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In chapter 9 of Spivak's Calculus, on derivatives, he mentions the "Leibnizian Notation" for the derivative of a function $f$, $\frac{df(x)}{dx}$. In a footnote on page 155, he writes

Leibniz was led to this symbol by his intuitive notion of the derivative, which he considered to be, not the limit of quotients $\frac{f(x+h)-f(x)}{h}$, but the "value" of this quotient when $h$ is an "infinitely small" number. This "infinitely small" quantity was denoted $dx$ and the corresponding "infinitely small" difference $f(x+dx)-f(x)$ by $df(x)$. Although this point of view is impossible to reconcile with properties (P1)-(P13) of the real numbers, some people find this notion of the derivative congenial.

The bold section has been highlighted by me. What does he mean with that?

Rodrigo de Azevedo
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xoux
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    He means that you can’t make sense of infinitesimals in the ordinary real numbers. – Randall Apr 03 '22 at 13:46
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    https://math.stackexchange.com/questions/3105978/does-every-complete-ordered-field-mathbbf-hold-the-archimedean-property – Ethan Bolker Apr 03 '22 at 14:17
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    There are different axiomatizations of real numbers. The one describe in Spivak's is the one that is adopted by most Mathematicians. Other axiomatizations, yield to different types of reals (hyperreals, etc) and is the subject of non Standard Analysis. Within that framework, the infinitesimals" dy, dx, etc are well define entities. – Mittens Apr 03 '22 at 14:35
  • So the infinitesimals are not well-defined entities in standard analysis? If the axiomatization described by Spivak is the most common one, and if that is used in standard analysis, and since infinitesimals are everywhere in Calculus, how can standard analysis get by without well-defined infinitesimals? – xoux Apr 03 '22 at 14:51
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    @evianpring: right. At issue is the fact that on the one hand, $dx$, $dy$, etc behave like small quantities that are smaller than any positive number but are not zero. This is not possible in the "standard" axiomatization of the reals. Further extensions of reals (also via equivalent classes as in standard analysis) make sense of such "infinitesimals". The article has a nice non-technical overview of history and facts about infinitesimals. – Mittens Apr 03 '22 at 15:39
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    The point of treating $dy/dx$ as a "limit of quotients" (in Spivak's words) rather than just simply a quotient (as Leibniz considered it) is that we don't need to use infinitesimals for anything in calculus. The only vestige of infinitesimals is the Leibniz notation, which we have to keep reminding ourselves is "not a ratio." Or you can go the non-standard route; but you don't have to. – David K Apr 22 '22 at 22:48
  • @DavidK not sure I understand what you mean about not needing to use infinitesimals in calculus. Take for example linear approximation of functions, or certain proofs of certain theorems (e.g. divergence theorem, fundamental theorem for gradients, stokes' theorem) that involve integrals. Don't these concepts all use the idea of infinitesimals? – xoux Apr 22 '22 at 23:31
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    No, in the standard model it's all limits or error terms, none of which is an infinitesimal. You might like to use infinitesimal notions as a kind of mnemonic or guide to intuition in some cases, but formally they don't come into play unless you make the considerable effort to develop non-standard analysis. – David K Apr 22 '22 at 23:35

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Within that axiomatic, there are no infinitesimal numbers, that is, there is no number $\mu>0$ such that $(\forall n\in\Bbb N):\mu<\frac1n$. That's so because the Archimedean property follows from those axioms. And that property states that $\Bbb N$ has no upper bound. But if such a number $\mu$ existed, we would have $(\forall n\in\Bbb N):n<\frac1\mu$.

Mittens
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José Carlos Santos
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  • It seems that you have defined an infinitesimal number as $\mu>0$ such that $forall n \in \mathbb{N}$, $\mu <\frac{1}{n}$. When calculus books speak of infinitesimals, is this the definition that they are using? I have taken a standard Calculus course, and every single treatment of infinitesimals that I have ever encountered has seemed incomplete, ambiguous, and confusing. Under what axiomatic, then, is there such a thing as an infinitesimal number? – xoux Apr 03 '22 at 14:19
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    @evianpring there are precise treatments, though I never studied them. Taking a look at a book such as Lectures on the Hyperreals: An Introduction to Nonstandard Analysis by Robert Goldblatt might be interesting to you. – Steven Apr 03 '22 at 14:32
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    @evianpring There are many versions of your comment's question, and most of them have been answered on this site across one or more questions (e.g. how to make one argument or another about infinitesimals rigorous, the simplest way to construct an infinitesimal, understanding how nonstandard analysis can be used for single variable calculus, understanding the formal setup of nonstandard analysis). "Elementary Calculus: An Infinitesimal Approach" is probably the most accessible book, but might not answer many questions. Feel free to post a new question if you can't find what you're looking for. – Mark S. Apr 04 '22 at 11:25
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    @evianpring "When calculus books speak of infinitesimals, is this the definition that they are using?" Not typically, no. But it is the sort of issue Spivak was alluding to in the quote, which is why I upvoted this answer. – Mark S. Apr 04 '22 at 11:27
  • @Steven: There is no real benefit of using hyperreals. Firstly, constructing the hyperreals needs philosophically dubious foundational assumptions. Secondly, asymptotic analysis suffices to provide all the benefits of thinking in terms of 'vanishing quantities' without them being zero. Thirdly, we can rigorously get Leibniz-style real analysis by treating each variable as a function rather than a fixed number, and treating arithmetic on them as pointwise, as sketched in this post, and it is compatible with asymptotic analysis. – user21820 May 03 '22 at 18:09
  • @evianpring: See my above comment. Note that of course these variables do not satisfy the same properties as real numbers. But it is a rigorous treatment of Leibniz-style real analysis. Unfortunately, most people are unaware that it can be done that way. – user21820 May 03 '22 at 18:14
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The precise meaning of "infinitesimal" depends on context, but it common to define an infinitesimal number as a positive number $x$ such that $x<1/n$ for every natural number $n$; similarly, an infinite number satisfies $x>n$ for every natural number $n$. With this definition, it can be proven that $\mathbb R$ does not contain infinite or infinitesimal numbers.

To prove that infinite numbers do not exist in $\mathbb R$, assume for the sake of contradiction that there is an infinite number $x\in\mathbb R$. Then, $x$ is an upper bound of $\mathbb N$, and so by the axiom of completeness, $\mathbb N$ has a least upper bound $\alpha$. Since $\alpha-1<\alpha$, it follows that $\alpha-1$ is not an upper bound of $\mathbb N$; in particular, there is an $n_0\in\mathbb N$ such that $n_0>\alpha_0-1$. But then $n_0+1>\alpha$ and $n_0+1\in\mathbb N$, contradicting the fact that $\alpha$ is an upper bound of $\mathbb N$. Hence, $\mathbb N$ is not bounded above in $\mathbb R$.

To prove that infinitesimal numbers do not exist in $\mathbb R$, assume for the sake of contradiction that there is an infinitesimal $y\in\mathbb R$. Then, $n<1/y$ for every natural number $n$, and so $1/y$ is an upper bound of $\mathbb N$, contradicting the fact that $\mathbb N$ is not bounded above in $\mathbb R$.

Joe
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