Q. 25 B5.0( 3 Votes )

Let A = Q × Q and

Answer :

(i) Let (e, f) be the identity element for *

for (a, b) Q × Q


We have-


(a, b) * (e, f) = (a, b) = (e, f) * (a, b)


(ae, af+b) = (a, b) = (ea, eb+f)


From 1st & 2nd part-


ae = a e = 1 ...(1)


and, af+b = b af = 0 ...(2)


From 2nd & 3rd part-


a = ea e = 1 ...(3)


and, b = ef+b


b = (1)f + b [using (3)]


b = f + b


f = 0 ...(4)


From (2) & (4), we find that


a need not to be '0'.


e = 1, f = 0


(e, f) = (1, 0) Q × Q


(1, 0) is the identity element of A.


(ii) Let (a, b) Q × Q


Let (c, d) Q × Q


such that


(a, b) * (c, d) = (1, 0) = (c, d) * (a, b)


(ac, ad + b) = (1, 0) = (ca, cb + d)


ac = 1, ad + b = 0, ca = 1, cb + d = 0


c = (1/a), d = (-b/a), (1/a)b + d = 0 (a ≠ 0)


(c, d) = (1/a, -b/a) (a ≠ 0)


for a ≠ 0, (a, b)-1 = (1/a, -b/a)


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