# How to add infinitesimal to the real number system

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.)

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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 at 20:30

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.

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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.
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. –  Hurkyl Apr 28 at 10:04
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 $x=0$. –  Hurkyl Apr 28 at 11:33