f(x+y)=f(x)*f(y), f(1)=q. find the followings.

There are 5 questions on this problem. The one I have difficult with is the last one, I think I got the first four.

Given a function f(x+y)=f(x)*f(y) and f(1)=q.

1. Find f(0).

f(1+0)=f(0)*f(1)=q

f(0)*q=q

f(0)=1

2. express f(-x) in terms of f(x)

f(-x)=f(x-2x)=f(x)*f(-2x)=f(x)*f(-x)*f(-x)

1=f(x)*f(-x)

f(-x)=1/f(x)

3. find f(n), when n is positive number.

since f(1)=q

f(2)=f(1)*f(1)=q^2

f(3)=f(2)*f(1)=q^3

f(n)=q^n

4. find f(n), when n is all integer (Z)

since f(0)=1=q^0

f(-1)=1/f(1)=1/q=q^-1

f(-2)=1/f(2)=1/(q^2)=q^-2

...

f(n)=q^n

5. find f(n), when n is all rational number.

I know I need to set n=a/b to take account n is a rational number. But I don't know how to proceed. Looking for help here :)

Re: f(x+y)=f(x)*f(y), f(1)=q. find the followings.

1-4 is correct.

For 5, you should mention that b has to be non-zero and without loss of generality b>0

f(a) = f((a/b)*b) = f((a/b) + (a/b) + (a/b) ... + (a/b)) [adding b times] = f(a/b)f(a/b)--f(a/b) [multiplying the function by itself b times] = [f(a/b)]^b

Hence f(a/b) = [f(a)]^(1/b) = (q^a)^(1/b) [by 4] = q^(a/b) [by exponent rules] which is what we expected. :)