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 a * b = b * a. Therefore, the operation * is commutative.
(iii) (2 * 3) = 1 and (4 * 5) = 1
∴(2 * 3) * (4 * 5) = 1 * 1 = 1