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?
 A: Since $S = \emptyset$ does not satisfy (3), $S$ is nonempty, say $x \in S$. Now (2) implies that $1 = x/x \in S$. It then follows that $S$ is closed under reciprocals, and hence also under multiplication. So it suffices to show that $1$ can be written as $y + z$ for some $y,z \in S$, because then each $x = xy + xz$ is a sum of elements of $S$.
Suppose for contradiction that $1 - y \not \in S$ for every $y \in S$. Since $\{1, -1\}$ does not satisfy (3), we can choose $x \in S$ with $|x| \not= 1$. Since $S$ is closed under multiplication, $x^2 \in S$, and so $1 - x^2 \not\in S$ by our assumption. 
Condition (3) implies that $Q \setminus S \subseteq qS \cup \{0\}$. This is actually equality because we are given $0 \not \in S$, and if $x \in S \cap (qS \setminus \{0\})$ then $x = qy$ for some $y \in S$, and then $q = y/x \in S$, which violates (3).
So $1 - x$ and $1 - x^2$ both belong to $qS$, and therefore $\frac{1}{1 + x} = (1 - x)/(1-x^2) \in qS/qS = S/S \subseteq S$. Rewriting $\frac{1}{1+x} = 1 - \frac{x}{1+x}$, we obtain $1 - \frac{x}{1+x} \in S$. Since $\frac{x}{1+x} \in S$ as well, this contradicts our assumption.
A: 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.
