In mathematics, if you squint, the word “real” has a kind of eldritch quality. Although calculus students work confidently with them, and they enable all kinds of nice behavior in topology, the definition of a real number is, well, surreal. The title of this blog post is a pun. You might ask why we need the real numbers, and one answer is “for completeness”. For the same reason, I thought a blog post would be useful.

## 1. Constructing numbers

I learned a little of how to prove things and the culture of mathematics by taking a class taught out of Apostol’s Calculus. As a first-year in college, I remember being thoroughly at sea in that class, whose professor apologetically translated the symbols “$$\forall$$” and “$$\exists$$” into English a couple of days into the class—they had appeared on the board within the first week.

One of the aspects of that class that I appreciated was the construction of the real numbers. I’d like to take us on a whirlwind tour through that construction.

We seek a mathematical object $$\mathbb{R}$$ called the “real numbers” with the following reasonable properties of numbers.

• Counting numbers $$1,2,\ldots$$ are real numbers.
• Real numbers may be added to produce other real numbers. Addition is associative and commutative.
• Real numbers may also be subtracted. Subtraction “undoes” addition. There is a unique number $$0$$ which is equal to $$x - x$$ for every real number $$x$$. This number $$0$$ is an identity for addition.
• Real numbers may be multiplied. Multiplication is associative and commutative and distributes over addition. It follows that $$x * 0 = 0$$ for every real number $$x$$.
• Multiplication can be “undone” by division in the sense that $$x \div y$$ is the unique number $$z$$ such that $$z * y = x$$—except that because $$x * 0 = 0$$ for every real number $$x$$, there is no sensible way to define division by $$0$$, so we don’t.
• Real numbers are ordered in a way that extends the usual ordering of the counting numbers and is compatible with addition and multiplication in the way that you expect.
• A set $$S$$ of real numbers has an upper bound when there is a real number $$B$$ such that every $$x$$ in $$S$$ satisfies $$x \le B$$. A least upper bound (or supremum) for $$S$$ is an upper bound $$\sup S$$ such that every upper bound $$B$$ satisfies $$\sup S \le B$$. Put another way, if $$y < \sup S$$, then $$y$$ is not an upper bound for $$S$$, so there must be some actual element $$x$$ of $$s$$ satisfying $$y < x$$. Every set of real numbers that has an upper bound has a least upper bound.

While “reasonable”, as I’ll attempt to argue in a second, this last property is a good deal more complex than the previous ones. Indeed, this behavior of real numbers is what makes them eldritch (and different from the rational numbers).

Here’s why this property is reasonable: suppose you have a way of approximating a quantity by real numbers. Like for example, suppose you have a method of computing more and more digits of this number in a decimal expansion. I claim that a reasonable definition of numbers should have an actual number that is what is being approximated. That is, given that we are able to compute (theoretically) $$\pi$$ to an arbitrary number of decimal places, $$\pi$$ is a real number. Notice that decimal expansions of $$\pi$$ have an upper bound ($$4$$ will do, for instance), and that each decimal expansion is larger than the last. Thus, since the set of decimal expansions of $$\pi$$ has a least upper bound, we can define that least upper bound to “be” $$\pi$$. It turns out to have all the properties of $$\pi$$ that we can compute analytically, so this “synthetic” definition makes sense.

So let’s go through our whirlwind tour.

## 2. The counting numbers

We’re going to suppose that we already understand the counting numbers: they have the defining property that you start somewhere and always know what the next one is. That is, after $$1$$ comes $$2$$, after $$2$$ you have $$3$$ and so on.

In other simulated confusions, ahem, constructions of various fields of mathematics, one starts with the idea of a “set” and ways of forming new sets from old ones and from there defines the counting numbers and their successor function by some scheme. Let’s not do things this way: it makes possible certain silly statements, like, “what are the elements of $$6$$” which don’t seem reasonable to me.

There are people who are skeptical about the idea of infinity in mathematics. I don’t want to poo-poo or misrepresent these people’s intellectual commitments, which can be really interesting to read defended. I’ll just confine myself to noticing that there’s a real psychological difference between the claim “I will never run out of counting numbers, since after any counting number I know which one comes next, and my scheme for counting never reuses numbers” and “I can conceive of the collection of counting numbers together in its total infinity”.

## 3. The integers

Addition of counting numbers is another something we understand. We know how to compute $$2 + 3$$ and that it is $$3 + 2$$. I feel compelled to note the long history of the statement $$2 + 2 = 5$$, which dates back to at least Orwell’s 1984 but is older, and had a recent featured appearance in the “culture wars”, which caused well-intentioned progressive mathematicians a certain amount of angst and bending-over-backwards to affirm that $$2 + 2 = 5$$ can be a sensical mathematical statement; precisely the kind of thing that Orwell famously decried. Part of what makes this construction exercise sort of exhilerating and deeply silly is that what we are doing is agreeing amongst ourselves that we know what the symbols $$2$$, $$+$$, $$5$$ and $$=$$ mean. In so doing, it must be noted that this agreement and the meanings agreed upon have a certain arbitrariness to them. Changing our agreements about meaning should have nothing to do with doublethink, newspeak, etc. but are instead an opportunity to invite creativity, a necessary part of the practice of mathematics, back into what can be an intensely conservative field.

Anyway, our goal here is to follow our common sense here, a goal that has dubious outcomes in many situations, and indeed the goal of this blog post is to point up the dubiousness of the real numbers.

Subtraction is also something we understand, but it leads us to our first hint of the eldritch: negative numbers. Since $$2 - 3$$ is a sensible formulation: we would like to be able to subtract numbers willy-nilly, it forces us to define numbers which we did not have previously.

It’s fun to play with $$0$$ in particular: the rules of algebra tell us that since $$x + y = x + y$$, we have $$x - x = y - y$$ for every pair of real numbers. Similarly, $$0$$ is the only number for which $$x + 0 = x$$. If there were another such number, say $$0'$$, adding it to $$0$$ allows us to deduce that $$0 + 0' = 0$$ on the one hand, but also $$0'$$ on the other, so these are the same number by the transitive nature of equality.

## 4. The rational numbers

Integers may be freely multiplied, and we argued above that division, whatever it is, should make sense only for nonzero numbers. It’s fun to see that our definitions force the existence of at most one number $$1$$ for which $$x * 1 = x$$ for all real numbers $$x$$, that $$x * 0 = 0$$ for all real numbers $$x$$ and so $$1 \ne 0$$.

But how to define numbers that allow division? We have already from school the idea of fractions $$\frac{a}{b}$$. This generalizes in a fun way: a ring in mathematics, is a place where you can play the two games of addition and multiplication. Addition, like in school, is associative and commutative, and has an identity element $$0$$ and negatives, but mathematicians are happy to merely require multiplication to be associative and distribute over addition. Usually we require the existence of a $$1$$ in our rings, but there are times that mathematicians forego even that.

Anyway, if you have a set $$S$$ in a ring $$R$$ (with $$1$$) which is multiplicatively closed in the sense that $$s * t$$ is in $$S$$ when $$s$$ and $$t$$ are and contains $$1$$, you can form a ring of fractions $$S^{-1}R$$ whose elements are formal fractions with denominators in $$S$$. When $$R$$ is commutative (that is, multiplication in $$R$$ is commutative) and $$S$$ does not contain $$0$$ nor any element $$s$$ for which there exists a nonzero element $$t$$ such that $$st = 0$$, the ring of fractions behaves essentially like you would expect: $$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$$, $$\frac{a}{b} * \frac{c}{d} = \frac{ac}{bd}$$, and so on. The cool thing I wanted to mention is that this construction actually makes sense more generally!

In our situation, the set $$S$$ is just all of the integers which are nonzero, which implies that the ring of fractions is actually a field, in the sense that we have a commutative ring in which every nonzero element of the field has a multiplicative inverse.

## 5. Ordering

There’s a little to say here, but not too much. One wants an ordering on the real numbers that extends the usual one on the counting numbers and is compatible with addition in the sense that if $$a < b$$ implies that $$a + c < b + c$$. This actually implies that negation reverses order since $$a < b$$ implies that $$a + -(a + b) < b + -(a + b)$$, and these numbers are equal to $$-b$$ on the left and $$-a$$ on the right. Since for every pair of counting numbers $$a$$ and $$b$$, we have the trichotomy: either $$a = b$$, $$a < b$$ or $$a > b$$, it seems reasonable that this should be true for real numbers as well. It cannot be true that $$a < b$$ implies that $$a * c < b * c$$ for all numbers, because of the existence of negation and zero, but we can say that we would like this to be true for all real numbers $$c$$ which are positive (that is, for which $$c > 0$$).

Actually, although the ordering stuff is pretty straightforward, (in particular, we didn’t have to invent new numbers) it’s worth pointing out that having such a total order is one of the properties which characterize the real numbers.

## 6. Upper bounds

This is where things really get eldritch. First of all, the rational numbers, $$\mathbb{Q}$$, do not have least upper bounds. Firstly, note that $$\sqrt{2}$$ is not a rational number. If you’ve never proved this, it’s so cute: suppose you could write $$\sqrt{2}$$ as $$\frac{p}{q}$$ for integers $$p$$ and $$q$$ which are in “lowest terms”. Then you have to ask whether $$p$$ and $$q$$ are even or odd. It turns out neither answer will do, so it cannot be possible to write $$\sqrt{2}$$ as $$\frac{p}{q}$$. But, we can clearly ask, of any rational number $$r = \frac{p}{q}$$, whether $$r^2$$ is greater than or less than $$2$$. The collection of all rational numbers for which the answer is “less than” has an upper bound but no least upper bound. This is a fun exercise: suppose you have a rational number $$r$$ whose square is greater than $$2$$. Can you find a general way of constructing a rational number $$r'$$ which also satisfies $$r'^2 > 2$$ but which is smaller than $$r$$?

So to satisfy the least upper bound condition, we are going to have to add real numbers. A lot of them, actually. Formally, define a cut $$r$$ in $$\mathbb{Q}$$ to be a partition of $$\mathbb{Q}$$ into two nonempty pieces, $$r^-$$ and $$r^+$$, which do not overlap (this is what I mean by a partition) and are closed under ordering, in the sense that if $$s$$ is an element of $$r^-$$ and $$t < s$$, then $$t \in r^-$$, and similarly if $$s \in r^+$$ and $$t > s$$, then $$t \in r^+$$.

Now, if $$S$$ is a subset of $$\mathbb{Q}$$ which is bounded above, then $$S$$ generates a cut $$r$$ by the rule that an element of $$\mathbb{Q}$$ is in $$r^-$$ if it is less than or equal to some element of $$S$$. Because $$S$$ is bounded above, the complement of this set $$r^+$$ is nonempty. So if we can turn the collection of Dedekind cuts into a field, we’ll win.

Defining the rules of algebra on these cuts is not so bad, but not fun enough to want to do here. I’ll confine myself to saying that $$(r + r')^-$$ should just be the set of elements of $$\mathbb{Q}$$ of the form $$q + q'$$, where $$q \in r^-$$ and $$q' \in r'^-$$.

This is a strange definition! It’s super weird! It works great!

## 7. Completeness

The definitions above have a couple interesting and useful consequences. The first one I want to mention is that the real numbers are Archimedean in the sense that every real number $$r$$ has a counting number $$n$$ larger than it. Indeed, if $$r$$ is negative this is clear, and if $$r$$ is positive, the collection of counting numbers less than $$r$$ is bounded above (by $$r$$), has a least upper bound (which is actually a counting number, it turns out), and we can take $$n$$ to be that supremum plus one.

Here is the other. A sequence $$x_1,x_2,\ldots$$ of real numbers converges to a real number $$x$$ when for every $$\epsilon > 0$$ (which you are supposed to suppose is given to you by an adversary) you can find a large number $$N > 0$$ such that for all $$n > N$$, the statement $$|x_n - x| < \epsilon$$ is true. In English, the elements of the sequence are getting closer and closer to $$x$$, even if it may be the case that none of them is precisely equal to $$x$$.

A priori weaker than the notion of convergence is the notion of being Cauchy. (Augustin-Louis Cauchy, possibly mathematics’ most useful pedant, went around poking holes in the thinking of his contemporaries and predecessors, or so I’m told. As a result of his efforts, mathematics is more rigorous, which given its unreasonable applications to reality, is probably for the best. I have a stuffed animal given to me by a high school boyfriend; in college I named him Cauchy.)

A Cauchy sequence is one for which for any $$\epsilon > 0$$, there exists $$N$$ big enough so that whenever $$n$$ and $$m$$ are chosen greater than $$N$$, we have $$|x_n - x_m| < \epsilon$$. In English, the elements of the sequence get closer and closer to each other as the sequence goes along.

> Theorem. Every Cauchy sequence of real numbers converges.

More generally, in a metric space (more about which soon), the notions of convergence and being Cauchy make sense by using the notion of distance. A metric space is complete if the above theorem holds. Thus $$\mathbb{R}$$ is complete as a metric space, while its subset $$\mathbb{Q}$$ is not.