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

Check the binary operation * is commutative :

We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

L.H.S =(a, b) * (c, d)

=(a + c, b + d)

R. H. S = (c, d) * (a, b)

=(a + c, b + d)

Hence, L.H.S = R. H. S

Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R

* is commutative (a, b) * (c, d) = (a + c, b + d)

Check the binary operation * is associative :

We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

L.H.S =  (a, b) * ( (c, d) * (x, y) ) =  (a+c+x, b+d+y)

R.H.S =  ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)

Thus, L.H.S = R.H.S

Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R

Thus, the binary operation * is associative

Checking for Identity Element:

e is identity of * if (a, b) * e = e * (a, b) = (a, b)

where e = (x, y)

Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)

= (x + a , b + y) = (a, b)

Now, (a + x, b + y) = (a, b)

Now comparing these, we get:

a+x = a

x = a -a = 0

Next compare: b +y = b

y = b-b = 0

Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.

Therefore, the operation * does not have any identity element.

Answer 2

Given A: N×N with binary operation ∗ definrd by (a,b)∗(c,d)=(a+c,c+d).
Step: 1-Checking if the operation is commutative:
An opertion ∗ on A is commutative if
a∗b=b∗a∀a,bϵA
∴(a,b)∗(c,d)=((a+c),(c+d))
Similarly, (c,d)∗(a,b)=((c+a),(d+c))
=((a+c),(c+d)) as addition is commutaitive in N.
⇒(a,b)∗(c,d)=(c,d)∗(a,b)
the operation ∗ is commutative.
Checking the operation is associative:
An operation ∗ on A is associative if
a∗(b∗c)=(a∗b)∗c∀a,b,cϵA
∴((a,b)∗(c,d))∗(e,f)=(a+c,b+d)∗(e,f)
Similarly, =(a+c+e,b+d+f)
(a∗b)∗((c,d)∗(e,f))=(a,b)∗(c+e,d,f)
=(a+c+e,b+d+f)
⇒((a,b)∗(c,d))∗(e,f)=(a∗b)∗((c,d)∗(e,f))
the operation ∗ is associative.
Step:-2 Checking if the operation has an identity,
We know that the element eϵN an identity element for operation ∗
if a∗e=e∗a for all aϵN
Lete=(e1,e2)ϵA,a=(a1,a2)ϵA
∴a∗e=(a1,a2)∗(e1∗e2)=(a1+e1,a2+e2)
however,this is not equal to a=(a1,a2) which for example would imply that 
a1=a1+e1→e1=0,which is not possible.
hence no identity element (e1,e2) exists in N for the operation ∗.