∈ Q0.
So, we can say that o is associative on Q0.

(ii) Let x be the identify element in Qwith respect to * such that
a o x = a x o a ,∀ a ∈ Q0
⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q0
x = 2 ∈ Q0,∀ a ∈ Q
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 *.

### 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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