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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?

it's a hire car baby
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1 Answers1

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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.

So the evidence points to "no."

Noah Schweber
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  • This seems a bit complicated. There is a much more accessible notion of “ordered field” (and indeed $\mathbb{Z}_2$ is a field), which basically means “can you order it so it makes sense with the field operations?” The reals and rationals are an ordered field, while finite fields are not. – BRSTCohomology Jan 08 '18 at 16:29
  • Very interesting, thanks. The challenge to your answer that springs to mind is that the notion of an ordering could be simplified to the notion of a partitioning of the set. Perhaps this is a partitioning into elements having a non-negative (suitably defined) logarithm versus the other elements? – it's a hire car baby Jan 08 '18 at 16:59
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    @MarcusAurelius That's exactly what I'm talking about. I'm using "orderable" instead of "ordered" because the phrase "ordered field" (etc.) really ought to refer to a field with a particular order, etc. – Noah Schweber Jan 08 '18 at 17:05
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    @MarcusAurelius: $\Bbb{Z}_p$ refers to the $p$-adic integers here, which do not form a field. – Torsten Schoeneberg Jan 08 '18 at 17:52
  • Oh whoops, my bad! – BRSTCohomology Jan 08 '18 at 17:57
  • How do we square this answer off with the statement $\frac{1}{3}\leq1$? I mean, how is it that the question even has meaning? There must be some boundary where the concept as stated in the question (albeit vaguely, but nevertheless clearly enough to be understood), breaks down. It seems clear that $\mathbb{Z}$ considered as a subset of $\mathbb{Z}_2$ contains elements $\geq 1$, so even if not all elements in $\mathbb{Z}_2$ can be ordered... – it's a hire car baby Jan 08 '18 at 19:31
  • ...correct me if I'm wrong but does your answer assume $\geq$ is a partial or total order on $\mathbb{Z}_2$, and is it necessary to do so? – it's a hire car baby Jan 08 '18 at 19:34
  • @RobertFrost I'm assuming that it's a total ordering. – Noah Schweber Jan 08 '18 at 19:48
  • Thanks. I guess complex numbers can be $\geq 1$ provided their imaginary part is zero, and they can also be $\lvert\cdot\rvert\geq 1$ so I guess I wonder if this analogy can be continued into $\mathbb{Q}_2$ in other than the $\lvert\cdot\rvert_2$ way – it's a hire car baby Jan 08 '18 at 19:48
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    @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 08 '18 at 21:51
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    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 08 '18 at 21:55
  • @TorstenSchoeneberg thanks. Does "no natural choice" mean "provably no natural choice" or "no apparent natural choice"? I presume the latter, given my vague question and vague term 'natural'. – it's a hire car baby Jan 08 '18 at 22:28
  • @TorstenSchoeneberg ...so interpreting your answer I may be looking for an ordering on a subfield of Q2 which preserves the natural ordering of Q. If it we're possible I would want an embedding in which all square roots of positive numbers were positive. – it's a hire car baby Jan 08 '18 at 22:36
  • @TorstenSchoeneberg I guess I'm imagining we already have an ordering on Z which might be preserved and extended. – it's a hire car baby Jan 09 '18 at 04:48
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    "... 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 09 '18 at 05:33
  • @TorstenSchoeneberg my immediate desire to investigate what you say is to apply the logarithm function to those square roots of $17$ and see if it can discriminate positive versus negative in a way analogous to the way the logarithm of a negative real number is a complex number. e.g. perhaps one logarithm is convergent over $\mathbb{Q}_2$ while the other isn't. Unfortunately this simple suggestion would be a substantial undertaking for me!... – it's a hire car baby Jan 09 '18 at 20:10
  • ...The difference between logs of positive and negative square roots in $\mathbb{C}$ is $i\pi$. It turns out I have seen arguments that the obvious analogy for $i\pi$ in $\mathbb{Q}_2$ is $0$, which makes me all the more curious whether the logs of these two square roots are both convergent in $\mathbb{Q}_2$ and if so, what is the difference between them. – it's a hire car baby Jan 09 '18 at 20:15
  • @RobertFrost: Nice idea, but if I'm not mistaken, the $2$-adic logarithm has the properties $log(-1) = 0$ (maybe you mean this by "analogy of $i\pi$ is $0$" -- generally, one sets $log(\xi)=0$ for every root of unity $\xi$ in the $p$-adic world, but for $p=2$ this is actually what the series $log(-1) = log(1-2) = -\sum_{n=1}^\infty 2^n/n$ converges too), and $log(-x)=log(x)$ for all $x\in \Bbb{Z}_2^*$. – Torsten Schoeneberg Jan 15 '18 at 04:27
  • It is true that for the square roots $±x$ of an element $x^2 \equiv 1$ mod $8$ in $\Bbb{Z}_2$, one is $\equiv 1$ and one is $\equiv 3$ mod $4$. That is a good enough statement, which IMHO would only lead to confusion if one mixes it up with what one traditionally calls "positive" or "negative". This is a general philosophy in the p-adic world: one loses the linear order of positive and negative, but what one has instead is all these nested congruences. This is how the p-adic value "orders" numbers: http://www.nt.th-koeln.de/fachgebiete/mathe/knospe/p-adic/ – Torsten Schoeneberg Jan 15 '18 at 04:57
  • @TorstenSchoeneberg thanks; very interesting and helpful. You say it leads to confusion but I'm trying to unravel very complicated confusion (the Collatz ordering of the odd integers) so in a sense anything other than confusion/complexity/chaos is a sign of going in the wrong direction. – it's a hire car baby Jan 15 '18 at 11:22
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    This is unnecessary confusion. The term "positive" has a kind of clear meaning (pos. cones ...) incl. the properties "every element is either positive or its add. inverse is" and "sum of positives is positive". Whereas in your proposal, a sum of three "positive" numbers is "negative". We should avoid this confusion by giving your subset another name, not "positive". But wait, there is already a name for it: It's "the elements $\equiv 1$ mod 4". Oh yes. That's clear, and everyone understands it's closed under multiplication but not addition, how its additive inverses are $\equiv 3$ mod 4 etc. – Torsten Schoeneberg Jan 16 '18 at 05:56