How to use induction on $p+q$ in functional equations

Question

Question -

Find all $f : \Bbb Q^+\to\Bbb Q^+$ such that

a) $f(x)+f(1/x)=1$

b) $f(f(x))=f(x+1)/f(x)$.

My try -

By checking some values I get $f(1)=1/2$ $f(2)=1/3$ .....

and using induction I proved for all natural numbers... Now i don't know how to extend it to positive rational numbers.... For extending something like these we have to first prove non-negativity,additivity,monotonicity , etc ....so that I can prove by standard methods but none of them is working here...

In the hint they wrote prove using induction on p+q Where $r=p/q$ is positive rationaI .... I did not able to see how to do it.....

Any help will be helpful... Thankyou

Answer 1

Claim: $ f( \frac{p}{q}) = \frac{q}{p+q}$.
We can check that this satisfies both conditions.

We will prove this by Strong Induction on $ n = p + q$.

Base case: $n = 2$.
This forces $ p = q = 1$, and we need to show that $ f(1) = \frac{1}{2}$.
This follows directly from condition 1, setting $x = 1$.

Induction step: Suppose that for all $ p + q \leq n$, we have $ f( \frac{p}{q}) = \frac{q}{p+q}$.
Now, for a given $ x^* = \frac{ p^* }{ q^*}, p^* + q^* = n+1$.
From condition 1 (and knowing $f(1)$), we may assume that $ p^* > q^*$.
Letting $ x = x^* - 1 = \frac{ p^* - q^* } { q^*}$, and applying condition 2, we get

$$f( \frac{q^*}{p^*} ) = \frac{ f( \frac{p^*}{q^*}) } { f( \frac{ p^* - q^* } { q^* }) } $$

Hint: Apply condition 1, and we have a system of linear equations.

Can you take it from here to show that $f( \frac{p^*}{q^*}) = \frac{q^* } { p^* + q^* } $, thereby completing the induction step ?