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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

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Solution

(i)

The table of the operation is shown below:

* 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

The value of the function ( 2*3 )*4 from the table is,

( 2*3 )*4=1*4 =1

The value of the function 2*( 3*4 ) from the table is,

2*( 3*4 )=2*1 =1

Thus, the value of ( 2*3 )*4 is 1 and the value of 2*( 3*4 ) is 1 .

(ii)

For the function to be commutative a*b=b*a .

The function for any random value of a and b be a=4 and b=2 is,

4*2=2 2*4=2

So for every value of a,b{ 1,2,3,4,5 } , the function is commutative.

Thus, the function is commutative.

(iii)

The value of the function ( 2*3 )*( 4*5 ) from the table is,

( 2*3 )*( 4*5 )=1*1 =1

Thus, the value of ( 2*3 )*( 4*5 ) is 1 .


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