f(xy)=f(x)f(y)-f(x+y)+1when x=0 & y=1 then
f(0)=f(0)f(1)-f(1)+1
f(0)[1-f(1)]=1[1-f(1)]
so there r 2 possibilities : f(0)=1 ; or f(1)=1
Case 1 : when f(0)=1
then for any p element of Q, we've
f(0)=f(0)f(p)-f(p+1)+1
f(p)=f(p+1)
so f is a constant function = 1
Case 2: when f(1)=1 then
for x=1 & y=1
f(1)=f(1)f(1)-f(2)+1
f(2)=1
so f(3)=1,f(4)=1..........
similarly for any p element of Q f(p)=f(p+1)
so again f is a constant func. =1
So f (x)=1
M i correct ????
Moderation Team
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