Question

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

Answer 1

(i) according to table , 2*3 = 1
so, (2*3)*4 = 1*4 = 1 [ according to table ]
according to table , 3*4 = 1
so, 2*(3*4) = 2*1 = 1 [ according to table ]

(ii) we can see 2*1 = 1 [ from table ]
1*2 = 1 [ from table ]
e.g., 2*1 = 1*2 = 1 , where 1,2 belongs to {1,2,3,4,5}
therefore, * is commutative.

(iii) according to table , (2*3) = 1 and (4*5) = 1
so, (2*3)*(4*5) = 1*1 = 1 [ according to table ]

Answer 2

Answer:

(i) according to table , 2*3 = 1

so, (2*3)*4 = 1*4 = 1 [ according to table ]

according to table , 3*4 = 1

so, 2*(3*4) = 2*1 = 1 [ according to table ]

(ii) we can see 2*1 = 1 [ from table ]

1*2 = 1 [ from table ]

e.g., 2*1 = 1*2 = 1 , where 1,2 belongs to {1,2,3,4,5}

therefore, * is commutative.

(iii) according to table , (2*3) = 1 and (4*5) = 1

so, (2*3)*(4*5) = 1*1 = 1 [ according to table ]

Step-by-step explanation: