## 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 = a ^{b} for all a, b ∈ N.**

**(ii) ‘O’ on Z defined by a O b = a ^{b} 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= a ^{b} + b^{a} 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 [∵ a^{b}≠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}

= 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,

X_{6} | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 4 | 0 | 2 | 4 |

3 | 3 | 0 | 3 | 0 | 3 |

4 | 4 | 2 | 0 | 4 | 2 |

5 | 5 | 4 | 3 | 2 | 1 |

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,

+_{6} | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 | 0 |

2 | 2 | 3 | 4 | 5 | 0 | 1 |

3 | 3 | 4 | 5 | 0 | 1 | 2 |

4 | 4 | 5 | 0 | 1 | 2 | 3 |

5 | 5 | 0 | 1 | 2 | 3 | 4 |

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,

a^{b}, b^{a} ∈ 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 = (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 = ab ^{2}**

**(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 + 4b ^{2}**

**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 = ab^{2}

Let a, b ∈ R

⇒ a, b^{2} ∈ R

⇒ ab^{2} ∈ 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 + 4b^{2}

Let a, b ∈ R

⇒ a, 4b^{2} ∈ R

⇒ a + 4b^{2} ∈ 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* = 2*a* + *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:**

LCM | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 2 | 6 | 4 | 10 |

3 | 3 | 5 | 3 | 12 | 15 |

4 | 4 | 4 | 12 | 4 | 20 |

5 | 5 | 10 | 15 | 20 | 5 |

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 = 2 ^{ab} 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 = a ^{2} + b^{2} 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 = ab ^{2} 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 = a ^{b} 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 = 2^{ab}

= 2^{ba}

= 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 * (2^{bc})

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

(a * b) * c = (2^{ab}) * 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 = a^{2} + b^{2}

= b^{2} + a^{2}

= 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 ⊙ (b^{2} + c^{2})

= a^{2} + (b^{2} + c^{2})^{2}

= a^{2} + b^{4} + c^{4} + 2b^{2}c^{2}

(a ⊙ b) ⊙ c = (a^{2} + b^{2}) ⊙ c

= (a^{2} + b^{2})^{2} + c^{2}

= a^{4} + b^{4} + 2a^{2}b^{2} + c^{2}

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 = ab^{2}

b * a = ba^{2}

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 * (bc^{2})

= a (bc^{2})^{2}

= ab^{2} c^{4}

(a * b) * c = (ab^{2}) * c

= ab^{2}c^{2}

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 * (b^{2} + c^{2} – 2 b c)

= (a – b^{2} – c^{2} + 2bc)^{2}

(a * b) * c = (a – b)^{2} * c

= (a^{2} + b^{2} – 2ab) * c

= (a^{2} + b^{2} – 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 = a^{b}

b * a = b^{a}

Therefore, a * b ≠ b * a

Thus, * is not commutative on N.

Now we have to check associativity of *

a * (b * c) = a * (b^{c})

=

(a * b) * c = (a^{b}) * c

= (a^{b})^{c}

= a^{bc}

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 – [1].**

**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 Q _{0} (set of non-zero rational numbers) defined by a * b = (3ab/5) for all a, b ∈ Q_{0}. 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 ∈ Q_{0}

a * b = (3ab/5)

= (3ba/5)

= b * a

Therefore, a * b = b * a, for all a, b ∈ Q_{0}

Now we have to prove associativity of *

Let a, b, c ∈ Q_{0}

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 ∈ Q_{0}

Thus * is associative on Q_{0}

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 ∈ Q_{0}

a * e = a and e * a = a, ∀ a ∈ Q_{0}

3ae/5 = a and 3ea/5 = a, ∀ a ∈ Q_{0}

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

Thus, 5/3 is the identity element in Q_{0} 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 = R _{0} × R, where R_{0} 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) ∈ R_{0} × 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 ∈ R_{0 }and 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 ∈ R_{0 }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)

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 ∈ R_{0 }and 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 ∈ R_{0 }and 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:

×_{4} | 0 | 1 | 2 | 3 |

0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 |

2 | 0 | 2 | 0 | 2 |

3 | 0 | 3 | 2 | 1 |

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

**Solution:**

1 +_{5 }1 = remainder obtained by dividing 1 + 1 by 5

= 2

3 +_{5 }1 = remainder obtained by dividing 3 + 1 by 5

= 2

4 +_{5 }1 = remainder obtained by dividing 4 + 1 by 5

= 3

So, the composition table is as follows:

+_{5} | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 1 | 2 | 3 | 4 |

1 | 1 | 2 | 3 | 4 | 0 |

2 | 2 | 3 | 4 | 0 | 1 |

3 | 3 | 4 | 0 | 1 | 2 |

4 | 4 | 0 | 1 | 2 | 3 |

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

**Solution:**

Here,

1 ×_{6 }1 = remainder obtained by dividing 1 × 1 by 6

= 1

3 ×_{6 }4 = remainder obtained by dividing 3 × 4 by 6

= 0

4 ×_{6 }5 = remainder obtained by dividing 4 × 5 by 6

= 2

So, the composition table is as follows:

×_{6} | 0 | 1 | 2 | 3 | 4 | 5 |

0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 | 5 |

2 | 0 | 2 | 4 | 0 | 2 | 4 |

3 | 0 | 3 | 0 | 3 | 0 | 3 |

4 | 0 | 4 | 2 | 0 | 4 | 2 |

5 | 0 | 5 | 4 | 3 | 2 | 1 |

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

**Solution:**

Here,

1 ×_{5 }1 = remainder obtained by dividing 1 × 1 by 5

= 1

3 ×_{5 }4 = remainder obtained by dividing 3 × 4 by 5

= 2

4 ×_{5 }4 = remainder obtained by dividing 4 × 4 by 5

= 1

So, the composition table is as follows:

×_{5} | 0 | 1 | 2 | 3 | 4 |

0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | 3 | 4 |

2 | 0 | 2 | 4 | 1 | 3 |

3 | 0 | 3 | 1 | 4 | 2 |

4 | 0 | 4 | 3 | 2 | 1 |

**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:

×_{10} | 1 | 3 | 7 | 9 |

1 | 1 | 3 | 7 | 9 |

3 | 3 | 9 | 1 | 7 |

7 | 7 | 1 | 9 | 3 |

9 | 9 | 7 | 3 | 1 |

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.