What properties of "numbers" are used to assert that for given numbers $a$ and $b$, $a≠b$, there exists a number $x$ such that $a < x < b$ ?

In the texts I've read, this seems to be assumed without explanation in discussions that are otherwise quite careful about such things.

For example, in Spivak's Calculus (4E, p. 123) this fact is used to demonstrate that the function $f(x) = x^2$ does not take on its maximum on the interval (0,1) because for any $0 ≤ y < 1$ there is an $x$ such that $y < x < 1$. Up to that point, the only properties of "numbers" (whatever they may be) that have been defined are those of an ordered field, and it is not clear to me that this property can be derived from those.

I gather this amounts to the "numbers" in question having "dense order". (Correct me if I'm wrong.) But I'm unclear where that comes from.

Also note: up to the point where this question arises, there has been no mention of what "numbers" are (i.e., whether they are $\mathbb Q$ or $\mathbb R$), only that they have the properties of an ordered field.


2 Answers 2


It is easy to show that in any ordered field if $a < b$, then $a < \frac{a + b}{2} < b$. In any ordered field 2 is invertible, so this intermediate value exists.

  • $\begingroup$ Indeed! I think that's the answer I was looking for. So just to break this down further the be sure I'm not missing something: the reason I can divide by 2 is that the multiplicative inverse of 2 exists if 2 does. Similar reasoning leads to $\mathbb Q$. Correct? $\endgroup$
    – orome
    Sep 8, 2011 at 13:45
  • $\begingroup$ And so that I don't appear to be to much of an idiot: In the texts I'm reading, at the point where my question arises, there's no specification of what "numbers" mean, other than a specification of the properties they have. Hence my question (and the appropriateness of this answer). $\endgroup$
    – orome
    Sep 8, 2011 at 14:24
  • 1
    $\begingroup$ yes; even more "broken down" from the very beginning: 1)Big Bang 2)1 is in $\mathbb R$, since $\mathbb R$ is a field. 3)By closure of operations, 1+1 is also in $\mathbb R$ 3)As you said, in a field**, every nonzero element is invertible, and **$ \mathbb R$ does not have characteristic 2, i.e., the above result is not true in fields of characteristic 2; and trivially not true for the field of 2 elements by cardinality reasons alone--there is no third element that could be between 0 and 1. $\endgroup$
    – gary
    Sep 8, 2011 at 16:40
  • $\begingroup$ Aren't there ordered fields of characteristic 2? If so, 2 would not be invertible there. $\endgroup$
    – gary
    Sep 8, 2011 at 16:43
  • 3
    $\begingroup$ @gary: No, there are no ordered fields of finite characteristic. If $F$ is a field, then $1\neq 0$, so either $1\lt 0$ or $1\gt 0$. But $1=1^2$ and squares are always positive in an ordered field, so $1\gt 0$. Adding $1$ to both we get $2\gt 1 \gt 0$; add $1$ again to get $3\gt 2\gt 1\gt 0$. Inductively, for every $n\in\mathbb{N}-\{0\}$, $n\gt 0$, hence $n\neq 0$; thus, $F$ does not have positive characteristic. $\endgroup$ Sep 8, 2011 at 17:00

If we just assume connectedness of the reals, if there were no numbers between y and 1, we would be lead to the contradiction that we could then express $\mathbb R= (-\infty,y]\cup[1,\infty)$, i.e., $\mathbb R$ is the union of two sets with disjoint closure, i.e., we get a disconnection of $\mathbb R$, as the union of two non-disjoint sets, or as the union of non-empty separated sets.

Alternatively,using field properties, if we assume $\frac {1}{2}$ is in $\mathbb R$ ($1$ is in $\mathbb R$, then so is $1+1$, then so is $(1+1)^{-1})$, then so is $\frac{1}{2^n}$ for all natural n. By the Archimedean property , there is some integer $m$ such that $\frac{1}{2^m}<1-y$ by field closure properties then, $y+\frac {1}{2^m}$ is a real number, but $y+\frac{1}{2^m}<1$.

EDIT: 3rd paragraph removed, since I have no good proof for it.

  • 4
    $\begingroup$ But it is much harder to prove that $\mathbb R$ is connected than it is to show that there is an element between $a<b$ in an ordered field. I hope this doesn't sound too negative, because this is a good point in the case of $\mathbb R$, but this method assumes more and gives less than taking $\frac{a+b}{2}$. E.g., it wouldn't show that there is a rational between two rationals. $\endgroup$ Sep 8, 2011 at 5:35
  • $\begingroup$ I had a proof similar to Oliver's in mind, but when I saw his, I ended up using something else; maybe the lub property. No problem, tho, I welcome input. $\endgroup$
    – gary
    Sep 8, 2011 at 5:46
  • $\begingroup$ That's understandable, as there's no point covering the same ground. I took the question to be about a way to show this before the notion of completeness is introduced (and the l.u.b. property is equivalent to connectedness for ordered fields). If you do assume the l.u.b. property, then you get the Archimedean property, from which you can deduce the stronger conclusion that there is a rational between $a<b$. $\endgroup$ Sep 8, 2011 at 5:49
  • $\begingroup$ Jonas: I added an argument from the perspective of the field properties of $\mathbb R$, which uses the $LUB$, I guess in the background. $\endgroup$
    – gary
    Sep 8, 2011 at 6:01
  • $\begingroup$ gary: I see. I'm not really sure why you used $1/2^n$ rather than $1/n$ (or $(1-y)/2$), but that argument using the Archimedean property makes sense. This is a nitpick, but it could be mentioned that the fact that $(1+1)^{-1}$ exists is a consequence of the fact that $1+1\neq 0$ (because $1>0\implies 1+1>0$). $\endgroup$ Sep 8, 2011 at 6:07

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