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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 3 |

Exercise | 1.1 |

Category | RD Sharma Solutions |

Table of Contents

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

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

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

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

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

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

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

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

**Solution:**

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

Let a, b ∈ N. Then,

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

⇒ a * b ∈ N

Therefore,

a * b ∈ N, ∀ a, b ∈ N

Thus, * is a binary operation on N.

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

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

⇒ a * b = 3^{-1}

= ∉ Z

Thus, * is not a binary operation on Z.

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

If a = 1 and b = 1,

a * b = a + b – 2

= 1 + 1 – 2

= 0 ∉ N

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

So, * is not a binary operation on N.

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

Consider the composition table,

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,

ab, ba ∈ N

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

⇒ a ⊙ b ∈ N

Thus, ⊙ is a binary operation on N.

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

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

a * b =

=

= [which is not defined]

For a = 2 and b = -1

a * b does not belongs to Q

So, * is not a binary operation in Q.

**Question 2. Determine whether or not the definition of * given below gives a binary operation. In the event that * is not a binary operation give justification of this.**

**(i) On Z ^{+}, defined * by a * b = a – b**

**(ii) On Z ^{+}, define * by a*b = ab**

**(iii) On R, define * by a*b = 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 – bIf a = 1 and b = 2 in Z

^{+}, thena * 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 bLet 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 = aLet 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.

**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:**

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}.

**Question 5. Let S = {a, b, c}. Find the total number of binary operations on S.**

**Solution:**

Number of binary operations on a set with n elements is

Here, S = {a, b, c}

Number of elements in S = 3

Number of binary operations on a set with 3 elements is

**Question 6. Find the total number of binary operations on {a, b}.**

**Solution:**

We have,

S = {a, b}

The total number of binary operation on S = {a, b} in

**Question 7. Prove that the operation * on the set**

**M=**** defined by A + B = AB is a binary operation.**

**Solution: **

We have,

and

A + B = AB for all A, B ∈ M

Let A =\ and B =

Now, AB =

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

⇒ ac ∈ R and bd ∈ R

⇒

⇒ A * B ∈ M

Hence, the operator * defines a binary operation on M

**Question 8. Let S be the set of all rational numbers of the form **** where m ∈ Z and n = 1, 2, 3. Prove that * on S defined by a * b = ab is not a binary operation**

**Solution: **

S = set of rational numbers of the form where m ∈ Z and n = 1, 2, 3

Also, a * b = ab

Let a ∈ S and b ∈ S

⇒ ab =

Therefore, a * b ∉ S

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

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

**Solution:**

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

Now,

(2*3) = 2 × 2 + 3

= 4 + 3

(2*3)*4 = 7*4 = 2 × 7 + 4

= 14 + 4

= 18

**Question 10. Let * be a binary operation on N given by a*b = LCM(a, b) for all a, b ∈ N. Find 5*7.**

**Solution:**

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

Now,

5*7 = LCM (5, 7)

= 35

### (i) Find 2 * 4, 3 * 5, 1 * 6

**Solution:**

We are given that a * b = L.C.M. (a, b)

⇒ 2 * 4 = L.C.M. (2, 4) = 4

and, 3 * 5 = L.C.M. (3, 5) = 15

now, 1 * 6 = L.C.M. (1, 6) = 6

Hence, 2 * 4 = 4, 3 * 5 = 15 and 1 * 6 = 6.

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

**Solution:**

For Commutativity:

Let a, b ∈ N

a * b = L.C.M. (a, b) = L.C.M. (b, a) = b * a

Therefore, a * b = b * a ∀ a, b ∈ N

Thus * is commutative on N.

For Associativity:

Let a, b, c ∈ N

⇒ a * (b * c) = a * L.C.M. (b, c) = L.C.M. (a, (b, c)) = L.C.M. (a, b, c)

And, (a * b) * c = L.C.M. (a, b) * c = L.C.M. ((a, b), c) = L.C.M. (a, b, c)

Therefore, (a * (b * c) = (a * b) * c, ∀ a, b, c ∈ N

Thus, * is associative on N.

**Solution:**

For Commutativity:

Let a, b ∈ Z

Then a * b = a + b + ab = b + a + ba = b * a

Therefore, a * b = b * a, ∀ a, b ∈ Z

Hence, * is commutative on Z.For Associativity:

Let a, b, c ∈ Z, Then,

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

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

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

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

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

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

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

Thus, * is associative on Z.

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

**Solution:**

For Commutativity:

Let a, b ∈ N

a * b = 2

^{ab}= 2^{ba}= b * aTherefore, a * b = b * a, ∀ a, b ∈ N

Thus, * is commutative on NFor Associativity:

Let a, b, c ∈ N

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

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

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

Thus, * is not associative on N.

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

**Solution:**

For Commutativity:

Let a, b ∈ Q, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Q.For Associativity:

Let a, b, c ∈ Q, then

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

and, (a * b) * c = (a – b) * c = a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

### (iv) ‘⊙’ on Q defined by a ⊙ b = a^{2} + b^{2} for all a, b ∈ Q

**Solution:**

For Commutativity:

Let a, b ∈ Q, then

a ⊙ b = a

^{2}+ b^{2}= b^{2}+ a^{2}= b ⊙ aClearly, a ⊙ b = b ⊙ a, ∀ a, b ∈ Q

Thus, ⊙ is commutative on Q.For Associativity:

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}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 = ab^{2} for all a, b ∈ Q

**Solution:**

For Commutativity:

Let a, b ∈ Q, then

a * b = ab

^{2}b * a = ba

^{2}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 * (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) ‘*’ 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 * (b

^{2}+ c^{2}– 2bc)= (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}Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

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

**Solution:**

For Commutativity:

Let a, b ∈ Q, then

a * b = ab + 1

= ba + 1

= b * a

Clearly, a * b = b * a, for all a, b ∈ Q

Thus, * is commutative on Q.For Associativity:

Let a, b, c ∈ Q, then

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

= a (bc + 1) + 1

= abc + a + 1

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

= (ab + 1) c + 1

= abc + c + 1

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Q.

### (xi) ‘*’ on N defined by a * b = a^{b }for all a, b ∈ N

**Solution:**

For Commutativity:

Let a, b ∈ N, then

a * b = a

^{b}b * a = b

^{a}Clearly, a * b ≠ b * a

Thus, * is not commutative on N.For Associativity:

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

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

^{b})^{c }= a^{bc}Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on N.

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

**Solution:**

Let a, b ∈ Z, then

a * b = a – b

b * a = b – a

Clearly, a * b ≠ b * a

Thus, * is not commutative on Z.For Associativity:

Let a, b, c ∈ Z, then

a * (b * c) = a * (b – c)

= a – (b – c)

= a – (b + c)

(a * b) * c = (a – b) – c

= a – b – c

Clearly, a * (b * c) ≠ (a * b) * c

Thus, * is not associative on Z.

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

**Solution:**

For Commutativity:

Let a, b ∈ Q, then

a * b = (ab/4)

= (ba/4)

= b * a

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

Thus, * is commutative on QFor Associativity:

Let a, b, c ∈ Q, then

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

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

= (a b c/16)

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

= [(ab/4) c]/4

= abc/16

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

Thus, * is associative on Q.

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

**Solution:**

For Commutativity:

Let a, b ∈ Z, then

a * b = a + b – ab

= b + a – ba

= b * a

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

Thus, * is commutative on Z.For Associativity:

Let a, b, c ∈ Z

a * (b * c) = a * (b + c – bc)

= a + b + c- b c – ab – ac + abc

(a * b) * c = (a + b – ab) c

= a + b – ab + c – (a + b – ab)

= a + b + c – ab – ac – bc + a b c

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

Thus, * is associative on Z.

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

**Solution:**

For Commutativity:

Let a, b ∈ N, then

a * b = gcd (a, b)

= gcd (b, a)

= b * a

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

Thus, * is commutative on N.

Now we have to check associativity of *

Let a, b, c ∈ N

a * (b * c) = a * [gcd (a, b)]

= gcd (a, b, c)

(a * b) * c = [gcd (a, b)] * c

= gcd (a, b, c)

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

Thus, * is associative on N.

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

**Solution:**

Let a, b ∈ Q – {-1}.

Then aob = a + b – ab

= b+ a – b = boa

Therefore,

aob = boa for all a, b ∈ Q – {-1}

Thus, o is commutative on Q – {-1}.

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

**Solution:**

Let a, b ∈ Z

a * b = 3a + 7b

and, b * a = 3b + 7a

Clearly, a * b ≠ b * a for all a, b ∈ Z.

Example, Let a = 1 and b = 2

1 * 2 = 3 × 1 + 7 × 2 = 3 + 14 = 17

2 * 1 = 3 × 2 + 7 × 1 = 6 + 7 = 13

Therefore, there exist a = 1, b = 2 ∈ Z such that a * b ≠ b * a

Thus, * is not commutative on Z.

### Question 7. On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.

**Solution:**

Let a, b, c ∈ Z

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

= a (bc + 1) + 1

= a b c + a + 1

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

= (ab + 1) c + 1

= a b c + c + 1

Clearly, a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Z

Thus, * is not associative on Z.

### Question 8. Let S be the sum of all real numbers except −1 and let * be an operation defined by a * b = a + b + ab for all a,b ∈ S. Determine whether * is a binary operation on S. If yes, check its commutativity and associativity.

**Solution:**

Given: a * b = a + b + ab, a, b ∈ S = R − {−1}

Let a, b ∈ S.

Thus, ab ∈ S and hence, a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.For Commutativity:

a * b = a + b + ab = b +a + ba = b * a

Hence, * is commutative.For Associativity:

Let a, b, c ∈ Z, Then,

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

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

= a + b + c + ab + ac + bc + abc …..(a)

Now, a * (b * c) = a * (b + c + bc)

= a + b + c + bc + ac +ab +abc …..(b)

From (a) and (b), it is clear that a * (b * c) = (a * b) * c for all a, b, c ∈ Q

Hence, * is associative on Q.

### Question 9. On Q, the set of rational numbers, * is defined by a * b = (a – b)/2, show that * is not associative.

**Solution:**

Let a, b, c ∈ Q. Then,

(a * b) * c = * c = = …….(a)

Now, a * (b * c) = a * = ……….(b)

From (a) and (b), it is clear that a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

### Question 10. Let binary operation * : R×R⇥R is defined as a * b = 2a + b. Find (2 * 3) * 4

**Solution:**

Given, a * b = 2a + b

⇒ (2 * 3) * 4 = (2 × 2 + 3) * 4 = 7 * 4 = (2 × 7 + 4) = 18

Hence, (2 * 3) * 4 = 18.

### Question 11. On Z, the set of integers, a binary operation * is defined as a * b = a + 3b − 4. Prove that * is neither commutative nor associative on Z.

**Solution:**

For Commutativity:

a * b = a + 3b − 4 ≠ b + 3a − 4 = b * a

⇒ a * b ≠ b * a

Hence * is not commutative on Z.For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + 3b − 4) * c

= a + 3b − 4 + 3c − 4

= a + 3b + 3c − 8 …….(a)

Now, a * (b * c) = a + 3(b + 3c − 4) − 4

= a + 3b + 9c − 16 ……(b)

From (a) and (b), we get a * (b * c) ≠ (a * b) * c for all a, b, c ∈ Q

Hence, * is not associative on Q.

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

### a * b = ab/5, prove that * is associative on Q.

**Solution:**

Let a, b, c ∈ Z, then,

(a * b) * c = ab/5 * c = abc/25 …..(a)

and, a * (b * c) = a * bc/5 = abc/25 ….(b)

From eq (a) and (b), we have

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

Hence, * is associative on Q.

### Question 13. The binary operation * is defined as a * b = ab/7 on the set Q of rational numbers. Prove that * is associative on Q.

**Solution:**

Let a, b, c ∈ Z, then,

(a * b) * c = ab/7 * c = abc/49 …..(a)

and, a * (b * c) = a * bc/7 = abc/49 ….(b)

From eq(a) and (b), we have

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

Hence, * is associative on Q.

### Question 14. On Q, the set of all rational numbers, a binary operation * is defined as (a + b)/2 . Show that * is not associative on Q.

**Solution:**

Let a, b, c ∈ Z, then,

(a * b) * c = * c = = …(a)

a * (b * c) = a * = = …(b)

From eq(a) and (b), we have,

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

Hence, * is not associative on Q.

### Question 15. Let S be the sum of all real numbers except 1 and let * be an operation defined by a * b = a + b − ab for all a, b ∈ S. Prove that:

### (i) * is a binary operation on S.

**Solution:**

Let a, b ∈ S

Thus, ab ∈ S and hence,

a + b − ab ∈ S or a * b ∈ S

Hence, a * b S is a binary operation.

### (ii) is commutative and associative.

**Solution:**

For Commutativity:

a * b = a + b − ab = b + a − ba = b * a

Hence, * is commutative.For Associativity:

Let a, b, c ∈ Z, Then,

(a * b) * c = (a + b − ab) * c

= a + b − ab + c + (a + b − ab)c

= a + b + c − ab − ac − bc + abc …..(a)

Now, a * (b * c) = a * (b + c − bc)

= a + b + c − bc − ac − ab +abc …..(b)

From eq(a) and (b), it is clear that

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

Hence, * is associative on Q.

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

^{+}.

**Solution:**

Let e be the identity element in I

^{+}with respect to * such thata * 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 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.

_{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:**

Firstly we will prove commutativity of *

Let a, b ∈ Q_{0}

a * b = (3ab/5)

= (3ba/5)

= b * a

⇒ a * b = b * a, for all a, b ∈ Q_{0.}Now we will prove associativity of *

Let a, b, c ∈ 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

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

So we can say that * is associative on Q_{0}.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 ∈ Q_{0}

a * x = a and x * a = a, ∀ a ∈ Q_{0}

3ax/5 = a and 3xa/5 = a, ∀ a ∈ Q_{0}

x = 5/3 ∀ a ∈ Q_{0}[a ≠ 0]

So we can say that, 5/3 is the identity element in Q_{0}with respect to *.

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

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

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

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

**Solution:**

(i)First we will check commutativity of *

Let us assume that a, b ∈ Q – {-1}

a * b = a + b + ab

= b + a + ba

= b * a

⇒

a * b = b * a, ∀ a, b ∈ Q – {-1}Now we will prove associativity of *

Let us assume that a, b, c ∈ Q – {-1}, Then,

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

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

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

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

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

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

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

So we can say that, * is associative on Q – {-1}

(ii)Let us assume that x be the identity element in I+ with respect to * such that

a * x = a = x * a, ∀ a ∈ Q – {-1}

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

a + x + ax = a and x + a + xa = a, ∀ a ∈ Q – {-1}

x + ax = 0 and x + xa = 0, ∀ a ∈ Q – {-1}

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

x = 0, ∀ a ∈ Q – {-1} [a ≠ -1]

so we can say that , 0 is the identity element in Q – {-1} with respect to *.

(iii)Let us assume that a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. Then,

a * b = e = b * a

a * b = e and b * a = e

a + b + ab = 0 and b + a + ba = 0

b (1 + a) = – a Q – {-1}

b = -a/1 + a Q – {-1} [a ≠ -1]

So we can say that, -a/1 + a is the inverse of a ∈ Q – {-1}.

**Question 4. Let A = 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.

_{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 us assume that X = (a, b) and Y = (c, d) ∈ A, ∀ a, c ∈ R_{0}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 ∈ R_{0}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 ∈ 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)

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 Q**_{0} of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q_{0}

_{0}of all non zero rational numbers defined by a * b = ab/2 for all a, b ∈ Q

_{0}

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

**(ii) Find the identity element in Q _{0 }.**

**(iii) Find the invertible element of Q _{0}.**

**Solution:**

(i)We have to show,‘*’ is commutative.

Let a, b ∈ Q_{0}.

a o b = ab/2 = ba/2

⇒ b o a

⇒ a o b = b o a, ∀ a, b ∈ Q_{0}.

So, o is commutative on Q_{0.}Now, we will show, ‘*’ is Associative.

Let a, b, c ∈ Q_{0}

a o (b 0 c) = a o (bc/2)

= (a(bc/2))/2

= abc /4

⇒ (a o b) o c = (ab/2) o c

= abc/4

⇒ a o (b o c) = (a o b) o c ∀ a, b, c ∈ Q_{0.}

So, we can say that o is associative on Q_{0.}

(ii)Let x be the identify element in Q_{0 }with respect to * such that

a o x = a x o a ,∀ a ∈ Q_{0}

⇒ ax /2 = a and xa /2 = a, ∀ a ∈ Q_{0}

x = 2 ∈ Q_{0,}∀ a ∈_{ }Q_{0 }

So, we can say that, 2 is the identity element in Q_{0}with respect to o.

(iii)Let us assume that a ∈ Q_{0}and b ∈ Q_{0}be the inverse of a.

⇒ a o b = e = b o a = e

⇒ ab/2 = 2 and ba/2 = 2

⇒ b = 4/a ∈ Q_{0}

So, we can say that, 4/a is the inverse of a∈ Q_{0}.

**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 R**_{0} denote the set of all non zero real number and let A = R_{0} x R_{0} . If ‘*’ is a binary operation on A defined by ( a, b) * (c ,d) = (ac , bd) for all (a , b)(c , d) ∈ A.

_{0}denote the set of all non zero real number and let A = R

_{0}x R

_{0}. If ‘*’ is a binary operation on A defined by ( a, b) * (c ,d) = (ac , bd) for all (a , b)(c , d) ∈ A.

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

**(ii) Find the identity element in A.**

**(iii) Find the invertible element in A.**

**Solution:**

In the question we have given (a, b) * (c ,d) = (ac , bd) for all (a,b)(c,d) ∈ A.

(i) Let us assume that , (a,b)(c,d) ∈ A. So,

(a, b) * (c ,d) = (ac , bd)

=(ca , bd) [ ac = ca and bd = db ]

=(c , d)*(a , b)

⇒ (a, b) * (c,d) = (ac,bd)

So we can say that , ‘*’ is commutative on A.⇒ Now we will find associativity on A.

Let us assume that , (a,b),(c,d),(e,f) ∈ A.

⇒ ((a,b)*(c,d))*(e,f) = (ac , bd)*(e,f)

=(ace , bdf) –(i)

Now (a,b)*((c,d)*(e,f)) =(a,b)*(ce,df)

=(ace , bdf) –(ii)

From equation (i) and (ii).

((a,b)*(c,d))*(e,f) = (a,b)*((c,d)*(e,f))

So we can say that , ‘*’ is associative on A.(ii) Let find identity element in A.

Let assume that (x,y) ∈ A be the identity element with respect to *.

(a,b) * (x,y) = (x,y)*(a,b) = (a,b) for all (a,b) ∈ A.

⇒ (ax , by) = (a,b)

⇒ ax = a & by = b

⇒ x = 1 & y = 1

So we can say that (1,1) will be identity element.(iii) Now we will find invertible element in A.

Let assume that (c,d) ∈ A be the inverse of (a,b) ∈ A

(a,b)*(c,d) = (c,d)*(a,b) = x

(ac , bd) = (1,1) [e = (1,1) ]

ac = 1 & bd = 1

c = 1/a & d = 1/b

So we can say that (1/a ,1/b) will be the inverse of (a,b) with respect to *.

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

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

**Solution:**

The binary operation * on N can be defined as:

a*b = H.C.F of a and b

And we also know that , HCF(a,b) = HCF(b,a) . a,b ∈ N.

So we can say that , a * b = b * a

So , the operation * is commutative.For a,b,c ∈ N. So we have.

(a * b) * c = (HCF(a,b))*c = HCF(a,b,c)

a * (b * c) = a * (HCF(a,b)) = HCF(a,b,c)

So it can be said that (a * b) * c = a * (b * c)

So we can say that , the operation * is associative.Now , an element e ∈ N will be the identity for the operation.

* if a * e = a = e * a ,∀ a ∈ N.

But we can say that , this relation is not true for any a ∈ N.

So we can say that , the operation * does not have any identity in N.

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