Can we find the average of three numbers in this way? Let $f_n$ denote the "averaging" function $[0,1]^n \rightarrow [0,1]$ defined as follows. $$f_n(x_0,\cdots,x_{n-1}) = \frac{x_0 + \cdots + x_{n-1}}{n}$$
Then clearly, it is possible to express $f_4$ in terms of $f_2$.
$$f_4(a,b,c,d) = \frac{a+b+c+d}{4} = \frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} = f_2(f_2(a,b),f_2(c,d)).$$
Question. Can we express $f_3$ in terms of $f_2$ in this way, using $0$ and $1$ as well if necessary? Perhaps this is trivial, but I cannot see the answer.
A more precise statement of the problem.
Let $A$ denote the following set of "primitive symbols." $$\{a,b,c,0,1\}.$$ We define a class of "valid expressions" $B$ as the least superset of $A$ such that if $\alpha,\beta \in B$, then $f_2(\alpha,\beta) \in B$.
The problem can be solved in one of two ways, listed here below.


*

*Option 1. Find an expression in $B$ that correctly defines $f_3(a,b,c).$

*Option 2. Show that no such expression exists.

 A: You cannot build $f_3$ in such a way.
A first solution, not the most rigorous but it gets the idea across without using too much algebra: 
Let $R$ be any ring where $2$ is invertible. Then $\frac{x+y}{2}$ is defined on elements of $R$. 
This allows us to define a mapping $\varphi: B \times R^3 \to R$, where $\varphi(\beta,a,b,c)$ is the result of evaluating the symbolic expression $\beta$, with the values $a,b,c \in R$ substituted in. 
(For example, if $\beta = f_2(f_2(a,b),c)$, then $\varphi(\beta,a,b,c) = \frac{a+b}{4} + \frac{c}2$.)
Suppose there is an expression $\beta \in B$ such that $\varphi(\beta,a,b,c) = \frac{a+b+c}{3}$. Then $\varphi(\beta,1,0,0) = \frac{1}3$, and so $3$ is invertible in $R$. 
(This part bothers me a bit, because it's a bit presumptuous to even write down the expression $\frac{a+b+c}{3}$. Perhaps it would be better stated that $3 \varphi(\beta,a,b,c) = a+b+c.$ Either way, $3$ ends up being invertible so long as 2 is, if we suppose that we can build $f_3$ by iterating $f_2$, and allowing ourselves to substitute in 1 and 0.)
And, so, it suffices to note there are rings $R$ where 3 is not invertible, but 2 is. One such ring is $\mathbb{Z}/3$.

I intended the above solution to avoid possibly confusing abstraction, but in response to comments below I will be super rigorous: 
Let $S = \mathbb{Z}[\frac{1}2]$ denote the ring of rational numbers with denominator a power of two. We can define a mapping of sets $\tau : B \to S[a,b,c]$ by sending $a,b,c,0,1$ to themselves, and $f_2(x,y)$ to $\frac{1}2 (\tau(x) + \tau(y))$. A simple induction proves that the image of $\tau$ consists solely of linear polynomials.
The map $\tau$ extends linearly to a map of $S-$modules $\varphi: SB \to S[a,b,c]$. We have also the change of rings morphism $\theta : S[a,b,c] \to \mathbb{R}[a,b,c]$, and evaluation maps $ev_{(x,y,z)} : \mathbb{R}[a,b,c] \to \mathbb{R}$. 
This gives chains of $S$-module morphisms 
$$SB \xrightarrow{\varphi} S[a,b,c] \xrightarrow{\theta} \mathbb{R}[a,b,c] \xrightarrow{ev} \mathbb{R}.$$
where $ev$ is any one of the many evaluation maps we might choose from.
The map $\varphi$ also has image solely among the linear polynomials, while $\theta$ is injective and degree-preserving.
If $ev_{(x,y,z)}P = ev_{(x,y,z)} Q$ for every $x,y,z \in [0,1]$, and $P,Q,$ are linear polynomials, then $P = Q$.
Therefore, it suffices to show that there is no $\beta \in SB$ such that $\theta (\varphi(\beta)) = \frac{a+b+c}{3}.$ 
Proof of claim: Suppose otherwise. Then $\theta (3 \varphi(\beta)) = 3\theta (\varphi (\beta)) = a+b+c$. The map $\theta$ is injective, so $3\varphi(\beta) = a+b+c$. Evaluating at $(a,b,c) = (1,0,0)$ gives that $3\varphi(\beta)(1,0,0) = 1 \in S$; yet, $3$ is not invertible in $S$ (*). This gives a contradiction.  
(*) For various reasons: to stick to my guns a bit, one being that, since $2$ is a unit in $\mathbb{Z}/3$, there is a unique ring morphism $S \to \mathbb{Z}/3$ extending the canonical quotient map $\mathbb{Z} \to \mathbb{Z}/3$. As $3$ is not a unit in $\mathbb{Z}/3$, it cannot be a unit of $S$.
A: We will prove something a bit more general. We prove that no composition can ever take the value $1/3$ when supplied $0$'s and $1$'s as argument. We define $B'$ to be all functions $[0,1]^n \to \mathbb{R}$ which can be written as a composition of $f_2$, $0$ and $1$. 
More precisely, $B'$ is the smallest superset of $A' = \{f_2,0,1\}$, where $0,1$ are the constant functions, such that $B'$ is closed under composition.
It's clear that the $B$ defined in the problem, interpreted appropriately, is a subset of this.


*

*$B'$ is given by finite compositions of elements of $A'$, since this is a superset of $A'$ that is closed under composition of $f_2$, and any such superset must contain finite compositions. This is a standard argument in recursive definitions. More precisely, let $A_0' =A'$ and $A_{k+1}' = A_k' \cup \{f_2 \circ (\beta_1,\beta_2) \mid \beta_1,\beta_2 \in A_k'\}$. Then $B' = \cup_k A_k'$. To be even more precise, we should include arity $ar(\beta)$ of elements $\beta \in A_k'$ and the tuple $(\beta_1,\beta_2)$ will have different choices of arity between $\min(ar(\beta_1),ar(\beta_2))$ and $ar(\beta_1)+ar(\beta_2)$ depending on how we mix the arguments.

*For any $\beta \in B'$ of arity $n$, and any $n$-vector $v$ of $0$'s and $1$'s, $\beta(v)$ is rational, and in the reduced form, the denominator is a power of $2$. This is easily proved via induction on $k$ above, since $\beta \in A_k'$ for some $k$.
