Let a set S have two numbers, 0 and 1.
S = {0, 1}
We may choose any two numbers of this set, calculate the average and then add the result to the set. We may repeat this process.
For example
S = {0, 1}
S = {0, 0.5, 1} 0.5 = (0 + 1)/2
S = {0, 0.25, 0.5, 1} 0.25 = (0 + 0.5)/2
Now, given a number k, ex. k = 0.6, is it possible to show whether or not this number will ever appear on this set ?
Starting with a set $S_0$ and letting each $S_{n+1}$ be the union of $S_n$ with the set of averages of pairs of members of $S_n,$ we show by induction on $n$ that every member of any $S_n$ is of the form $2^{-n}\sum_{j=1}^KA_js_j$ for some selection $s_1,...,s_K$ of members of $S_0,$ where $A_1,...,A_K$ are non-negative positive integers that satisfy $\sum_{j=1}^KA_j=2^n.$
( $K$ can be anything).
If $S_0=\{0,1\}$ then any $x\in S_n$ must be $2^{-n}(A_1\cdot 0+A_2\cdot 1)$ with $A_1+A_2=2^n,$ that is, $x=2^{-n}A_2$ with $0\le A_2 \in\Bbb Z.$
Now if $0.6=3/5=2^{-n}A_2$ then $3\cdot 2^n=5A_2,$ which requires $5$ to be a divisor of $3\cdot 2^n,$ which is impossible.