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Why we can analytically augment the algebraic definition of $ε$ in dual numbers so to distinguish $ε$ from $-ε$ but cannot do so in complex and split-complex numbers? Is there a theorem on this?

Algebraically one cannot distinguish between $\varepsilon$ and $-\varepsilon$. But it is possible to augment their definition analytically so to distinguish them.

In dual numbers there is a common equality: for differentiable at $x=a$ function $f(x)$, $f(a+b\varepsilon)=f(a)+b\varepsilon f'(a)$.

Now, one can define that if at point $x=a$ $f(x)$ has right and left derivatives $f'_r(a)$ and $f'_l(a)$, and they are not equal, then $f(a+\varepsilon)=f(a)+\varepsilon f'_r(a)$ and $f(a-\varepsilon)=f(a)-\varepsilon f'_l(a)$. In other words, $\varepsilon$ is defined as a positive infinitesimal, and $-\varepsilon$ is defined as a negative infinitesimal. This provides an optional analytic definition distinguishing $\varepsilon$ from $-\varepsilon$, but the algebraic structure can work just well without such additional analytic property (but with it one can evaluate more functions at more dual numbers).

Still, in dual numbers one cannot distinguish $\varepsilon$ from $a \varepsilon$ when $a>0$ even with this analytic addition.

So, my question is: why is it possible to define dual numbers in such a way so to distinguish the sign of dual unity, but the same cannot (?) be done with sign of imaginary and hyperbolic unities and with scale in dual numbers?

Is there any strong argument, why?

Is it because the lexicographical ordering in dual numbers naturally embeds into ordering of reals, while in complex and split-complex numbers it does not?

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  • $\begingroup$ For a differentiable function $f$ of a real variable, you seem to take the formula for $f(a+b\varepsilon)$ as a (provable) fact. So how do you define $f(a+b\varepsilon)$? If $f$ is given by a formula (such as a polynomial) you can just plug substitute $a+b\varepsilon$ for $$x$ in the formula and see if the result makes sense (and is independent of the formula!). But in general? $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 2, 2022 at 6:37
  • $\begingroup$ @LaurentMoret-Bailly as stated in the post, if $f$ is not differentiable, but has right or left derivative then we can define $f(a+b\varepsilon)$ via right derivative and $f(a-b\varepsilon)$ via left derivative. $\endgroup$
    – Anixx
    Commented Jun 2, 2022 at 6:44
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    $\begingroup$ Well, my point is that you can define it any way you please. What would make such a definition relevant? $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 2, 2022 at 11:42
  • $\begingroup$ And by the way, with your definition, what is $f(a+b\varepsilon)$ if $b=-1$? $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 3, 2022 at 18:28
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    $\begingroup$ In your first comment you wrote "we can define $f(a+b\varepsilon)$ via right derivative". What else does this mean? $\endgroup$
    – Laurent Moret-Bailly
    Commented Jun 5, 2022 at 15:59

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The short answer is "because you are considering $\mathbb{C}$ and $\mathbb{R}[\epsilon]$ them with different structure", which is an artificial choice.

Maybe to illustrate the point : If I want to work with $\mathbb{Q}[\sqrt{2}]$ would you consider that I can tell $\sqrt{2}$ and $-\sqrt{2}$ apart ? I'm not sure you can come up with an an argument for one answer over the other everybody would agree on.

Whether you can distinguish or not these depends on the type of structure you include. As you pointed out, "algebraically" - that is only using the ring structure - you can't distinguish $\epsilon$ and $-\epsilon$, to distinguish them you added new structure, for example the lexicographical order on $\mathbb{R}[\epsilon]$, or the way it acts differentiable functions can be evaluated on dual numbers.

Similarly, If you only consider $\mathbb{C}$ as a field, then you can't distinguish between $i$ and $-i$ because there is a field automorphism exchanging them, but you could add structure on $\mathbb{C}$ that would allow to distinguish them. For example, one could completely arbitrarily consider $\mathbb{C}$ as a field with a marked element "i", or put a "lexicographical order" on it, so that $i> -i$; that would make $i$ and $-i$ distinguished. Of course this structure is not very interesting, so you don't want to do this, and most people will not accept this as an answer.

But what about the structure of "being oriented" as a 2-dimensional real vector space (in addition to its field structure) ? That's a structure on the complex numbers that I would consider relevant, especially if you are using complex numbers to do geometry, and that allows to distinguish between $i$ and $-i$.

And on the contrary, if you consider $\mathbb{C}$ with less structure, for example as just a real vector space, then you can't tell $i$ and $-i$ apart but, you can't tell $i$ and $2i$ or $2i+3$ apart either.

For the question of $\mathbb{Q}[\sqrt{2}]$ the answer could depend on whether you are considering it as a field or as an ordered field.

So at the end of the day, the only question is whether you would consider that a certain structure that could be used to distinguish two (or more) elements is relevant or not. And that's a purely "sociological" question that has nothing to do with the mathematics, only with what we chose to consider relevant or not.

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    $\begingroup$ The exemple of the lexicographical order wasn't meant to be natural. But the point is that the notion of what is "natural" or not, is fairly arbitrary. By orientation of a vecteur space I mean as in: en.wikipedia.org/wiki/Orientation_(vector_space) This is an additional structure you put on $\mathbb{C}$ so it won't be definable in terms of the algebraic structure nor its analytical properties (given these are all preserved by complex conjugation). But this is no different from the fact that the way you distinguished $\epsilon$ from $-\epsilon$ depended on other structures. $\endgroup$
    – Simon Henry
    Commented Jun 1, 2022 at 23:26
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    $\begingroup$ The kind of objection you voice against the structure I added to distinguish $i$ and $-i$ can be said voiced about how you chose to define $f(a+b\epsilon)$ : you are chosing to use the left derivative if $b>0$ and the right derivative if $b<0$, but you could have made the other choice. By making this choice you are adding structure, which can then be used to distinguishes element that were not previously disinguishable. That is no different than chosing what will be considered a "postive orientation" for curve in the complex plan. $\endgroup$
    – Simon Henry
    Commented Jun 2, 2022 at 16:17
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    $\begingroup$ Note that my point isn't that these additional structure are "natural" or not, but that deciding what is natural or not is arbitrary. And you won't be able to distinguishes $i$ and $-i$ using only algebraic properties and "analytical properties" (in the sense of using analytical functions) as these are all preserved/equivariant under the conjugation automorphisms. $\endgroup$
    – Simon Henry
    Commented Jun 2, 2022 at 16:20
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    $\begingroup$ Ok, This is both not true (draw an arrow on a window and try it) and not relevant (the structure you are adding to $\mathbb{R}[\epsilon]$ is not the orientation of the line, but the definition of $f(a+b\epsilon)$. You are choosing to make $\epsilon$ corresponds to right, but you could exchange the role of $+\epsilon$ and $-\epsilon$). To give an example closer to what you do, I could say that I distinguish $i$ from $-i$ because the imaginary part of $i$ is $1$. Whatever you want to answer to this, my only point is that you can say the same thing about your definition. $\endgroup$
    – Simon Henry
    Commented Jun 3, 2022 at 15:16
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    $\begingroup$ well I don't know what you did, but you probably did it wrong, if you're arrow point right on one side, it will point left on the other side... But anyway, as I said that's not even relevant to what we were saying. I think the difference you are seeing between $\mathbb{R}$ and $\mathbb{C}$ is that the orientation of $\mathbb{R}$ is determined by its algebraic structure (we have a convention that "1" is right of "0" that gives the orientation). While the orientation of $\mathbb{C}$ is not determined by the algebraic structure (conjugation change the order but preserve the algebraic structure). $\endgroup$
    – Simon Henry
    Commented Jun 6, 2022 at 14:17

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