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## Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4

### 2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.4

Question 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^{+}, * define by a * b = a – b

Answer:

2 ∈ Z^{+}, 5 ∈ Z^{+}, 2 – 5 = -3 ∉ Z^{+
}hence not a binary operation.

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

Answer:

∀ a ∈ Z^{+}, b ∈ Z^{+} ab ∈ Z^{+}, hence a * b ∈ Z^{+
}∴ * is a binary operation.

(iii) On R, define * by a*b = ab^{2
}Answer:

a ∈ R, b ∈ R then ab^{2} ∈ R

∴ a*b ∈ R

∴ * is a binary operation.

(iv) On Z^{+}, define * by a * 6 = |a – b|

Answer:

a ∈ Z^{+}, b ∈ Z^{+}, |a – b| ∈ Z^{+
}∴ a * b ∈ Z^{+
∴ }* is a binary operation.

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

Answer:

a ∈ Z^{+}, b ∈ Z^{+} then a ∈ Z^{+
}∴ a * b ∈ Z^{+
}∴ * is a binary operation.

Question 2.

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

(i) On Z, define a * b = a – b

Answer:

a = 3, b = 2, a – b = 3 – 2= 1

b-a = 2-3 = -1

a – b ≠ b – a, hence not commutative

a = 3, b = 2, c = 1

(a * b) * c = (3 – 2) * 1 = 1 * 1 = 0

a*(b * c) = 3* (2 * 1) = 3*(2 – 1) = 3 * 1 = 3 – 1 = 2

(a * b) * c ≠ a * (b * c), hence * is not associative

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

Ans:

a ∈ Q, b ∈ Q

a * b = ab + 1

b * a = ab + 1

∴ a * b = b * a, hence commutative ( a * b )*c = (ab+1) *c

= (ab + 1) c + 1 = abc + c + 1

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

= a (bc + 1) + 1 = abc + a + 1

∴ (a * b) *c * a* (b * c)

hence * is not associative

(iii) On Q,define \(a * b=\frac{a b}{2}\)

Answer:

(iv) On Z^{+}, define a*b = 2“*

Answer:

a g Z^{+}, be Z^{+},

a*b = 2^{ab}

b * a = 2^{ba} = 2^{ab} a * b = b * a

hence * is commutative

(a * b) * c = 2^{ab} * c = 2^{ab} * c = 2^{2al}’^{x}‘

a*(b*c) = a* 2^{ac} = 2^{a}–^{2}“

hence (a * b) * c * a * (b * c)

.: * is not Associative.

(v) On Z^{+}, define a * b = a^{b
}Answer:

a ∈ Z^{+}, b ∈ Z^{+}, a * b = a^{b
}b * a = b^{a
}a * b ≠ b * a, hence * is not commutative

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

hence * is not associative.

(vi) On R – {- 1}, define \(a * b=\frac{a}{b+1}\)

Answer:

Question 3.

Consider the binary ∧ operation on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}. Write the operation table of the operation ∧

Answer:

∧ | 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 |

Question 4.

Consider a binary * operation on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).

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

(ii) * Is commutative?

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

(Hint: use the following table)

Table 1.2

* | 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 |

Answer:

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

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

(ii) Since the table is symmetric with main diagonal it is commutative.

(iii) (2 * 3) * (4 * 5) = 1 * 1 = 1.

Question 5.

*’Let be the binary operation on the set {1, 2, 3, 4, 5} defined by a *’ b = H.C.F. of a and b. Is the *’ operation same as the * operation defined in Exercise 4 above? Justify your answer.

Answer:

a, b ∈ {1,2, 3,4, 5}, a* b = H.C.F. of a and b we composition table for *’

*’ | 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 |

Both the composition tables are exactly same hence the operation* and *’ are same.

Question 6.

Let* be the binary operation on N given by a * b = L.C.M. of a and Find

(i) 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* ?

Answer:

(i) 5 *7 = L.C.M. of 5, 7 = 35

20 *16 = L.C.M. of 20, 16 = 80.

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

= b * a

hence * is commutative.

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

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

hence * is associative.

(iv) Let ‘e’ be the identity element, than

a * e = e * a = a ∀ a ∈ N

L.C.M. (a, e) = L.C.M. (e, a) = a

⇒ e divides a ∀a ∈N

⇒ e = 1

∴ 1 is the identity element.

(v) Let a ∈ N be invertible.

∴ be N such that a * b = b * a = L.C.M. (a, b) = 1

⇒ a = 1, b = 1

∴ only invertible element in N is 1

Question 7.

Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.

Answer:

L.C.M. of 3, 5 = 15

3 * 5 = L.C.M. of (3, 5) = 15 g {1,2, 3,4,5}

hence * is not a binary operation.

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?

Answer:

a ∈ N, b ∈ N, a ∈ b = H.C.F. of a and b

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

hence * is commutative

a ∈ N, b ∈ N, e ∈ N.

a* (b * e) = H.C.F. a, b, c

= (a * b) * c * is associative.

Let e ∈ N be the identity element

a * e = e * a = a

H.C.F. a and e = a

∴ a divides e ∀ a ∈ N

which is not possible as there does not exist such a number e, which is divisible by every number.

∴ identity element does not exist.

Question 9.

Let * be a binary operation on the set Q of rational numbers as follows. Find which of the binary operations are commutative and which are associative.

(i) a * b = a – b

Answer:

a * b = a – b,∀ ^{a}> be Q

a * b = a – b * a = b – a a * b ≠ b * a

hence not commutative

(a * b) * c = (a – b) * c = a – b – c

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

hence ≠ is not associative

(ii) a * b = a^{2} + b^{2
}Answer:

a, b ∈ Q ⇒ a * b = a^{2} + b^{2
}= b^{2} + a^{2} = b * a

hence * is commutative

(a * b) * c = (a^{2} + b^{2}) * c^{2} = (a^{2} + b^{2})^{2} + c^{2
}a*(b * c) = a* (b^{2 }+ c^{2}) = a^{2} + (b^{2} + c^{2})^{2
}hence * is not associative

(iii) a * b = a + ab

Answer:

a * b = a + ab

b * a = b + ab ∴ a * b ≠ b * a,

hence * is not commutative

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

= a + ab + ac + abc

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

= a + a (b + bc)

= a + ab + bca

≠ (a * b) * c

hence * is not associative

(iv) a * b = (a – b)^{2
}Answer:

a * b = (a – b)^{2} = (b – a)^{2}= b * a

hence * is not commutative

(a * b) * c = (a – b)^{2} * c = ((a-b)^{2} – c)^{2} = (a – b)^{4}+ c^{2} – 2c (a – b)^{2
}a*(b * c) = a* (b – c)^{2} = [a -(b – c)]^{2} = a^{2} + (b – c)^{2} – 2a (b – c)^{2
}hence not associative

(v) \(a * b=\frac{a b}{4}\)

Answer:

(vi) a*b = ab^{2
}Answer:

a ∈ Q, be Q , a * b = ab^{2
}b * a = ba^{2} ∴ a * b≠ b * a

hence not commutative.

(a * b) * c = (ab^{2}) * c = ab^{2}c^{2
}a * (b * c) = a * (bc^{2}) = ab^{2}c^{4
}(a * b) * c = a * (b)

∴ * is not associative.

Question 10.

Show that none of the operations given above has identity.

(i) Let ‘e’ be the identity in Q such that

Answer:

a * e = e * a = a

a – e = a = e – a

⇒ e = 0 or e = 2a ∀ a ∈ Q

which is not possible. Hence no identity.

(ii) a * b = a^{2} + b^{2
}Answer:

Let e ∈ Q be the identity element

a * e = a^{2} + e^{2 }=a e * a = c^{2} + a^{2} = a

which is not possible as a^{2} ≠ a ∀ a ∈ Q

hence identity element doesn’t exist.

(iii) a*b = a + ab

Answer:

Let e ∈ Q be the identity element

a * e = a + ae = a ⇒ ae = 0, ⇒ e = 0

e * a = e + ae = a ⇒ \(e=\frac{a}{1+a}\)

∴\(\mathrm{e}=0 \text { and } \mathrm{e}=\frac{\mathrm{a}}{1+\mathrm{a}} \forall \mathrm{a} \in \mathrm{Q} \) which is not possible. Hence identity element does not exist.

(iv) a * b = (a – b)^{2
}Answer:

Let e be the identity element

a * e = (a – e)^{2} = a

e * a = (e – a)^{2} = a

∴ a – e = ≠ a, e – a = ± a

⇒ e = 0 and e = 2a ∀ a ∈ Q which is not possible.

Hence no identity element.

(v) \(a * b = \frac{a b}{4}\)

Answer:

Let e be the identity element

\(a * e=\frac{a e}{4}=a \text { and } e * a=\frac{e a}{4}=a\)

∴ e = 4

∴ 4 is the identity element.

(vi) a*b = ab^{2
}Answer:

Let e be the identity element

a * e = ae^{2} = a

e * a = ea^{2} = a

∴\( \mathrm{e}^{2}=1 \text { and } \mathrm{e}=\frac{1}{\mathrm{a}} \quad \forall \mathrm{a} \in \mathrm{Q}\)

which is not possible. Hence identity element does not exist.

Question 11.

Let A = N x N and * be the binary operation on A defined by (a, b)* (c, d) = (a + c, b + d) Show that is commutative and associative. Find the identity element for * on A, if any. Answer:

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

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

hence commutative

[(a,b) * (c,d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f) = (a, b) * [(c,d) * (e, f)] hence * is associative.

Let (e, f) be the identity element of A

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

(a + e, b + 0 = (a, b) ⇒ (e = 0, f = 0) ∉ N

hence no identity element.

Question 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 ∈ N.

Answer:

If a * b = a+ b, then

a * a = a + a = 2a ≠ a

∴ statement is false

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

Answer:

Since commutative a* (b*c) = (b*c) *a

= (c *b) * a

hence true.

(iii) For every binary operation defined on a set having exactly one element a is necessarily commutative and associative .with a as the identity element and a being the inverse of a.

Answer:

A = {a}

* : A x A → A defined by a* a = a ∀ a ∈ A Also from the table it is commutative and associative. Also a is the identity element and also inverse of a .

Question 13.

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

(A) *Is both associative and commutative?

(B) *Is commutative but not associative?

(c) *Is associative but not commutative?

(D) *Is neither commutative nor associative?

Answer:

a*b = a^{3} + b^{3} = b^{3} + a^{3} = b*a

hence commutative

(a*b) * c = (a^{3} + b^{3}) *c = (a^{3} + b^{3})^{3} + c^{3
}a * (b * c) = a * (b^{3} + c^{3}) = a^{3} + (b^{3} + c^{3})^{3
}hence not associative

“B “ is the correct answer.