Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wonder if there can be numbers (in some extended theory) for which all reals are either smaller or larger than this number, but no real number is equal to that number?!

Is there some extension of number which allows that? Under what conditions (axiom etc.) there is no such number.

share|cite|improve this question
Since we are trying to get rid of number tag, see meta, I've retagged the question to number-systems. If you have a better idea, feel free to replace it with a more appropriate tag. – Martin Sleziak Jun 18 '12 at 11:21
It's OK. I found it hard to find a keyword especially since I'm not an expert :) – Gerenuk Jun 18 '12 at 12:00
up vote 8 down vote accepted

Under the axioms of the real numbers this cannot occur. You must add new elements to the real numbers, note that if $\varepsilon$ is smaller than all $\frac1n$ but still positive then $\frac1\varepsilon$ is larger than any real number.

Such $\varepsilon$ is called infinitesimal and their existence is incompatible with the real numbers per se. There is a branch, however, called non-standard analysis in which these numbers play an important role.

One example to such field is called Hyperreal numbers.

share|cite|improve this answer
Thanks. What does it mean "under axioms of real numbers". Do I have to abondon one of the axioms to introduce "numbers inbetween"? Which axiom would that be? Do hyperreals follow the condition about order I required? – Gerenuk Jun 18 '12 at 7:37
Another example is called Surreal number, which I think is more valuable. – Frank Science Jun 18 '12 at 7:48
The axiom that has to be abandoned is what is usually called the Completeness Axiom. One standard version of it is that any non-empty set (of reals) which is bounded above has a least upper bound. – André Nicolas Jun 18 '12 at 7:48
@Gerenuk: Hyperreal numbers are ordered and form a field. However they do not have the Archemedian property of the real numbers (every number is smaller than a natural number). Andre gave another property which does not hold. – Asaf Karagila Jun 18 '12 at 7:53
@Frank: If you want proper classes, sure. The. You have to explain what is a class, what are ordinals, etc. – Asaf Karagila Jun 18 '12 at 7:54

You want a "non-archimedean" ordered field. Examples are hyperreals and surreals. My favorite one: the transseries (G. A. Edgar, "Transseries for Beginners," Also try the Levi-Civita numbers: . See here for many examples.

share|cite|improve this answer

Consider the field $\mathbb{R}(x)$ of all (formal) rational functions in one variable with real coefficients. While this is not an ordered field, it is an orderable field -- it is possible to define an ordering $<$ on rational functions that is consistent with the usual laws of arithmetic.

For any ordering $<$ of $\mathbb{R}(x)$, we can define sets $L = \{ a \in \mathbb{R} \mid a < x\}$ and $R = \{ a \in \mathbb{R} \mid a > x\}$, and we have $\mathbb{R} = L \cup R$ -- under this ordering, every real number is either less than or greater than the polynomial $x$.

It turns out the ordering $<$ is completely determined by $L$ and $R$, and conversely each way to choose $L$ and $R$ corresponds to an ordering of $\mathbb{R}(x)$.

The complete list of orderings are:

  • The ordering "$+\infty$" - $x$ is larger than every real number
  • The ordering "$-\infty$" - $x$ is smaller than every real number
  • The ordering "$a^+$" - $x$ is infinitesimally larger than $a$
  • The ordering "$a^-$" - $x$ is infinitesimally smaller than $a$

The labels I've chosen for the orderings refer to "where" $x$ is placed in relation to the real line.

Some good buzzwords that relate to this sort of topic are:

  • Real closed field
  • Formally real field
  • Real algebraic geometry
  • Semi-algebraic geometry

There is an easy way to write down a first-order theory whose models are examples of the sort of number system you ask for. For example,

  • Start with the language of ordered fields
  • Add a new constant symbol $\varepsilon$
  • Add in all of the ordered field axioms
  • Add in one axiom $0 < \varepsilon$
  • For every positive integer $n$, add in one axiom $\varepsilon < n$

Every model of this theory will have a number $\varepsilon$ with the property that it is larger than every non-positive real number, and smaller than every positive real number.

share|cite|improve this answer
Can I add multiple $\varepsilon$ this way? It won't break any field axioms? Will all the maths from university still work (integrals etc.), or do I need some exceptions (i.e. where does the archimedian matter)? Is which is the field which is more than real numbers, but leaves as much undergrad math valid as possible? – Gerenuk Jun 19 '12 at 11:18
If you stick to elementary algebra, any real closed field is indistinguishable from the real numbers in terms of what theorems/identities are true or what computations you can do. If you want to do real analysis (e.g. integrals), you need to set things up more along the lines of non-standard analysis. (i.e. hyperreals) – Hurkyl Jun 19 '12 at 20:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.