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

(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 a * b = a − b(ii) On Q, 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 (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 a ∨b = min {a, b}. Write the operation table of the operation∨.

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

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

(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 a *′ b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your 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 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(v) Which elements of N are invertible for the operation *?

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 *.