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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.
Solution
A = N × N
* is a binary operation on A and is defined by:
(a, b) * (c, d) = (a + c, b + d)
Let (a, b), (c, d) ∈ A
Then, a, b, c, d ∈ N
We have:
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
[Addition is commutative in the set of natural numbers]
∴(a, b) * (c, d) = (c, d) * (a, b)
Therefore, the operation * is commutative.
Now, let (a, b), (c, d), (e, f) ∈A
Then, a, b, c, d, e, f ∈ N
We have:
`((a, b)*(c,d)) * (e, f) = (a +c, b+d)*(e, f) = (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))`
Therefore, the operation * is associative.
An element `e = (e_1, e_2)` will be an identity element for the operation * if
`a"*"e = a = e"*"a "∀" a = (a_1,a_2) in A`
, i.e.,`(a_1 + e_1, a_2 + e_2) = (a_1, a_2) = (e_1 + a_1, e_2 + a_2)` which is not true for any element in A.
Therefore, the operation * does not have any identity element.
which is not true for any element in A.
Therefore, the operation * does not have any identity element.
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