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TOPICS
Exercise - 1.4

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
(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)	On  Z+, * is defined by a * b = a − b.
e.g., (a, b) = (2, 3) under * is a * b = a − b ⇒ 2 * 3 = 2 − 3 = −1 ∉ Z+.
It is not a binary operation.
```
```(ii) On Z+, * is defined by a * b = ab.
It is seen that for each a, b ∈ 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+.
e.g., (a, b) = (2, 4) under * is a * b = ab ⇒ 2 * 4 = 8 ∈ Z+.
Therefore, * is a binary operation.
```
```(iii) On R, * is defined by a * b = ab².
It is seen that for each a, b ∈ 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 .
e.g., (a, b) = (1.2, 3) under * is a * b = ab² ⇒ 1.2 * 3 = 3.6 ∈ R.
Therefore, * is a binary operation.
```
```(iv) On Z+, * is defined by a * b = |a − b|.
It is seen that for each a, b ∈ 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+.
e.g., (a, b) = (2, 3) under * is a * b = |a − b| ⇒ 2 * 3 = |2 − 3| = 1 ∈ Z+.
Therefore, * is a binary operation.
```
```(v)	 On Z+, * is defined by a * b = a.
It is seen that for each a, b ∈ Z+, there is a unique element a in Z+.
This means that * carries each pair (a, b) to a unique element a * b = a in Z+.
e.g., (a, b) = (3, 4) under * is a * b = a ⇒ 3 * 4 = 3 ∈ Z+.
Therefore, * is a binary operation.
```

Question-2 :-  For each 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
(v) On Z+, define a ∗ b = ab
(vi) On R – {– 1}, define a ∗ b = a/(b+1)

Solution :-
```(i) On Z, define a ∗ b = a – b
For commutative:        ∴ [a * b = b * a]
a ∗ b = a – b
b ∗ a = b – a
e.g., (a, b) = (2, 4) ∈ Z
2 * 4 = 2 - 4 = -2
4 * 2 = 4 - 2 = 2
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = (a–b) - c
a * (b * c) = a - (b-c)
e.g., (a, b, c) = (2, 4, 6) ∈ Z
(2 ∗ 4) * 6 = (2–4) - 6
= -2 - 6
= -8
2 * (4 * 6) = 2 - (4-6)
= 2 + 2
= 4
Hence, the operation * is not associative.
Therefore, it is neither commutative nor associative.
```
```(ii) On Q, define a ∗ b = ab + 1
For commutative:        ∴ [a * b = b * a]
a ∗ b = ab + 1
b ∗ a = ba + 1
e.g., (a, b) = (1/2, 2/3) ∈ Q
1/2 * 2/3 = 1/2 . 2/3 + 1 = 4/3
2/3 * 1/2 = 2/3 . 1/2 + 1 = 4/3
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = (ab + 1)c + 1
a * (b * c) = a(bc + 1) + 1
e.g., (a, b, c) = (1/2, 2/3, 3/4) ∈ Q
(1/2 * 2/3) * 3/4 = (1/2 . 2/3 + 1) x 3/4 + 1
= 4/3 . 3/4 + 1
= 1 + 1
= 2
1/2 * (2/3 * 3/4) = 1/2 x (2/3 . 3/4 + 1) + 1
= 1/2 . 3/2 + 1
= 3/4 + 1
= 7/3
Hence, the operation * is not associative.
Therefore, it is commutative but not associative.
```
```(iii) On Q, define a ∗ b = ab/2
For commutative:        ∴ [a * b = b * a]
a ∗ b = ab/2
b ∗ a = ba/2
e.g., (a, b) = (2/3, 4/3) ∈ Q
2/3 * 4/3 = (2/3 x 4/3)/2 = 8/18
4/3 * 2/3 = (2/3 x 4/3)/2 = 8/18
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = ((ab/2) x c)/2 = abc/4
a * (b * c) = (a x (bc/2))/2 = abc/4
e.g., (a, b, c) = (2/3, 4/3, 1/3) ∈ Q
(2/3 ∗ 4/3) * 1/3 = (2/3 x 4/3 x 1/3)/4
= 8/108
2/3 * (4/3 * 1/3) = (2/3 x 4/3 x 1/3)/4
= 8/108
Hence, the operation * is associative.
Therefore, it is commutative and associative.
```
```(iv) On Z+, define a ∗ b = 2ab
For commutative:        ∴ [a * b = b * a]
a ∗ b = 2ab
b ∗ a = 2ba
e.g., (a, b) = (2, 1) ∈ Z+
2 * 1 = 22 = 4
1 * 2 = 22 = 4
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = 2ab * c
a * (b * c) = a * 2bc
e.g., (a, b, c) = (2, 1, 3) ∈ Z+
(2 ∗ 1) * 3 = 22 * 3
= 4 * 3
= 212
2 * (1 * 3) = 2 * 23
= 2 * 8
= 216
Hence, the operation * is not associative.
Therefore, it is commutative but not associative.
```
```(v) On Z+, define a ∗ b = ab
For commutative:        ∴ [a * b = b * a]
a ∗ b = ab
b ∗ a = ba
e.g., (a, b) = (2, 1) ∈ Z+
2 * 1 = 2¹ = 2
1 * 2 = 1² = 1
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = ab * c
a * (b * c) = a * ba
e.g., (a, b, c) = (2, 1, 3) ∈ Z+
(2 ∗ 1) * 3 = 2¹ * 3
= 2 * 3
= 2³
= 8
2 * (1 * 3) = 2 * 1³
= 2 * 1
= 2¹
= 2
Hence, the operation * is not associative.
Therefore, it is neither commutative nor associative.
```
```(vi) On R – {– 1}, define a ∗ b = a/(b+1)
For commutative:        ∴ [a * b = b * a]
a ∗ b = a/(b+1)
b ∗ a = b/(a+1)
e.g., (a, b) = (3, -2) ∈ R – {– 1}
3 * -2 = 3/(-2+1) = -3
-2 * 3 = -2/(3+1) = -2/4 = -1/2
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a ∗ b) * c = a/(b+1) * c
a * (b * c) = a * b/(c+1)
e.g., (a, b, c) = (2, 1, 3) ∈ R – {– 1}
(2 ∗ 1) * 3 = 2/(1+1) * 3
= 1 * 3
= 1/(3+1)
= 1/4
2 * (1 * 3) = 2 * 1/(3+1)
= 2 * 1/4
= 2/(1/4 + 1)
= 8/5
Hence, the operation * is not associative.
Therefore, it is neither commutative nor 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 :-
```     Let A = {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.

```

Question-4 :-  Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).
(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) (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) For every a, b ∈{1, 2, 3, 4, 5}, we have a * 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 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 :-
```     The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b.
The operation table for the operation *′ can be given as:

```

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

Solution :-
```    The binary operation * on N is defined as a * 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 where a, b ∈ N.
∴a * b = b * a
Thus, the operation * is commutative.

(iii) For a, b, c ∈ N, we have:
(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
∴(a * b) * c = a * (b * 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 where a ∈ N
⇒ a * 1 = a = 1 * a, where 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 a * 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 a * b = L.C.M. of a and b a binary operation? Justify your answer.

Solution :-
```     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:

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

Solution :-
```    The binary operation * on N is defined as:
a * 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 where a, b ∈ N.
∴a * b = b * a
Thus, the operation * is commutative.

For a, b, c ∈ N, we have:
(a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c
a *(b * c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c
∴(a * b) * c = a * (b * c)
Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a where a ∈ N.
But this relation is not true for any a ∈ N.
Thus, the operation * does not have any identity in N.
```

Question-9 :- Let * be a binary operation on the set Q of rational numbers 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 operations are commutative and which are associative.

Solution :-
```(i) a * b = a − b
For Commutative :       ∴ [a * b = b * a]
a * b = a − b
b * a = b - a
e.g., (a, b) = (1/2, 1/3) ∈ Q
1/2 * 1/3 = 1/2 - 1/3 = 1/6
1/3 * 1/2 = 1/3 - 1/2 = -1/6
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (a − b) - c
a * (b * c) = a - (b - c)
e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q
(1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4
= 1/6 - 1/4
= -1/12
1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4)
= 1/2 - 1/12
= 5/12
Hence, the operation * is not associative.
```
```(ii) a * b = a² + b²
For Commutative :       ∴ [a * b = b * a]
a * b = a² + b²
b * a = b² + a²
e.g., (a, b) = (1/2, 1/3) ∈ Q
1/2 * 1/3 = (1/2)² + (1/3)²
= 1/4 + 1/9
= 13/36
1/3 * 1/2 = (1/3)² + (1/2)²
= 1/9 + 1/4
= 13/36
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (a² + b²) + c²
a * (b * c) = a² + (b² + c²)
e.g., (a, b, c) = (1, 2, 3) ∈ Q
(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
Hence, the operation * is associative.
```
```(iii) a * b = a + ab = a(1 + b)
For Commutative :       ∴ [a * b = b * a]
a * b = a(1 + b)
b * a = b(1 + a)
e.g., (a, b) = (1/2, 1/3) ∈ Q
1/2 * 1/3 = 1/2 x (1 + 1/3)
= 1/2 x 4/3
= 2/3
1/3 * 1/2 = 1/3 x (1 + 1/2)
= 1/3 x 3/2
= 1/2
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (a − b) - c
a * (b * c) = a - (b - c)
e.g., (a, b, c) = (1/2, 1/3, 1/4) ∈ Q
(1/2 * 1/3) * 1/4 = (1/2 - 1/3) - 1/4
= 1/6 - 1/4
= -1/12
1/2 * (1/3 * 1/4) = 1/2 - (1/3 - 1/4)
= 1/2 - 1/12
= 5/12
Hence, the operation * is not associative.
```
```(iv) a * b = (a − b)²
For Commutative :       ∴ [a * b = b * a]
a * b = (a − b)²
b * a = (b − a)²
e.g., (a, b) = (3, 2) ∈ Q
3 * 2 = (3 - 2)² = 1
2 * 3 = (2 - 3)² = 1
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (a − b)² * c
a * (b * c) = a * (b - c)²
e.g., (a, b, c) = (3, 2, 1) ∈ Q
(3 * 2) * 1 = (3 - 2)² * 1
= 1 * 1
= (1 - 1)²
= 0
3 * (2 * 1)  = 3 * (2 - 1)²
= 3 * 1
= (3 - 1)²
= 4
Hence, the operation * is not associative.
```
```(v)  a * b  = ab/4
For Commutative :       ∴ [a * b = b * a]
a * b = ab/4
b * a = ba/4
e.g., (a, b) = (4, 2) ∈ Q
4 * 2 = (4 x 2)/4 = 2
2 * 4 = (2 x 4)/4 = 2
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (ab/4) * c
a * (b * c) = a * (bc/4)
e.g., (a, b, c) = (4, 2, 1) ∈ Q
(4 * 2) * 1 = (8/4) * 1
= 2 * 1
= 2/4
= 1/2
4 * (2 * 1)  = 4 * (2/4)
= 4 * 1/2
= 4/8
= 1/2
Hence, the operation * is associative.
```
```(vi) a * b = ab²
For Commutative :       ∴ [a * b = b * a]
a * b = ab²
b * a = ba²
e.g., (a, b) = (3, 2) ∈ Q
3 * 2 = 3 x 2² = 12
2 * 3 = 2 x 3² = 18
Hence, the operation * is not commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = ab² * c
a * (b * c) = a * bc²
e.g., (a, b, c) = (3, 2, 1) ∈ Q
(3 * 2) * 4 = (3 x 2²) * 4
= 12 * 4
= 12 x 4²
= 192
3 * (2 * 4)  = 3 * (2 x 4²)
= 3 * 32
= 3 x (32)²
= 3072
Hence, the operation * is not associative.

```

Question-10 :- Find which of the operations given above has identity.

Solution :-
```     Let the identity be I.
(i) a * I = a - I
(ii) a * I = a² - I²
(iii) a * I = a + aI
(iv) a * I = (a - I)²
(v)  a * I = aI/4
(vi) a * I = aI²
Therefore, none of the operations given above has identity.
```

Question-11 :-  Let A = N × 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.

Solution :-
```    A = N x N and * is a binary operation defined on A.
(a, b) * (c, d) = (a+c, b+d)
= (c+a, d+b)
= (c, d) * (a, b)
Hence, The operation is commutative.
Again, [(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)
Here, [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)]
Hence, The operation is associative.
Let identity function be (e, f), then (a, b) * (e, f) = (a+e, b+f)
For identity function a = a + e
e = 0
and for b + f = b
f = 0
As 0 ≠ N, therefore, identity-element does not exist.
```

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

Solution :-
```(i)  * being a binary operation on N, is defined as a * a =  a∀a ∈ N
Hence operation * is not defined, therefore, the given statement is false.

(ii) * being a binary operation on N.
c * b = b * c and (c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * c) = (c * b) * a,  therefore the given statement is true.
```

Question-13 :-  Consider a binary operation ∗ on N defined as a ∗ b = 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?

Solution :-
```    Binary operation ∗ on N defined as a ∗ b = a³ + b³
For Commutative :       ∴ [a * b = b * a]
a * b = a³ + b³
b * a = b³ + a³
e.g., (a, b) = (2, 3) ∈ Q
2 * 3 = (2)³ + (3)³
= 8 + 27
= 35
3 * 2 = (3)³ + (2)³
= 27 + 8
= 35
Hence, the operation * is commutative.
For associative:        ∴ [(a * b) * c = a * (b * c)]
(a * b) * c = (a³ + b²) + c³
a * (b * c) = a³ + (b³ + c³)
e.g., (a, b, c) = (1, 2, 3) ∈ Q
(1 * 2) * 3 = (1³ + 2³) * 3
= (1 + 8) * 3
= 9 * 3
= 9³ + 3³
= 729 + 27
= 756
1 * (2 * 3) = 1 * (2³ + 3³)
= 1 * (8 + 27)
= 1 * 35
= 1³ + 35³
= 1 + 42875
= 42876
Hence, the operation * is not associative.
Therefore, option (B) is correct.
```
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