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

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 = ab2

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 = ab2

It is seen that for each a, b ϵ R, there is a unique element ab2 in R.

This means that * carries each pair (a, b) to a unique element a*b = ab2 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 = 2ab

iv. On Z+, define a * b = ab

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 = 2ab

2abϵ Z+, so the operation * is binary operation

It is known that

ab ba for a, b Z

Therefore, 2ab = 2ab 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 = 24×3 = 212

1*(2*3) = 1*22×3 =1×26= 1*64 = 264

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 = ab

ab ϵ Z+, so the operation * is binary operation

It can be observed that 1*2 = 12 = 1 and 2*1 = 21 = 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 = 23*4 = 8*4 = 84 = (3)4 = 212

2*(3*4) = 2*34 = 2*81 = 281

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

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

Q 5. Let * be the binary operation of the set {1, 2, 3, 4, 5} defined by a*b = HCF of a and b. Is the operation * is the same as the operation * defined in Q.4 above? Justify your answer.
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 *.

Q 7. Is * defined on the set A = {1, 2, 3, 4, 5} by a*b = LCM of a and b, a binary operation? Justify your answer.
Sol. The operation * on the set A = {1, 2, 3, 4, 5} is defined as a*b = LCM of a and b. Now, 2*3 = LCM of 2 and 3 = 6. But 6 does not belong to the given set. Hence, the given operation * is not a binary operation.
Q 8. Let * be the binary operation on N defined by a*b = HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
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 = a2+b2

iii. a*b = a + ab

iv. a*b = (a-b)2

v. a*b = ab/4

vi. a*b = ab2

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 = ab2

It can be observed that for 2, 3 ϵ Q

2*3 = 2.32 = 18 and 3*2 = 3.22 = 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 = (1.22)*3 = 4*3 = 4.32 = 36

1*(2*3) = 1*(2.32) = 1*18 = 1.182 = 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.

Q 10. Show that none of the operation given in Q.9 has identity.
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 = a2 +b2

If a*e = a, then a2 + e2 = a

For a = -2, (-2)4 +e2 = 4+e2 ≠-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 = ab2

If a*e = a → ae2 = a

→ e2 =1 → e±1

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

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

Q 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.
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 = (e1, e2) ϵ A will be an identity element for thye operation * if

a*e = a= e*a ∀ a =(a1, a2) i.e.,

(a1+e1, a2+e2)= (a1, a2) = (e1+a1, e2+a2)

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 = a3+b3. 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 = a3+b3

For a, b ϵ N, we have

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

Therefore, the operation * is commutative.

It can be observed that

(1*2)*3 = (13+23)*3 = 9*3 = 93+33 = 729 + 27 = 756

(1*(2*3) = 1*(23+33)= 1*(8+27) = 1*35

= 13+353 = 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).

Updated: October 9, 2020 — 2:01 pm