# Building the Real Numbers

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- 7979

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Skills to Develop

- Show why building real numbers is logically necessary

Contrary to the title of this section we will not be rigorously building the real numbers here. Instead our goal is to show why such a build is logically necessary, and to give a sense of some of the ways this has been accomplished in the past. This may seem odd given our uniform emphasis on mathematical rigor, especially in the third part of the text, but there are very good reasons for this.

One is simple practicality. The fact is that rigorously building the real numbers and then showing that they have the required properties is extraordinarily detailed work, even for mathematics. If we want to keep this text to a manageable size (we do), we simply don’t have the room.

The second reason is that there is, as far as we know, very little for you to gain by it. When we are done we will have the real numbers. The same real numbers you have been using all of your life. They have the same properties, and quirks, they’ve always had. To be sure, they will not have lost any of their charm; They will be the same delightful mix of the mundane and the bizarre, and they are still well worth exploring and getting to know better. But nothing we do in the course of building them up logically from simpler ideas will help with that exploration.

A reasonable question then, is, “*Why bother?*” If the process is overwhelmingly, tediously detailed (it is) and gives us nothing new for our eﬀorts, why do it at all?

Doing mathematics has been compared^{1} to entering a dark room. At ﬁrst you are lost. The layout of the room and furniture are unknown so you fumble about for a bit and slowly get a sense of your immediate environs, perhaps a vague notion of the organization of the room as a whole. Eventually, after much, often tedious exploration, you become quite comfortable in your room. But always there will be dark corners; hidden areas you have not yet explored. Such a dark area may hide anything; the latches to unopened doors you didn’t know were there; a clamp whose presence explains why you couldn’t move that little desk in the corner; even the light switch that would allow you to illuminate an area more clearly than you would have imagined possible.

But, and this is the point, there is no way to know what you will ﬁnd there until you walk into that dark corner and begin exploring. Perhaps nothing. But perhaps something wonderful.

This is what happened in the late nineteenth century. The real numbers had been used since the Pythagoreans learned that \(\sqrt{2}\) was irrational. But really, most calculuations were (and still are) done with just the rational numbers. Moreover, since \(\mathbb{Q}\) forms a “*set of measure zero*,” it is clear that most of the real numbers had gone completely unused. The set of real numbers was thus one of those “*dark corners*” of mathematics. It had to be explored.

“*But even if that is true*,” you might ask, “I have no interest in the logical foundations of the real numbers, especially if such knowledge won’t tell me anything I don’t already know. Why do I need to know all of the details of constructing \(\mathbb{R}\) from \(\mathbb{Q}\)?"

The answer to this is very simple: You don’t.

That’s the other reason we’re not covering all of the details of this material. We will explain enough to light up, dimly perhaps, this little corner of mathematics. Later, should you need (or want) to come back to this and explore further you will have a foundation to start with. Nothing more.

Until the nineteenth century the geometry of Euclid, as given in his book The Elements, was universally regarded as the touchstone of mathematical perfection. This belief was so deeply embedded in Western culture that as recently as 1923, Edna St. Vincent Millay opened one of the poems in her book The Harp Weaver and Other Poems with the line “*Euclid alone has looked on beauty bare.*”

Euclid begins his book by stating \(5\) simple axioms and proceeds, step by logical step, to build up his geometry. Although far from actual perfection, his methods are clean, precise and eﬃcient – he arrives at the Pythagorean Theorem in only \(47\) steps (theorems) – and even today Euclid’s Elements still sets a very high standard of mathematical exposition and parsimony.

The goal of starting with what is clear and simple and proceeding logically, rigorously, to what is complex is still a guiding principle of all mathematics for a variety of reasons. In the late nineteenth century, this principle was brought to bear on the real numbers. That is, some properties of the real numbers that at ﬁrst seem simple and intuitively clear turn out on closer examination, as we have seen, to be rather counter-intuitive. This alone is not really a problem. We can have counter-intuitive properties in our mathematics – indeed, this is a big part of what makes mathematics interesting – as long as we arrive at them logically, starting from simple assumptions the same way Euclid did.

Having arrived at a view of the real numbers which is comparable to that of our nineteenth century colleagues, it should now be clear that the real numbers and their properties must be built up from simpler concepts as suggested by our Italian friends in the previous section.

In addition to those properties we have discovered so far, both \(\mathbb{Q}\) and \(\mathbb{R}\) share another property which will be useful. We have used it throughout this text but have not heretofore made it explicit. They are both linearly ordered. We will now make this property explicit.

Definition \(\PageIndex{1}\)

A number ﬁeld is said to be linearly ordered if there is a relation, denoted “\(<\),” on the elements of the ﬁeld which satisﬁes all of the following for all \(x\), \(y\), and \(z\) in the ﬁeld.

- For all numbers \(x\) and \(y\) in the ﬁeld, exactly one of the following holds:
- \(x < y\)
- \(x = y\)
- \(y < x\)

- If \(x < y\), then \(x + z < y + z\) for all \(z\) in the ﬁeld.
- If \(x < y\), and \(0 < z\), then \(x \cdot z < y \cdot z\).
- If \(x < y\) and \(y < z\) then \(x < z\).

Any number ﬁeld with such a relation is called a linearly ordered number ﬁeld and as the following problem shows, not every number ﬁeld is linearly ordered.

Exercise \(\PageIndex{1}\)

- Prove that the following must hold in any linearly ordered number ﬁeld.
- \(0 < x\) if and only if \(-x < 0\).
- If \(x < y\) and \(z < 0\) then \(y \cdot z < x \cdot z\).
- For all \(x \neq 0\), \(0 < x^2\).
- \(0 < 1\).

- Show that the set of complex numbers (\(\mathbb{C}\)) is not a linearly ordered ﬁeld.

In a thorough, rigorous presentation we would now assume the existence of the natural numbers (\(\mathbb{N}\)), and their properties and use these to deﬁne the integers, (\(\mathbb{Z}\)). We would then use the integers to deﬁne the rational numbers, (\(\mathbb{Q}\)). We could then show that the rationals satisfy the ﬁeld axioms worked out in the previous section, and that they are linearly ordered.

Then – at last – we would use \(\mathbb{Q}\) to deﬁne the real numbers (\(\mathbb{R}\)), show that these also satisfy the ﬁeld axioms and also have the other properties we expect: Continuity, the Nested Interval Property, the Least Upper Bound Property, the Bolzano-Weierstrass Theorem, the convergence of all Cauchy sequences, and linear ordering.

We would start with the natural numbers because they seem to be simple enough that we can simply assume their properties. As Leopold Kronecker (1823-1891) said: “*God made the natural numbers, all else is the work of man.*”

Unfortunately this is rather a lot to ﬁt into this epilogue so we will have to abbreviate the process rather severely.

We will assume the existence and properties of the rational numbers. Building \(\mathbb{Q}\) from the integers is not especially hard and it is easy to show that they satisfy the axioms worked out by Salviati, Sagredo and Simplicio in the previous section. But the level of detail required for rigor quickly becomes onerous.

Even starting at this fairly advanced position in the chain of logic there is still a considerable level of detail needed to complete the process. Therefore our exposition will necessarily be incomplete.

Rather than display, in full rigor, how the real numbers can be built up from the rationals we will show, in fairly broad terms, three ways this has been done in the past. We will give references later in case you’d like to follow up and learn more.

### The Decimal Expansion

This is by far the most straightforward method we will examine. Since we begin with \(\mathbb{Q}\), we already have some numbers whose decimal expansion is inﬁnite. For example, \(\frac{1}{3} = 0.333....\) We also know that if \(x ∈ \mathbb{Q}\)then expressing \(x\) as a decimal gives either a ﬁnite or a repeating inﬁnite decimal.

More simply, we can say that \(\mathbb{Q}\) consists of the set of all decimal expressions which eventually repeat. (If it eventually repeats zeros then it is what we’ve called a ﬁnite decimal.)

We then deﬁne the real numbers to be the set of all inﬁnite decimals, repeating or not.

It may feel as if all we have to do is deﬁne addition and multiplication in the obvious fashion and we are ﬁnished. This set with these deﬁnitions obviously satisfy all of the ﬁeld axioms worked out by our Italian friends in the previous section. Moreover it seems clear that all of our equivalent completeness axioms are satisﬁed.

However, things are not quite as clear cut as they seem.

The primary diﬃculty in this approach is that the decimal representation of the real numbers is so familiar that everything we need to show seems obvious. But stop and think for a moment. Is it really obvious how to deﬁne addition and multiplication of inﬁnite decimals? Consider the addition algorithm we were all taught in grade school. That algorithm requires that we line up two numbers at their decimal points:

\[d_1d_2 \cdot d_3d_4 \\ + \delta _1 \delta _2 \cdot \delta _3 \delta _4\]

We then begin adding in the rightmost column and proceed to the left. But if our decimals are inﬁnite we can’t get started because there is no rightmost column!

A similar problem occurs with multiplication.

So our ﬁrst problem is to deﬁne addition and multiplication in \(\mathbb{R}\) in a manner that re-captures addition and multiplication in \(\mathbb{Q}\).

This is not a trivial task.

One way to proceed is to recognize that the decimal notation we’ve used all of our lives is really shorthand for the sum of an inﬁnite series. That is, if \(x = 0 \cdot d_1d_2d_3 ...\) where \(0 ≤ d_i ≤ 9\) for all \(i ∈ N\) then

\[x = \sum_{i=1}^{\infty } \frac{d_i}{10^i}\]

Addition is now apparently easy to deﬁne: If \(x = \sum_{i=1}^{\infty } \frac{d_i}{10^i}\) and \(y = \sum_{i=1}^{\infty } \frac{\delta _i}{10^i}\) then

\[x + y = \sum_{i=1}^{\infty } \frac{e_i}{10^i} \text{ where } e_i = d_i + \delta _i\]

But there is a problem. Suppose for some \(j ∈ \mathbb{N}\), \(e_j = d_i + δ_i > 10\). In that case our sum does not satisfy the condition \(0 ≤ e_i ≤ 9\) so it is not even clear that the expression \(\sum_{i=1}^{\infty } \frac{e_i}{10^i}\) represents a real number. That is, we may not have the closure property of a number ﬁeld. We will have to deﬁne some sort of “*carrying*” operation to handle this.

Exercise \(\PageIndex{2}\)

Deﬁne addition on inﬁnite decimals in a manner that is closed.

**Hint**-
Find an appropriate “

*carry*” operation for our deﬁnition.

A similar diﬃculty arises when we try to deﬁne multiplication. Once we have a notion of carrying in place, we could deﬁne multiplication as just the multiplication of series. Speciﬁcally, we could deﬁne

\[\begin{align*} (0\cdot a_1a_2a_3\cdots )\cdot (0\cdot b_1b_2b_3\cdots ) &= \left ( \frac{a_1}{10} + \frac{a_2}{10^2} + \cdots \right )\cdot \left ( \frac{b_1}{10} + \frac{b_2}{10^2} + \cdots \right )\\ &= \frac{a_1b_1}{10^2} + \frac{a_1b_2 + a_2b_1}{10^3} + \frac{a_1b_3 + a_2b_2 + a_3b_1}{10^4} + \cdots \end{align*}\]

We could then convert this to a “*proper*” decimal using our carrying operation.

Again the devil is in the details to show that such algebraic operations satisfy everything we want them to. Even then, we need to worry about linearly ordering these numbers and our completeness axiom.

Another way of looking at this is to think of an inﬁnite decimal representation as a (Cauchy) sequence of ﬁnite decimal approximations. Since we know how to add and multiply ﬁnite decimal representations, we can just add and multiply the individual terms in the sequences. Of course, there is no reason to restrict ourselves to only these speciﬁc types of Cauchy sequences, as we see in our next approach.

### Cauchy Sequences

As we’ve seen, Georg Cantor began his career studying Fourier series and quickly moved on to more foundational matters in the theory of inﬁnite sets.

But he did not lose his fascination with real analysis when he moved on. Like many mathematicians of his time, he realized the need to build \(\mathbb{R}\) from \(\mathbb{Q}\). He and his friend and mentor Richard Dedekind (who’s approach we will see in the next section) both found diﬀerent ways to build \(\mathbb{R}\) from \(\mathbb{Q}\).

Cantor started with Cauchy sequences in \(\mathbb{Q}\).

That is, we consider the set of all Cauchy sequences of rational numbers. We would like to deﬁne each such sequence to be a real number. The goal should be clear. If \(\left ( s_n \right )_{n=1}^{\infty }\) is a sequence in \(\mathbb{Q}\) which converges to \(\sqrt{2}\) then we will call (\(s_n\)) the real number \(\sqrt{2}\).

This probably seems a bit startling at ﬁrst. There are a lot of numbers in (\(s_n\)) (countably inﬁnitely many, to be precise) and we are proposing putting all of them into a big bag, tying it up in a ribbon, and calling the whole thing \(\sqrt{2}\). It seems a very odd thing to propose, but recall from the discusion in the previous section that we left the concept of “*number*” undeﬁned. Thus if we can take any set of objects and deﬁne addition and multiplication in such a way that the ﬁeld axioms are satisﬁed, then those objects are legitimately numbers. To show that they are, in fact, the real numbers we will also need the completeness property.

A bag full of rational numbers works as well as anything if we can deﬁne addition and multiplication appropriately.

Our immediate problem though is not addition or multiplication but uniqueness. If we take one sequence (\(s_n\)) which converges to \(\sqrt{2}\) and deﬁne it to be \(\sqrt{2}\), what will we do with all of the other sequences that converge to \(\sqrt{2}\)?

Also, we have to be careful not to refer to any real numbers, like the square root of two for example, as we deﬁne the real numbers. This would be a circular – and thus useless – deﬁnition. Obviously though, we can refer to rational numbers, since these are the tools we’ll be using.

The solution is clear. We take all sequences of rational numbers that converge to \(\sqrt{2}\), throw them into our bag and call that \(\sqrt{2}\). Our bag is getting pretty full now.

But we need to do this without using \(\sqrt{2}\) because it is a real number. The following two deﬁnitions satisfy all of our needs.

Definition \(\PageIndex{2}\)

Let \(x = \left ( s_n \right )_{k=1}^{\infty }\) and \(y = \left (\sigma _n \right )_{k=1}^{\infty }\) be Cauchy sequences in \(\mathbb{Q}\). \(x\) and \(y\) are said to be equivalent if they satisfy the following property: For every \(ε > 0\), \(ε ∈ \mathbb{Q}\), there is a rational number \(\mathbb{N}\) such that for all \(n > N\), \(n ∈ \mathbb{N}\),

\[|s_n - σ_n| < ε\]

We will denote equivalence by writing, \(x ≡ y\).

Exercise \(\PageIndex{3}\)

Show that:

- \(x ≡ x\)
- \(x ≡ y ⇒ y ≡ x\)
- \(x ≡ y\) and \(y ≡ z ⇒ x ≡ z\)

Definition \(\PageIndex{3}\)

Every set of all equivalent Cauchy sequences deﬁnes a real number.

A very nice feature of Cantor’s method is that it is very clear how addition and multiplication should be deﬁned.

Definition \(\PageIndex{4}\)

If

\[x = \left \{ \left ( s_n \right )_{k=1}^{\infty } \mid \left ( s_n \right )_{k=1}^{\infty } \text { is Cauchy in } \mathbb{Q} \right \}\]

and

\[y = \left \{ \left ( \sigma _n \right )_{k=1}^{\infty } \mid \left ( \sigma _n \right )_{k=1}^{\infty } \text { is Cauchy in } \mathbb{Q} \right \}\]

then we deﬁne the following:

**Addition**:

\[x + y = \left \{ \left ( t_n \right )_{k=1}^{\infty } \mid t_k = s_k + \sigma _k, \forall (s_n)\epsilon \; x, \text{ and } (\sigma _n)\epsilon \; y \right \}\]

**Multiplication**:

\[x \cdot y = \left \{ \left ( t_n \right )_{k=1}^{\infty } \mid t_k = s_k \sigma _k, \forall (s_n)\epsilon \; x, \text{ and } (\sigma _n)\epsilon \; y \right \}\]

The notation used in Deﬁnition \(\PageIndex{3}\) can be diﬃcult to read at ﬁrst, but basically it says that addition and multiplication are done component-wise. However since \(x\) and \(y\) consist of all equivalent sequences we have to take every possible choice of \((s_n) ∈ x\) and \((σ_n) ∈ y\), form the \(\text {sum (product)} \left (s_n + \sigma _n \right )_{n=1}^{\infty } (\left (s_n \sigma _n \right )_{n=1}^{\infty })\) and then show that all such \text{sums (products)}\) are equivalent. Otherwise \text{addition (multiplication)}\) is not well-deﬁned: It would depend on which sequence we choose to represent \(x\) and \(y\).

Exercise \(\PageIndex{4}\)

Let \(x\) and \(y\) be real numbers in \(\mathbb{Q}\) (that is, let them be sets of equivalent Cauchy sequences). If (\(s_n\)) and (\(t_n\)) are in \(x\) and (\(σ_n\)) and (\(τ_n\)) are in \(y\) then

\[\left (s_n + t_n \right )_{n=1}^{\infty } \equiv \left (\sigma _n + \tau _n\right )_{n=1}^{\infty }\]

Theorem \(\PageIndex{1}\)

Let \(0∗\) be the set of Cauchy sequences in \(\mathbb{Q}\) which are all equivalent to the sequence \((0, 0, 0, ...)\). Then

\[0∗ + x = x\]

**Proof**-
From Problem \(\PageIndex{4}\) it is clear that in forming \(0∗ +x\) we can choose any sequence in \(0∗\) to represent \(0∗\) and any sequence in \(x\) to represent \(x\). (This is because any other choice will yield a sequence equivalent to \(0∗ + x\).)

Thus we choose \((0,0,0,...)\) to represent \(0∗\) and any element of \(x\), say \((x_1,x_2,...)\), to represent \(x\). Then

\[\begin{align*} (0,0,0,...) + (x_1,x_2,x_3,...) &= (x_1,x_2,x_3,...)\\ &= x \end{align*}\]

Since any other sequences taken from \(0∗\) and \(x\) respectively, will yield a sum equivalent to \(x\) (see Problem \(\PageIndex{3}\)) we conclude that

\[0∗ + x = x\]

Exercise \(\PageIndex{5}\)

Identify the set of equivalent Cauchy sequences, \(1∗\), such that \(1∗ \cdot x = x\).

Exercise \(\PageIndex{6}\)

Let \(x\), \(y\), and \(z\) be real numbers (equivalent sets of Cauchy sequences). Show that with addition and multiplication deﬁned as above we have:

- \(x + y = y + x\)
- \((x + y) + z = x + (y + z)\)
- \(x \cdot y = y \cdot x\)
- \((x \cdot y) \cdot z = x \cdot (y \cdot z)\)
- \(x \cdot (y + z) = x \cdot y + x \cdot z\)

Once the existence of additive and multiplicative inverses is established ^{2} the collection of all sets of equivalent Cauchy sequences, with addition and mulitiplication deﬁned as above satisfy all of the ﬁeld axioms. It is clear that they form a number ﬁeld and thus deserve to be called numbers.

However this does not necessarily show that they form \(\mathbb{R}\). We also need to show that they are complete in the sense of Chapter 7. It is perhaps not too surprising that when we build the real numbers using equivalent Cauchy sequences the most natural completeness property we can show is that if a sequence of real numbers is Cauchy then it converges.

However we are not in a position to show that Cauchy sequences in \(\mathbb{R}\) converge. To do this we would ﬁrst need to show that these sets of equivalence classes of Cauchy sequences (real numbers) are linearly ordered.

Unfortunately showing the linear ordering, while not especially hard, is time consuming. So we will again invoke the prerogatives of the teacher and brush all of the diﬃculties aside with the assertion that it is straightforward to show that the real numbers as we have constructed them in this section are linearly ordered and are complete. If you would like to see this construction in full rigor we recommend the book, *The Number System by H. A. Thurston* [16].^{3}

### Dedekind Cuts

An advantage of building the reals via Cauchy sequences in the previous section is that once we’ve identiﬁed equivalent sequences with real numbers it is very clear how addition and multiplication should be deﬁned.

On the other hand, before we can even start to understand that construction, we need a fairly strong sense of what it means for a sequence to converge and enough experience with sequences to be comfortable with the notion of a Cauchy sequence. Thus a good deal of high level mathematics must be mastered before we can even begin.

The method of “*Dedekind cuts*” ﬁrst developed by Richard Dedekind (though he just called them “*cuts*”) in his 1872 book, Continuity and the Irrational Numbers shares the advantage of the Cauchy sequence method in that, once the candidates for the real numbers have been identiﬁed, it is very clear^{4} how addition and multiplication should be deﬁned. It is also straightforward to show that most of the ﬁeld axioms are satisﬁed.

In addition, Dedekind’s method also has the advantage that very little mathematical knowledge is required to get started. This is intentional. In the preface to the ﬁrst edition of his book, Dedekind states:

This memoir can be understood by anyone possessing what is usually called common sense; no technical philosophic, or mathematical, knowledge is in the least degree required. (quoted in [5])

While he may have overstated his case a bit, it is clear that his intention was to argue from very simple ﬁrst principles just as Euclid did.

His starting point was the observation we made in Chapter 1: The rational number line is full of holes. More precisely we can “*cut*” the rational line in two distinct ways:

- We can pick a rational number, \(r\). This choice divides all other rational numbers into two classes: Those greater than \(r\) and those less than \(r\).
- We can pick one of the holes in the rational number line. In this case all of the rational fall into two classes: Those greater than the hole and those less.

But to speak of rational numbers as less than or greater than something that is not there is utter nonsense. We’ll need a better (that is, a rigorous) deﬁnition.

As before we will develop an overall sense of this construction rather than a fully detailed presentation, as the latter would be far too long to include.

Our presentation will closely follow that of Edmund Landau’s in his classic 1951 text Foundations of Analysis [7]. We do this so that if you choose to pursue this construction in more detail you will be able to follow Landau’s presentation more easily.

Definition \(\PageIndex{5}\): Dedekind Cut

A set of positive^{5} rational numbers is called a cut if

**Property:** It contains a positive rational number but does not contain all positive rational numbers.

**Property II:** Every positive rational number in the set is less than every positive rational number not in the set.

**Property III:** There is no element of the set which is greater than every other element of the set.

Given their intended audiences, Dedekind and Landau shied away from using too much notation. However, we will include the following for those who are more comfortable with the symbolism as it may help provide more perspective. Speciﬁcally the properties deﬁning a Dedekind cut \(α\) can be written as follows.

**Property I:** \(α \neq ∅\) and \(\mathbb{Q}^+ - α \neq ∅\).

**Property II:** If \(x ∈ α\) and \(y ∈ \mathbb{Q}^+ - α\), then \(x < y\). (Alternatively, if \(x ∈ α\) and \(y < x\), then \(y ∈ α\).)

**Property III:** If \(x ∈ α\), then \(∃ z ∈ α\) such that \(x < z\).

**Properties I-III** really say that Dedekind cuts are bounded open intervals of rational numbers starting at \(0\). For example, \((0,3) ∩ \mathbb{Q}^+\) is a Dedekind cut (which will eventually be the real number \(3\)). Likewise, \(\{x|x^2 < 2\} ∩ \mathbb{Q}^+\) is a Dedekind cut (which will eventually be the real number \(\sqrt{2}\)). Notice that care must be taken not to actually refer to irrational numbers in the properties as the purpose is to construct them from rational numbers, but it might help to ground you to anticipate what will happen.

Take particular notice of the following three facts:

- Very little mathematical knowledge is required to understand this deﬁnition. We need to know what a set is, we need to know what a rational number is, and we need to know that given two positive rational numbers either they are equal or one is greater.
- The language Landau uses is very precise. This is necessary in order to avoid such nonsense as trying to compare something with nothing like we did a couple of paragraphs up.
- We are only using the positive rational numbers for our construction. The reason for this will become clear shortly. As a practical matter for now, this means that the cuts we have just deﬁned will (eventually) correspond to the positive real numbers.

Definition \(\PageIndex{6}\)

Let \(α\) and \(β\) be cuts. Then we say that \(α\) is less than \(β\), and write \[α < β\] if there is a rational number in \(β\) which is not in \(α\).

Note that, in light of what we said prior to Deﬁnition \(\PageIndex{1}\) (which is taken directly from Landau), we notice the following.

Theorem \(\PageIndex{2}\)

Let \(α\) and \(β\) be cuts. Then \(α < β\) if and only if \(α ⊂ β\).

Exercise \(\PageIndex{7}\)

Prove Theorem \(\PageIndex{2}\) and use this to conclude that if \(α\) and \(β\) are cuts then exactly one of the following is true:

- \(α = β\)
- \(α < β\)
- \(β < α\)

We will need ﬁrst to deﬁne addition and multiplication for our cuts and eventually these will need to be extended of \(\mathbb{R}\) (once the non-positive reals have also been constructed). It will be necessary to show that the extended deﬁnitions satisfy the ﬁeld axioms. As you can see there is a lot to do.

As we did with Cauchy sequences and with inﬁnite decimals, we will stop well short of the full construction. If you are interested in exploring the details of Dedekind’s construction, Landau’s book [7] is very thorough and was written with the explicit intention that it would be accessible to students. In his “*Preface for the Teacher*” he says

I hope that I have written this book, after a preparation stretching over decades, in such a way that a normal student can read it in two days.

This may be stretching things. Give yourself at least a week and make sure you have nothing else to do that week.

Addition and multiplication are deﬁned in the obvious way.

Definition \(\PageIndex{7}\): Addition on cuts

Let \(α\) and \(β\) be cuts. We will denote the set \(\{x + y|x ∈ α,y ∈ β\}\) by \(α + β\).

Definition \(\PageIndex{8}\): Multiplication on cuts

Let \(α\) and \(β\) be cuts. We will denote the set \(\{xy|x ∈ α,y ∈ β\}\) by \(αβ\) or \(α \cdot β\).

If we are to have a hope that these objects will serve as our real numbers we must have closure with respect to addition and multiplication. We will show closure with respect to addition.

Theorem \(\PageIndex{3}\): Closure with Respect to Addition

If \(α\) and \(β\) are cuts then \(α + β\) is a cut.

**Proof**-
We need to show that the set \(α + β\) satisﬁes all three of the properties of a cut.

**Proof of Property I**Let \(x\) be any rational number in \(α\) and let \(x_1\) be a rational number not in \(α\). Then by Property II \(x < x_1\).

Let \(y\) be any rational number in \(β\) and let \(y_1\) be a rational number not in \(β\). Then by Property II \(y < y_1\).

Thus since \(x + y\) represents a generic element of \(α + β\) and \(x + y < x_1 + y_1\), it follows that \(x_1 + y_1 \not{∈} α + β\).

**Proof of Property II**We will show that the contrapositive of Property II is true: If \(x ∈ α + β\) and \(y < x\) then \(y ∈ α + β\).

First, let \(x ∈ α + β\). Then there are \(x_α ∈ α\) and \(x_β ∈ β\) such that \(y < x = x_α + x_β\). Therefore \(\frac{y}{x_\alpha + x_\beta } < 1\), so that

\[x_\alpha \left (\frac{y}{x_\alpha + x_\beta } \right ) < x_\alpha\]

and

\[x_\beta \left (\frac{y}{x_\alpha + x_\beta } \right ) < x_\beta\]

Therefore \(x_\alpha \left (\frac{y}{x_\alpha + x_\beta } \right ) \epsilon \; \alpha\) and \(x_\beta \left (\frac{y}{x_\alpha + x_\beta } \right ) \epsilon \; \beta\).

Therefore

\[y = x_\alpha \left (\frac{y}{x_\alpha + x_\beta } \right ) + x_\beta \left (\frac{y}{x_\alpha + x_\beta } \right ) \epsilon \; \alpha +\beta\]

**Proof of Property III**Let \(z ∈ α + β\). We need to ﬁnd \(w > z\), \(w ∈ α + β\). Observe that for some \(x ∈ α\) and \(y ∈ β\)

\[z = x + y\]

Since \(α\) is a cut, there is a rational number \(x_1 ∈ α\) such that \(x_1 > x\). Take \(w = x_1 + y ∈ α + β\). Then

\[w = x1 + y > x + y = z\]

This completes the proof of this theorem.

Exercise \(\PageIndex{8}\)

Show that if \(α\) and \(β\) are cuts then \(α \cdot β\) is also a cut.

At this point we have built our cuts and we have deﬁned addition and multiplication for cuts. However, as observed earlier the cuts we have will (very soon) correspond only to the positive real numbers. This may appear to be a problem but it really isn’t because the non-positive real numbers can be deﬁned in terms of the positives, that is, in terms of our cuts. We quote from Landau [7]:

These cuts will henceforth be called the “*positive numbers;*” .

We create a new number \(0\) (to be read “*zero*”), distinct from the positive numbers.

We also create numbers which are distinct from the positive numbers as well as distinct from zero, and which we will call negative numbers, in such a way that to each \(ξ\) (I.e. to each positive number) we assign a negative number denoted by \(-ξ\) (− to be read “*minus*”). In this, \(-ξ\) and \(-ν\) will be considered as the same number (as equal) if and only if \(ξ\) and \(ν\) are the same number.

The totality consisting of all positive numbers, of \(0\), and of all negative numbers, will be called the real numbers.

Of course it is not nearly enough to simply postulate the existence of the non-negative real numbers. All we have so far is a set of objects we’re calling the real numbers.

For some of them (the positive reals^{6}) we have deﬁned addition and multiplication. These deﬁnitions will eventually turn out to correspond to the addition and multiplication we are familiar with.

However we do not have either operation for our entire set of proposed real numbers. Before we do this we need ﬁrst to deﬁne the absolute value of a real number. This is a concept you are very familiar with and you have probably seen the following deﬁnition: Let \(α ∈ \mathbb{R}\). Then

\[\left | \alpha \right | = \begin{cases} \alpha & \text{ if } \alpha \geq 0\\ -\alpha & \text{ if } \alpha < 0 \end{cases}\]

Unfortunately we cannot use this deﬁnition because we do not yet have a linear ordering on \(\mathbb{R}\) so the statement \(α ≥ 0\) is meaningless. Indeed, it will be our deﬁnition of absolute value that orders the real numbers. We must be careful.

Notice that by deﬁnition a negative real number is denoted with the dash (’**-**’) in front. That is \(χ\) is positive while \(-χ\) is negative. Thus if \(A\) is any real number then one of the following is true:

- \(A = χ\) for some \(χ ∈ \mathbb{R}\) (\(A\) is positive)
- \(A = -χ\) for some \(χ ∈ \mathbb{R}\) (\(A\) is negative)
- \(A = 0\)

We deﬁne absolute value as follows:

Definition \(\PageIndex{9}\)

Let \(A ∈ \mathbb{R}\) as above. Then

\[\left | A \right | = \begin{cases} \chi & \text{ if } A = \chi \\ 0 & \text{ if } A = 0\\ \chi & \text{ if } A = -\chi \end{cases}\]

With this deﬁnition in place it is possible to show that \(mathbb{R}\) is linearly ordered. We will not do this explicitly. Instead we will simply assume that the symbols “\(<\)” “\(>\),” and “\(=\)” have been deﬁned and have all of the properties we have learned to expect from them.

We now extend our deﬁnitions of addition and multiplication from the positive real numbers (cuts) to all of them. Curiously, multiplication is the simpler of the two.

Definition \(\PageIndex{10}\): Multiplication

Let \(α\), \(β ∈ \mathbb{R}\). Then

\[\alpha \cdot \beta = \begin{cases} -\left | \alpha \right | \left | \beta \right | & \text{ if } \alpha > 0, \beta < 0 \text{ or } \alpha < 0, \beta > 0 \\ \left | \alpha \right | \left | \beta \right | & \text{ if } \alpha < 0, \beta > 0\\ 0 & \text{ if } \alpha = 0 \text{ or } \beta = 0 \end{cases}\]

Notice that the case where \(α\) and \(β\) are both positive was already handled by Deﬁnition \(\PageIndex{8}\) because in that case they are both cuts.

Next we deﬁne addition.

Definition \(\PageIndex{11}\): Addition

Let \(α\), \(β ∈ \mathbb{R}\). Then

\[\alpha + \beta = \begin{cases} -(\left | \alpha \right | + \left | \beta \right |) & \text{ if } \alpha < 0, \beta < 0 \\ \left | \alpha \right | - \left | \beta \right | & \text{ if } \alpha > 0, \beta < 0, \left | \alpha \right | > \left | \beta \right |\\ 0 & \text{ if } \alpha > 0, \beta < 0, \left | \alpha \right | = \left | \beta \right |\\ -(\left | \alpha \right | - \left | \beta \right |) & \text{ if } \alpha > 0, \beta < 0,\left | \alpha \right | < \left | \beta \right | \\ \beta +\alpha & \text{ if } \alpha < 0, \beta > 0 \\ \beta & \text{ if } \alpha = 0 \\ \alpha & \text{ if } \beta = 0 \end{cases}\]

But wait! In the second and fourth cases of our deﬁnition we’ve actually deﬁned addition in terms of subtraction. ^{7} But we haven’t deﬁned subtraction yet! Oops!

This is handled with the deﬁnition below, but it illuminates very clearly the care that must be taken in these constructions. The real numbers are so familiar to us that it is extraordinarily easy to make unjustiﬁed assumptions.

Since the subtractions in the second and fourth cases above are done with positive numbers we only need to give meaning to the subtraction of cuts.

Definition \(\PageIndex{12}\)

If \(α\), \(β\) and \(δ\) are cuts then the expression

\[α−β = δ\]

is deﬁned to mean

\[α = δ + β\]

Of course, there is the detail of showing that there is such a cut \(δ\). (We warned you of the tediousness of all this.) Landau goes through the details of showing that such a cut exists. We will present an alternative by deﬁning the cut \(α - β\) directly (assuming \(β < α\)). To motivate this deﬁnition, consider something we are familiar with: \(3 - 2 = 1\). In terms of cuts, we want to say that the open interval from \(0\) to \(3\) “*minus*” the open interval from \(0\) to \(2\) should give us the open interval from \(0\) to \(1\). Taking elements from \((0,3)\) and subtracting elements from \((0,2)\) won’t do it as we would have diﬀerences such as \(2.9 - 0.9 = 2\) which is not in the cut \((0,1)\). A moment’s thought tells us that what we need to do is take all the elements from \((0,3)\) and subtract all the elements from \((2,∞)\), restricting ourselves only to those which are positive rational numbers. This prompts the following deﬁnition.

Definition \(\PageIndex{13}\)

Let \(α\) and \(β\) be cuts with \(β < α\). Deﬁne \(α - β\) as follows:

\[α - β = \{x - y|x ∈ α \text{ and } y \not{∈} β\} ∩ Q^+\]

To show that, in fact, \(β + (α - β) = α\), the following technical lemma will be helpful.

Lemma \(\PageIndex{1}\)

Let \(β\) be a cut, \(y\) and \(z\) be positive rational numbers not in \(β\) with \(y < z\), and let \(ε > 0\) be any rational number. Then there exist positive rational numbers \(r\) and \(s\) with \(r ∈ β\), and \(s \not{∈} β\), such that \(s < z\), and \(s - r < ε\).

Exercise \(\PageIndex{9}\)

Prove Lemma \(\PageIndex{1}\).

**Hint**-
Since \(β\) is a cut there exists \(r_1 ∈ β\). Let \(s_1 = y \not{∈} β\). We know that \(r_1 < s_1 < z\). Consider the midpoint \(\frac{s_1+r_1}{2}\). If this is in \(β\) then relabel it as \(r_2\) and relabel \(s_1\) as \(s_2\). If it is not in \(β\) then relabel it as \(s_2\) and relabel \(r_1\) as \(r_2\), etc.

Exercise \(\PageIndex{10}\)

Let \(α\) and \(β\) be cuts with \(β < α\). Prove that \(β + (α - β) = α\).

**Hint**-
It is pretty straightforward to show that \(β + (α - β) ⊆ α\). To show that \(α ⊆ β + (α - β)\), we let \(x ∈ α\). Since \(β < α\), we have \(y ∈ α\) with \(y \not{∈} β\). We can assume without loss of generality that \(x < y\). (Why?) Choose \(z ∈ α\) with \(y < z\). By the Lemma \(\PageIndex{1}\), there exists positive rational numbers \(r\) and \(s\) with \(r ∈ β\), \(s ∈ β\), \(s < z\), and \(s - r < z - x\). Show that \(x < r + (z - s)\).

We will end by saying that no matter how you construct the real number system, there is really only one. More precisely we have the following theorem which we state without proof.^{8}

Theorem \(\PageIndex{4}\)

Any complete, linearly ordered ﬁeld is isomorphic^{9} to \(\mathbb{R}\).

Remember that we warned you that these constructions were fraught with technical details that are not necessarily illuminating. Nonetheless, at this point, you have everything you need to show that the set of all real numbers as deﬁned above is linearly ordered and satisﬁes the Least Upper Bound property.

But we will stop here in order, to paraphrase Descartes, to leave for you the joy of further discovery.

### References

^{1 }By Andrew Wiles, the man who proved Fermat’s Last Theorem.

^{2 }We will not address this issue here, but you should give some thought to how this might be accomplished.

^{3 }Thurston ﬁrst builds R as we’ve indicated in this section. Then as a ﬁnal remark he shows that the real numbers must be exactly the inﬁnite decimals we saw in the previous section.

^{4 }“Clear” does not mean “easy to do” as we will see.

^{5 }Take special notice that we are not using the negative rational numbers or zero to build our cuts. The reason for this will become clear shortly.

^{6 }That is, the cuts.

^{7 }Notice also that the ﬁfth case refers to the addition as deﬁned in the second case.

^{8 }In fact, not proving this result seems to be standard in real analysis references. Most often it is simply stated as we’ve done here. However a proof can be found at http://math.ucr.edu/ res/math205A/uniqreals.pdf.

^{9 }Two linearly ordered number ﬁelds are said to be isomorphic if there is a one-to-one, onto mapping between them (such a mapping is called a bijection) which preserves addition, multiplication, and order. More precisely, if \(\mathcal{F}_1\) and \(\mathcal{F}_2\) are both linearly ordered ﬁelds, \(x\), \(y ∈ \mathcal{F}_1\) and \(\varphi : \mathcal{F}_1 \rightarrow \mathcal{F}_2\) is the mapping then

- \(φ(x + y) = φ(x) + φ(y)\)
- \(φ(x \cdot y) = φ(x) \cdot φ(y)\)
- \(x < y ⇒ φ(x) < φ(y)\)

### Contributor

Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)