Show that $x=y+z$ for all $x \in S$ We are given a set $S$ as a subset of the rational numbers defined by:


*

*$0 \notin S$  

*If $s_1 , s_2 \in S$, then $\frac {s_1}{s_2} \in S$  

*There exists a nonzero rational number $q \notin S$ such that every nonzero number in $Q \setminus S$ is of the form $qs$ for some $s \in S$.


Prove that if $x \in S$, then there exist $y$ and $z$ in S such that $x=y+z$.
Taking $s_1=s_2$ tells us that $1 \in S$, and thus $ \frac{1}{s} \in S $, and $s^k \in S$ for $k$ an integer and $s \in S$. Perhaps it can be proved by taking $y=x^a, z=x^b$ and proving integers $a$ and $b$ always exist, but I can't figure it out. Can anyone help me?

First notice that we only need to find one pair $s_1$, $s_2 \in S$ such that $s_1+s_2 \in S$. Then for any $x \in S$, we have
$$x=x.1=x\Bigg(\frac{s_1}{s_1+s_2}+\frac{s_2}{s_1+s_2}\Bigg) = x\frac{ s_1}{s_1+s_2}+x\frac{s_2}{s_1+s_2} $$
and we may then take $y = x\frac{ s_1}{s_1+s_2}$, $z = x\frac{s_2}{s_1+s_2}$.
Now $S$ is a subgroup of $\mathbb{Q}^{*}$, and condition 3. implies that 
$\mathbb{Q}^{*} = S \cup qS$ for some $q \not\in S$. Since $\mathbb{Q}^{*}$ is the union of two cosets of $S$, the index of $S$ in $\mathbb{Q}^{*}$ is $2$.
Therefore, the square of any non-zero rational belongs to $S$, so picking $s_1 = 9$ and $s_2 = 16$ does the job.
