Question

Let A = Q * Q, where Q is the set of rational numbers, and * be a binary operation on
A defined by (a, b) * (c,d) = (ac, b + ad) for all (a, b), (c,d) € A, then find
(i) the identity element of * in A.
(ii) invertible elements of A, and hence write the inverse of elements (5, 3) and (1/2,4)​

Answer 1

Answer:

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Step-by-step explanation:

Let A = Q * Q, where Q is the set of rational numbers, and * be a binary operation on

A defined by (a, b) * (c,d) = (ac, b + ad) for all (a, b), (c,d) € A, then find

(i) the identity element of * in A.

(ii) invertible elements of A, and hence write the inverse of elements (5, 3) and (1/2,4)

Answer 2

$\huge \sf {\orange {\underline {\pink{\underline{Answer :-}}}}}$

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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 ∈ Q

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 ∈ Q

* 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 ∈ Q

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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