In the book Understanding analysis, by Abbot, when discussing the Archimedean property, the author states that there are ordered field extensions of $\mathbb{Q}$ that include "numbers" bigger than every natural number.

Could someone provide examples and and explanation why this could be the case?

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    $\begingroup$ Consider the field of rational functions over $\mathbb Q$, ordered by: $f>g$ if for sufficiently large $x$, $f(x)>g(x)$. $\endgroup$
    – Wojowu
    Jan 29, 2019 at 16:22
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    $\begingroup$ Sounds as though he might be talking about something related to the hyperreal system, which has infinitesimals which (i) are smaller than any real number while being bigger than $0$ and (ii) have reciprocals bigger than any real number but smaller than $\infty$? $\endgroup$
    – timtfj
    Jan 29, 2019 at 16:24

4 Answers 4


A concrete example

Consider the ring $\mathbb{Q}[x]$ of polynomials over $\mathbb{Q}$. These can be linearly ordered as in this answer, in a way that preserves the ordering of the rationals. Now we have the "regular" naturals in this field: 1, 2,3, and so on. But the polynomial $x$ is larger than all of these, and $x^2$ is larger than $x$, and so on. So in this ring there are elements that are greater than all the ordinary natural numbers.

By standard abstract algebra, the ordering on $\mathbb{Q}[x]$ can be extended to an ordering on the field of fractions of the ring. That will give an ordered field that contains $\mathbb{Q}[x]$ as a subring, and as such contains elements larger than all the ordinary natural numbers.

There are other algebraic constructions that give other non-Archimedean ordered fields, as well.

An example from logic

In logic, the compactness theorem says that if we have a set of axioms so that every finite subset of the axioms is consistent on its own, then the whole set is consistent. Take our axioms to include the axioms of an ordered ring, and also the axiom $z > n$ for each $n$. For any finite subset of these axioms, we can choose a value of $z$ large enough to make that finite subset true in $\mathbb{Q}$. So there is a model in which all the axioms are true at once. In that model, whatever $z$ is must be an element larger than every natural number.

In nonstandard analysis

In nonstandard analysis, we work with a field that extends of the reals in which there are infinitesimals. For this purpose, an infinitesimal is an element $\delta$ so that $0 < \delta$ but $\delta < 1/n$ for each ordinary natural number $n$. Then if $\delta$ is infinitesimal, $1/\delta$ exists (because $\delta \not = 0$, and $1/\delta$ is larger than $n$ for all $n$.


Consider the field of rational functions over $\mathbb{Q}$. I don't know what the standard notation for this is, but let's denote it by $\mathbb{Q}(x)$.

$$\mathbb{Q}(x)=\left\{\frac{p(x)}{q(x)}: p(x),q(x)\in\mathbb{Q}[x] \text{ and } q(x)\neq 0 \text{ is a monic polynomial.}\right\}$$

To define an order on it, define $f>g$ if and only if $f-g>0$ where a rational function is positive if and only if its leading coefficient of the numerator is positive. I leave the process of checking the axioms to you.

Now compare $\frac{x}{1}$ and $\frac{n}{1}$. We see that $\frac{x}{1}-\frac{n}{1}=\frac{x-n}{1}>0$. So, $x$ is bigger than any natural number $n$.

Edit: After Carl Mummert's wonderful comment, it is worthy to add that in order to have a well-defined order, we should first set the leading coefficient in the denominator to be $1$ by multiplying the whole fraction by the inverse of this coefficient if necessary. So, we can assume that $q(x)$ is monic. Now, we can see that the order we have defined is also an extension of the order on $\mathbb{Q}$ as well.

  • $\begingroup$ This is a good example but the order is hard to define; $4/(-1)$ is not a positive element, but the leading coefficient of the numerator is positive. $\endgroup$ Jan 29, 2019 at 17:17
  • $\begingroup$ @CarlMummert You're right. But the OP hasn't said that the order too must be an extension of the order on rational numbers. I mean it's open to interpretation. The OP's question reads "ordered field extensions of $\mathbb{Q}$". This is an ordered field and it's an extension of $\mathbb{Q}$. But you're absolutely right that the orders seem not to be compatible which is a very good point. I think if we add an extra hypothesis that in cases like what you mentioned, $4/(-1)$ should be seen as $(-4)/1$, the issue might be resolved but I'm not sure. What do you think? $\endgroup$ Jan 29, 2019 at 17:20
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    $\begingroup$ I see your point; I was thinking of extensions that preserve the ordering. There is a way to do this by looking at cross products, the same way we extend the order of the integers to the order of the rationals: $a/b < c/d$ if $ad < bc$. So that reduces to the problem to ordering $\mathbb{Q}[x]$. as in math.stackexchange.com/questions/1920165/… $\endgroup$ Jan 29, 2019 at 17:24
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    $\begingroup$ Yes, I think that resolves it. $\endgroup$ Jan 29, 2019 at 18:22
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    $\begingroup$ You could also say $f>0$ whenever $f(x)>0$ for all sufficiently large $x$. For your specific field, this is equivalent to your definition, but I think it generalizes better (e.g. to fields of algebraic functions). $\endgroup$
    – Micah
    Jan 29, 2019 at 19:37

Here's a well known construction of an ordered field extension of $\mathbb Q$ that comes from logic and from nonstandard analysis, as mentioned in the answer of @CarlMummert.

Choose a nonprincipal ultrafilter on the natural numbers $\mathbb N$: an ultrafilter is a finitely additive, $0,1$-valued measure defined on all subsets $A \subset \mathbb N$, such that $\mathbb N$ has measure $1$; and $\mu$ is nonprincipal if $\{i\}$ has measure zero for every $i \in \mathbb N$. The existence of a nonprincipal ultrafilter is an exercise in applying the axiom of choice.

Consider the set $\mathbb Q^{\mathbb N}$ which is the set of all sequences of rational numbers. Define an equivalence relation on this set: given $\underline x = (x_i)$ and $\underline y = (y_i) \in \mathbb Q^{\mathbb N}$, define $\underline x \approx \underline y$ if the set of indices $i$ for which $x_i = y_i$ has measure $1$. Let $[\underline x] = [x_1,x_2,x_3,...]$ denote the equivalence class of $\underline x = (x_1,x_2,x_3,...)$. Define addition and multiplication in the obvious way: $[\underline x] + [\underline y] = [\underline z]$ means that the set of indices for which $x_i + y_i = z_i$ has measure $1$, and similarly for multiplication. Define inequality similarly: $[\underline x] < [\underline y]$ if the set of indices for which $x_i < y_i$ has measure $1$. Check that everything is well-defined, and that you get an ordered field.

To embed $\mathbb Q$ into this field, map the rational number $q$ to $[q,q,q,q,q,q,q,q,q,q,...]$. Check that this is an embedding of ordered fields.

To identify a number greater than any natural number, take $[1,2,3,4,5,6,7,8,9,...]$.


Conway’s surreal numbers form a totally ordered Field (where the capital ‘F’ means that they are a proper class (not just a set) containing (a field isomorphic to) the real (and hence rational) numbers. They include many transfinite numbers, including the ordinals, i.e. numbers larger than all reals, rationals or natural numbers. Wikipedia also tells us that they form a universal ordered field in the sense that any ordered field may be modelled using them.

You ask for an “explanation why this could be the case” — that is hard to provide when one does not know why you think it might not be possible, as you apparently do.

  • $\begingroup$ It may be that the ordinals are included in Conway's "Field", but it probably isn't fair to "just" say that the ordinals are "numbers larger than all the reals" when cardinals and ordinals count different things, and they "agree" on the finite sets. $\endgroup$
    – nomen
    Jan 29, 2019 at 23:24
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    $\begingroup$ @nomen - it is true that ordinals, cardinals, and real numbers are distinctly different things by definition. But it is also true that there are natural isomorphisms between the countable cardinals, the countable ordinals, and the minimal subset of $\Bbb R$ containing $0$ and $1$ and closed under addition. As such, we generally identify these three sets as being the "Natural numbers" and don't balk over the different definitions. Similarly the ordinals, cardinals and reals all isomorphically embed in the surreals, where PJTraill's description is true. $\endgroup$ Jan 30, 2019 at 0:23
  • $\begingroup$ @nomen: As Paul Sinclair says, the surreals contain isomorphs of the reals and the ordinals. Since the images of the (transfinite) ordinals come out larger than those of the reals, that seems good enough to me. The questioner gave no indication that they meant a specific sort of number, mentioning $ \mathbb Q $ and natural numbers, but neither cardinals nor ordinals. That the latter give notions of size of sets and well-ordered sets respectively does not matter; what matters is the order in the ordered field given in an answer. I take it that by ‘“numbers”’ they meant elements of that field. $\endgroup$
    – PJTraill
    Jan 31, 2019 at 0:38

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