This question arose today at Yevgeny Gordon's talk, *"Will nonstandard analysis be
the analysis of the future?"* at the CUNY Logic
Workshop. Here is my way of asking it.


Consider the ordered real field $\newcommand\R{\mathbb{R}}\R$ with a predicate for the natural numbers, and a
nonstandard version of it $\R^\*$, the hyper-reals, which an ordered
field extending $\R$ and having the transfer property, a map
$a\mapsto a^\*$, which preserves the truth of any statement in the
language of ordered fields, allowing also the predicate for the
natural numbers. The hyper-real numbers of the form $a^\*$ are
referred to as the *standard* elements of $\R^\*$.


**Question.** Suppose that a hyper-real number $a\in\R^\*$ is
definable in $\R^\*$ by a formula $\varphi$ in the language of
ordered fields with a predicate for natural numbers, but where the
scope of the quantifiers is only over the standard elements. Must
$a$ be algebraic?


It is easy to see that every algebraic hyper-real number is
definable in this way. For example, the hyper-real $\sqrt{2}$ is
definable in $\R^\*$ as the unique $x$ for which $x^2=2$ and $0<x$.
Similarly, any algebraic hyper-real (algebraic over the standard
integers) is the unique solution in $\R^\*$ in a certain standard
rational interval of a polynomial equation over the standard
integers. So in fact, every algebraic hyper-real is quantifier-free
definable in $\R^\*$, even without the predicate for the natural
numbers. So the question is equivalent to asking:


**Question.** If a hyper-real is definable by a formula whose
quantifiers have scope restricted to the standard reals (allowing a
predicate for the natural numbers), then is it quantifier-free
definable?


Meanwhile, if you think about numbers like $e$ and $\pi$, it is not
clear how to define them in $\R^\*$ without quantifying over all the
(possibly nonstandard) natural numbers. For example, $e$ is the
limit of $(1+\frac 1n)^n$, and so $e^\*$ is the unique $x$ in $\R^\*$
such that $$\forall \epsilon>0\ \exists N\ \forall n\geq N \
|(1+\frac 1n)^n-x|<\epsilon.$$ This definition can be undertaken in $\R$, using the predicate for $\mathbb{N}$, which allows one to define exponentiation and so on. But in $\R^\*$, these quantifiers are not only
over the standard numbers; we have to quantify also over the
nonstandard numbers. If you restrict to standard $\epsilon$ and
standard $N$ and $n$ only, then there will be an entire interval of
hyper-reals that are that close to those numbers---anything
infinitesimally close to $e$ will do. So the restricted-scope version of the definition will not succeed as a definition.


Similarly, it is not clear how to define $\pi$ or indeed any other
transcendental hyper-real number while quantifying only over
standard numbers.


I believe that [Tarski's theorem on real-closed
fields](https://en.wikipedia.org/wiki/Real_closed_field) will prove
the positive result for the special case of the question, where
$\varphi$ does not use the predicate for the natural numbers (but
still has the scope of all quantifiers restricted to the standard
hyper-reals). My reason for this expectation is that I believe we
can apply Tarski's elimination of quantifiers procedure to such a
$\varphi$ and thereby prove that $\varphi(x)$ is equivalent to a
quantifier-free assertion in the language of ordered fields. Then,
using the fact the algebraic numbers form an elementary
substructure of the reals, as ordered fields, it follows that the
existence of a solution in $\R$ is equivalent to the existence of a
solution in the algebraic numbers. And so the given number must be
algebraic.


But I am a little fuzzy on the details of how the scope-restriction
affects this argument, and so if you can affirm or refute it, I
would be grateful.



Are the definable hyper-reals, using quantifiers only over the standard reals and natural numbers, the same as the algebraic numbers?
$\newcommand{\st}{\textrm{st}}\newcommand{\bR}{{\bf R}}\newcommand{\bN}{{\bf N}}$
I think the limit of any definable convergent sequence $(a\_n)$ (including $e$) is definable by the formula
$$\varphi(x)=(\exists x'\in {\bR}^\st\, x'=x)\land \forall N\in\bN^\st\exists n\_0\in \bN^\st\forall n\in \bN^\st\_{>n\_0}(a\_n-x)^2<N^{-1}.$$


Since the language includes natural numbers, we have all of Peano arithmetic, so I think by a similar argument, every computable (and I guess every definable in PA) real should be definable. The point is that if you allow quantifiers which range over a given set, then you automatically make that very set definable, so you have at least as much strength as you would have in the smaller structure.


**Edit**: The idea was first mentioned by Ramiro de la Vega in comments, which I neglected to read before writing this answer.


