# 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:

1. $0 \notin S$
2. If $s_1 , s_2 \in S$, then $\frac {s_1}{s_2} \in S$
3. 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?

• My suspicion is that $\mathbb Q_{>0}$ is the only such $S$ (with $q =$ any negative rational). Can anybody think of any other possibilities? – TonyK Apr 3 '16 at 11:58
• If I'm not mistaken, then conditions in the question state precisely that $S$ is a subgroup of multiplicative group of rational numbers of index $2$. – Wojowu Apr 3 '16 at 12:48
• @TonyK Numbers with even $p$-adic valuation. – Wojowu Apr 3 '16 at 12:57

## 2 Answers

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.

• How are you writing $\frac{1}{1+x} = 1 - \frac{1}{1+x}$? – Seven Apr 3 '16 at 13:04
• Sorry, I'm an idiot. I meant $\frac{1}{1+x} = 1 - \frac{x}{1+x}$, but that works the same. – Aidan Sims Apr 3 '16 at 13:05
• Okay. But now how does $\frac{x}{1+x} \in S$? – Seven Apr 3 '16 at 13:13
• Well, we started with x \in S. We showed that $\frac{1}{1+x} \in S$. And we know that $S$ is closed under multiplication. – Aidan Sims Apr 3 '16 at 13:23
• Yeah. I realized that. I'm an idiot too :) – Seven Apr 3 '16 at 13:25

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.