# Is there a natural way for elements of $\mathbb{Z}_2$ to be $\geq1$?

Is there a natural way to define elements $x$ of $\mathbb{Z}_2$ as being $\geq1$ in the conventional sense?

By which I mean e.g. the following numbers are $<1$:

$\frac{-1}{3}=\sum2^nk_n:$ $k_n=0$ for even $n,$ otherwise $k_n=1$.

$\frac{2}{3}=\overline{01}10<1$

Can this unambiguously be written $x\geq1$ or is there scope for that to be misunderstood?

Note that any notion of "$\ge 1$" is equivalent to an ordering of the ring in question - namely, view $a$ as less than $b$ if $b-a+1\ge 1$. So really you're asking about whether there is a natural way to order $\mathbb{Z}_2$. Although it is of course impossible to prove that there is no "natural" ordering on $\mathbb{Z}_2$, since "natural" is an informal term, we can show that any ordering of $\mathbb{Z}_2$ would have to be algebraically pathological:
If $\mathbb{Z}_2$ were an orderable ring, then $\mathbb{Q}_2$ would be an orderable field - any ordering of a ring induces an ordering on its field of fractions, and if the former is compatible with the ring structure then the latter is compatible with the field structure. But there is no ordering of $\mathbb{Q}_2$ which is compatible with the field structure. This can be seen by noting that $\mathbb{Q}_2$ contains an element whose square is $-7$, and hence $-1$ can be written (in $\mathbb{Q}_2$) as the sum of finitely many squares; this means that there is no real closed field containing $\mathbb{Q}_2$, hence by the Artin-Shreier theorem $\mathbb{Q}_2$ is not an orderable field.
• @MarcusAurelius: $\Bbb{Z}_p$ refers to the $p$-adic integers here, which do not form a field. – Torsten Schoeneberg Jan 8 '18 at 17:52
• @RobertFrost: For every field embedding $i:K\rightarrow \Bbb{R}$, pulling back the order on $\mathbb{R}$ via $i$ gives a total ordering $<_i$ on $K$ (dep. on $i$). $K=\Bbb{Q}$ has a unique embedding into $\Bbb{R}$. Other subfields $K$ of $\Bbb{Q}_2$ have (several) embeddings into $\Bbb{R}$. But beware, e.g. there are two elements $\pm x$ in $\Bbb{Q}_2$ with $x^2 =17$, but depending on which $i: \mathbb{Q}(x) \rightarrow \Bbb{R}$ you choose, $x$ is positive or negative. In $\Bbb{Q}_2$, there is no natural choice of whether $x$ or $-x$ is "positive". – Torsten Schoeneberg Jan 8 '18 at 21:51
• More generally, be aware that when you speak of "negative" numbers, you are assuming implicitly that you already have an ordering. So another way to express Noah Schweber's answer is that there is no meaningful way to partition all of $\mathbb{Q}_2$ into "negative" and "positive" numbers. – Torsten Schoeneberg Jan 8 '18 at 21:55
• "... in which all square roots of positive numbers were positive." That can't be, since square roots come in pairs $\pm x$. E.g. the square roots of $17$ in $\Bbb{Q}_2$, according to this, start $...01001$ and $...0111$. Each is the additive inverse of the other. Which one of them would you want to make "positive"? -- Re last comment: As said, $\Bbb{Q}$ embeds uniquely to $\Bbb{R}$, hence all those orders automatically extend the usual ones on $\Bbb{Q} \supset \Bbb{Z}$. The problem is the arbitrariness for everything that is $\notin \Bbb{Q}$. – Torsten Schoeneberg Jan 9 '18 at 5:33