Does a subset $S$ of $\mathbb{Q}$ with the following conditions, equal $\mathbb{Q}$?

Question

Let $S$ be a subset of $\mathbb{Q}$ such that:

1. $0 \in S$;
2. If $x\in S$, then $(x-1) \in S$ and $(x+1) \in S$;
3. If $x \in S \setminus\{0,1\}$, then $\frac{1}{x(x-1)} \in S$.

Does $S = \mathbb{Q}$?

Answer 1

No, $S$ need not be $\Bbb{Q}$.

The goal here will be to prove this by proving a stronger result, and then ask 2 natural followup questions. We will start with some preamble to get some notation in place.

Define $f:\Bbb{Q}\backslash\{0,1\}\rightarrow\Bbb{Q}$ to be the map from $\left(3\right)$, that is

$$f\left(x\right)=\frac{1}{x\left(x-1\right)}$$

Let $A$ be any subset of $\Bbb{Q}$, define $S\left[A\right]$ inductively as follows.

Set $S_0\left[A\right]=A+\Bbb{Z}$

Now let $S_{n+1}\left[A\right]=S_n\left[A\right]\cup \left(f\left(S_n\left[A\right]\backslash\{0,1\}\right)+\Bbb{Z}\right)$

Finally define $S\left[A\right]$ to be the ascending union, i.e.

$$S\left[A\right]=\bigcup\limits_{n=0}^{\infty}S_n\left[A\right]$$

From this it is clear that $S\left[A\right]$ is the smallest set satisfying $\left(2\right)$ and $\left(3\right)$ which contains $A$ (We ignore $\left(1\right)$ as we can simply choose $0\in A$ additionally we will use the shorthand $S\left[0\right]=S\left[\{0\}\right]$ ).

Let us consider a class of elements not in $S\left[A\right]$. First suppose that $q\in S\left[A\right]$, then $q\in S_n\left[A\right]$ for some $n$. Choose the smallest such $n$.

If $n=0$ then $q\in A+\Bbb{Z}$

If $n>0$ then $q\in S_{n}\left[A\right]=S_{n-1}\left[A\right]\cup \left(f\left(S_{n-1}\left[A\right]\backslash\{0,1\}\right)+\Bbb{Z}\right)$

But $q\not\in S_{n-1}\left[A\right]$ so $q\in f\left(S_{n-1}\left[A\right]\backslash\{0,1\}\right)+\Bbb{Z} \subseteq f\left(\Bbb{Q}\backslash\{0,1\}\right)+\Bbb{Z}$

Now we know that if $q\not\in A+\Bbb{Z}$ and $q\not\in f\left(\Bbb{Q}\backslash\{0,1\}\right)+\Bbb{Z}$ then $q\not\in S\left[A\right]$.

We now examine elements not in $f\left(\Bbb{Q}\backslash\{0,1\}\right)+\Bbb{Z}$.

Set $U=\Bbb{Q}\backslash\left(f\left(\Bbb{Q}\backslash\{0,1\}\right)+\Bbb{Z}\right)$ and call elements of $U$ nongeneratable.

$\textbf{Now comes the main proof}$

We will show that $\frac{2}{p}\in U$ for all primes $p>3$.

Suppose $\frac{2}{p}=f\left(\frac{a}{b}\right)+z$. Additionally assume $\frac{a}{b}$ is reduced.

$$f\left(\frac{a}{b}\right)=\frac{1}{\frac{a}{b}\left(\frac{a}{b}-1\right)}=\frac{1}{\frac{a}{b}\cdot\frac{a-b}{b}}=\frac{b^2}{a\left(a-b\right)}$$

Define $\tilde{n}=a$,$m=a-b$, then

$$f\left(\frac{a}{b}\right)=\frac{\left(\tilde{n}-m\right)^2}{\tilde{n}m}$$

As $z\in\Bbb{Z}$ for $\frac{\left(\tilde{n}-m\right)^2}{\tilde{n}m}+z$ to be $\frac{2}{p}$ we require that $p|\tilde{n}m$.

As everything is symmetric in $\tilde{n},m$ and due to the defining property of primes we may without loss of generality choose $\tilde{n}=pn$. Now,

$$\frac{\left(\tilde{n}-m\right)^2}{\tilde{n}m}+z=\frac{\left(pn-m\right)^2}{pnm}+z=\frac{\left(pn-m\right)^2+pnmz}{pnm}=\frac{2nm}{pnm}$$

Therefore,

$$\left(pn-m\right)^2+pnmz=2nm$$

and so,

$$\left(pn-m\right)^2\equiv2nm\mod pnm\tag{4}$$

Now $\frac{a}{b}\neq0,1$ implies that $n,m\neq0$. Additionally the fact that $\frac{a}{b}$ is reduced tells us that $\tilde{n}$ and $m$ are coprime, which is equivalent to $n$ and $m$ coprime and $p\nmid m$.

Now let $q$ be any prime. First assume $q|n$.

Now $q|pnm$ so the equivalence $\left(4\right)$ holds $\bmod q$. This gives us,

$$\left(pn-m\right)^2\equiv m^2\equiv 2nm \equiv 0 \mod q$$

Thus $m^2\equiv 0 \pmod q$ and as $q$ is prime we conclude $m\equiv0 \pmod q$.

This implies $n,m$ not coprime, a contradiction. Therefore $q\nmid n$.

Now instead assume $q|m$, again we can consider $\left(4\right)$ $\bmod q$. This gives,

$$\left(pn-m\right)^2\equiv \left(pn\right)^2\equiv 2nm \equiv 0 \mod q$$

From this we conclude either $p\equiv0$ or $n\equiv0 \pmod q$.

The second is a contradiction so instead we assume $p\equiv0 \pmod q$.

But now $p$ and $q$ are primes so $p=q$.

Therefore $p|m$ which is again a contradiction.

Putting this together (along with F.T.A.) we conclude $\left|n\right|=\left|m\right|=1$. Now we can calculate that the $4$ choices for $n,m$ give us

$$f\left(\frac{a}{b}\right)=\frac{\left(p-1\right)^2}{p} \text{ or } \frac{-\left(p+1\right)^2}{p}$$

Choosing $z$ so that $f\left(\frac{a}{b}\right)+z\in \left[0,1\right]$ we get

$$f\left(\frac{a}{b}\right)+z=\frac{1}{p} \text{ or } \frac{p-1}{p}$$

As $p>3$ we conclude $f\left(\frac{a}{b}\right)+z\neq\frac{2}{p}$ which is a contradiction.

Therefore $\frac{2}{p}\in U$.

From this we deduce that $\frac{2}{p} \in S\left[A\right]$ if and only if $\frac{2}{p} \in A+\Bbb{Z}$.

Now if $A$ is finite there is some maximum denominator among the fractions in $A$ (when in reduced form), therefore we may choose some prime $p$ larger than that and conclude that $\frac{2}{p} \not\in A+\Bbb{Z}$.

Therefore $\frac{2}{p} \not\in S\left[A\right]$ and so $S\left[A\right]\neq\Bbb{Q}$.

The question asks about the special case with $0\in A$. $S\left[0\right]$ is the smallest such set, and we see from the above that $S\left[0\right]\neq\Bbb{Q}$.

$ \square $

We now pose $2$ immediate followups.

1. Is $S\left[U\right]=\Bbb{Q}$?

2. Is $S\left[0\right]=\Bbb{Q}\backslash U$?

Answer 2

Consider the minimal closure $S$ of $\{0\}$ under the two operations. From the first operation, we know that $S$ is a union of cosets of $\mathbb Z$. Only the second operation can generate new cosets.

We claim $\frac{3}{5}\not\in S$. Otherwise $\frac{1}{(a/b)(a/b-1)}+\mathbb Z=\frac{3}{5}+\mathbb Z$ where $\gcd(a,b)=1$.

Observe $\frac{1}{(a/b)(a/b-1)}=\frac{b^2}{a(a-b)}$ and $\gcd(b, a(a-b))=\gcd(b, a^2)=1$, hence $\gcd(b^2, a(a-b))=1$ as well.

And if $\frac{b^2}{a(a-b)}=\frac{3}{5} + k = \frac{3+5k}{5}$, since $\gcd(3+5k, 5)=1$, we must have $b^2=\pm (3+5k)$, thus $b^2\equiv \pm 3 \mod 5$. But neither $3$ nor $-3$ is a quadratic residue mod $5$.

Note that there is a general pattern. Using the same method, for any prime $p\equiv 1\mod 4$ and the Legendre symbol $(\frac{a}{p})=-1$, we conclude $b^2\equiv \pm a \mod p$, and since $(\frac{-1}{p})=1$, we always have $(\frac{a}{p})=1$, a contradiction.

Note that if we replace the second condition by the stronger $\frac{1}{x(x-n)}\in S$ for any $x\not=0,n$ and $n\in\mathbb Z$, the above argument is still valid.