# Lecture 19

For centuries, the real numbers were considered in an informal way: nobody knew exactly how to define $\mathbb{R}$, but they knew what it ought to look like.

For the time being, and *for the time being only* we’ll investigate the reals in a similar, informal way. For now, you can regard the real numbers $\mathbb{R}$ as being built out of decimals (as you did at school). In the last lecture of the course, we’ll sort this out, and consider a modern construction of the reals.

Our mental picture of the reals should be a picture of a numberline. Here’s a numberline with some interesting points marked on:

I’ve marked on the integers $-1$, $0$, $1$, $2$ and $3$, which are all in $\mathbb{Z}$ and hence in $\mathbb{Q}$.

I’ve also marked on $\sqrt{2}$, which we now know to be irrational, and $\pi$, which I’ve claimed to you is irrational: these things are in the set $\mathbb{R}\backslash\mathbb{Q}$ of irrational numbers.

In my mind, I think of the real numbers $\mathbb{R}$ as a solid line, and the rational numbers $\mathbb{Q}$ as a very fine gauze net stretched out within it.

Bear in mind that that the rationals $\mathbb{Q}$ are a lovely system of numbers: we can add and subtract and multiply and divide rationals and remain inside the rationals. Formally: if $x\in\mathbb{Q}$ and $y\in\mathbb{Q}$, then $x+y$, $x-y$, $xy$ and $x/y$ (if $y$ is nonzero) are all elements of $\mathbb{Q}$. We say that $\mathbb{Q}$ is *closed* under addition, subtraction, multiplication and division.

The reals $\mathbb{R}$ are also a lovely system of numbers, closed not just those four operations but many others: square roots (of positive numbers), sines, cosines, and so on.

The irrational numbers $\mathbb{R}\backslash\mathbb{Q}$ are not a lovely system of numbers: they are not closed under any of these things.

For example, can we think of two irrational numbers whose sum is rational?

$\sqrt{2} + (1-\sqrt{2}) = 1$.

Can we think of two irrational numbers whose product is rational?

$\left(\sqrt{2}\right)\left(\sqrt{2}\right) = 2$.

So, the irrational numbers $\mathbb{R}\backslash\mathbb{Q}$ really are just the big messy clump left over in $\mathbb{R}$ when you remove $\mathbb{Q}$.

However, at least the following is true:

#### Proposition

Let $x$ be irrational, and $y$ be rational. Then $x+y$ is irrational.

Also, if $y$ is nonzero, then $xy$ is irrational.

#### Proof

We prove the first one by contradiction. Suppose that $x+y$ is rational. Then $(x+y)-y$ is also rational, being obtained by subtracting two rational numbers, but it’s equal to $x$ which we know to be irrational. That’s the contradiction we wanted.

We prove the second one by contradiction too. Suppose that $xy$ is rational. Then $(xy)/y$ is also rational, as it’s obtained by dividing two rational numbers (with the latter nonzero), but it’s equal to $x$ which we know to be irrational. That’s the contradiction we wanted.

## Convergent sequences

Now our mission is to study the real numbers. When we were studying the integers, the main theme running through it all was to do with divisibility. Divisibility is, of course, not a very sensible thing to ask about over the reals.

As a result, real analysis (the study of $\mathbb{R}$), and the questions which are interesting and helpful to ask, is very different to number theory.

It turns out that the most interesting things you can ask about are to do with *approximation*. Why is the notion of approximation so important?

When we write that $\pi = 3.1415926535897932384626433\cdots,$ the point is that the digits give a kind of address telling you how to find $\pi$ on the numberline. The number $\pi$ is close to $3$, closer to $3.1$, closer still to $3.14$, even closer still to $3.141$, and so on.

The notion of *convergence*, which I’ll define shortly, is a way of encoding this concept of increasingly good approximation. We will say that the sequence of rational numbers $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots$ “converges to $\pi$”. That’s supposed to mean that if you follow the address, you’ll end up homing in on $\pi$.

The definition will seem complicated, and probably harder to get your head around than other definitions in the course. However, that’s because it really is a subtle concept: all the simpler approaches you might think of are wrong.

The most obvious wrong definition is this:

Why is this completely wrong? Well, for example, the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots$ also gets closer and closer to $1000$: $\begin{aligned} {}1000 - 3 &= 997\\ {}1000 - 3.1 &= 996.9\\ {}1000 - 3.14 &= 996.86\\ {}1000 - 3.141 &= 996.859\\ {}1000 - 3.1415 &= 996.8585\end{aligned}$ Of course, this sequence never gets particularly close to $1000$ (the sequence never goes above $4$, so it never gets within $996$ of $1000$), but it’s always getting closer!

But this means that if our definition of “converging to $x$” were the completely wrong definition “gets closer and closer to $x$”, then the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots$ would “converge to $\pi$”, but it would also “converge to $1000$”.

But that’s not what we want: this sequence is a terrible way of getting to $1000$, but a good way of getting to $\pi$.

Here’s a slightly better idea:

More precisely, this definition is trying to say that for any positive distance, this sequence comes within this distance of $x$.

In symbols, this can be written: $\forall\epsilon>0,\quad \exists n\in\mathbb{N}\quad\mathrm{s.t.}\quad\left|a_n-x\right|<\epsilon.$ Before we understand why this is wrong, we ought to make sure we know what this is supposed to mean.

I like to think of it as an argument with a very dangerous and unpleasant *evil opponent*. The evil opponent gets to choose a (positive real) distance, and we win if the sequence gets within that distance of $x$, and we lose if it doesn’t.

In order to be *sure* of winning, we have to know how to beat the evil opponent whatever they say.

So, in investigating how close the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots,$ gets to $1000$, then if the evil opponent is stupid enough to ask “does the sequence get within distance $100000$?” we’ll win. But we can’t just hope they’ll ask that. Instead they might ask “does the sequence get within distance $0.001$ of $1000$?”, and then we’d lose, because it never gets anywhere near there.

On the other hand, if we’re investigating how close the sequence gets to $\pi$, we win no matter what they say. If they ask “does the sequence get within distance $1000000$ of $\pi$?”, then we say “yes, $3$ is within $1000000$ of $\pi$”, and win. If they ask “does the sequence get within distance $0.0001$ of $\pi$, then we say “yes, $3.14159$ is within $0.0001$ of $\pi$”, and win. If they ask a question with an even tinier positive number in, we just take more digits and use that and say “yes”. We’re happy: the evil opponent can’t beat us.

This is a much better concept. But it’s still not right.

Here’s an example of what could go wrong. Consider the sequence $1.1,\quad 2.01,\quad 1.001,\quad 2.0001,\quad 1.00001,\quad 2.000001,\quad \ldots$ Does it converge to $1$? Does it converge to $2$?

It’s lousy as an address for either: if we give these instructions to the numberline’s village postal worker, they’ll get very annoyed as they keep walking from somewhere near $1$ to somewhere near $2$ and back again.

But according to the definition above it would “converge” to both, because it gets as close as you like to $1$ and it also gets as close as you like to $2$.

So we need to find some way of saying that it has to make its mind up eventually. The obvious thing to do is to say is that (for any $\epsilon>0$) it has to get within $\epsilon$ of $x$, and then stay within $\epsilon$ of $x$ forever.

This leads us to our final definition:

**Definition:** Let $x$ be a real number. A sequence of real numbers $a_0, a_1,
a_2,\ldots$ is said to *converge to $x$* if we have $\forall \epsilon>0,\quad \exists N\in\mathbb{N}\quad\mathrm{s.t.}\quad\forall n>N,\quad \left|a_n-x\right|<\epsilon.$

So that says “no matter what positive real $\epsilon$ our evil opponent gives us, we can point out some $N$, such that all the terms $a_{N+1}, a_{N+2}, a_{N+3}, \ldots$ are all within $\epsilon$ of $x$”.

That does an excellent job of making precise the concept of “gets close and stays close forever”, and it’s the right definition!

Now, suppose we ask whether the sequence $3,\quad 3.1,\quad 3.14,\quad 3.141,\quad 3.1415,\quad \ldots$ converges to $\pi$. It does, because no matter what $\epsilon$ our evil opponent asks about, we can find some term of the sequence beyond which all terms are within $\epsilon$ of $\pi$. For example, all terms after the $(N+1)$st term are within $10^{-N}$ of $\pi$.

Does that converge to $1000$? No, it never comes within $1$ of $1000$ (for example), so it certainly doesn’t stay within $1$ of $1000$ forever.

What about the sequence $a_0 = 1.1,\quad a_1 = 2.01,\quad a_2 = 1.001,\quad a_3 = 2.0001,\quad \ldots?$ Does that converge to anything?

No, it doesn’t. In particular, it doesn’t converge to $1$, because while it’s sometimes close to $1$, it’s also sometimes close to $2$. So there is no $N$ where $a_n$ is always within $0.1$ of $1$ for all $n>N$: all the odd-numbered $a_n$ aren’t in that range.

Similarly, it doesn’t converge to $2$, because while it’s sometimes close to $2$, it’s sometimes close to $1$. So there is no $N$ where $a_n$ is always within $0.1$ of $2$ for all $n>N$: all the even-numbered $a_n$ aren’t in that range.

So, given the difficulties we’ve had in finding the right definition, perhaps you’ll have some sympathy for the fact that it took about two centuries to sort real analysis out properly. In what remains of the course I’ll try to make you like this definition.