Why are the rational numbers not continuous?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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, $\sqrt2$ and $π$ are probably numbers they want on their number line. However 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 $f(a) < 0 < f(b)$, thereby 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 $\vert a_m-a_n\vert$ gets arbitrary small for $m$ and $n$ big enough). But only in the real numbers there will always be some limit $x$ contained in all the intervals $(a_n, b_n)$. This property is called completeness.

Answer 5

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 !

Answer 6

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.)

Answer 7

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 $\dfrac{1}{n}<\varepsilon$.

Answer 8

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.