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NCERT solution class 12 chapter 1 Relations and Functions exercise 1.4 mathematics part 1

EXERCISE 1.4


Page No 24:

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 − b

(ii) On Z+, define * by ab

(iii) On R, define * by ab2

(iv) On Z+, define * by = |− b|

(v) On Z+, define * by a

Answer:

(i) On Z+, * is defined by * 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 ab ∈ Z+, there is a unique element ab in Z+.

This means that * carries each pair (ab) to a unique element * b ab in Z+.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab2.

It is seen that for each ab ∈ R, there is a unique element ab2 in R.

This means that * carries each pair (ab) to a unique element * b abin R.

Therefore, * is a binary operation.

(iv) On Z+, * is defined by * b = |a − b|.

It is seen that for each ab ∈ Z+, there is a unique element |a − b| in Z+.

This means that * carries each pair (ab) to a unique element * b = |a − b| in Z+.

Therefore, * is a binary operation.

(v) On Z+, * is defined by a * b = a.

* carries each pair (ab) to a unique element * b a in Z+.

Therefore, * is a binary operation.

Question 2:

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

(i) On Z, define − b

(ii) On Q, define ab + 1

(iii) On Q, define 

(iv) On Z+, define = 2ab

(v) On Z+, define ab

(vi) On − {−1}, define 

Answer:

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

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ 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

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z

Hence, the operation * is not associative.

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

It is known that:

ab = ba &mnForE; a, b ∈ Q

⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q

⇒ * b = * b &mnForE; a, b ∈ 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 × 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Q

Therefore, the operation * is not associative.

(iii) On Q, * is defined by * b 

It is known that:

ab = ba &mnForE; a, b ∈ Q

⇒ &mnForE; a, b ∈ Q

⇒ * b = * a &mnForE; a, b ∈ Q

Therefore, the operation * is commutative.

For all a, b, c ∈ Q, we have:

Therefore, the operation * is associative.

(iv) On Z+, * is defined by * b = 2ab.

It is known that:

ab = ba &mnForE; a, b ∈ Z+

⇒ 2ab = 2ba &mnForE; a, b ∈ Z+

⇒ * b = * a &mnForE; a, b ∈ Z+

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ Z+

Therefore, the operation * is not associative.

(v) On Z+, * is defined by * b = ab.

It can be observed that:

 and 

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ Z+

Therefore, the operation * is not commutative.

It can also be observed that:

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ Z+

Therefore, the 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 can also be observed that:

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

Therefore, the operation * is not associative.

Question 3:

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

Answer:

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as  b = min {ab}

&mnForE; ab ∈ {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

Page No 25:

Question 4:

Consider a binary operation * on the 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).

(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

Answer:

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

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

(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have * b = b * a. Therefore, the operation * is commutative.

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

∴(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5:

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

Answer:

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *′ b = H.C.F 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 tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

Question 6:

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

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

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

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

Answer:

The binary operation * on N is defined as * b = L.C.M. of a and b.

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

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

(ii) It is known that:

L.C.M of a and b = L.C.M of b and a &mnForE; a, b ∈ N.

a * b = * a

Thus, the operation * is commutative.

(iii) For a, b∈ N, we have:

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

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

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N

⇒ a * 1 = a = 1 * a &mnForE; a ∈ 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 * b = e = b * a.

Here, e = 1

This means that:

L.C.M of a and b = 1 = L.C.M 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 *.

Question 7:

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

Answer:

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

a * b = L.C.M. of a and b.

Then, the operation table for the given operation * can be given as:

*

1

2

3

4

5

1

1

2

3

4

5

2

2

2

6

4

10

3

3

6

3

12

15

4

4

4

12

4

20

5

5

10

15

20

5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

Question 8:

Let * be the binary operation on defined by = H.C.F. of and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

Answer:

The binary operation * on N is defined as:

* b = H.C.F. of a and b

It is known that:

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

a * b = * a

Thus, the operation * is commutative.

For ab∈ N, we have:

(* b)* c = (H.C.F. of a and b) * c = H.C.F. of ab, and c

*(* c)= *(H.C.F. of b and c) = H.C.F. of ab, and c

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

Now, an element ∈ N will be the identity for the operation * if * e = a = e* a ∈ N.

But this relation is not true for any ∈ N.

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

Question 9:

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

(i) − (ii) a2 + b2

(iii) ab (iv) = (− b)2

(v) (vi) ab2

Find which of the binary operations are commutative and which are associative.

Answer:

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

It can be observed that:

and 

 ; where

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(ii) On Q, the operation * is defined as * b = a2 + b2.

For a, b ∈ Q, we have:

a * b = b * a

Thus, the operation * is commutative.

It can be observed that:

(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34

1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 170∴ (1*2)*3 ≠ 1*(2*3) , where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

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

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(iv) On Q, the operation * is defined by a * b = (a − b)2.

For ab ∈ Q, we have:

* b = (a − b)2

* a = (b − a)2 = [− (a − b)]2 = (a − b)2

∴ * b = b * a

Thus, the operation * is commutative.

It can be observed that:

Thus, the operation * is not associative.

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

For ab ∈ Q, we have:

∴ * b = * a

Thus, the operation * is commutative.

For a, b, c ∈ Q, we have:

∴(* b) * c = a * (* c)

Thus, the operation * is associative.

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

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

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

Question 10:

Find which of the operations given above has identity.

Answer:

An element ∈ Q will be the identity element for the operation * if

* e = a = e * aa ∈ Q.

We are given

* b =

ab4

⇒ a*e = a⇒ae4=a⇒ e=4Similarly, it can be checked for e*a=a, we get e=4Thus, e = 4 is the identity.

Question 11:

Let A = × and * be the binary operation on A defined by

(ab) * (cd) = (cd)

Show that * is commutative and associative. Find the identity element for * on A, if any.

Answer:

A = N × N

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

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

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

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

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

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e∈ N

We have:

Therefore, the operation * is associative.

An element will be an identity element for the operation * if

, i.e., which is not true for any element in A.

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

Page No 26:

Question 12:

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set N N.

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

Answer:

(i) Define an operation * on N as:

* b = a + b a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (* b) * a

= (* c) * [* is commutative]

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

= L.H.S.

∴ a * (* c) = (* b) * a

Therefore, statement (ii) is true.

Question 13:

Consider a binary operation * on defined as a3 + b3. Choose the correct answer.

(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:

On N, the operation * is defined as * b = a3 + b3.

For, ab, ∈ N, we have:

* b = a3 + b3 = b3 + a3 = * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ N

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.


 

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