Visual representation of the fact that there are more irrational than rational numbers.

Would anybody know of a visual or even (preferably) geometric representation of this?

To make it more specific: Text, symbols and written numbers are predominantly used as labels, and and less to represent the (ir)rationals themselves, or relations between the two groups. Like, if you need an image or a picture for your answer, that's probably it.

For example:

It does not have to be a rigorous proof, just a visual approach—if possible.

At the same time, a (somewhat elementary) explanation of why it does not make sense to try to visualize this relation (geometrically) is just as welcome.

• It definitely falls short of being a convincing argument, but looking at Thomae's function is at least moderately enlightening; I can see the rationals are all separated out, then a whole line of irrationals piled at the bottom! Jun 15, 2016 at 1:51
• @MiloBrandt It becomes convincing after an infinite amount of time spent zooming in. Jun 15, 2016 at 1:53
• Why do you even need a visual representation that will most definitely fail to be a rigorous proof?
– Vim
Jun 15, 2016 at 2:34
• @Vim Whatever the answer may be, I have to ask to find out. Jun 15, 2016 at 2:54
• @KarloGrba that's of course absolutely fine. But, just in my opinion, the reality might let you down this time because you can't always expect intuition to inspire proof. By the way, I find you accepted an answer using Lebesgue measure and claiming to be rigorous, but I don't think it can qualify as a real proof. See my comment below, if you are interested.
– Vim
Jun 15, 2016 at 3:03

The rationals can be mapped into the lattice points $(n, m)$, which are an infinite set of isolated points in the plane.

The irrationals, by any of the standard ways of mapping two reals into one (such as zipping the digits of a pair of reals) fill the whole plane (with quibbles about two representations or points missed being disregarded).

• Could you eventually explain how or what fills the whole plane, I'm missing that part. Jun 15, 2016 at 2:01
• Take the base $b$ representation of a real number, where $b \ge 3$. Make two real numbers from it by taking the base $b$ digits in the even positions for the first ordinate, and the base $b$ digits in the odd positions for the second ordinate. This is similar to the construction that Cantor used to show that the cardinality of points in k-dimensional space is the same as the reals Jun 15, 2016 at 2:31
• To be a bit more concrete, but using the same idea. Think of rational numbers as "the lines of the plane which intersect $\mathbb{Z} \times \mathbb{Z}$ in a proper extension of $\{(0,0)\}$"; and think of irrational numbers as "the lines of the plane which intersect $\mathbb{Z} \times \mathbb{Z}$ exactly in $\{ (0,0)\}$". Indeed, this is essentially what it is done in the construction explained in arxiv.org/abs/math/0301015 (for the technicalities one only considers lines through the origin of coordinates and avoids the vertical line) Jun 15, 2016 at 9:51
• In your second example, the most well-known such function maps both the irrationals and the rationals to dense sets in the plane. Just because the rationals also have a discrete injection doesn't imply that the irrationals don't similarly have a discrete injection. Of course we know they can't, but I fail to see how this example elucidates that fact. Jun 15, 2016 at 17:57

Here is a geometric difference between the rationals and the irrationals: the length of the rationals equals zero, whereas the length of the irrationals equals infinity.

Here's why the rationals have zero length. Start with an enumeration of the rationals: $$p_1,p_2,p_3,...$$

Pick your favorite tiny positive number $\epsilon>0$.

For each $k=1,2,3,...$, let $I_k$ be the interval centered on $p_k$ of radius $\frac{\epsilon}{2^k}$. Now take the union of these intervals: $$X = I_1 \cup I_2 \cup I_3 \cup \cdots$$ The total length of $X$ is no more than the sum $$\text{Length}(I_1) + \text{Length}(I_2) + \text{Length}(I_3) + \cdots = \frac{\epsilon}{2} + \frac{\epsilon}{2^2} + \frac{\epsilon}{2^3} + \cdots = \epsilon$$ But the rationals are contained in $X$, so the total length of the rationals is at most $\epsilon$.

But you can repeat this argument for tinier and tinier values of $\epsilon$, approaching zero.

So the total length of the rationals is zero.

But the total length of the irrationals is infinity, because it equals the the total length of the whole real line (which is infinity) minus the total length of the rationals (which is zero).

Thus, there are more irrationals than rationals.

Now this argument may sound fishy, but it turns out to be completely rigorous. Once you have developed the Lebesgue measure of the real line, substitute the phrase "Lebesgue measure" for "length", and you've got a proof.

• Although correct, this may raise the question "why can't I do the opposite"? Of course, you can't because the irrationals cannot be enumareted, but that's only if you already know that there are more irrationals than rationals... Jun 15, 2016 at 2:49
• Lebesgue measure isn't really a good measure of cardinals. For instance, the Cantor set has Lebesgue measure zero too, but it has the same cardinal as $\Bbb R$, which has $\infty$ Lebesgue measure. So I'm very suspicious of the rigour of your "proof".
– Vim
Jun 15, 2016 at 2:54
• While I agree with these comments regarding cardinality, I also note that the question does not mention cardinality, whereas it does mention geometry. Jun 15, 2016 at 11:13
• Also, although you do need to know that the rationals are denumerable in order to carry out this proof, you do not need to know that the reals (or the irrationals) are nondenumerable; in fact this proof shows that the irrationals are non denumerable. Cantor's first proof that the reals are non denumerable was also a simple argument with intervals (simpler than the one in my answer), and was discovered by him before his diagonalization argument and before he instituted the formal study of cardinality. en.wikipedia.org/wiki/Georg_Cantor%27s_first_set_theory_article Jun 15, 2016 at 11:25

If we make a rectangular grid with integer coordinates, it's possible to assign a unique angle to any rational number, using the definition $\tan \phi=y/x$ for $\phi \in (-\pi/2, \pi/2)$.

For positive rationals it would look something like this:

It's intuitively clear to me that the lines corresponding to the rational numbers can't fill all the space here (the gaps are especially noticeable around the numbers with small denominators / numerators).

This can be considered an illustration to marty cohen's answer.

• Why is it obvious that the lines can't fill all the space? (Without assuming the result we're trying to illustrate.) In particular, the density of the rationals means that, between any two lines (or any line and the axis), you can always fit another line. Jun 15, 2016 at 19:27
• @DavidRicherby, I do not consider it proof, only an illustration to marty's answer. There is no pictures in any other answer, so I wanted to add one. Jun 15, 2016 at 19:39
• Thanks for the awesome image! Jun 15, 2016 at 20:01
• @YuriyS I understand that it's supposed to be an illustration, rather than a proof, but I'm taking issue with your claim that something is obvious. Regardless of levels of proof, I don't think it's obvious at all. Also, Marty's answer doesn't use the lines at all: it just points out that the rationals correspond to the grid points and the irrationals to everything else. Jun 15, 2016 at 20:15
• @YuriyS Insofar as you're now only saying that it's obvious to you, rather than that it should be obvious to me, I suppose I'm happy. Nonetheless, I'd encourage you to add some explanation of why it's true. It seems to me to be that you're essentially saying, "It's obvious because most reals aren't rationals" but that's exactly the thing you're supposed to be illustrating! Jun 15, 2016 at 22:04

In general, my favorite way to think about this is to think about a countable set (rationals) as the set of all finite strings of $0$'s and $1$'s and an uncountable set (reals) as the set of all infinite sequences of $0$'s and $1$'s.

I think this makes it "geometrically" obvious, at least for me.

EDIT 1: Any countable set is in bijection with $\mathbb N$, each element of which has a finite binary representation. Similarly any element of $\mathbb R$ is (by definition of as the completion of $\mathbb Q$) an infinite sequence of rational numbers, where for example you could fix the first $n$ terms of the $n$'th element (ie. $1,10,100,1001,10010,\ldots$).

Although the diagonal argument is at first unappealing, it is in fact geometric in nature, and simpler than it appears. The idea is the following:

Suppose $\mathbb R$ is countable, then the interval $[0,1]$ is also countable (due to the bijection with $\tan^{-1}$) and there is a bijection $\phi:\mathbb N\to\mathbb [0,1]$ (we denote $\phi(n)$ by $\phi_n$, and the $k$'th binary decimal of $\phi_n$ as $\phi_n^k$), and we can list all the elements of $\mathbb R$

$$\label{arr}\tag{1}\phi_1=\phi_1^1\phi_1^2\phi_1^3\ldots\\ \phi_2=\phi_2^1\phi_2^2\phi_2^3\ldots\\ \phi_3=\phi_3^1\phi_3^2\phi_3^3\ldots\\ \vdots$$

now consider the real number $$x=\tilde\phi_1^1\tilde\phi_2^2\tilde\phi_3^3\ldots$$ where $\tilde\phi_i^i=1-\phi_i^i$, then clearly it will follow that $x\neq \phi_1$, $x\neq \phi_2$, $x\neq \phi_3$, etc... because $x$ always disagrees with $\phi_i$ on the $i$'th decimal, and thus $x\not\in[0,1]$ because if it were, then there would exist some $i$ such that $\phi_i=x$, which we showed impossible. Contradiction.

The "geometric" part of this argument is the fact that you choose $x$ to be the opposite of the diagonal of the array \ref{arr}.

EDIT 2: I am not sure thinking of this geometrically is necessary, as it obscures the fact that countability vs uncountability is not really a property of the "geometry" in the real line. In fact I like that Lee Mosher pointed out the intuitive property I am guessing you were looking for (and in your drawing) that the rationals have "measure" zero while irrationals have "measure" one in the interval, but as some people pointed out in the comments, this is not a characterization of cardinality. It is true that every countable subset of the real line has "measure" zero, but there are also uncountable sets (1/3-Cantor set) also with "measure" zero.

So here is a nice proof I like that shows that this is really not a geometric property in the sense that you were thinking of, but a property of sets.

Again we take a countable set as all finite binary sequences (which we call $C$), and an uncountable set as the set of all functions $C\to \bf 2$ (where $\bf 2$ denotes the set $\{0,1\}$) which we denote $\bf 2^C$. We see that $\bf 2^C$ is the same as the set of all infinite binary sequences, since the element $\phi\in\bf 2^C$ represents the sequence $\phi_1\phi_2\phi_3\ldots$.

We now show the following lemma (where $B^A$ is the set of all functions $A\to B$):

Lemma 1: (Lawvere) If there exists a surjective function $\Theta:A\to B^A$, then any function $f:B\to B$ has a fixed point.

Given $f:B\to B$, consider $g:A\to B$ given by $g(a)=f(\Theta(a)(a))$, then since $\Theta$ is surjective there exists $a_f$ such that $\Theta(a_f)=g$, and thus $\Theta(a_f)(a_f)=g(a_f)=f(\Theta(a_f)(a_f))$, so $\Theta(a_f)(a_f)$ is a fixed point of $f$.

Now, suppose there was a bijection $\Theta:C\to\bf 2^C$ (and hence surjection), then the map $$0\mapsto 1\\1\mapsto 0$$ has a fixed point, which is clearly not the case. Contradiction.

This is a somewhat more "geometric" argument, somewhat analogous to Cantor's. Consider any sequence of distinct real numbers $x_n$. I will construct a real number $y$ that is not in the sequence, as the limit of a sequence $y_n$. Start by choosing $y_1 \ne x_1$. At stage $n$, we will have a real number $y_n$ that is not $x_1, \ldots, x_n$. Let $a_1, \ldots, a_n$ be $x_1, \ldots, x_n$ sorted in increasing order. For convenience we take $a_0 = -\infty$ and $a_{n+1} = +\infty$. $x_{n+1}$ will be in one of the $n+1$ open intervals $(a_j, a_{j+1})$. If we happen to have $x_{n+1} = y_n$, let $y_{n+1}$ be some point of $(x_{n+1}, a_{j+1})$, otherwise $y_{n+1} = y_n$. In fact I'll take $$y_{n+1} = \cases{y_n + 10^{-n} & if a_{j+1} = \infty\cr y_n + 10^{-n}(a_{j+1}-y_n) & otherwise}$$ It's easy to show that this is a nondecreasing sequence and bounded above, therefore has a limit as $n \to \infty$, and that this limit can't be one of the $x_n$.
Visually, think of $y_n$ as sitting at a point on a number line, with the $x_n$ as (single-point) hailstones coming down from above, and sticking to the number line as they land. If a hailstone hits $y_n$, it hops a tiny bit to the right, but by such a small distance that it will never get much closer to any of the existing hailstones.