# RD Sharma Solutions for Class 12 Maths Chapter 3 Binary Operations

## RD Sharma Solutions for Class 12 Maths Chapter 3 – Free PDF Download

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### Access RD Sharma Solutions For Class 12 Maths Chapter 3 – Binary Operatio

Exercise 3.1

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,

a∈ 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

= 1/3 ∉ 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 = (a – 1)/ (b + 1)

= (2 – 1)/ (- 1 + 1)

= 1/0 [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.

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 Zdefine * 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 Zdefine * 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.

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

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}.

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 [latex]n^{n^{2}}[/latex]

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is [latex]3^{3^{2}}[/latex]

Exercise 3.2 Page No: 3.12

1. Let ‘*’ be a binary operation on N defined by a * b = l.c.m. (a, b) for all a, b ∈ N
(i) Find 2 * 4, 3 * 5, 1 * 6.

(ii) Check the commutativity and associativity of ‘*’ on N.

Solution:

(i) Given a * b = 1.c.m. (a, b)

2 * 4 = l.c.m. (2, 4)

= 4

3 * 5 = l.c.m. (3, 5)

= 15

1 * 6 = l.c.m. (1, 6)

= 6

(ii) We have to prove commutativity of *

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.

Now we have to prove associativity of *

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)

(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.

2. Determine which of the following binary operation is associative and which is commutative:

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

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

Solution:

(i) We have to prove commutativity of *

Let a, b ∈ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b ∈ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c ∈ N

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

= 1

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

= 1

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

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

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.

Now we have to prove associativity of *

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

(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.

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:

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

4. Check the commutativity and associativity of each of the following binary operations:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution:

(i) First we have to check commutativity of *

Let a, b ∈ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Now we have to prove associativity of *

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

(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

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Z

Thus, * is associative on Z.

(ii) First we have to check commutativity of *

Let a, b ∈ N

a * b = 2ab

= 2ba

= b * a

Therefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on N

Now we have to check associativity of *

Let a, b, c ∈ N

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

=[latex]2^{a*2^{bc}}[/latex]

(a * b) * c = (2ab) * c

=[latex]2^{ab*2^{c}}[/latex]

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(iii) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Therefore, a * 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 * (b – c)

= a – (b – c)

= a – b + c

(a * b) * c = (a – b) * c

= a – b – c

Therefore,

a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q

(iv) First we have to check commutativity of ⊙

Let a, b ∈ Q, then

a ⊙ b = a2 + b2

= b2 + a2

= b ⊙ a

Therefore, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ on Q

Now we have to check associativity of ⊙

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

Therefore,

(a ⊙ b) ⊙ c ≠ a ⊙ (b ⊙ c)

Thus, ⊙ is not associative on Q.

(v) First we have to check commutativity of o

Let a, b ∈ Q, then

a o b = (ab/2)

= (b a/2)

= b o a

Therefore, a o b = b o a, ∀ a, b ∈ Q

Thus, o is commutative on Q

Now we have to check associativity of o

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

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

= [(ab/2) c] /2

= (a b c)/4

Therefore a o (b o c) = (a o b) o c, ∀ a, b, c ∈ Q

Thus, o is associative on Q.

(vi) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab2

b * a = ba2

Therefore,

a * 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) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

Therefore, a * b ≠ b * a

Thus, * is not commutative on Q.

Now we have to prove associativity on Q.

Let a, b, c ∈ Q, then

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

= a + a (b + b c)

= a + ab + a b c

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

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

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

Therefore a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(viii) First we have to check commutativity of *

Let a, b ∈ R, then

a * b = a + b – 7

= b + a – 7

= b * a

Therefore,

a * b = b * a, for all a, b ∈ R

Thus, * is commutative on R

Now we have to prove associativity of * on R.

Let a, b, c ∈ R, then

a * (b * c) = a * (b + c – 7)

= a + b + c -7 -7

= a + b + c – 14

(a * b) * c = (a + b – 7) * c

= a + b – 7 + c – 7

= a + b + c – 14

Therefore,

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

Thus, * is associative on R.

(ix) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = (a – b)2

= (b – a)2

= b * a

Therefore,

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

a * (b * c) = a * (b – c)2

= a * (b2 + c2 – 2 b c)

= (a – b2 – c2 + 2bc)2

(a * b) * c = (a – b)2 * c

= (a2 + b2 – 2ab) * c

= (a2 + b2 – 2ab – c)2

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(x) First we have to check commutativity of *

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c ∈ Q, then

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

= a (b c + 1) + 1

= a b c + a + 1

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

= (ab + 1) c + 1

= a b c + c + 1

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

(xi) First we have to check commutativity of *

Let a, b ∈ N, then

a * b = ab

b * a = ba

Therefore, a * b ≠ b * a

Thus, * is not commutative on N.

Now we have to check associativity of *

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

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

= (ab)c

= abc

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N

(xii) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Therefore,

a * b ≠ b * a

Thus, * is not commutative on Z.

Now we have to check associativity of *

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

Therefore, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z

(xiii) First we have to check commutativity of *

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

Now we have to check associativity of *

Let a, b, c ∈ Q, then

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

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

= (a b c/16)

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

= [(ab/4) c]/4

= a b c/16

Therefore,

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

Thus, * is associative on Q.

(xiv) First we have to check commutativity of *

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

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

Thus, * is commutative on Z.

Now we have to check associativity of *

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – b c)

= a + b + c- b c – ab – ac + a b c

(a * b) * c = (a + b – a b) c

= a + b – ab + c – (a + b – ab) c

= a + b + c – ab – ac – bc + a b c

Therefore,

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

Thus, * is associative on Z.

(xv) First we have to check commutativity of *

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)

Therefore,

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

Thus, * is associative on N.

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 – .

Solution:

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – ba

= boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}

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

b * a = 3b + 7a

Thus, a * b ≠ b * a

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.

7. On the set Z of integers a binary operation * is defined by a 8 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

Thus, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

Exercise 3.3 Page No: 3.15

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+

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

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]

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

Exercise 3.4 Page No: 3.25

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 have to prove commutativity of *

Let a, b ∈ Z. then,

a * b = a + b – 4

= b + a – 4

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Z

Thus, * is commutative on Z.

Now we have to prove associativity of Z.

Let a, b, c ∈ Z. then,

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

Therefore,

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

Thus, * is associative on Z.

(ii) Let e be the identity element in Z with respect to * such that

a * e = a = e * a ∀ a ∈ Z

a * e = a and e * a = a, ∀ a ∈ Z

a + e – 4 = a and e + a – 4 = a, ∀ a ∈ Z

e = 4, ∀ a ∈ Z

Thus, 4 is the identity element in Z with respect to *.

(iii) Let a ∈ Z and b ∈ Z be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b – 4 = 4 and b + a – 4 = 4

b = 8 – a ∈ Z

Thus, 8 – a is the inverse of a ∈ Z

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:

First we have to prove commutativity of *

Let a, b ∈ Q0

a * b = (3ab/5)

= (3ba/5)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q0

Now we have to 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

Therefore a * (b * c) = (a * b) * c, for all a, b, c ∈ Q0

Thus * is associative on Q0

Now we have to find the identity element

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

a * e = a = e * a ∀ a ∈ Q0

a * e = a and e * a = a, ∀ a ∈ Q0

3ae/5 = a and 3ea/5 = a, ∀ a ∈ Q0

e = 5/3 ∀ a ∈ Q[because a is not equal to 0]

Thus, 5/3 is the identity element in Q0 with respect to *.

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 have to check commutativity of *

Let a, b ∈ Q – {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, ∀ a, b ∈ Q – {-1}

Now we have to prove associativity of *

Let 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

Therefore,

a * (b * c) = (a * b) * c, ∀ a, b, c ∈ Q – {-1}

Thus, * is associative on Q – {-1}.

(ii) 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]

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

(iii) Let 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} [because a not equal to -1]

Thus, -a/1 + a is the inverse of a ∈ Q – {-1}

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 X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ Rand b, d ∈ R

Then, X O Y = (ac, bc + d)

And Y O X = (ca, da + b)

Therefore,

X O Y = Y O X, ∀ X, Y ∈ A

Thus, O is not 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 ∈ Rand 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)

Therefore, X O (Y O Z) = (X O Y) O Z, ∀ X, Y, Z ∈ A

(ii) Let E = (x, y) be the identity element in A with respect to O, ∀ x ∈ Rand y ∈ R

Such that,

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)

Considering (ax, bx + y) = (a, b)

ax = a

x = 1

And bx + y = b

y = 0 [since x = 1]

Considering (xa, ya + b) = (a, b)

xa = a

x = 1

And ya + b = b

y = 0 [since x = 1]

Therefore (1, 0) is the identity element in A with respect to O.

(iii) Let F = (m, n) be the inverse in A ∀ m ∈ Rand n ∈ R

X O F = E and F O X = E

(am, bm + n) = (1, 0) and (ma, na + b) = (1, 0)

Considering (am, bm + n) = (1, 0)

am = 1

m = 1/a

And bm + n = 0

n = -b/a [since m = 1/a]

Considering (ma, na + b) = (1, 0)

ma = 1

m = 1/a

And na + b = 0

n = -b/a

Therefore the inverse of (a, b) ∈ A with respect to O is (1/a, -b/a)

Exercise 3.5 Page No: 3.33

1. Construct the composition table for ×4 on set S = {0, 1, 2, 3}.

Solution:

Given that ×4 on set S = {0, 1, 2, 3}

Here,

1 ×4 1 = remainder obtained by dividing 1 × 1 by 4

= 1

0 ×4 1 = remainder obtained by dividing 0 × 1 by 4

= 0

2 ×4 3 = remainder obtained by dividing 2 × 3 by 4

= 2

3 ×4 3 = remainder obtained by dividing 3 × 3 by 4

= 1

So, the composition table is as follows:

2. Construct the composition table for +5 on set S = {0, 1, 2, 3, 4}

Solution:

1 +1 = remainder obtained by dividing 1 + 1 by 5

= 2

3 +1 = remainder obtained by dividing 3 + 1 by 5

= 2

4 +1 = remainder obtained by dividing 4 + 1 by 5

= 3

So, the composition table is as follows:

3. Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.

Solution:

Here,

1 ×1 = remainder obtained by dividing 1 × 1 by 6

= 1

3 ×4 = remainder obtained by dividing 3 × 4 by 6

= 0

4 ×5 = remainder obtained by dividing 4 × 5 by 6

= 2

So, the composition table is as follows:

4. Construct the composition table for ×5 on set Z5 = {0, 1, 2, 3, 4}

Solution:

Here,

1 ×1 = remainder obtained by dividing 1 × 1 by 5

= 1

3 ×4 = remainder obtained by dividing 3 × 4 by 5

= 2

4 ×4 = remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

5. For the binary operation ×10 set S = {1, 3, 7, 9}, find the inverse of 3.

Solution:

Here,

1 ×10 1 = remainder obtained by dividing 1 × 1 by 10

= 1

3 ×10 7 = remainder obtained by dividing 3 × 7 by 10

= 1

7 ×10 9 = remainder obtained by dividing 7 × 9 by 10

= 3

So, the composition table is as follows:

From the table we can observe that elements of first row as same as the top-most row.

So, 1 ∈ S is the identity element with respect to ×10

Now we have to find inverse of 3

3 ×10 7 = 1

So the inverse of 3 is 7.