Question

Let * be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a − b
(ii) a * b = a^2 + b^2
(iii) a * b = a + ab
(iv) a * b = (a − b)^2
(v) a * b = ab / 4
(vi) a * b = ab^2

Find which of the binary operations are commutative and which are associative.

Answer 1

* be a binary operation on the set Q of rational numbers as follows :

(i) we can see that 1 * 2 = 1 - 2 = -1
2 * 1 = 2 - 1 = 1
e.g., 1 * 2 ≠ 2 * 1
therefore, * is commutative.

Let 1 , 2 , 3 ∈ ℚ
(1 * 2 ) * 3 = (1 - 2) * 3 = -1 * 3 = -1 - 3 = -4
1 * (2 * 3) = 1 * (2 - 3) = 1 * -1 = 1 - (-1) = 2
e.g., (1 * 2) * 3 ≠ 1 * (2 * 3)
therefore, * is associative .

(ii) ∗ be a binary operation on the set Q of rational numbers is defined as
a ∗ b = a² + b²
For a, b ϵ Q, we get,
a * b = a² + b² = b² + a² = b * a
therefore, * is commutative.
We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q
Therefore, the operation * is not associative.

(iii) ∗ be a binary operation on the set Q of rational numbers is defined as
a ∗ b = a + ab
We can see that 1 * 2 = 1 + 1 × 2 = 1 + 2 = 3
and 2 * 1 = 2 + 2 × 1 = 2+2 = 4
⇒ 1 *2 ≠ 2 * 1: where 1,2 ϵ Q.
therefore, * is not commutative.
Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q
Therefore, the operation * is not associative.

(iv)  ∗ be a binary operation on the set Q of rational numbers is defined as
a ∗ b = (a – b)²
For a, b ϵ Q, we have,
a ∗ b = (a – b)²
b ∗ a = (b – a)² = [-(a – b)]² = (a –b)²
⇒ a * b = b * a
⇒ the operation * is commutative.
Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q
Therefore, the operation * is not associative.

(v) ∗ be a binary operation on the set Q of rational numbers is defined as a*b = ab/4
we can see that , 1 * 2 = 1 × 2/4 = 1/2
2 * 1 = 2 × 1/4 = 1/2
e.g., 1 * 2 = 2 * 1
therefore, * is commutative.

(1 * 2 ) * 3 = (1 × 2/2) * 3 = 1 * 3 = 3/2
1 * (2 * 3) = 1 * (2 × 3/2) = 1 * 3 = 3/2
e.g., (1 * 2 ) * 3 = 1 * (2 * 3)
therefore , * is associative.

(vi)  ∗ be a binary operation on the set Q of rational numbers is defined as
a ∗ b = ab²
We can see that for 2, 3 ϵ Q
2 * 3 = 2. 3² = 18 and 3 * 2 = 3. 2² = 12
⇒ 2 * 3 ≠ 3 * 2
Therefore, the operation * is not commutative.
Also, We can see that (1 * 2) * 3 ≠ 1 *(2 * 3), where 1,2,3 ϵ Q
Therefore, the operation * is not associative