Question

8. LetA=NxN and be a binary operation on A defined by
(a, b) (c,d) = (a + c, b + d)
a) find (1, 2) (2, 3)
b) Prove that is commutative
c) Prove that is associative​

Answer 1

Answer:

Let (a, b),(c, d) in N×N be two elements

then (a,b)*(c, d)=(a+c, b+d) and (c,d)*(a, b)=(c+a,d+b)

therefore (a,b)*(c,d)=(c,d)*(a,b)

therefore * is commutative

Let (a,b),(c,d),(e,f) be three elements in N×N

Then ((a,b)*(c,d))*(e,f)=(a+c,b+d)*(e,f)=(a+c+e,b+d+f)

and (a,b)*((c,d)*(e,f))=(a,b)*(c+e,d+f)=(a+c+e, b+d+f)

Therefore ((a,b)*(c,d))*(e,f)=(a,b)*((c,d)*(e,f))

Therefore * is associative on N×N

Answer 2

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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.

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Hope It's Helpful....:)