The question says like this:
Let $S \subset [0,1]$ be a set satisfying the following two properties.(1)$0,1 \in S$;(2)For any $n \in \mathbb{N}$ and pairwise distinct numbers $s_{1},...,s_{n} \in S$ we have $\frac{s_{1}+...+s_{n}}{n} \in S$. Show that $S=\mathbb{Q} \cap [0,1]$.
My thought just like this: I first write down $0,1$ and they generate $\frac{1}{2}$ and then ganerate $\frac{1+\frac{1}{2}}{2}=\frac{3}{4}$ and so on. I try to construct a triangel just like the following:
On nth row are numbers like $\frac{i}{n}$, $i$ from $0$ to $n$.
I want to use induction method. For the first line, we know that it is in the set $S$, I wany to prove that if nth row are in the set $s$,then the $(n+1)th$ row are also in $S$. I am stuck in this step. Can anyone help me with this stuck?
ps: I have another kind of induction proof now.I find that it firstly correct for all $2^{k}$ and once we had proved it, we can try to prove that if it works for $x$,then it also works for $x−1$, and finally we can prove taht it works for all natural numbers.
For $k=1$ we can know that it works for $2^{1}$, suppose that it works for $2^{k}$ then we can find that $\frac{0+\frac{1}{2^{k}}}{2}=\frac{1}{2^{k+1}}$ is in $S$ and do similar operations, we can prove that it works for $2^{k+1}$. What we left now is trying to prove that if it works for $x$,then it also works for $x−1$. And now, I stuck here.
ps:anyone who is also interested in this question can click into the "linked" question on the right side. And thanks to Mr. or Mrs. @SarveshRavichandranlyer 's comment which remind me that we can only prove that $\mathbb{Q} \cap [0,1] \subset S$.