**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**

**Solution: **

If a, b belongs to Z+

a * b = a – b which may not belong to Z+

For eg: 1 – 3 = -2 which doesn’t belongs to Z+

Therefore, * is not a Binary Operation on Z+

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

**Solution: **

If a, b belongs to Z+

a * b = ab which belongs to Z+

Therefore, * is Binary Operation on Z+

**(iii) On R, define * by a * b = ab²**

**Solution:**

If a, b belongs to R

a * b = ab

^{2 }which belongs to RTherefore, * is Binary Operation on R

**(iv) On Z+, define * by a * b = |a – b|**

**Solution:**

If a, b belongs to Z+

a * b = |a – b| which belongs to Z+

Therefore, * is Binary Operation on Z+

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

**Solution:**

If a, b belongs to Z+

a * b = a which belongs to Z+

Therefore, * is Binary Operation on Z+

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

**(i) On Z, define a * b = a – b **

**Solution:**

a) Binary:If a, b belongs to Z

a * b = a – b which belongs to Z

Therefore, * is Binary Operation on Z

b) Commutative:If a, b belongs to Z, a * b = b * a

LHS = a * b = a – b

RHS = b * a = b – a

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:If a, b, c belongs to Z, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a – b + c

RHS = (a – b) * c = a – b- c

Since, LHS is not equal to RHS

Therefore, * is not Associative

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

**Solution:**

a) Binary:If a, b belongs to Q, a * b = ab + 1 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:If a, b belongs to Q, a * b = b * a

LHS = a * b = ab + 1

RHS = b * a = ba + 1 = ab + 1

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:If a, b, c belongs to Q, a * (b * c) = (a * b) * c

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

RHS = (a * b) * c = abc + c + 1

Since, LHS is not equal to RHS

Therefore, * is not Associative

**(iii) On Q, define a ∗ b = ab/2**

**Solution :**

a) Binary:If a, b belongs to Q, a * b = ab/2 which belongs to Q

Therefore, * is Binary Operation on Q

b) Commutative:If a, b belongs to Q, a * b = b * a

LHS = a * b = ab/2

RHS = b * a = ba/2

Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:If a, b, c belongs to Q, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * (bc/2) = (abc)/2

RHS = (a * b) * c = (ab/2) * c = (abc)/2

Since, LHS is equal to RHS

Therefore, * is Associative

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

^{ab}

**Solution:**

a) Binary:If a, b belongs to Z+, a * b = 2

^{ab}which belongs to Z+Therefore, * is Binary Operation on Z+

b) Commutative:If a, b belongs to Z+, a * b = b * a

LHS = a * b = 2

^{ab}RHS = b * a = 2

^{ba}= 2^{ab}Since, LHS is equal to RHS

Therefore, * is Commutative

c) Associative:If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * 2

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

^{ab}* c = 2^{2abc}Since, LHS is not equal to RHS

Therefore, * is not Associative

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

^{b}

**Solution:**

a) Binary:If a, b belongs to Z+, a * b = a

^{b}which belongs to Z+Therefore, * is Binary Operation on Z+

b) Commutative:If a, b belongs to Z+, a * b = b * a

LHS = a * b = a

^{b}RHS = b * a = b

^{a}Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:If a, b, c belongs to Z+, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * b

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

^{b}* c = a^{bc}Since, LHS is not equal to RHS

Therefore, * is not Associative

**(vi) On R – {– 1}, define a ∗ b = a / (b + 1)**

**Solution: **

a) Binary:If a, b belongs to R, a * b = a / (b+1) which belongs to R

Therefore, * is Binary Operation on R

b) Commutative:If a, b belongs to R, a * b = b * a

LHS = a * b = a / (b + 1)

RHS = b * a = b / (a + 1)

Since, LHS is not equal to RHS

Therefore, * is not Commutative

c) Associative:If a, b, c belongs to A, a * (b * c) = (a * b) * c

LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1

RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)

Since, LHS is not equal to RHS

Therefore, * is not Associative

**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 ∧. **

**Solution: **

^ 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.**

**(Hint: use the following table) **

* | 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 |

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

**Solution:**

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

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

**(ii) Is ∗ commutative?**

**Solution:**

The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.

**(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).**

**Solution:**

(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.**

**Solution:**

Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.

*’ 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 see that the operation *’ is the same as the operation * in Exercise 4 above.

**Question 6: Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find**

**(i) 5 ∗ 7, 20 ∗ 16**

**Solution:**

If a, b belongs to N

a * b = LCM of a and b

5 * 7 = 35

20 * 16 = 80

**(ii) Is ∗ commutative?**

**Solution:**

If a, b belongs to N

LCM of a * b = ab

LCM of b * a = ba = ab

a*b = b*a

Thus, * binary operation is commutative.

**(iii) Is ∗ associative?**

**Solution:**

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

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

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

Thus, * binary operation is associative.

**(iv) Find the identity of ∗ in N**

**Solution:**

Let ‘e’ is an identity

a * e = e * a, for a belonging to N

LCM of a * e = a, for a belonging to N

LCM of e * a = a, for a belonging to N

e divides a

e divides 1

Thus, e = 1

Hence, 1 is an identity element

**(v) Which elements of N are invertible for the operation ∗? **

**Solution:**

a * b = b * a = identity element

LCM of a and b = 1

a = b = 1

only ‘1’ is invertible element in N.

### Chapter 1 Relations and Functions – Excercise 1.4 | Set 2

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