Where are the irrational numbers?

Question

I’m new to real analysis and have recently encountered the fact that there are more irrational numbers than rational. However, I can’t seem to reconcile the fact that I can easily think of rational numbers but not irrationals. So, my question is, where are the irrationals? To illustrate what I mean: Where do I have to go in the real number line to find this great amount of irrationals? Do I have to look deep into small numbers? In other words, there’s not a lot of irrationals in easily-identifiable numbers (like whole numbers or their square roots) but once we go deep to where numbers are separated by an arbitrarily small number, say 1/100000000, we find lots of irrationals. Or maybe the irrationals are hiding somewhere in the millions and billions, but not in numbers that a human would easily count (like 1-100).

In a sense this question is a bit vague, and my image of the irrational number line is vague, but hopefully it’s clear what I mean. Where are the irrationals!?

Answer 1

Finding irrationals has been dubbed to be like "finding hay in a haystack". They literally are everywhere in the reals, but it's very hard to explicitly list many of them (outside of roots).

Perhaps the best perspective I can offer is the decimal one. Real number can be expressed as decimal expansions. Which decimal expansions correspond to rational numbers? Those that are "eventually periodic", i.e., their decimal expansion eventually begins looping with some period.

At least to me, the idea of there being far fewer numbers that are eventually periodic than numbers that aren't is a much easier bullet to bite than abstractly thinking about the sizes of irrational and rational sets.

Answer 2

A slightly different take on Rushabh's answer that might make it even easier to appreciate: rather than speaking of the analytic rationals and reals, I'm going to speak of some topological 'rationals' and 'reals' — that is, a countable set of values and an uncountable set that is the completion of the countable ones.

To do this, I'm going to throw away the arithmetic structure of numbers and look only at the metric structure of who's close to who, and I'm going to replace the rationals with the dyadic rationals (between 0 and 1). This structure has an 'easier' isomorphic description: the rationals in this world correspond to all finite-length binary sequences $\{a_1, a_2, \ldots, a_n\}$, and the reals correspond to all$^*$ infinite-length binary sequences $\{a_1, a_2, \ldots, a_n, a_{n+1}, \ldots\}$. We can talk about how close one sequence is to another: if we have two sequences $\mathbf{a}=\{a_i\}$ and $\mathbf{b}=\{b_i\}$, then we define the distance between them as $2^{-n}$, with $n$ being the first index where they disagree. Likewise, we can put an order on these sequences: again letting $n$ be the first index where $\{a_i\}$ and $\{b_i\}$ disagree, we say $\mathbf{a}\gt\mathbf{b}$ iff $a_n\gt b_n$ — in other words, if $a_n=1$ and $b_n=0$.

I find this model a lot easier to think of in terms of completions and density: it's easy to see that there are infinitely many rationals between any two rationals, but that there are 'even more' reals — every infinite binary sequence, not just the finite ones.

$^*$ To be a little more proper about respecting the distance, we need to work with an equivalence relation that identifies any finite sequence $\{a_i: i\leq m\}$ with all the 'zero-terminated' sequences $\{b_i: i\leq n\}$ that have $b_i=a_i$ for $1\leq i\leq m$ and $b_i=0$ for $m\lt i\leq n$, as well as the infinitely long zero-terminated $n\to infty$ limit of the various $\mathbf{b}$s. But these are IMHO technical details, and for a raw understanding I find it a little better to just ignore them.