The strange behaviour of rational and irrational numbers on the number line

A few years ago I was studying from “The Foundations of Mathematics” by Ian Stewart and David Tall (an awesome book by the way) and started reading about the real number line. A “real” number is a number which has no imaginary component (hence real) and can be a rational or an irrational number.

A rational number can be expressed as a fraction of integers for example

$2,\, \frac{8}{10}, \, \frac{57}{117}$

are rational numbers. we can write these as decimals (eg $\frac{8}{10}= 0.8$ ) but what characterises decimal expansions as rational numbers is the fact that they are either finite or, if they do go on forever, they end up in some cycling pattern.
Irrational numbers such as

$e,\, \pi,\, \sqrt{2}$

have decimal expansions which go on forever and never repeat. For example $\pi = 3.14159265 \dots$

OK but that isn’t the weird thing i was talking about (just wanted to introduce some terminology). The weird thing is how the rational and irrational numbers appear on the number line.

Between any two rational numbers there’s an irrational number

Lets take two distinct rational numbers $a$ and $b$ . Could be any you like. We can always show that there is an irrational number somewhere between them. First note that adding anything to $a$ that is smaller than $(b-a)$ will give you a number between $a$ and $b$ (although it won’t necessarily be irrational). For example
$a + \frac{1}{2} (b-a)$ will be half way between $a$ and $b$ .
That is

$a < a + \frac{1}{2} (b-a) < b$

All we need to do now is make it irrational. An easy way is swapping the $\frac{1}{2}$ with $\frac{\sqrt{2}}{2}$ (note that an irrational number multiplied or divided by a rational number is itself irrational).

$a < a + \frac{\sqrt{2}}{2} (b-a) < b$

The number $a + \frac{\sqrt{2}}{2} (b-a)$ is always irrational. Therefore, we have a way of finding a number in between $a$ and $b$ which is irrational!

Between any two irrational numbers theres a rational number

This time let $a$ and $b$ be distinct irrational numbers. Let’s assume that $a < b$ .
Both will have infinitely long decimal expansions and even if they are really close to one another at some point you’ll reach a digit in $b$ that is higher than the one in $a$ at the same point.
For example:

$a = 0.1183280734217940 \dots$

and

$b = 0.1183280765645311 \dots$

Here, you have to get to the tenth significant figure before you see that $b$ is indeed greater than $a$ . Now you might have realised already how we can make a rational number between $a$ and $b$ ! You truncate $b$ directly after this digit!
So in this case let

$c = 0.118328076$

( $b$ truncated to 10 significant figures). We find that $c$ is less than $b$ because we’ve chopped a bunch of numbers off the end and it is greater than $a$ because the digit at the final decimal place in $c$ is larger than the one in the same place in $a$ . Not only is $c$ between $a$ and $b$ but, as it has a finite decimal sequence, it is rational.

How to have an infinite number of neighbours and no friends

OK, so we’ve established that between any two rational numbers there’s an irrational one and between any two irrational numbers there’s a rational one. If, like me hearing this for a first time, you now have a picture of the real number line where rationals and irrationals alternate then i’m sorry but that’s not how things are. You see there isn’t just one irrational number between any two rationals but an infinite number of them. I invite you to go back and think about how to prove this. Likewise, between any two irrationals there are an infinite number of rationals. Why does this mean they don’t alternate? Well there’s no concept of one number being “next to” another.

Imagine you move into a house and you go next door to meet your neighbour. On the way you find another smaller house in between your house and theirs. “OK I guess this is my real neighbour then. I’ll go say hi.” But on the way to that house you find another even smaller house between… and so on… infinitely.
You never actually meet your neighbour because you always find someone lives closer and closer.
The same is true of numbers. There is no concept of a “closest” number to another because no matter how close two numbers are there will always be a number in between them (actually there will be an infinite number of numbers in between them). This stems from the completeness of the real numbers: the concept that no gaps exist on the real number line.

The “no neighbour problem” above bears direct resemblance to the Zeno’s “Achilles and the tortoise” paradox:
Achilles and the tortoise agree to having a race but knowing the tortoise doesn’t stand a chance against his much faster self, Achilles gives the tortoise a head start. So the race begins and Achilles runs towards the tortoise. The problem is he will never actually reach the tortoise. Lets say that the tortoise is 50 m ahead and can run half the speed of Achilles. Then by the time Achilles reaches the 50 m mark the tortoise has reached 75 m. A short time later Achilles reaches 75 m but the tortoise has by now reached 87.5 m. When Achilles gets to 87.5 m the tortoise has moved even further. Thats the root of the paradox. It doesn’t matter how much faster Achilles is, any amount of getting closer to the tortoise is counteracted by the relentless plodding of the tortoise. Taking to it’s extreme this way of thinking can be used to argue that any movement at all is impossible. In order to take one step forward you first need to take half a step forward. But before you manage to do that you first must’ve taken a quarter of a step etc. etc. So any movement whatsoever requires an infinite number of smaller movements and of course its impossible to do an infinite number of things in an finite amount of time. Good look with that.
I must say however at this point that this was a paradox and now it isn’t. That’s because we now have methods of adding together an infinite number of numbers (as well as a grasp of the infinitesimally small) which helps make more sense of the situation (so yes movement is possible after all).
Zeno would have us believe that moving forward 1 m is impossible because first you need to get halfway, before that you must have reached a quarter of the way and so on ad infinitum. Because each of these tasks takes some amount of time, and there an infinite number of tasks, surely it takes an infinite amount of time to carry them out?
Well let’s see, lets add $\frac{1}{2} +\frac{1}{4} +\frac{1}{8}+ \dots$ and so on with each new fraction half the size of the last. What does this summation get you? Thinking about it: every new fraction you add makes the total bigger, and you keep on going… maybe it’s just infinitely large?
Let’s call the summation $x$ .

$x = \frac{1}{2} +\frac{1}{4} +\frac{1}{8}+\frac{1}{16}+ \dots$

we can find what $x$ is with some algebraic manipulation. Consider $2x$ .

$2x = 2 \times \left( \frac{1}{2} +\frac{1}{4} +\frac{1}{8}+\frac{1}{16}+ \dots \right)$

$= 1 + \frac{1}{2} +\frac{1}{4} +\frac{1}{8}+\frac{1}{16}+ \dots$

notice that $2x$ starts with 1 but the rest ( $\frac{1}{2} +\frac{1}{4}+ \dots$ etc) is just $x$ .
So whatever $x$ actually is we now know that

$2x = 1 + x$

which we can solve, in the usually way of rearranging equations, to find $x$ of course is one!
So we can add an infinite number of numbers together and get a finite number. As long as the numbers get smaller and smaller quickly enough. We say that the series of numbers we have added together converge to a finite number, in this case 1.
So yes Achilles will eventually catch the tortoise (bang on the 100 m line actually) but that doesn’t help me find my neighbour.

There are many, many more irrational numbers then rational numbers

Anyway back to numbers. Trying to show that rationals and irrationals don’t alternate led to an infinite regress. But it wasn’t entirely convincing that they don’t alternate in some weird infinite way was it? Here’s the real evidence that they don’t alternate: there are more irrational numbers than rational numbers. Not in the way that if you have 6 apples and 11 bananas there are more bananas, there are an infinite number of rationals and an infinite number of irrationals.
It just so happens that the infinity of irrationals is much, much bigger than that of the rationals.
We say that something is countably infinite if there exists a one-to-one correspondence between it and the natural numbers. ie. We can count them.
Take even numbers for example: we can line them up with the natural numbers like so

$2 \rightarrow 1$
$4 \rightarrow 2$
$6 \rightarrow 3$
and so on…

There are an infinite number of even numbers but for each even we can assign a number and therefore we say the evens are countably infinite. The rationals share this countable infinity (yes there are as many fractions as there are numbers!) but the irrationals don’t.
You can count an infinite number of irrationals and still find you’ve “missed some out”. We therefore say there are an uncountable number of irrational numbers. So although you could tell me what the ${112}^{th}$ even number is it’s impossible to say what the ${1}^{st}$ irrational number is.

That’s it. So, there you go. Between any two irrationals you have an infinite number of rationals (no matter how close they are) and between any two rationals you have an even more infinite number of irrationals.

The strange behaviour of rational and irrational numbers on the number line

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