Question

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 = $\frac{ab}{2}$
(iv) On Z+, define a ∗ b = $2^{ab}$
(v) On Z+, define a ∗ b = $a^b$
(vi) On R - {- 1}, define a ∗ b = $\frac{a}{b+1}$

Answer 1

i} binary because it is using two variables, not commutative because a-b is not equal to b-a, not associative since a-(b-c) is not equal to (a-b)-c

ii} it is commutative since ab-1=ba-1

iii} commutative

iv}u commutative

v}r associative

vi} binary

Answer 2

Hello sir ☺☺☺☺☺☺☺


(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 a * b = ab + 1.

It is known that:

ab = ba &mnForE; a, b ∈ Q

⇒ ab + 1 = ba + 1 &mnForE; a, b ∈ Q

⇒ a * b = a * 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.

3 , 4 , 5 ,6 in page I solve


Hope you understand