Question

Let A = Z × Z and * be a binary operation on A defined by
(a,b)*(c,d) = (ad + bc, bd).

Find the identity element for * in the set A.​

Answer 1

Answer:

An element (e, f) ϵ Z × Z be the identity element, if

(a, b) * (e, f) = (a, b) = (e, f) * (a, b) ∀ (a, b) ϵ Z × Z

i.e., if, (af + be, bf) = (a, b) = (eb + fa, fb)

i.e., if, af + be = a = eb + fa and bf = b = fb …(1)

i.e., if, f = 1, e = 0 …(2)

Hence, (0, 1) is the identity element.

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 ∈ Z

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 ∈ Z

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 = Z×Z, 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....:)