I'm teaching myself about hyperreal numbers. My main motivation for doing so is that they include infinite numbers, whose existence I hear disputed & doubted often as "quantifying the unquantifiable".
In this YouTube video, around twelve minutes in, it states that
$$\Bbb R(x)=\left\{ \frac{f(x)}{g(x)}\,\middle|\, f(x), g(x)\in \Bbb R[x], g(x)\neq 0\right\},$$
where $\Bbb R[x]$ is the set of polynomials in $x$ with real coefficients, is an instantiation of the hyperreal numbers ${}^*\Bbb R$.
What is meant by "instantiation" here?
Please note that hyperreal numbers are not rigorously defined in the video.
I have only a rudimentary understanding of hyperreal numbers; I'm not sure about what they are exactly. Perhaps that is where my confusion lies.
The Wikipedia article on hyperreals doesn't contain a definition either.
For questions on what hyperreals are, see the following:
Their answers make sense but I wouldn't say I have a firm grasp of them.
I'm given to understand that hyperreals are of use to model theory. That might explain where the terminology in question comes from. However, I don't remember where I heard this claim. (I think it was in a Numberphile video.)
A suspicion: I think $\Bbb R(x)$ is to ${}^*\Bbb R$ what the quotient $\Bbb R[x]/(x^2+1)$ is to $\Bbb C$.
Please help 🙂
Best Answer
The claim in the video you link to is incorrect; to be honest, no video making such a claim should be used as a reference for technical topics. There is no substantive sense in which $\mathbb{R}(x)$ is related to hyperreals.
The most that can be said is that $\mathbb{R}(x)$ is an ordered (well, fine, orderable) field with infinitesimal elements - that is, a non-Archimedean field. Hyperreal fields, however, are much more than merely non-Archimedean fields. A hyperreal field must have a "high degree of similarity" to $\mathbb{R}$ itself, in a particular technical sense. $\mathbb{R}(x)$ demonstrably lacks this, most obviously since it does not satisfy the property "$\forall u\ge 0\exists v(v^2=u)$." So we don't even have $\mathbb{R}(x)\equiv\mathbb{R}$, let alone that $\mathbb{R}(x)$ is a hyperreal field (see below)!
To someone new to the subject of nonstandard analysis, this may feel like uncharitable hair-splitting. I want to emphasize as strongly as possible that the difference between mere non-Archimedeanness and hyperrealness is huge. Remember that the whole point of hyperrealness is that a hyperreal field should let you "transfer" results to the standard real numbers $\mathbb{R}$, so that infinitesimal methods can still provide real results. The fact that this is possible at all is really neat and awesome, and ignoring precisely what makes hyperrealness special in the interest of a superficial gain in accessibility is a pedagogical disservice.
It may help at this point to see the (most common in my experience) precise definition of hyperreal-ness, in order to contrast it with $\mathbb{R}(x)$. Let $\Sigma$ be the expansion of the language of fields by a symbol for every finite-arity relation on the reals, and let $\mathcal{R}$ be the reals construed as a $\Sigma$-structure in the obvious way. A hyperreal field is then just a proper elementary extension $\mathcal{F}$ of $\mathcal{R}$.
Note that strictly speaking a "hyperreal field" isn't a field but rather an expansion of a field (compare $\mathcal{R}$ with the much-less-complicated $(\mathbb{R};+,\times,0,1)$. This is a bit messy, so we describe $\mathcal{F}$ as a field equipped with a third-order operation ${}^*$ which assigns to each finite-arity relation $A$ on $\mathbb{R}$ an extension $^*A$ on the underlying set of $F$ which satisfies some simple properties. This amounts to the same thing as the above, but is arguably more convenient in two significant ways: pdagogically this approach is more accessible to non-logicians, and technically it lets us pay more careful attention to exactly how much transfer is needed for a given application.