# Are we allowed, by the definition of induction, to assume that $f_{1}(x)$ still has the same formula, even though we switched $n$?

by ArielK   Last Updated November 18, 2018 11:20 AM

I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that

For all natural number $$n$$, $$f_{0}(x) = \frac{x}{x+1} \& f_{n+1}(x)=(f_{0}(f_{n}(x))) \implies f_{n}(x) = \frac{x}{(n+1)x + 1}$$

So if we first verify it for n=0 then the explicit formula for $$f_{n}$$ is given. But then, if we try to verify it for any other $$n$$, let's say $$n=1$$, we would have the formula for $$f_{0}(x)$$ as always, and we would also have that $$f_{2}(x) = f_{0}(f_{1}(x))$$ but how is that suppose to lead us to conclude the explicit formula for $$f_{1}$$ if we don't know what $$f_{2}(x)$$ looks like?

Did I get induction wrong?

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