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I read in the book "Calculus with infinitesimals" (Efrain Soto Apolinar) that $dx=1/N$ and $N$ is the number of elements of the set of the natural numbers (letter $N$ is used to indicate the cardinality of the set of natural numbers).

In other source I read that for hyperreal numbers "$\varepsilon = 1/\omega$" and $\omega$ is number greater than any real number.

Why do we use 2 different types of infinity to define the same infinitesimals?

Thanks.

Mike_bb
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    Any system for dealing rigorously with infinitesimals will have its own internally consistent definitions. Since these two systems have different axioms, they will not have "the same infinitesimals". – Ethan Bolker Jun 10 '22 at 14:58
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    It comes down to $\aleph_0$ vs $\omega$. – J.G. Jun 10 '22 at 15:00
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    @J.G. It really shouldn't - neither ordinals nor cardinals are directly relevant to infinitesimals, at least in any rigorous presentation I've ever seen. – Noah Schweber Jun 10 '22 at 15:25
  • @NoahSchweber Me neither, but the OP has seen some strange things. – J.G. Jun 10 '22 at 15:50
  • @EthanBolker +1 What is difference between $dx$ and $\varepsilon$? Is $dx$ more natural (for use in physics) than $\varepsilon$? Thanks. – Mike_bb Jun 14 '22 at 14:30
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    Until you get to the formal study of infinitesimals you will see $dx$ in informal (but essentially correct) applications in physics and mathematics and $\epsilon$ in mathematical proofs. See https://math.stackexchange.com/questions/1991575/why-cant-the-second-fundamental-theorem-of-calculus-be-proved-in-just-two-lines/1991585#1991585 – Ethan Bolker Jun 14 '22 at 15:22

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It's a good question, I think. There is a formulation of this, which skirts around the issue, called Nonstandard analysis, and instead concentrates on the properties of infinitesimals, rather than what they are.

However, I think your question is more about why it doesn't matter which kind of infinity you use, when you work with functions that converge to $0$ - perhaps an intuitive way to look at it would be to say that because $\aleph_0$, the countable infinity, is smaller than $\aleph_1$, the uncountable one, you might argue that $\frac{1}{\aleph_0} \ge \frac{1}{\aleph_1}$; but $\frac{1}{\aleph_0}$ is already $0$, and we don't have another zero that is somehow smaller, so there isn't any difference between the outcome of using one kind of infinity rather than another.

The really important thing about infinitesimals, IMO, is the observation that the value of $\frac{0}{0}$ depends entirely on how you arrive at the $0$ on top and the one at the bottom - that is, how you converge to $0$.

Now, I'm sure there are many people with a more firm grasp of things, who will have felt their toes curl up on reading my explanation, but I don't think is entirely wrong. Sometimes a good lie is better than a bad truth ;-)

j4nd3r53n
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