What is the hyperreal multiplicative inverse of $1 + \epsilon$, and how do we show it exists?

Question

What is the multiplicative inverse of $1 + \epsilon$, in the ordered field of hyperreals or surreals?

Simple algebra shows it must be equal $1-\epsilon+\epsilon^2-\epsilon^3+\epsilon^4...$ But how do we prove that that number exists as a hyperreal or surreal?

If it does exist, is it realizable as a finite set of arithmetic operations on $\mathbb{R} + \epsilon$?

If not, how do we characterize the hyperreal or surreal numbers that are not realizable as such? And what symbols are customarily used for them (or why do they not have common symbols)?

If we simply take the ordered field of reals and introduce $\epsilon$, all the standard surreal numbers follow (e.g. $\omega$, e.g. $2\epsilon$, e.g. $1+\epsilon$, etc.), but I don't see how to prove that the infinite sum above exists as a hyperreal or surreal.

A similar question can be asked for $\frac{1}{1-\epsilon} = 1 + \epsilon + \epsilon^2 + \epsilon^3...$

---

UPDATE $\epsilon$ is the most basic infinitesimal, greater than zero but less than any real. In surreal numbers, it's $(\{0\},\{1,1/2,1/4,1/8/1/16,1/32...\})$.

$\frac{1}{1 +\epsilon} \ne 1 - \epsilon$, because $(1+\epsilon) * (1 - \epsilon) = 1 - \epsilon^2 $. It must be slightly greater than $(1 - \epsilon)$ , but the difference is less than $\epsilon^2$.

Another way to ask this is: $d = 1 - \frac{1}{1 +\epsilon}$.

$\epsilon^2 < d < \epsilon$. Is there a way to characterize $d$ and it's size compared to $\epsilon$?

Answer 1

Hyperreals

What is the multiplicative inverse of $1+\epsilon$ in the ordered field of hyperreals?

I assume we're talking about Robinson's Hyperreals (not more general hyperreal fields) being created via an ultrapower construction.

For those unfamiliar, the basic idea behind of the construction isn't too complicated. I like Terry Tao's voting analogy. A hyperreal is a sequence of reals (well, an equivalence class of sequences) that vote each time you ask about a property (like "are you bigger than 5?"). How to determine which infinite collections of voters count as good majorities is handled by some technical stuff, but you don't have to worry about that to get the idea.

I assume that by $\epsilon$, the OP meant the following equivalence class: $[(1,\frac12,\frac13,\frac14,\ldots)]$. Then $1+\epsilon$ is $[(2,\frac32,\frac43,\frac54,\ldots)]$ and its reciprocal $\dfrac{1}{1+\epsilon}$ is $[(\frac12,\frac23,\frac34,\ldots)]$. It is "realizable as a finite set of arithmetic operations" in that it equals $1/(1+\epsilon)$.

---

Surreals

What is the multiplicative inverse of $1+\epsilon$ in the ordered field of surreals?

The surreals don't fit in a set (the collection of surreals is a proper class) so in some contexts people would not call the surreals an ordered field.

I assume that by $\epsilon$, the OP meant the following surreal: $\{0\mid 1,\frac12,\frac13,\frac14,\ldots\}$. Then $1+\epsilon$ is $\{1\mid 2,\frac32,\frac43,\frac54,\ldots\}$. Its reciprocal $\dfrac{1}{1+\epsilon}$ is not $\{\frac12,\frac23,\frac34,\ldots\mid1\}=1-\epsilon$. It's actually a bit of a pain to write down from first principles (see (5.28) of Surreal Numbers - An Introduction), but I think it might be equal to $\{1-\epsilon,1-\epsilon+\epsilon^2-\epsilon^3\ldots\mid1,1-\epsilon+\epsilon^2\ldots\}$.

It is "realizable as a finite set of arithmetic operations" in that it equals $1/(1+\epsilon)$.

---

Other Comments

If we simply take the ordered field of reals and introduce $\epsilon$, I don't see how to prove that the infinite sum above exists as a hyperreal or surreal.

What do you mean by "introduce"? The hyperreals and surreals, whatever your definitions, are known to act like fields, so division exists. As discussed above, there are even explicit ways to build a representative of a reciprocal for both. No infinite sum is required.

If we simply take the ordered field of reals and introduce $\epsilon$, all the standard surreal numbers follow (e.g. $\omega$, e.g. $2\omega$, e.g. $1+\epsilon$, etc.)

If all you do is "introduce" $\epsilon$ and allow arithmetic operations including division, then you only get what one might call the ordered field of rational functions, which doesn't have enough weird elements to be an example of Robinson's hyperreals, let alone the not-set-sized surreals.

Answer 2

A few comments:

In the field $\mathbf{No}$ of surreal numbers, there is one simplest, or first born number $\epsilon$ which is positive and smaller than all positive real numbers (i.e. positive and infinitesimal), and it is presented as you wrote.

In a field of hyperreal numbers, there are no distinguished positive infinitesimals, and there is nothing special about any hyperreal number represented by this or that sequence in a ultrapower of $\mathbb{R}$.

---

In fields of hyperreal numbers and in $\mathbf{No}$, for any positive infinitesimal $\varepsilon$, the sequence $(\sum \limits_{k=0}^n (-\varepsilon)^k)_{n \in \mathbb{N}}$ does not converge in the order topology, so the expression $1-\varepsilon + \varepsilon^2+\cdot\cdot\cdot$ does not make sense as a limit. In those fields, there are infinitely many numbers $x$ which satisfy $1-\varepsilon,1-\varepsilon+\varepsilon^2-\varepsilon^3,\cdot\cdot\cdot<x<\cdot\cdot\cdot,1-\varepsilon+\varepsilon^2,1$. Again in fields of hyperreal numbers, there is no reason to distinguish between two such numbers. However $(1+\varepsilon)^{-1}$ is one of those. In $\mathbf{No}$, one can consider the simplest one $x_{\varepsilon}$. As Mark S. wrote, we have $x_{\epsilon}=(1+\epsilon)^{-1}$. But in general, say for $\varepsilon=\epsilon +\frac{1}{\omega^{\omega}}$, you have $x_{\varepsilon}=x_{\epsilon}$, so $x_{\varepsilon}\neq(1+\varepsilon)^{-1}$.

---

Now there is a more intricate notion of sum of infinite families in $\mathbf{No}$, and this notion makes sense of $1-\varepsilon + \varepsilon^2+\cdot\cdot\cdot$ whenever $\varepsilon$ is infinitesimal. With that definition of sum, you have indeed $1-\varepsilon + \varepsilon^2+\cdot\cdot\cdot=(1+\varepsilon)^{-1}$. For more information, look up Hahn series fields and summable families in Hahn series fields, knowing that $\mathbf{No}$ is a Hahn series field.

---

Finally, if $F$ is an ordered field containing $\mathbb{R}$ and $\iota \in F$ is positive and infinitesimal, then there is no finite algebraic expression for $(1+\varepsilon)^{-1}$ in $F$ if by that you mean a polynomial $P \in \mathbb{R}[X]$ with $P(\iota)=(1+\iota)^{-1}$. In fact $\iota$ is transcendent over $\mathbb{R}$, because given $Q \in \mathbb{R}[X]$ irreducible, you have $Q(\iota)=Q(0)+\iota'$ for some infinitesimal $\iota'$. By irreducibility $Q(0)$ is a non zero real number, so $Q(\iota)\neq 0$.