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Is it possible in the system of dual numbers ($a+\epsilon b$; $\epsilon^2=0$) to calculate $\epsilon/\epsilon =1$? How then does one deal with $\epsilon^2/\epsilon^2=1$ versus $\epsilon^2/\epsilon^2=0/0$?

The same question for infitesimal calculus using hyperreal numbers where: $\epsilon \neq 0$ but $\epsilon^2=0$?

I probably did not use the correct formulation w.r.t. hyperreal numbers. I meant the axiom (?) in smooth infinitesimal analysis where it is assumed: $\epsilon \neq 0$ but $\epsilon^2=0$.
I am not quite sure how this analysis is related to nonstandard-analysis and hypercomplex numbers. I came across this topic in the book: A Primer of infinitesimal analysis (John L. Bell).

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    $\begingroup$ Uhm. I don't know about "dual numbers", but in the hyperreal numbers you certainly don't have $\varepsilon^2=0$. The hyperreal numbers are an ordered field, so $x^2=0\iff x=0$. $\endgroup$ – Asaf Karagila Mar 15 '14 at 19:44
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In the dual numbers, ${\mathbb R}[\epsilon]$ ($={\mathbb R}[X]/(X^2)$), $\epsilon$ is not invertible, so the expression $\epsilon / \epsilon$ ($= \epsilon \epsilon^{-1})$ is undefined.

In hyperreals, as Asaf Karagila mentions in the comments, $\epsilon^2 \neq 0$. There you do have $\epsilon / \epsilon = \epsilon^2 / \epsilon^2 = 1$ (as the hyperreals are a field and $\epsilon$ is a non-zero element).

I had a very quick look at the book by Bell. That's not only using infinitesimals, but also a different kind of logic (no law of excluded middle!). That's not for the faint-of-heart :-): for a given $x$, the statement "$x = 0 \lor x \neq 0$" is not necessarily true in that setting.

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    $\begingroup$ I was going to write the same (!) Just to observe that if you had $\epsilon^2=0$ and $\frac{\epsilon}{\epsilon}=1$ you would have $$\epsilon=\epsilon\frac{\epsilon}{\epsilon}=\frac{\epsilon^2}{\epsilon}=0$$If you want to invert a zero-divisor you have to lose something - so if you want addition and multiplication to behave normally, you have to leave some expressions undefined. $\endgroup$ – Mark Bennet Mar 15 '14 at 20:18
  • $\begingroup$ W.r.t. hyperreals, I probably misformulated, I meant nilsquare infinitesimals, edited my question. $\endgroup$ – Gerard Mar 16 '14 at 10:31
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"Multiplication by $\epsilon$" gives an isomorphism $\mathbf{R} \to \epsilon \mathbf{R}[\epsilon]$ (i.e. the ideal of multiples of $\epsilon$ in the dual numbers). Thus it's not terribly abusive to write "division by $\epsilon$" as the inverse of this isomorphism. If you do so, then $\epsilon / \epsilon$ is the real number 1 (and not the dual number $1$).

More generally, in any one-dimensional vector space over $\mathbf{R}$, it is not terrible to define division of vectors to be the unique real number that satisfies the equation $(v/w)w = v$.

Generalizing in another direction, if $I$ is a principal ideal of a ring $R$ generated by $t$, then "multiplication by $t$" gives a surjective $R$-module homomorphism $\mu_t : R \mapsto I : r \mapsto tr$ (this is not a ring homomorphism!). Correspondingly, it is not terrible to define "division by $t$" on $I$ to be the inverse isomorphism $I \mapsto R / \ker \mu_t$ that maps $tr \mapsto r$.

e.g. if I write $f(x) + O(x^n)$ to be the equivalence class of $f(x)$ modulo $(x^n)$ in the ring $\mathbf{R}[x]$, then we would have the appealing fact

$$ \frac{xf(x) + O(x^n)}{x} = f(x) + O(x^{n-1}) $$

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  • $\begingroup$ Sorry, I am not currently able to understand your answer ... $\endgroup$ – Gerard Mar 16 '14 at 10:23
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As far as smooth infinitesimal analysis (SIA) is concerned, it is not correct that it assumes that $\epsilon \neq 0$. It cannot be proven that $\epsilon \neq 0$. What is true is that infinitesimals in SIA satisfy $\neg\neg(\epsilon = 0)$.

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  • $\begingroup$ I see. What is the difference between 1) smooth infinitesimal analysis (SIA) 2) nonstandard analysis and 3) hyperreal numbers? $\endgroup$ – Gerard Mar 26 '14 at 11:11
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    $\begingroup$ Nonstandard analysis (NSA) and hyperreal numbers is roughly speaking the same thing. The important point is that NSA uses classical logic, including the law of excluded middle (LEM). Meanwhile SIA is a different theory that also is able to account for infinitesimals but in a way very different from NSA. In particular, SIA uses intuitionistic logic rather than classical logic. In particular, LEM is not allowed. Thus, all proofs by contradiction become suspect, because the LEM is the basic ingredient in any proof by contradiction. $\endgroup$ – Mikhail Katz Mar 26 '14 at 12:14
  • $\begingroup$ Thank you for the clarification. Does NSA use an infinitesimal $\epsilon$ with $\epsilon^2 = 0$? And how would SIA look at $\epsilon^2 / \epsilon^2$ ? $\endgroup$ – Gerard Mar 26 '14 at 12:18
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    $\begingroup$ In NSA all infinitesimals are invertible, because they are nonzero elements in a hyperreal field, and all nonzero elements in a field are invertible. Here the situation $\epsilon^2=0$ does not arise, unless of course $\epsilon=0$. Nilsquare infinitesimals in SIA are not invertible (nilpotent elements can never be invertible). Therefore the expression $\epsilon^2/\epsilon^2$ here is undefined. $\endgroup$ – Mikhail Katz Mar 26 '14 at 12:44

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