## 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 Operations

Exercise 3.1 Page No: 3.4

**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}Â for all a, b âˆˆ N.**

**(ii) ‘O’ on Z defined by a O b = a ^{b}Â for all a, b âˆˆ Z.**

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

**(iv) ‘Ã— _{6}‘Â onÂ S = {1,Â 2,Â 3,Â 4,Â 5}Â definedÂ by a Ã—_{6}Â b = RemainderÂ whenÂ a bÂ isÂ dividedÂ byÂ 6.**

**(v) â€˜+ _{6}â€™ on S = {0, 1, 2, 3, 4, 5} defined by a +_{6} b **

**(vi) ‘âŠ™’ on N defined by a âŠ™ b= a ^{b}Â + b^{a}Â for all a, b âˆˆ N**

**(vii) â€˜*â€™ on Q defined by a * b = (a â€“ 1)/ (b + 1) for all a, b âˆˆ Q**

Solution:

(i) Given ‘*’ on N defined by a * b = a^{b}Â for all a, b âˆˆ N.

LetÂ a,Â bÂ âˆˆÂ N.Â Then,

a^{bÂ }âˆˆÂ NÂ Â Â Â Â Â [âˆµÂ a^{b}â‰ 0Â andÂ a, bÂ isÂ positiveÂ integer]

â‡’Â aÂ *Â bÂ âˆˆÂ N

Therefore,

aÂ *Â bÂ âˆˆÂ N,Â âˆ€Â a,Â bÂ âˆˆÂ N

Thus, * is a binary operation onÂ N.

(ii) Given ‘O’ on Z defined by a O b = a^{b}Â for all a, b âˆˆ Z.

BothÂ a = 3Â andÂ b = -1Â belongÂ toÂ Z.

â‡’ aÂ *Â b = 3^{-1}

= 1/3 âˆ‰ Z

Thus, * is not a binary operation onÂ Z.

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

IfÂ aÂ = 1 andÂ b = 1,

a * b = a + b –Â 2

= 1 + 1 –Â 2

= 0Â âˆ‰ N

Thus, there exist a = 1 and b = 1 such that a * bÂ âˆ‰ N

So, * is not a binary operation onÂ N.

(iv) Given ‘Ã—_{6}‘Â onÂ S = {1,Â 2,Â 3,Â 4,Â 5}Â definedÂ by a Ã—_{6}Â b = RemainderÂ whenÂ a bÂ isÂ dividedÂ byÂ 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.

â‡’ ForÂ a = 2Â andÂ b = 3,

a Ã—_{6}Â b = 2 Ã—_{6}Â 3 = remainderÂ whenÂ 6Â dividedÂ byÂ 6 = 0 â‰ S

Thus,Â Ã—_{6}Â isÂ notÂ aÂ binaryÂ operationÂ onÂ S.

(v) Given â€˜+_{6}â€™ 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.

â‡’ ForÂ a = 2Â andÂ b = 3,

a Ã—_{6}Â b = 2 Ã—_{6}Â 3 = remainderÂ whenÂ 6Â dividedÂ byÂ 6 = 0 â‰ Thus,Â Ã—_{6}Â isÂ notÂ aÂ binaryÂ operationÂ onÂ S.

(vi) Given ‘âŠ™’ on N defined by a âŠ™ b= a^{b}Â + b^{a}Â for all a, b âˆˆ N

LetÂ a,Â b âˆˆ N.Â Then,

a^{b},Â b^{a}Â âˆˆ N

â‡’ a^{b}Â + b^{a}Â âˆˆ NÂ Â Â Â Â Â [âˆµAddition is binary operation on N]

â‡’ a âŠ™ b âˆˆ N

Thus,Â âŠ™Â isÂ aÂ binaryÂ operationÂ onÂ N.

(vii) Given â€˜*â€™ on Q defined by a * b = (a â€“ 1)/ (b + 1) for all a, b âˆˆ Q

If a = 2 and b = -1 in Q,

a * b = (a â€“ 1)/ (b + 1)

= (2 â€“ 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) OnÂ Z ^{+}, defined * byÂ aÂ *Â bÂ =Â aÂ â€“Â b**

**(ii) On Z ^{+}, define * byÂ a*bÂ =Â ab**

**(iii) OnÂ R, define * byÂ a*bÂ =Â ab ^{2}**

**(iv) OnÂ Z ^{+}Â define * byÂ aÂ *Â bÂ = |aÂ âˆ’Â b|**

**(v) On Z ^{+ }define * by a * b = a**

**(vi) On R, define * by a * b = a + 4b ^{2}**

**Here,Â Z^{+}Â denotes the set of all non-negative integers.**

**Solution:**

(i) Given OnÂ *Z*^{+}, defined * byÂ aÂ *Â bÂ =Â aÂ â€“Â b

If a = 1 and b = 2 in Z^{+}, then

a * b = a â€“ b

= 1 â€“ 2

= -1 âˆ‰ Z^{+ }[because Z^{+} is the set of non-negative integers]

For a = 1 and b = 2,

a * b âˆ‰ Z^{+}

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

(ii) Given Z^{+}, define * byÂ a*bÂ =Â a b

Let a, b âˆˆ Z^{+}

â‡’ a, b âˆˆ Z^{+}

â‡’ a * b âˆˆ Z^{+}

Thus, * is a binary operation on R.

(iii) Given onÂ R, define byÂ a*bÂ =Â ab^{2}

Let a, b âˆˆ R

â‡’ a, b^{2} âˆˆ R

â‡’ ab^{2} âˆˆ R

â‡’ a * b âˆˆ R

Thus, * is a binary operation on R.

(iv) Given onÂ Z^{+}Â define * byÂ aÂ *Â bÂ = |aÂ âˆ’Â b|

Let a, b âˆˆ Z^{+}

â‡’ | a â€“ b | âˆˆ Z^{+}

â‡’ a * b âˆˆ Z^{+}

Therefore,

a * b âˆˆ Z^{+}, âˆ€ a, b âˆˆ Z^{+}

Thus, * is a binary operation on Z^{+}.

(v) Given on Z^{+ }define * by a * b = a

Let a, b âˆˆ Z^{+}

â‡’ a âˆˆ Z^{+}

â‡’ a * b âˆˆ Z^{+}

Therefore, a * b âˆˆ Z^{+} âˆ€ a, b âˆˆ Z^{+}

Thus, * is a binary operation on Z^{+}.

(vi) Given On R, define * by a * b = a + 4b^{2}

Let a, b âˆˆ R

â‡’ a, 4b^{2} âˆˆ R

â‡’ a + 4b^{2} âˆˆ R

â‡’ a * b âˆˆ R

Therefore, a *b âˆˆ R, âˆ€ a, b âˆˆ R

Thus, * is a binary operation on R.

**3. Let * be a binary operation on the set I of integers, defined byÂ aÂ *Â bÂ = 2aÂ +Â bÂ âˆ’ 3. Find the value of 3 * 4.**

**Solution:**

Given *a*Â *Â *b*Â = 2*a*Â +Â *b*Â â€“ 3

3 * 4 = 2 (3) + 4 â€“ 3

= 6 + 4 â€“ 3

= 7

**4. Is * defined on the set {1, 2, 3, 4, 5} byÂ aÂ *Â bÂ = LCM ofÂ aÂ andÂ bÂ 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 considerÂ aÂ = 2 andÂ bÂ = 3,Â a * b =Â LCM ofÂ aÂ andÂ bÂ = 6 âˆ‰Â {1, 2, 3, 4, 5}.

Thus, * is not a binary operation on {1, 2, 3, 4, 5}.

**5. LetÂ SÂ = {a,Â b,Â c}. Find the total number of binary operations onÂ S.**

**Solution:**

Number of binary operations on a set withÂ nÂ elements is \(n^{n^{2}}\)

Here,Â SÂ = {a,Â b,Â c}

Number of elements inÂ SÂ = 3

Number of binary operations on a set with 3 elements is \(3^{3^{2}}\)

Exercise 3.2 Page No: 3.12

**1. Let ‘*’ be a binary operation onÂ NÂ defined by aÂ *Â bÂ = l.c.m. (a,Â b) for allÂ a,Â bÂ âˆˆÂ N
(i) Find 2 * 4, 3 * 5, 1 * 6.**

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

**Solution:**

(i) Given aÂ *Â bÂ = 1.c.m. (a,Â 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 âˆˆ N

a * b = l.c.m (a, b)

= l.c.m (b, a)

= b * a

Therefore

a * b = b * a âˆ€ a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ 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, âˆ€ a, b , c âˆˆ N

Thus, * is associative on N.

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

**(i) * onÂ NÂ defined byÂ aÂ *Â bÂ = 1 for allÂ a,Â bÂ âˆˆÂ N**

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

**Solution:**

(i) We have to prove commutativity of *

Let a, b âˆˆ N

a * b = 1

b * a = 1

Therefore,

a * b = b * a, for all a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ 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 âˆˆ N

Thus, * is associative on N.

(ii) First we have to prove commutativity of *

Let a, b âˆˆ N

a * b = (a + b)/2

= (b + a)/2

= b * a

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus * is commutative on N.

Now we have to prove associativity of *

Let a, b, c âˆˆ 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) â‰ (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 âˆˆ N such that a * (b * c) â‰ (a * b) * c

Thus, * is not associative on N.

**3. LetÂ AÂ be any set containing more than one element. Let ‘*’ be a binary operation onÂ A defined byÂ aÂ *Â bÂ =Â bÂ for allÂ a,Â bÂ âˆˆÂ AÂ Is ‘*’ commutative or associative onÂ A?**

**Solution:**

Let a, b âˆˆ A

Then, a * b = b

b * a = a

Therefore a * b â‰ b * a

Thus, * is not commutative on A

Now we have to check associativity:

Let a, b, c âˆˆ A

a * (b * c) = a * c

= c

Therefore

a * (b * c) = (a * b) * c, âˆ€ a, b, c âˆˆ A

Thus, * is associative on A

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

**(i) ‘*’ onÂ ZÂ defined byÂ aÂ *Â bÂ =Â aÂ +Â bÂ +Â a bÂ for allÂ a,Â bÂ âˆˆÂ ZÂ **

**(ii) â€˜*â€™ on N defined by a * b = 2 ^{ab} for all a,Â bÂ âˆˆ N**

**(iii) â€˜*â€™ on Q defined by a * b = a â€“ b for all a, b âˆˆ Q**

**(iv) â€˜âŠ™â€™ on Q defined by a âŠ™ b = a ^{2} + b^{2} for all a, b âˆˆ Q**

**(v) â€˜oâ€™ on Q defined by a o b = (ab/2) for all a, b âˆˆ Q**

**(vi) â€˜*â€™ on Q defined by a * b = ab ^{2} for all a, b âˆˆ Q**

**(vii) â€˜*â€™ on Q defined by a * b = a + a b for all a, b âˆˆ Q**

**(viii) â€˜*â€™ on R defined by a * b = a + b -7 for all a, b âˆˆ R**

**(ix) â€˜*â€™ on Q defined by a * b = (a â€“ b) ^{2} for all a, b âˆˆ Q**

**(x) â€˜*â€™ on Q defined by a * b = a b + 1 for all a, b âˆˆ Q**

**(xi) â€˜*â€™ on N defined by a * b = a ^{b} for all a, b âˆˆ N**

**(xii) â€˜*â€™ on Z defined by a * b = a â€“ b for all a, b âˆˆ Z**

**(xiii) â€˜*â€™ on Q defined by a * b = (ab/4) for all a, b âˆˆ Q**

**(xiv) â€˜*â€™ on Z defined by a * b = a + b â€“ ab for all a, b âˆˆ Z**

**(xv) â€˜*â€™ on Q defined by a * b = gcd (a, b) for all a, b âˆˆ Q**

**Solution:**

(i) First we have to check commutativity of *

Let a, b âˆˆ Z

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Z

Now we have to prove associativity of *

Let a, b, c âˆˆ 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, âˆ€ a, b, c âˆˆ Z

Thus, * is associative on Z.

(ii) First we have to check commutativity of *

Let a, b âˆˆ N

a * b = 2^{ab}

= 2^{ba}

= b * a

Therefore, a * b = b * a, âˆ€ a, b âˆˆ N

Thus, * is commutative on N

Now we have to check associativity of *

Let a, b, c âˆˆ N

Then, a * (b * c) = a * (2^{bc})

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

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

=\(2^{ab*2^{c}}\)

Therefore, a * (b * c) â‰ (a * b) * c

Thus, * is not associative on N

(iii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = a â€“ b

b * a = b â€“ a

Therefore, a * b â‰ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c)

= a â€“ (b â€“ c)

= a â€“ b + c

(a * b) * c = (a â€“ b) * c

= a â€“ b â€“ c

Therefore,

a * (b * c) â‰ (a * b) * c

Thus, * is not associative on Q

(iv) First we have to check commutativity of âŠ™

Let a, b âˆˆ Q, then

a âŠ™ b = a^{2} + b^{2}

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

= b âŠ™ a

Therefore, a âŠ™ b = b âŠ™ a, âˆ€ a, b âˆˆ Q

Thus, âŠ™ on Q

Now we have to check associativity of âŠ™

Let a, b, c âˆˆ Q, then

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

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

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

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

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

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

Therefore,

(a âŠ™ b) âŠ™ c â‰ a âŠ™ (b âŠ™ c)

Thus, âŠ™ is not associative on Q.

(v) First we have to check commutativity of o

Let a, b âˆˆ Q, then

a o b = (ab/2)

= (b a/2)

= b o a

Therefore, a o b = b o a, âˆ€ a, b âˆˆ Q

Thus, o is commutative on Q

Now we have to check associativity of o

Let a, b, c âˆˆ 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, âˆ€ a, b, c âˆˆ Q

Thus, o is associative on Q.

(vi) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = ab^{2}

b * a = ba^{2}

Therefore,

a * b â‰ b * a

Thus, * is not commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ 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) â‰ (a * b) * c

Thus, * is not associative on Q.

(vii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = a + ab

b * a = b + ba

= b + ab

Therefore, a * b â‰ b * a

Thus, * is not commutative on Q.

Now we have to prove associativity on Q.

Let a, b, c âˆˆ 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) â‰ (a * b) * c

Thus, * is not associative on Q.

(viii) First we have to check commutativity of *

Let a, b âˆˆ R, then

a * b = a + b â€“ 7

= b + a â€“ 7

= b * a

Therefore,

a * b = b * a, for all a, b âˆˆ R

Thus, * is commutative on R

Now we have to prove associativity of * on R.

Let a, b, c âˆˆ R, then

a * (b * c) = a * (b + c â€“ 7)

= a + b + c -7 -7

= a + b + c â€“ 14

(a * b) * c = (a + b â€“ 7) * c

= a + b â€“ 7 + c – 7

= a + b + c â€“ 14

Therefore,

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

Thus, * is associative on R.

(ix) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = (a â€“ b)^{2}

= (b â€“ a)^{2}

= b * a

Therefore,

a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c âˆˆ Q, then

a * (b * c) = a * (b â€“ c)^{2}

= a * (b^{2} + c^{2} â€“ 2 b c)

= (a â€“ b^{2} â€“ c^{2} + 2bc)^{2}

(a * b) * c = (a â€“ b)^{2} * c

= (a^{2} + b^{2} â€“ 2ab) * c

= (a^{2} + b^{2} â€“ 2ab â€“ c)^{2}

Therefore, a * (b * c) â‰ (a * b) * c

Thus, * is not associative on Q.

(x) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = ab + 1

= ba + 1

= b * a

Therefore

a * b = b * a, for all a, b âˆˆ Q

Thus, * is commutative on Q

Now we have to prove associativity of * on Q

Let a, b, c âˆˆ 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) â‰ (a * b) * c

Thus, * is not associative on Q.

(xi) First we have to check commutativity of *

Let a, b âˆˆ N, then

a * b = a^{b}

b * a = b^{a}

Therefore, a * b â‰ 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) â‰ (a * b) * c

Thus, * is not associative on N

(xii) First we have to check commutativity of *

Let a, b âˆˆ Z, then

a * b = a â€“ b

b * a = b â€“ a

Therefore,

a * b â‰ b * a

Thus, * is not commutative on Z.

Now we have to check associativity of *

Let a, b, c âˆˆ Z, then

a * (b * c) = a * (b â€“ c)

= a â€“ (b â€“ c)

= a â€“ (b + c)

(a * b) * c = (a â€“ b) â€“ c

= a â€“ b â€“ c

Therefore, a * (b * c) â‰ (a * b) * c

Thus, * is not associative on Z

(xiii) First we have to check commutativity of *

Let a, b âˆˆ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

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

Thus, * is commutative on Q

Now we have to check associativity of *

Let a, b, c âˆˆ 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 âˆˆ Q

Thus, * is associative on Q.

(xiv) First we have to check commutativity of *

Let a, b âˆˆ Z, then

a * b = a + b â€“ ab

= b + a â€“ ba

= b * a

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

Thus, * is commutative on Z.

Now we have to check associativity of *

Let a, b, c âˆˆ Z

a * (b * c) = a * (b + c â€“ b c)

= a + b + c- b c â€“ ab â€“ ac + a b c

(a * b) * c = (a + b â€“ a b) c

= a + b â€“ ab + c â€“ (a + b â€“ ab) c

= a + b + c â€“ ab â€“ ac â€“ bc + a b c

Therefore,

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

Thus, * is associative on Z.

(xv) First we have to check commutativity of *

Let a, b âˆˆ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

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

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c âˆˆ 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 âˆˆ N

Thus, * is associative on N.

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

**Solution:**

Let a, b âˆˆ Q â€“ {-1}.

Then aob = a + b â€“ ab

= b+ a â€“ ba

= boa

Therefore,

aob = boa for all a, b âˆˆ Q â€“ {-1}

Thus, o is commutative on Q â€“ {-1}

**6. Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative?**

**Solution:**

Let a, b âˆˆ Z

a * b = 3a + 7b

b * a = 3b + 7a

Thus, a * b â‰ b * a

Let a = 1 and b = 2

1 * 2 = 3 Ã— 1 + 7 Ã— 2

= 3 + 14

= 17

2 * 1 = 3 Ã— 2 + 7 Ã— 1

= 6 + 7

= 13

Therefore, there exist a = 1, b = 2 âˆˆ Z such that a * b â‰ 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 âˆˆ Z. Prove that * is not associative on Z.**

**Solution:**

Let a, b, c âˆˆ 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) â‰ (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 âˆˆ I^{+}.**

**Solution:**

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

a * e = a = e * a, âˆ€ a âˆˆ I^{+}

a * e = a and e * a = a, âˆ€ a âˆˆ I^{+}

a + e = a and e + a = a, âˆ€ a âˆˆ I^{+}

e = 0, âˆ€ a âˆˆ 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 â€“ 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, âˆ€ a âˆˆ Q â€“ {-1}

a * e = a and e * a = a, âˆ€ a âˆˆ Q â€“ {-1}

a + e + ae = a and e + a + ea = a, âˆ€ a âˆˆ Q â€“ {-1}

e + ae = 0 and e + ea = 0, âˆ€ a âˆˆ Q â€“ {-1}

e (1 + a) = 0 and e (1 + a) = 0, âˆ€ a âˆˆ Q â€“ {-1}

e = 0, âˆ€ a âˆˆ Q â€“ {-1} [because a not equal to -1]

Thus, 0 is the identity element in Q â€“ {-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 â€“ 4 for all a, b âˆˆ 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 âˆˆ Z. then,

a * b = a + b â€“ 4

= b + a â€“ 4

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Z

Thus, * is commutative on Z.

Now we have to prove associativity of Z.

Let a, b, c âˆˆ Z. then,

a * (b * c) = a * (b + c – 4)

= a + b + c -4 â€“ 4

= a + b + c â€“ 8

(a * b) * c = (a + b â€“ 4) * c

= a + b â€“ 4 + c â€“ 4

= a + b + c â€“ 8

Therefore,

a * (b * c) = (a * b) * c, for all a, b, c âˆˆ 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 âˆ€ a âˆˆ Z

a * e = a and e * a = a, âˆ€ a âˆˆ Z

a + e â€“ 4 = a and e + a â€“ 4 = a, âˆ€ a âˆˆ Z

e = 4, âˆ€ a âˆˆ Z

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

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

a * b = e = b * a

a * b = e and b * a = e

a + b â€“ 4 = 4 and b + a â€“ 4 = 4

b = 8 â€“ a âˆˆ Z

Thus, 8 â€“ a is the inverse of a âˆˆ 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 âˆˆ 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 âˆˆ Q_{0}

a * b = (3ab/5)

= (3ba/5)

= b * a

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

Now we have to prove associativity of *

Let a, b, c âˆˆ 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 âˆˆ 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 âˆ€ a âˆˆ Q_{0}

a * e = a and e * a = a, âˆ€ a âˆˆ Q_{0}

3ae/5 = a and 3ea/5 = a, âˆ€ a âˆˆ Q_{0}

e = 5/3 âˆ€ a âˆˆ 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 â€“ {-1} defined by a * b = a + b + ab for all a, b âˆˆ Q â€“ {-1}. Then,**

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

**(ii) Find the identity element in Q â€“ {-1}**

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

**Solution:**

(i) First we have to check commutativity of *

Let a, b âˆˆ Q â€“ {-1}

Then a * b = a + b + ab

= b + a + ba

= b * a

Therefore,

a * b = b * a, âˆ€ a, b âˆˆ Q â€“ {-1}

Now we have to prove associativity of *

Let a, b, c âˆˆ Q â€“ {-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, âˆ€ a, b, c âˆˆ Q â€“ {-1}

Thus, * is associative on Q â€“ {-1}.

(ii) Let e be the identity element in I^{+} with respect to * such that

a * e = a = e * a, âˆ€ a âˆˆ Q â€“ {-1}

a * e = a and e * a = a, âˆ€ a âˆˆ Q â€“ {-1}

a + e + ae = a and e + a + ea = a, âˆ€ a âˆˆ Q â€“ {-1}

e + ae = 0 and e + ea = 0, âˆ€ a âˆˆ Q â€“ {-1}

e (1 + a) = 0 and e (1 + a) = 0, âˆ€ a âˆˆ Q â€“ {-1}

e = 0, âˆ€ a âˆˆ Q â€“ {-1} [because a not equal to -1]

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

(iii) Let a âˆˆ Q â€“ {-1} and b âˆˆ Q â€“ {-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 â€“ {-1}

b = -a/1 + a Q â€“ {-1} [because a not equal to -1]

Thus, -a/1 + a is the inverse of a âˆˆ Q â€“ {-1}

**4. LetÂ AÂ =Â R _{0}Â Ã—Â R, whereÂ R_{0}Â denote the set of all non-zero real numbers.Â AÂ binary operation ‘O’ is defined onÂ AÂ as follows: (a,Â b) O (c,Â d) = (ac,Â bcÂ +Â d) for all (a,Â b), (c,Â d) âˆˆÂ R_{0}Â Ã—Â R.**

**(i) Show that ‘O’ is commutative and associative onÂ 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) âˆˆ A, âˆ€ a, c âˆˆ R_{0 } and b, d âˆˆ R

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

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

Therefore,

X O Y = Y O X, âˆ€ X, Y âˆˆ 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), âˆ€ a, c, e âˆˆ R_{0 } and b, d, f âˆˆ 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, âˆ€ X, Y, Z âˆˆ A

(ii) Let E = (x, y) be the identity element in A with respect to O, âˆ€ x âˆˆ R_{0 } and y âˆˆ R

Such that,

X O E = X = E O X, âˆ€ X âˆˆ 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 âˆ€ m âˆˆ R_{0 } and n âˆˆ 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) âˆˆ A with respect to O is (1/a, -b/a)

Exercise 3.5 Page No: 3.33

**1. Construct the composition table for Ã— _{4}Â on setÂ SÂ = {0, 1, 2, 3}.**

**Solution:**

Given that Ã—_{4}Â on setÂ *S*Â = {0, 1, 2, 3}

Here,

1 Ã—_{4} 1 = remainder obtained by dividing 1 Ã— 1 by 4

= 1

0 Ã—_{4} 1 = remainder obtained by dividing 0 Ã— 1 by 4

= 0

2 Ã—_{4} 3 = remainder obtained by dividing 2 Ã— 3 by 4

= 2

3 Ã—_{4} 3 = remainder obtained by dividing 3 Ã— 3 by 4

= 1

So, the composition table is as follows:

Ã—_{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 Ã— _{6}Â on setÂ SÂ = {0, 1, 2, 3, 4, 5}.**

**Solution:**

Here,

1 Ã—_{6 }1 = remainder obtained by dividing 1 Ã— 1 by 6

= 1

3 Ã—_{6 }4 = remainder obtained by dividing 3 Ã— 4 by 6

= 0

4 Ã—_{6 }5 = remainder obtained by dividing 4 Ã— 5 by 6

= 2

So, the composition table is as follows:

Ã—_{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 Ã— _{5}Â on set Z_{5}Â = {0, 1, 2, 3, 4}**

**Solution:**

Here,

1 Ã—_{5 }1 = remainder obtained by dividing 1 Ã— 1 by 5

= 1

3 Ã—_{5 }4 = remainder obtained by dividing 3 Ã— 4 by 5

= 2

4 Ã—_{5 }4 = remainder obtained by dividing 4 Ã— 4 by 5

= 1

So, the composition table is as follows:

Ã—_{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 Ã— _{10 }set S = {1, 3, 7, 9}, find the inverse of 3.**

**Solution:**

Here,

1 Ã—_{10 }1 = remainder obtained by dividing 1 Ã— 1 by 10

= 1

3 Ã—_{10 }7 = remainder obtained by dividing 3 Ã— 7 by 10

= 1

7 Ã—_{10 }9 = remainder obtained by dividing 7 Ã— 9 by 10

= 3

So, the composition table is as follows:

Ã—_{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 âˆˆ S is the identity element with respect to Ã—_{10}

Now we have to find inverse of 3

3 Ã—_{10} 7 = 1

So the inverse of 3 is 7.

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

## RD Sharma Class 12 Solutions Chapter 3 Binary Operations

Let us have a look at some of the important concepts that are discussed in this chapter.

- Definition and meaning of binary operations
- Number of binary operations
- Types of binary operations
- Commutativity
- Associativity
- Distributivity

- Identity element
- The inverse of an element
- Composition table
- Multiplication modulo