NCERT 12 Maths Solutions of Relations and Functions – Chapter 1 Ex 1.4

Q 1. Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

i. On Z^{+}, defined * by a* b = a- b

ii. On Z^{+}, defined * by a* b = ab

iii. On R, defined * by a* b = ab^{2}

iv. On Z^{+}, defined a* b = |a-b|

v. On Z^{+}, defined * by a* b = a

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

It is not a binary operation as the image of (1, 2) under * is 1*2 = 1-2 = -1∉ Z^{+}

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

It is seen that for each a, b ϵ Z^{+}, there is a unique element ab in Z^{+}. This means that * carries each pair (a, b) to a unique element

a*b = ab in Z^{+}

Therefore, * is a binary operation.

(iii) On R, * is defined by a*b = ab^{2}

It is seen that for each a, b ϵ R, there is a unique element ab^{2} in R.

This means that * carries each pair (a, b) to a unique element a*b = ab^{2} in R. Therefore, * is binary operation.

(iv) On Z^{+}, * is defined by a*b = |a-b|

It is seen that for each a, b ϵ Z^{+ }there is a unique element |a-b| in Z^{+}

This means that * carries each pair (a, b) to a unique element a*b = |a-b| in Z^{+}. Therefore, * is a binary operation.

(v) On Z^{+}, * is defined by a*b = a

It is seen that for each a, b ϵ Z^{+}, there is a unique element a ϵ Z^{+}

This means that * carries each pair (a, b) to a unique element

a*b = a in Z^{+}

Therefore, * is a binary operation

Q 2. For each binary operation * defined below, determine whether * is binary, commutative or associative.

i. On Z, define a * b = a- b

ii. On Q, define a * b = ab + 1

iii. On Q, define a * b = ab/2

iv. On Z^{+}, define a * b = 2^{ab}

iv. On Z^{+}, define a * b = a^{b}

v. On R – {-1}, define a * b = a/b+1

Sol. (i) On Z, * is defined by a*b = a-b

a-b ϵ Z, so the operation * is binary

It can be observed that 1*2 = 1-2 = -1 and 2*1 = 2-1 = 1

Therefore, 1*2 ≠ 2*1, where 1, 2 ϵ Z

Hence, the operation * is not commutative

Also, we have (1*2)*3 = (1-2)*3 = -1*3 = -1-3 = -4

1*(2*3) = 1*(2-3) = 1* -1 = 1-(-1) = 2

Therefore, (1*2)*3≠ 1*(2*3), where 1, 2, 3 ϵ Z

Hence, the operation * is not associative

(ii) On Q, * is defined by a*b = ab+1

ab+1 ϵ Q, so operation * is binary

It is known that

ab = ba for a, b ϵ Q

Therefore, ab + 1 = ba + 1 for a, b ϵ Q

Therefore, a*b = a*b for a, b ϵ Q

Therefore, the operation * is commutative. It can be observed that

(1*2)*3 = (1×2+1)*3 = 3*3 = 3×3 +1 = 10

1*(2*3) = 1*(2×3+1) = 1*7+1 = 8

Therefore, (1*2)*3 ≠1*(2*3), where 1, 2, 3 ϵ Q

Therefore, the operation * is not associative

(iii) On Q, * is defined by a* b = ab/2

ab/2 ϵ Q, so the operation * is binary. It is known that

ab= ba for a, bϵ Q

Therefore, ab/2 = ba/2 for a, b ϵ Q

Therefore, a*b = b*a for a, bϵ Q

Therefore, the operation * is commutative

For all a, b, c ϵ Q, we have

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

Therefore, the operation * is associative

(iv) On Z^{+}, * is defined by a*b = 2^{ab}

2^{ab}ϵ Z^{+}, so the operation * is binary operation

It is known that

ab ba for a, b Z

Therefore, 2^{ab} = 2^{ab} for a, b ϵ Z^{+}

Therefore, a*b = b*a for a, bϵ Z^{+}

Therefore, the operation * is commutative. It can be observed that

(1*2)*3 = 2^{(1×2)}*3 = 4*3 = 2^{4×3} = 2^{12}

1*(2*3) = 1*2^{2×3} =1×2^{6}= 1*64 = 2^{64}

Therefore, (1*2)*3≠ 1*(2*3), where 1, 2, 3 ϵ Z^{+}

Therefore, the operation * is not associative.

(v) On Z^{+}, * is defined by a*b = a^{b}

a^{b} ϵ Z^{+}, so the operation * is binary operation

It can be observed that 1*2 = 1^{2} = 1 and 2*1 = 2^{1} = 2

Therefore, 1*2 ≠ 2*1, where 1, 2,ϵ Z^{+}

Therefore, the operation * is not commutative

It can be observed that

(2*3)*4 = 2^{3}*4 = 8*4 = 8^{4} = (^{3})^{4} = 2^{12}

2*(3*4) = 2*3^{4} = 2*81 = 2^{81}

Therefore, (2*3)*4≠2*(3*4) ; where 2, 3, 4 ϵ Z^{+}

Therefore, the operation * is not associative

(vi) On R-(-1), * is defined by a*b = a/b+1

a/b+1ϵ R for b ≠-1, so that operation * is binary

It can be observed that

1*2 = 1/2+1=1/3 and 2*1 = 2/1+1=2/2=1

Therefore, 1*2 ≠ 2*1 where 1, 2 ϵ R-{-1}

Therefore, the operation * is not commutative

It can also be observed that

Therefore, (1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ R-{-1}

Therefore, the operation * is not associative

Sol. The binary operation ^ on the set {1, 2, 3, 4, 5} in defined as

a^b = min{a, b} for a, b ϵ {1, 2, 3, 4, 5}

Thus, the operation table for the given operation ^ can be given as

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

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 2 | 2 | 2 |

3 | 1 | 2 | 3 | 3 | 3 |

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

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

Q 4. Consider a binary operation * on set {1, 2, 3, 4, 5} given by the following multiplication table:

i .Compute (2*3)*4 and 2*(3*4)

ii. Is * commutative?

iii. Compute (2*3)*(4*5)

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

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

Sol. (i) We have (2*3)*4 = (1)*4=1

And 2*(3*4) = 2*1 = 1

(ii) For every a, b ϵ {1, 2, 3, 4, 5}, we have a*b=b*a

Therefore, the operation * is commutative

(iii) We have (2*3) = 1 and (4*5) = 1

Therefore, (2*3)*(4*5) = 1*1=1

Sol. The binary operation *, on the set {1, 2, 3, 4, 5} is defined as a*b = HCF of a and b, the operation table for the operation* can be given as

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

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

We observe that the operation table for the operation * and operation * in Q.4 are the same. Thus, the operation * is the same as the operation *.

Q 6. Let * be the binary operation on N given by a*b = LCM of a and b.

i. Find 5*7, 20*16

ii. Is * commutative?

iii. Is * associative?

iv. Find the identity of * in N

v. Which elements of N are invertible for the operation *?

Sol. The binary operation * on N is defined as a*b = LCM of a and b

i. We have 5*7 = LCM of 5 and 7 = 35

and 20*16 = LCM of 20 and 16 = 80

ii. It is known that

LCM of a and b = LCM of b and a for a, b ϵ N

Therefore, a*b = b*a. Thus, the operation * is commutative

iii. For a, b, c ϵ N, we have

(a*b)*c = (LCM of a and b)*c = LCM of a, b, and c

a*(b*c) = a*(LCM of b and c) = LCM of a, b, and c

Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative.

iv. It is known that

LCM of a and 1 = a = LCM of 1 and a, a ϵ N

a*1 = a = 1*a, a ϵ N

Thus, 1 is the identity of * in N

v. An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a*b = e = b*a

Here, e=1. This means that

LCM of a and b = 1 = LCM of b and a

This case is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

Sol. The binary operation * on N is defined as a*b = HCF of a and b.

It is known that HCF of a and b = HCF of b and a for a, b ϵ N.

Therefore, a*b = b*a. Thus, the operation * is commutative.

For a, b, c ϵ N, we have (a*b)*c = (HCF of a and b)* c = HCF of a, b and c

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

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

Thus, the operation * is associative

Now, an element e ϵ N will be the identity for the operation * if

a*e = a = e*a, for ∀ a ϵ N.

But this relation is not true for any a ϵ N

Thus, the operation * does not have identity in N.

Q 9. Let * be a binary operation on the set Q of rational number as follows:

i. a*b = a-b

ii. a*b = a^{2}+b^{2}

iii. a*b = a + ab

iv. a*b = (a-b)^{2}

v. a*b = ab/4

vi. a*b = ab^{2}

Find which of the binary operation are commutative and which are associative?

Sol. (i) On Q, the operation * is defined as a*b = a-b

It can be observed that for 2, 3, 4 ϵ Q, we have

2*3 = 2-3 = -1 and 3*2 = 3-2 = 1

2*3 ≠ 3*2

Thus, the operation * is not commutative

It can also be observed that

(2*3)*4 = (-1)*4 = -1-4 = -5

And 2*(3*4) = 2*(-1) = 2-(-1) = 3

2*(3*4) ≠2*(3*4)

Thus, the operation * is not associative

(ii) On Q, the operation * is defined as a*b =

For a, b ϵ Q, we have

a*b = = = b*a

Therefore, a*b = b*a

Thus, the operation * is commutative

It can be observed that

(1*2)*3 = ()*3 = (1+4)*4 = 5*4 = = 41

1*(2*3) = 1*() = 1*(4+9) = 1*13 = = 170

(1*2)*3 ≠1*(2*3) where 1, 2, 3 ϵ Q

Thus, the operation * is not associative

(iii) On Q, the operation * is defined as a*b = a+ ab

It can be observed that

1*2 = 1+1 x 2 = 1+2 = 3, 2*1 = 2 + 2 x 1 = 2+2 = 4

1*2 ≠ 2*1, where 1, 2ϵQ

Thus, the operation * is not commutative

It can be observed that

(1*2)*3 = (1+1×2)*3 = 3*3 = 3+3×3 = 3+9=12

1*(2*3) = 1*(2+2×3)= 1*8 =1+1×8 = 9

(1*2)*3 ≠1*(2*3) where 1, 2, 3 ϵ Q

Thus, the operation * is not associative

(iv) On Q, the operation * is defined by a*b = (a-b)^{2}

For a, b ϵ Q, we have

a*b = (a-b)^{2} and b*a = (b-a)^{2} = [-(a-b)]^{2} = (a-b)^{2}

Therefore, a*b = b*a

Thus, the operation * is commutative. It can be observed that

(1*2)*3 = (1-2)^{2}*3 = (-1)^{2}*3 = 1*3 = (1-3)^{2} = (-2)^{2}= 4

1*(2*3) = 1*(2-3)^{2} = 1*(-1)^{2} = 1*1 = (1-1)^{2} = 0

(1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ Q

Thus, the operation * is not associative

(v) On Q, the operation * is defined as a*b = ab/4

For a, b ϵ Q, we have a*b = ab/4 = ba/4 = b*a

Therefore, a*b = b*a

Thus, the operation * is commutative

For a, b, c ϵ Q, we have

Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative.

(vi) On Q, the operation * is defined as a*b = ab^{2}

It can be observed that for 2, 3 ϵ Q

2*3 = 2.3^{2} = 18 and 3*2 = 3.2^{2} = 12

Hence, 2*3 3*2

Also,

Thus, the operation * is not commutative.

It can also be observed that 1, 2, 3 ϵ Q

(1*2)*3 = (1.2^{2})*3 = 4*3 = 4.3^{2} = 36

1*(2*3) = 1*(2.3^{2}) = 1*18 = 1.18^{2} = 324

(1*2)*3 1*(2*3)

Also,

Thus, the operation * is not associative

Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative.

Sol. An element e ϵ Q will be the identity element for the operation * if

a*e = a = e*a, a ϵ Q

i. a*b = a-b

If a*e=a, a≠ 0 → a-e = a, a ≠0 → e = 0

Also, e*a =a →e-a =a →e=2a

e=0=2a, a≠0

But the identity is unique. Hence this operation has no identity.

ii. a*b = a^{2} +b^{2}

If a*e = a, then a^{2} + e^{2} = a

For a = -2, (-2)^{4} +e^{2} = 4+e^{2} ≠-2

Hence, there is no identity element.

iii. a*b = a+ab

If a*e =a → a+ae = a→ ae=0 →e=0, a≠

Also if

→ e*a=a → e+ea =a → e = , a≠0

But the identity is unique. Hence, this operation has no identity.

iv. a*b = (a-b)^{2}

If a*e = a, then (a-e)^{2} = a. A square is always positive, so for

a = -2, (-2, -e)^{2} ≠ -2

Hence, there is no identity element

v. a*b = ab/4

If a*e = a, then ae/4 = a. Hence, e=4 is the identity element

a*4 = 4*a = 4a/4 = a

vi. a*b = ab^{2}

If a*e = a → ae^{2} = a

→ e^{2 }=1 → e±1

But identity is unique. Hence this operation has no identity.

Therefore, only part (v) has an identity element.

Sol. Given that A = N x N

And * is a binary operation on A and is defined by

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

Let (a, b), (c, d)ϵ A

Then, a, b, c, d ϵ N

We have (a, b)*(c, d) = (a+c, b+d)

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

[Addition is commutative in the set of natural numbers]

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

Therefore, the operation * is commutative.

Now, let (a, b)(c, d), (e, f) ϵ A

Then, a, b, c, d, e, f ϵ N

We have, ((a, b)*(c, d))*(e, f) = (a+c, b+d)*(e, f) = (a+c+e, b+d+f)

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

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

Therefore, the operation * is associative.

An element e = (e_{1}, e_{2}) ϵ A will be an identity element for thye operation * if

a*e = a= e*a ∀ a =(a_{1}, a_{2}) i.e.,

(a_{1}+e_{1}, a_{2}+e_{2})= (a_{1}, a_{2}) = (e_{1}+a_{1}, e_{2}+a_{2})

Which is not true for any element in A.

Note that a+e = a for e = 0 but 0 does not belong to N.

Therefore, the operation * does not have any identity element.

Q 12. State whether the following statements are true or false? Justify

i. For an arbitrary binary operation * on a set N, a * a = a ∀ a ϵ

ii. If * is a commutative binary operation on N, then a*(b*c) = (c*b)*a

Sol. i. Define an operation * on N as a*b = a + b ∀ a, b ϵ N

Then, in particular, for b=a=3, we have 3*3 = 3+3 = 6≠3

Therefore, statement (i) is false

(ii) RHS = (c*b)*a = (b*c)*a [* is commutative]

= a*(b*c) [Again, as * is commutative]

=LHS

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

Therefore, statement (ii) is true.

Q 13. Consider a binary operation * on N defined as a*b = a^{3}+b^{3}. Choose the correct answer.

i. * is both associative and commutative

ii. * is commutative but not associative

iii. * is associative but not commutative

vi. * is neither commutative nor associative

Sol. On N, the operation * is defined as a*b = a^{3}+b^{3}

For a, b ϵ N, we have

a*b = b^{3} +b^{3} = b^{3} a^{3} = b*a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that

(1*2)*3 = (1^{3}+2^{3})*3 = 9*3 = 9^{3}+3^{3} = 729 + 27 = 756

(1*(2*3) = 1*(2^{3}+3^{3})= 1*(8+27) = 1*35

= 1^{3}+35^{3} = 1+(35)^{3} = 1 + 42875 = 42876

Therefore, (1*2)*3 ≠ 1*(2*3) where 1, 2, 3 ϵ N

Therefore, the operation * is not associative

Hence, the operation * is commutative but not associative

Thus, the correct answer is (b).