# Representing rational numbers on a number line

Though the cardinality of the set of natural numbers is the same as the cardinality of the set of rational numbers, when one looks at a number line this fact seems counterintuitive, since between every rational number there exist an infinite number of more rational numbers. Of course, it can be shown that $\mathbb{Q}$ is countable by simply ordering them on an infinite array (the "zig-zag" method). Thus there are the same "number" of rationals as there are natural numbers (they have the same size).

My question is as follows: if we "organize" the rational numbers in a certain way on a line -

$\\ \frac{1}{1}, -\frac{1}{1}, \frac{1}{2}, -\frac{1}{2}, \ldots \\ \frac{2}{1}, -\frac{2}{1}, \frac{2}{2}, -\frac{2}{2}, \ldots \\ \vdots \\ \frac{n}{1}, -\frac{n}{1}, \frac{n}{2}, -\frac{n}{2}, \ldots \\ \vdots$

where the "line" would be the "zig-zag" path ($0/1$, $1/1$, $-1/1$, $2/1$, $3/1$...; note that this line would not be straight geometrically, but assume it has been "straigtened" for sake of argument) the intervals between these numbers would have nothing in between them - they would be very similar to the natural numbers on a normal line. However, if we consider the rational numbers on a normal number line, there are an infinite number of rationals in between two arbitrary rational numbers, and hence the layman is mislead into thinking that rationals outnumber naturals. Why are these two "lines" so different? Why does the normal number line appear to be more "densely" populated with rationals than the "reorganized" number line? Lastly, I am curious whether or not, since $\mathbb{Q}$ and $\mathbb{N}$ are the same size, can the natural numbers be organized such that there are an infinite number of natural numbers in between two natural numbers, as can be done with rationals?

• Let complement the answers so far with an example. Instead of on a straight line, place the naturals on the unit circle in $\mathbb{R}^2$, mapping $n$ to $(\mathrm{cos} (n), \mathrm{sin} (n) )$. Then by a hand waving argument (I don't want to give a rigorous argument here), no point gets hit twice because $\pi$ is irrational, but the numbers are equally distributed (in some sense); in every segment, no matter how small, there still are such points. (For a rigorous proof, one should use some topological or number theoretic arguments.) – Ben Jun 15 '15 at 7:02

The difference is the topology you put on $\mathbb{Q}$. In these two cases you're generating an order topology, one based on the standard ordering, the other is an order inherited by the specific bijection you have with $\mathbb{N}$. For the last question, definitely yes. You take the inverse of the bijection you were using before. Whereas you were originally arranging $\mathbb{Q}$ to look like $\mathbb{N}$, the inverse is rearranging $\mathbb{N}$ to look like $\mathbb{Q}$.
• Can you kindly clarify what you mean by "the difference is the topology you put on $\mathbb{Q}$", as I have not yet studied topology? – MathematicsStudent1122 Jun 12 '15 at 14:52
You think about the rationals in two different ways. One the one hand, you look at it as a dense linear order without endpoints and on the other hand as another linear order without endpoints, which is order isomorphic to the integers. You can do this with any countable set $X$ (and in many different ways): Pick bijections $q: X \rightarrow \mathbb Q$, $i: X \rightarrow \mathbb Z$ and consider the linear orderings $x \le_q y :\Leftrightarrow q(x) \le q(y)$ and $x \le_i y :\Leftrightarrow i(x) \le i(y)$.
Let's take another example. Consider the set of natural numbers. Whenever we take a finite-length interval of the "real line", only finitely many natural numbers are in this segment. Consider the map $f: \mathbb N_{> 0} \rightarrow \mathbb N, \ n \mapsto \frac 1 n$. Now, every interval of length $> 0$ that contains $0$ also contains infinitely many points of the form $f(\frac 1 n)$ for natural numbers $n > 0$ (it even contains all, but finitely many, of them). But if we consider the new linear ordering $n \le_f m :\Leftrightarrow f(n) \le f(m)$, we basically get the same linear ordering (it is an order isomorphism). What is going on now, is that we changed the topological properties of the natural numbers, namely its metric.