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Relations and Functions Ex-1.4 Questions Answers

Subjects

Question 1 : Determinewhether or not each of the definition of * given below gives a binaryoperation. In the event that * is not a binary operation, give justificationfor this.


Answer 1 : (i) On Z_+,define * by a * b = a – b
(ii) On Z_+, define * by a * b = ab
(iii) On R, define * by a * b = ab²
(iv) On Z_+,define * by a * b = |a – b|
(v) On Z_+, define * by a * b = a


Solution
(i) If a > b, a * b = a – b > 0, which belongs to Z+.
But if a < b, a * b = a – b < 0, which does not belong to Z+
=> * given operation is not a binary operation.

(ii) For all a and b belonging to Z-1, ab also belongs to Z+.
The operation *, defined by a * b = ab is abinary operation.

(iii) For all a and b belonging to R, ab² also belongs to R.
The operation * defined by a * b = ab² isbinary operation.

(iv) For all a and b belonging to Z+, |a – b| also belongs to Z+1
The operation a * b = |a – b| is a binaryoperation.

(v) On Z+ defined by a * b = a
a, b
Z+ = a Z+
The operation * is a binary operation.

Question 2 : Foreach binary operation * defined below, determine whether * is commutative orassociative.

Answer 2 : (i) OnZ, define a * b = a – b
(ii) OnQ, define a * b = ab + 1
(iii) On Q, define a * b = 
(iv) On Z_+, define a * b = 2ab
(v) On Z_+, define a * b = ab
(vi) On R- {-1}, define a * b = 


Solution

(i)On Z, operation * is defined as
(a) a * b = a – b => b * a = b – a
But a – b ≠ b – a ==> a * b ≠ b * a
Defined operation is not commutative
(b) a – (b – c) ≠ (a – b) – c
Binary operation * as defined is not associative.

(ii)On Q, Operation * is defined as a * b = ab + 1
(a) ab + 1 = ba + 1, a * b = b * a
Defined binary operation is commutative.
(b) (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
Binary operation defined is not associative.

(iii)(a) On Q, operation * is defined as a * b = 
a * b = b * a
Operation binary defined is commutative.
be abc
(b) a * (b * c) = a * 
 =  and

(a * b) * c =  * c 

=> (a * b) * c =  * c 

Defined binary operation is associative.

(iv)On Z+ operation * is defined as a * b = 2ab
(a) a * b = 2ab, b * a = 2ba = 2ab
=> a * b = b * a .
Binary operation defined as commutative.
(b) a * (b * c) = a * 2ba = 2a.2bc
(a * b) * c = 2ab * c = 22ab
Thus (a * b) * c ≠ a * (b * c)
Binary operation * as defined as is notassociative.

(v)On Z+, a * b = ab
(a) b * a = ba .
ab ≠ ba = a* b ≠ b * a.
* is not commutative.
(b) (a*b)*c = ab*c = (ab)c = abc a*(b*c)
= a*bc = abc
Thus (a * b) * c ≠ (a * b * c)
Operation * is not associative.

(vi)On R, * −{−1} is defined by

It can be observed that and

1 * 2 ≠ 2 * 1 ; where 1, 2  − {−1}

Therefore,the operation * is not commutative.

It canalso be observed that:

(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3  −{−1}

Therefore,the operation * is not associative.

Question 3 : Considerthe 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 3 : Operation ^ table onthe set {1, 2, 3, 4, 5} is as follows.

Question 4 : Considera binary operation * on the set {1, 2, 3, 4, 5} given by the followingmultiplication table.


Answer 4 : (i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
Hint – use the following table)


Solution

(i) From the given table, we find
2*3 = 1, 1*4 = 1
(a) (2*3)*4 = 1 * 4 = 1
(b) 2*(3*4) = 2 * 1 = 1

(ii) Let a, b {1,2,3,4,5} From the given table, we find
a*a = a
a*b = b*a = 1 when a or b or are odd and a b.
2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b
Thus a * b = b * c
Binary operation * given is commutative.

(iii) Binary operation * given is commutative (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 aand b. Is the operation *’ same as the operation * defined in the exp no. 4above? Justify your answer.

Answer 5 : The set is {1,2,3,4, 5} and a * b = HCF of a and b.
Let us prepare the table of operation *.

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
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in N

Answer 6 :

Binaryoperation * defined as a * b = 1 cm. of a and b.
(i) 5 * 7 = 1 cm of 5 and 7 = 35
20 * 16= 1 cm of 20 and 16 = 80

(ii) a * b= 1 cm of a and b b * a = 1 cm of b and a
=> a * b = b * a, 1 cm of a, b and b, a are equal
Binary operation * is commutative.

(iii) a * (b * c) = 1 cm of a, b, c and (a * b) * c = 1 cm of a, b, c
=> a * (b * c) = (a * b) * c
=> Binary operation * given is associative.
(iv) Identity of * in N is 1
1 * a = a * 1 = a = 1 cm of 1 and a.

(v) Let * : N x N—> N defined as a * b = 1.com of (a, b)
For a = 1, b = 1, a * b = l b * a
Otherwise a * b ≠ 1
Binary operation * is not invertible
=> 1 is invertible for operation *

Question 7 : Is* defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binaryoperation? Justify your answer.

Answer 7 :

Thegiven set = {1,2,3,4,5} Binary operation is defined as a * b = 1 cm of a and b.4 * 5 = 20 which does not belong to the given set {1, 2, 3, 4, 5}.
It 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 binaryoperation on N?

Answer 8 : Binary operation on set N is defined as a * b = HCF of a and b
(a) We know HCF of a, b = HCF of b, a
a * b = b * a
Binary operation *is commutative.
(b) a*(b*c) = a* (HCF of b, c) = HCF of a and (HCF of b, c) = HCF of a, b and c
Similarly (a * b) * c = HCF of a, b, and c
=> (a * b) * c = a * (b * c)
Binary operation * as defined above is associative.
(c) 1 * a = a * 1 = 1 ≠ a
There does notexists any identity element.

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

Answer 9 : (i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = 
(vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.


Solution

Operation is on the set Q.,,
(i) defined as a * b = a – b
(a) Now b * a = b – a
But a – b ≠ b – a
a * b ≠ b * a
Operation * is not commutative.
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c (a * b) * c = (a – b) *c = a – b – c
Thus a * (b * c)¹ (a * b) * c = (a² + b²)² + c²
=> a * (b * c) ≠ (a * b) * c
The operation * as defined is notassociative.

(ii)(a) a * b = a² + b * a = b² + a² = a² + b².
a * b = b * a
This binary operation is commutative,
(b) a*(b*c) = a*(b² + c²) = a² + (b²)² + c²)²
=> (a*b)*c = (a² + b²)*c = (a² + b²) + c²
Thus a * (b*c) (a*b) * c
The operation * given is not associative.

(iii)Operation * is defined as a * b = a + ab
(a) b * a = b + ba
a * b ≠ b * a
The operation is not commutative.
(b) 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
=> a * (b * c) ≠ (a * b) * c
=> The binary operation is not associative.

(iv)The binary operation is defined as a * b = (a – b)²
(a) b*a = (b – a)² = (a – b)² => a*b = b*a
.’. This binary operation * is commutative.
(b) a*(b*c) = a*(b – c)² = [a – (b – c)²]² (a * b) * c
= (a – b)² * c
= [(a – b)² – c]²
=> a * (b * c) ≠ (a * b) * c
the operation * is not associative.

(v) On Q,the operation * is defined as 

For ab  Q, we have:

 * b = * a

Thus, theoperation * is commutative.

For a, b, c  Q, we have:

(* b) * c = a * (* c)

Thus, theoperation * is associative.


(vi) On Q, the operation * is defined as * b= ab2

Itcan be observed that:

Thus, the operation * is notcommutative.

It canalso be observed that:

Thus, the operation * is notassociative.

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

Question 10 : Showthat none of the operations given above has identity.

Answer 10 :

Thebinary operation * on set Q is
(i) defined as a*b = a – b
For identity element e, a*e = e*a = a
But a*e = a – e≠a and e*a = e – a≠a
There is no identity element for this operation

(ii) Binary operation * is defined as a * b = a² + b² ≠ a
This operation * has no identity.

(iii) The binary operation is defined as a*b = a+ab
Putting b = e, a + e = a + eb ≠ a
There is no identity element.

(iv) The binary operation is defined as a * b = (a – b)²
Put b = e, a * e = (a – e)² ≠ a for any value of
e
Q
=> there is no Identity Element.

(v) The operation is a * b = 
a * e =  ≠ a for any value of e Q

Operation * has no identity

(vi) The operation * is a * b = ab² Put b = e, a
*e = ae² and e * a = ea² ≠ a for any value of e
Q
=> There is no Identity Element. Thus, these operations have no Identity.


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