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I am trying to understand why the rational numbers are not continuous.

Given two rational numbers $a$ and $b$, I can always find a number $c = \frac{a+b}{2}$ between these two numbers. So when I plot the rational numbers as a line, this is a steady line (unlike natural numbers, which are obviously not continuous). Why are they not continuous?

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Between $a$ and $b$ you have also irrationals. –  Tom-Tom Feb 11 '14 at 9:37
Continuity is a property of functions. You seem to ask why the rational numbers are not connected (or path connected). –  Asaf Karagila Feb 11 '14 at 9:37
@iblue, define formally what is "continuous" for you. –  Martín-Blas Pérez Pinilla Feb 11 '14 at 9:47
OK! Not very formal, but your book says that there is a bijective correspondence between points of the line (the geometric thing) and real numbers. –  Martín-Blas Pérez Pinilla Feb 11 '14 at 9:55
This winds up being a question about infinities and infinitesmals. The rational numbers can come arbitrarily close to an irrational but can never equal it -- so the "line" can look arbitrarily continuous, but will never actually be so; there will always be a gap at the irrational. No matter how close you look, rationals can't equal an irrational until the number of digits becomes infinite... at which point you've just reinvented irrationals. –  keshlam Feb 11 '14 at 16:25

8 Answers 8

up vote 12 down vote accepted

You argue that the integers are "not continuous", which means that they are a disconnected set. By the same reasoning the rational numbers are disconnected, see http://www.proofwiki.org/wiki/Rational_Numbers_are_Totally_Disconnected/Proof_1.

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If you just want to point to a proof that the rationals are disconnected, shouldn't you link to proofwiki.org/wiki/Rational_Numbers_are_not_Connected instead? –  Rahul Feb 11 '14 at 23:34
@Rahul Yes, you are right. But I thought the link with totally disconnected gives even more information. –  Dietrich Burde Feb 12 '14 at 10:14

You are completely right in that the notion of holes depends on what you can put between your numbers. There is always a bigger set. The integers are contained in the rationals, which are contained in the real numbers, but those themselves can be viewed to have many holes, called non-standard real numbers (e.g. infinitesimals).

In order to give the ideas of “holes” and “being without holes” a meaning, you have to define what the total set of your numbers should be. If you say you want all rationals, that is fine. But there are properties which are only fulfilled by bigger sets of numbers.

For example you can easily construct a right triangle with legs of length $1$. Then the length of the hypotenuse will not be a natural number like $1$ nor even a rational number. It is $\sqrt2$, a hole in the rational numbers. The same happens if you want to measure the circumference of a circle with rational radius.

As for the statement in your book, $\sqrt2$ an $π$ are probably numbers they want on their number line, but what exactly the set of numbers on the number line is, depends on how you define it. There is not a right or wrong way. Just more and less useful ones for doing math with.

In analysis now there are still more things you would like to describe by numbers. For example you want to find zeros of a function. For the real numbers the following property holds:

Let $f: ℝ→ℝ$ be a continuous function and $a<b$ real numbers such that $f(a) < 0 < f(b)$. Then there is a $x$ between $a$ and $b$ such that $f(x) = 0$.

This is not true for the rational numbers. The reason is, you can always make your interval $(a, b)$ smaller and smaller, keeping $f(a) < 0 < f(b)$, so closing in on the zero of $f$. That way you get a sequence of (rational or real) numbers $(a_n)_{n\inℕ}$ which is Cauchy, meaning the distance $\vert a_{n+1}-a_n\vert$ gets arbitrary small. But only in the real numbers there will some limit $x$ contained in all the intervals $(a_n, b_n)$. This property is called completeness.

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I would prefer to keep two pieces of standard jargon separate, and distinguish being a continuum and being continuous.

Continuity is a property of functions. There's a standard "epsilon-delta" explication of the notion of continuous function. And the function $x^2\colon \mathbb{Q} \to \mathbb{Q}$ (for example) defined over the rationals is just as good a continuous function as its counterpart defined over the reals.

So you can have continuous functions defined over rationals; but the rationals are not a continuum. Why not? The notion of a continuum has its roots in geometry, and -- in the barest headline terms -- the fundamental thought is that certain geometric constructions, even repeatedly indefinitely, will have determinate results. In particular, if we "sum" finite lines by laying them end to end, then so long as the result is bounded, then it is indeed another finite line (how trite that sounds!).

But when we arithmetize geometry, this turns into the thought that for numbers apt for measuring lines in a true continuum, any increasing sequence of numbers-apt-for-measuring-a-line which is bounded above must converge to a limit which is another number-apt-for-measuring-a-line. Now, a bounded above sequence of rationals need not converge to a rational limit (it might converge to $\sqrt{2}$). By contrast, a bounded above sequence of reals will converge to a real limit. So the reals form the analogue of a geometrical continuum and the rationals don't.

(Sometimes it is said that forming a continuum is a matter of not being gappy. But the notion of gappiness is too blunt an instrument here. After all, the rationals are of course dense, i.e. between any two there is another -- so in one good sense they are not gappy. So we need to distinguish lack-of-gaps meaning denseness, and lack-of-gaps meaning being a continuum.)

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+1 for giving such a cogent, precise, yet non-technical explanation that nails the OP's question. –  user86418 Feb 11 '14 at 12:26
Just a note on the terminology, the OP cites a German title (Taschenbuch der Mathematik), so that will have used the word "stetig", which usually is used for "continuous", hence the literal translation. It has also a (related but) different meaning, old-fashioned, but honoured by none less than Dedekind: "... der Art, daß das Gebiet der Zahlen dieselbe Vollständigkeit oder, wie wir gleich sagen wollen, dieselbe Stetigkeit gewinnt, wie die gerade Linie." (Stetigkeit und Irrationale Zahlen, R. Dedekind, 1872) –  Daniel Fischer Feb 11 '14 at 13:20

Maybe I can help to tidy up some of the terminology. A linearly ordered set is called a linear continuum if it has no gaps or jumps. A jump is a cut where both sides have endpoints; an ordering without jumps is called dense. A gap is a cut where neither side has an endpoint; an ordering without gaps is called complete.

The natural numbers have jumps, but no gaps; they are complete, but not dense. The rational numbers have gaps, but no jumps; they are dense, but not complete. The real numbers have neither gaps nor jumps; they are a continuum.

Some authors (and translators from other languages) use the English adjective "continuous" to describe a continuum. As some of the comments and other answers point out, this usage may be confusing, since the adjective "continuous" is also used for functions, and the meaning isn't quite the same.

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You bring up some important things, but this would be a much better answer if you defined "cut" (and the "sides" thereof) in the text of your answer (if not also "linearly ordered set"). I know what you mean, but I think a reader who is not already familiar with these concepts would not know what you're talking about. –  Mark S. Feb 12 '14 at 0:21

We have here two "intuitive" concepts of "having no gaps", but they are not the same, when defined rigorously.

(1) A function $f$ is continuous at some point $c$ of its domain if the limit of $f(x)$ as $x$ approaches $c$ exists and is equal to $f(c)$ [usual mathematical formula ...].

The function $f$ is said to be continuous if it is continuous at every point of its domain. If the point $c$ in the domain of $f$ is not a limit point of the domain, then this condition is vacuously true, since $x$ cannot approach $c$ through values not equal $c$. Thus, for example, every function whose domain is the set of all integers is continuous.

As said above, you can define a function $f : \mathbb{Q} \rightarrow \mathbb{Q}$ (for example, the $id$ function, where $id(x)=x$) and there is no problem in applying the usual definition of continuity to that function.

(2) A linear continuum is a linearly ordered set $S$ that is densely ordered, i.e., between any two members there is another, and which "lacks gaps" in the sense that every non-empty subset with an upper bound has a least upper bound.

The real line is a linear continuum under the standard < ordering. Specifically, the real line is linearly ordered by <, and this ordering is dense and has the least-upper-bound property.

The set $\mathbb{Q}$ of rational numbers, which is a countable dense subset of $\mathbb{R}$, has not the least-upper-bound property; in this "view" it has gaps !

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You have many holes: all the irrationals. In fact, more holes than points!

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But the holes are not included in the rational numbers. If I just define "apple" as $1 < \text{apple} < 2$, there is also a hole in the irrationals. –  iblue Feb 11 '14 at 9:41
(A) Obviously! (B) Define where exactly is "apple". Between 1 and 2 isn't enough. (C) See my comment. –  Martín-Blas Pérez Pinilla Feb 11 '14 at 9:50
$\text{apple} < \sqrt{2}$ but if $x$ is a real number and $x < \sqrt{2}$ then $x < \text{apple}$. Okay? –  jwg Feb 11 '14 at 15:27
@jwg I don't know if you're aware, but that's a particularly good demonstration that the reals (in any formulation) are not the final ordered field. That is, $\rm{apple} = \sqrt{2} - \frac{1}{\omega}$ is a perfectly good definition of $\rm{apple}$ that satisfies all the above properties. Generally, when the reals aren't enough, look to the surreals. :) –  Ptharien's Flame Feb 11 '14 at 18:19
@Ptharien, the funny thing is the surreals aren't complete: "on the other hand, they fail the completeness property of the reals in the most extreme way: every nonempty set of Surreals has an upper bound and no least upper bound." (math.stackexchange.com/questions/221334/…) –  Martín-Blas Pérez Pinilla Feb 11 '14 at 19:10

Here's one way to think about it: The rationals are 'full of holes', so to speak, because there are some lengths out there in the world for which there is no corresponding rational number. For example, you probably know that the diagonal of a unit square has length equal to the square root of 2, which cannot be written as a quotient of integers. Intuitively speaking, the idea of the real line is to 'fill in the gaps' so that for every length there exists are corresponding real number.

The property of the rational numbers you've noticed, that you can always find a rational number between any two real numbers, is referred to as the density of the rational numbers in the reals.

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The big difference between $\Bbb Q$ and $\Bbb R$ stays only in one simple property.

Without considering any particular construction, we can define $\Bbb Q$ to be the $\bf field$ $\bf of$ $\bf fraction$ $\bf of$ $\bf the$ $\bf integers$ $\Bbb Z$: I say 'the', because all fields of fractions are equal up to isomorphisms. So our $\Bbb Q$ happens to be an (infinite) ordered field with the Archimedean property;

Now, leave apart the Archimedean property, and consider an ordered field with the following additional property: $$\text{every non-empty bounded above set has a least upper bound.}$$ You've so just obtained $\Bbb R$! In fact, it can be shown that there indeed exists one such field. And all $\bf complete$ (with complete we refer to this last property) $\bf ordered$ $\bf field$ are equal up to isomorphism. The Archimedean property now follows automatically from the completeness. We can then think of $\Bbb Q$ to be a subset of $\Bbb R$, since $\Bbb Q$ is isomorphic to a subset of $\Bbb R$.

Archimedean/Archimedian property:

The Archimedean (or Archimedian) property basically says that the set $\Bbb N$ is not bounded above in $\Bbb R$. Surely, if we consider $\Bbb N$ by itself (i.e., not embedded in $\Bbb R$) as an ordered set, then it is unbounded above by definition (in fact, for each natural $n\in\Bbb N$, there exists also the number $n+1 \in\Bbb N$, which is defined to be greater than $n$). Now, let's embed $\Bbb N$ in our $\Bbb R$. It can be done by induction. Let $\phi:\Bbb N\to \Bbb R$ be a map such that \begin{equation} \begin{cases} \phi(1)=1\qquad\qquad\qquad\text{to 1 we associate the multiplicative identity element of the field $\Bbb R$};\\ \phi(n+1)=\phi(n)+1\qquad\text{the number 2 goes in $1+1$, the number 3 goes in $1+1+1$, $\dots$)} \end{cases} \end{equation} (since now then we simply denote with $n$ the number $\phi(n)$ and we consider $\Bbb N$ only as a special subset of $\Bbb R$.) As before, we know that there is no natural $m$ such that $m>=n$ for all $n\in \Bbb N$, in fact nothing has changed in the immersion of $\Bbb N$ into $\Bbb R$. But, how do we know that there isn't any $r\in \Bbb R$ such that $r>=n$ for all $n\in \Bbb N$? In other word, how can we be sure the set $\Bbb N$ in unbounded above in $\Bbb R$? Well, this is simply a consequence of the completeness property we assumed for $\Bbb R$.

*proof:*$\,\,\,\,\,$suppose $\Bbb N$ were bounded above. Since $\Bbb N\ne \emptyset$, there would be a least upper bound $\alpha$ for $\Bbb N$. Then $$\alpha\ge n\qquad\text{for all $n$ in $\Bbb N$}.$$ Consequently $$\alpha\ge n+1\qquad\text{for all $n$ in $\Bbb N$},$$ since $n+1$ is in $\Bbb N$ if $n$ is in $\Bbb N$. But this mean that $\alpha -1 \ge n$ for all $n$ in $\Bbb N$, and this mean that $\alpha -1$ is also an upper bound for $\Bbb N$, contradicting the fact that $\alpha$ is the least upper bound. $\blacksquare$

One can show that this property is equivalent to the fact that, given an $\varepsilon>0$ it exist a natural $n$ such that $\dfrac{1}{n}<\varepsilon$.

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Can you state briefly what the Archimedian property exactly is for completeness' sake? –  Trevor Alexander Feb 12 '14 at 10:29
I've just edited my answer: see above :) –  Matteo Feb 12 '14 at 11:39
one more question: when you say, "You've so just obtained ℝ!" here, does your argument assert that $\mathbb{Q}$ does not contain non-empty bounded above sets with a least upper bound? I'm having trouble understanding why not. Couldn't you just choose the largest element of a set contained in $\mathbb{Q}$ as its upper bound? I guess I'm really asking what situation yields a supremum not in $\mathbb{Q}$ but in $\mathbb{R}$. –  Trevor Alexander Feb 13 '14 at 0:46
I sort of get it from the wikipedia article: "An example of a set that lacks the least-upper-bound property is ℚ, the set of rational numbers. Let S be the set of all rational numbers q such that $q^2 < 2$. Then S has an upper bound but no least upper bound in ℚ: If we suppose p ∈ ℚ is the least upper bound, a contradiction is immediately deduced because between any two reals x and y (including √2 and p) there exists some rational p', which itself would have to be the least upper bound (if p > √2) or a member of S greater than p (if p < √2)." –  Trevor Alexander Feb 13 '14 at 0:49
No, I'm not saying that "$\Bbb Q$ does not contain non-empty bounded above sets with a least upper bound", I'm only saying that not every bounded above subset of $\Bbb Q$ has a least upper bound in $\Bbb Q$. "Couldn't you just choose the largest element of a set contained in $\Bbb Q$ as its upper bound?" Well, no: how do you know such 'largest' element exists? From your example you see that indeed there isn't a largest $q$ such that $q^2<2$. –  Matteo Feb 13 '14 at 9:53

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