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I found a solution to this problem but I am not sure of one of the steps.

First let $a,b,c \in A$ then $a+b=2c$ thus $a/2+b/2=c$. We can continue this process to show that $\frac{1}{2^n}a+(1-\frac{1}{2^n})b \in A$ However, I am not sure how to conclude that $\frac{k}{2^n}a+(1-\frac{k}{2^n})b \in A$ for $k \in \mathbb{N}$ $0\leq k\leq 2^n$. I was thinking induction but i am not sure how. It is easy to show it is true for $n=1,2$ then assume it is true for $n$. I can show that $0\leq k\leq2^n$ works for $n+1$ but I do not know how to extend $k$ to $2^{n+1}$

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    I assume you get to $\frac{1}{2^n}a+(1-\frac{1}{2^n})b$ by dividing by $2$ and repeatedly adding $b$ each time? Try switching it up and add $a$ some of the time. – halrankard Jul 17 '20 at 19:06
  • A construction by dichotomy https://math.stackexchange.com/q/1371147/399263, look also to the linked post to this post with functions, this is very similar. – zwim Jul 17 '20 at 19:12

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