Question

Consider the binary operation table on set A ={1,2,3,6}

* 1 2 3 6
13 1 2 6
21 2 3 6
32 3 1 6
61 6 6 3
i) Find 2*6 , 6*1,2*(1*6), 6*(2*3)
ii) Is * commutative ? justify
iii) Find the identity element if exists.
iv) Find the elements which has inverse
*
PLEASE ANSWER THIS FASSSTT!!
I need this now as this is my assignment question!!

Answer 1

(i)

2 * 6 = 6

6 * 1 = 1

2 * (1 * 6) = 2 * 6 = 6

6 * (2 * 3) = 6 * 3 = 6

(ii) * is commutative if for any a, b ∈ A, we have a * b = b * a. You can verify this visually by checking for symmetry in the table along the main diagonal (going from top left to bottom right).

* is not commutative because

6 * 1 = 1

while

1 * 6 = 6

(iii) If e is the identity, then for any a ∈ A, we have a * e = e * a = a.

Here, 2 is the identity element because 1 * 2 = 1 and 2 * 1 = 1, and the same is true if we replace 1 with 2, 3, or 6.

(iv) Let a ∈ A. a has an inverse a ⁻¹ if a * a ⁻¹ = a ⁻¹ * a = e.

We know that e = 2, so look for any pairs that map to 2. The table shows that

1 * 3 = 2 and 3 * 1 = 2

2 * 2 = 2

so 1 has an inverse of 3, 3 has an inverse of 1, and 2 has itself as its inverse (which is to be expected for the identity).