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