AL Pure Maths - Function

(a) when n=1 f(x)=1+f(x)-1
when n=k, assume that f(kx)=(1+f(x))^k-1
f[(k+1)x]
=f(kx)+f(x)+f(kx)f(x)
=[1+f(x)]^k-1+f(x)+{[1+f(x)]^k-1}f(x)
=[1+f(x)]^(k+1)-1
So, by MI f(nx)=(1+f(x))^n-1
(b) if f(1)=-1,from definition f(x+1)=f(x)+f(1)-f(x)f(1)=f(1)=-1
So f is a constant function, which is a contradiction
(c) Part (i) f(n)=[1+f(1)]^(n)-1
Part (ii) f(m)=[1+f(m/n)]^(n)-1=>f(m/n)=[f(m)+1]^(1/n)-1
But f(m)=[1+f(1)]^m-1, So f(m/n)=[f(1)+1]^(m/n)-1
(d)
Since we can construct rational number r_1,r_2,...r_n such that lim n-> infinity r_n = x
On the other hand, f(r_n)=[1+f(1)]^(r_n)-1
lim n-> infinity f(r_n)= lim n-> infinity [1+f(1)]^(r_n)-1
Since f is a continous function
f(x)= [1+ f(1)]^x-1=a^x-1