Question

Let A = N × N and * be the binary operation on A defined by ( a , b ) * ( c , d ) = ( a + c , b + d ) Show that * is commutative and associative. Find the identity element for * on A, if any.

Answer 1

It is given that the domain of the function is A=N×N and ∗ is the binary operation in A that is defined as,

( a,b )∗( c,d )=( a+c,b+d )

Consider ( a,b ),( c,d )∈A, therefore all the variables are natural numbers.

Apply the given binary operation on ( a,b )∗( c,d ).

( a,b )∗( c,d )=( a+c,b+d )

Apply the given binary operation on ( c,d )∗( a,b ).

( c,d )∗( a,b )=( c+a,d+b ) =( a+c,b+d )

The value of ( a,b )∗( c,d ) is equal to ( c,d )∗( a,b ). So, the given operation is commutative.

Consider ( a,b ),( c,d ),( e,f )∈A, therefore all the variables are natural numbers.

Apply the given binary operation on ( ( a,b )∗( c,d ) )∗( e,f ).

( ( a,b )∗( c,d ) )∗( e,f )=( a+c,b+d )∗( e,f ) =( a+c+e,b+d+f )

Apply the given binary operation on ( a,b )∗( ( c,d )∗( e,f ) ).

( a,b )∗( ( c,d )∗( e,f ) )=( a,b )∗( c+e,d+f ) =( a+c+e,b+d+f )

The value of ( ( a,b )∗( c,d ) )∗( e,f ) is equal to ( a,b )∗( ( c,d )∗( e,f ) ). So, the given operation is associative.

An element e=( e 1 , e 2 )∈A is said to be an identity element of the operation ∗ if the following condition is satisfied.

a∗e=a e∗a=a

For every a=( a 1 , a 2 )∈A.

( a 1 + e 1 , a 2 + e 2 )=( a 1 , a 2 ) and ( e 1 + a 1 , e 2 + a 2 )=( a 1 , a 2 )

Which is not true for any element in A.

Thus, the operation ∗ does not have any identity element.