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My question may seem foolish, but I don't know that why it is hard to add infinitesimal to the real number system.

For example, if I add to the real number system two special symbols to denote infinitesimal and infinitely large quantity, and give an order relation and operations on the new system, then what problems do occur? (Of course such a number system may not satisfy the Dedekind completeness, but I hear existing real number systems with infinitesimal also don't.)

skymountain
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    Once you have the symbol $\varepsilon$, how do you define $\varepsilon^2$? What is $\sqrt{\varepsilon}$? You will probably want to define a whole lot of new infinitesimals. Robinson's construction of the hyperreals, which is a rather natural construction (as equivalence classes of real-valued sequences), requires the axiom of choice to prove that it can be ordered, so the matter is more difficult than it seems. – Samuel Apr 27 '13 at 20:30

2 Answers2

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One of the disadvantages that occurs by simply augmenting a formal symbol indicating infinitesimal quantity and infinitely large quantity is that many algebraic properties no longer hold. We expect that this augmentation results in a systematic way of dealing limits in a more intuitive way, and the breakdown of algebra is certainly not what we wanted in this blueprint.

And even if we succeed in enlarging $\Bbb{R}$ by adding such objects in some way that raises no algebraic issue somehow, it is very hard to relate the calculus on the resulting number system, say $F$, to the usual calculus on $\Bbb{R}$. For example, you may want to prove the continuity of $\sin x$ by arguing that

Assume $\epsilon \approx 0$. Then $\sin \epsilon \approx 0$ and $\cos \epsilon \approx 1$. Now, by the addition formula $$\sin (x + \epsilon) = \sin x \cos \epsilon + \cos x \sin \epsilon,$$ we have $\sin (x+\epsilon) \approx \sin x$.

This seemingly appealing argument, however, required us to define a sine function on $F$ which is consistent with the original one on $\Bbb{R}$. And as you may realize, this is quite a non-trivial job. Indeed, even the power series defining $\sin x$ is not guaranteed to converge in $F$ for $x = \epsilon$ as $F$ is not complete!

Historically, avoiding this catastrophe while achieving a sufficiently intuitive and powerful way of dealing with limits as infinitesimal calculus had been considered very hard until Abraham Robinson came up with his famous hyperreal number system. This number system is an ordered field $\Bbb{R}^{*}$ containing the real field $\Bbb{R}$ as a proper subset such that

  1. $\Bbb{R}^{*}$ contains both infinitesimally small numbers and infinitely large numbers, whose notions are intuitively well-behaving under the arithmetic operations and order relation.
  2. The transfer principle holds: For any reasonably simple statement on $\Bbb{R}$, its hyperreal version is true if and only if its original version is true.

Since any non-Archimedean ordered field that extends $\Bbb{R}$ satisfies the property 1 in some sense, it is the transfer principle that makes the notion of hyperreal numbers a powerful tool for infinitesimal calculus.

Sangchul Lee
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  • Hello , I found you have a very deep understanding of nonstandard analysis, I am eager to discuss some questions with you, could you send an email to redstone-cold@163.com so that we can get in touch ,thanks ! – iMath Sep 22 '16 at 15:53
  • @iMath, I have just touched the basics of hyperreal numbers a coupe of years ago, so I am by no means an expert of this topic. Please forgive me if I am reluctant to answering your questions. I am pretty sure that there are generous peoples who are experienced and are eager to answer to your questions. – Sangchul Lee Sep 27 '16 at 21:38
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To complement sos440's fine answer, I would mention that what you are proposing is in fact close to what is known in the literature as "dual numbers"; see Is the theory of dual numbers strong enough to develop real analysis, and does it resemble Newton's historical method for doing calculus? Here you have basically a single (up to a multiple) infinitesimal $\epsilon$ such that $\epsilon^2=0$. The resulting number system is helpful in certain applications in physics. However, to be able to extend some standard functions (beyond polynomials) to the new number system, one needs a more elaborate approach.

Community
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Mikhail Katz
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  • Did the OP talk about the dual numbers? Anyways, $\epsilon$ doesn't have a reciprocal in the dual numbers. Also, they fit naturally into the setting of differential geometry: $\epsilon$ is essentially just a tangent vector to $0$ on the differentiable manifold $\mathbb{R}$. And so you can plug them into differentiable functions too. –  Apr 28 '13 at 10:04
  • You can plug them into differentiable functions, but you don't get the expected properties, as soss440 pointed out. Thus, the addition formula for sine is problematic. – Mikhail Katz Apr 28 '13 at 11:27
  • The OP did not talk specifically about the dual numbers, but the intuition of having "only one infinitesimal" is best borne out by the dual numbers. That's why I mentioned them. – Mikhail Katz Apr 28 '13 at 11:28
  • I can't see what exactly you think is problematic. Calculus on differentiable manifolds is well-understood. And in this setting, the Taylor series for $\sin(\epsilon)$ is simply $\epsilon$ (because all of the higher terms are zero), and the addition formula gives the correct result $\sin(x+\epsilon) = \sin(x) + \epsilon \cos(x)$. The problem with the dual numbers is when things aren't differentiable -- e.g. you ask for something like $\sqrt{x}$ near $x=0$. –  Apr 28 '13 at 11:33
  • Apparently we agree then that what is "problematic" is that some useful functions like square root cannot be extended? – Mikhail Katz Apr 28 '13 at 11:57
  • In the dual numbers we have simply $f(x + \epsilon) = f(x) + \epsilon f^\prime (x)$. And for $ \sqrt{x}$ the derivative is undefined at x=0. Therefore, the exapnsion formula can't be applied. But this is a problem ofthe function and not of the dual numbef extension. – user48672 Dec 25 '14 at 15:44
  • The question is why do we have $f(x+\epsilon)=f(x)+\epsilon f'(x)$. If this is your definition of $f(x+\epsilon)$ then you haven't accomplished anything. Also, the problem with the square root function is not merely at the origin. The problem is again that square root is not defined at points of the form $x+\epsilon$. – Mikhail Katz Dec 25 '14 at 16:07