[SOLVED] functions on rational numbers

[ehrenfest]

1. The problem statement, all variables and given/known data
Find all functions from Q to Q which satisfy the following two conditions:
i)f(1)=2
ii)f(xy)=f(x)f(y)-f(x+y)+1 for all x,y in Q


2. Relevant equations



3. The attempt at a solution

I can show by integers that if x is an integer, then f(x)=x+1. However, I am having trouble getting the value of the function for rational numbers. I want to do induction on n to get the inverse integers 1/n, but I cannot get 1/2. There is probably something clever you can plug in for x and y to get f(1/2), but I can't think of it.

[Hurkyl]

You might have to solve for several rational numbers at once, rather than one at a time. Also, there might be more than one such function f, in which case all you can do is figure out where the degrees of freedom lie, and express all values of f in terms of them.

[Dick]

You know more than you think. You can also show f(q+1)=f(q)+1 for all rationals q. What's f(q+n) for n an integer? What's f((m/n)*n)?

[ehrenfest]

You know more than you think. You can also show f(q+1)=f(q)+1 for all rationals q. What's f(q+n) for n an integer? What's f((m/n)*n)?

What's f(q+n) for n an integer?

I can show that f(q+n) = f(q)+n when n is a positive integer. Any rational number has a representation m/n, where m is an integer and n is a positive integer.

Then

f(m/n*n) = f(m/n)*f(n)-f(m/n+n)+1

m+1 =f(m/n)*(n+1)-f(m/n)-n+1

f(m/n) = (m+n)/n

I believe that is the only function that satisfies the two conditions.

[Dick]

I believe you are right. Notice f(m/n)=m/n+1. So the function is really just f(q)=q+1.