A lemma for a new proof for the existence of a rational between two arbitrary real numbers

Question

I had in mind to prove the theorem "There exists a rational number between each two arbitrary Real numbers" and towards the end, I happened to need to prove this: $$$$ If we have the two numbers $b$ and $c$ ($b \neq c$), we can always create a number arbitrary close to say, $b$, using just $b$ and $c$. $$$$ Let us name $A_0 = \frac{b+c}{2}$. We take the average of $A_0$ and b, call it something like $A_1$ and continue this process. It is clear that each $A_{n+1}$ is closer to b than $A_{n}$. $$$$ To formalize the concept I defined the following, inductively: $$$$ Let $f(x)$ be an arbitrary function: We define $f_1(x)=f(x)$ and $f_n(x)=f_{n-1}(f(x))$ which is just a more detailewd writing of $f_n(x) = fofofo...of(x) $ n times $$$$ So, what we wish to prove now is this: If $b$ and $c$ are real numbers and $f(x)= \frac{b+x}{2}$, prove $$$$ $$\forall \epsilon>0 \hspace{0.5cm} \exists n (|f_n(c)-b|<\epsilon)$$ I'd be very thankful if someone provides a proof or a hint to the proof since this is of a very different form than the theorems I've seen before.

Answer 1

You have that $|f(x)-b|=\frac{|x-b|}{2}$ by the definition of $f(x)$. So by induction, $|f_{n}(x)-b|=\frac{|x-b|}{2^n}$. So in your last formula you just have to pick $n$ such that $\frac{|c-b|}{2^n}<\varepsilon$.