NCERT 12 Maths Solutions of Relations and Functions – Chapter 1 Ex 1.4
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
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
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
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
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 |
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 |
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
| * | 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 *.
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 *?
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 *.
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.
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?
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.
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.
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.
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
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.
i. * is both associative and commutative
ii. * is commutative but not associative
iii. * is associative but not commutative
vi. * is neither commutative nor associative
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).