Course 2 Book 2A: Numbers, Limits and the IntegralChapter 4

The Real Numbers

Completing ℚ with Cauchy sequences: the least upper bound property and the idea of completion.

37 min read · Updated Oct 2, 2026

Read with Tao, Analysis I, chapter "The real numbers" (Cauchy sequences, equivalent Cauchy sequences, the construction of the reals, ordering, the least upper bound property, real exponentiation part I).

In this chapter · 10 sections

4.1Approximations, ancient and modern
4.2Cauchy sequences
4.3When two sequences approximate the same number
4.4The real numbers
4.4.1Arithmetic
4.5Order
4.6The least upper bound property
4.6.1√2 exists
4.7Completion
4.8Computers don't use real numbers
4.9History: two ways to fill the gaps
4.10Exercises

The rationals are dense, but 2A.3 Integers and Rationals found a hole in them at $\sqrt 2$ . There are rationals whose squares come as close to $2$ as you like, yet none whose square is $2$ . This chapter fills every such hole at once and builds the real numbers $\mathbb{R}$ .

The idea is simple to say. A real number is what an endless process of better and better rational approximations is approximating. The work is in making that precise without referring to the thing being approximated, which doesn't exist yet. The answer combines the two tools of the last two chapters: sequences that settle down (Cauchy sequences), and the quotient construction, so that two processes approximating "the same number" count as the same real.

This construction is called completion. It is the single most reused construction in this guidebook, so the chapter ends by naming it explicitly. The function spaces in which the heat equation and Ricci flow are solved (3A.3 The Lebesgue Integral, 4A.9 Sobolev Spaces) are built in exactly the same way.

By the end of this chapter you will be able to:

define Cauchy sequences and equivalent sequences of rationals, and test examples;
explain how the real numbers are constructed and why the operations are well defined;
state and prove the least upper bound property, and use it to show that $\sqrt 2$ exists;
explain what "completion" means and recognise it later;
explain why the numbers inside a computer are not real numbers, with a documented case where the difference mattered.

Approximations, ancient and modern

People computed with irrational quantities long before anyone could say what they were. They did it the only way possible: with rational approximations good enough for the task.

In the world Data √2 on a clay tablet

The Babylonian tablet known as YBC 7289, in the Yale Babylonian Collection and dated to roughly 1800–1600 BCE, shows a square with its diagonals. Along a diagonal is written, in base-60 notation, the number $1;24,51,10$ , meaning

1 + \frac{24}{60} + \frac{51}{60^2} + \frac{10}{60^3} = \frac{30547}{21600} = 1.41421296\ldots

The true value is $\sqrt 2 = 1.41421356\ldots$ , so the scribe's figure is correct to about six decimal places, an error of less than one part in two million. It is the closest approximation to $\sqrt 2$ possible with three sexagesimal places after the point.

How might such an approximation be found? One method, which may well be the one the scribes used (the tablets don't say), is divide and average. If $x$ is a guess for $\sqrt 2$ that is too big, then $2/x$ is too small, and their average is a better guess:

x_{n+1} = \frac12\Big(x_n + \frac{2}{x_n}\Big).

Starting from $x_0 = 2$ :

x_1 = \tfrac32,\qquad x_2 = \tfrac{17}{12},\qquad x_3 = \tfrac{577}{408},\qquad x_4 = \tfrac{665857}{470832}.

Every one of these is rational, and their squares minus $2$ are $\tfrac14$ , about $0.0069$ , about $0.000006$ , and about $0.0000000000045$ . The number of correct digits roughly doubles at each step. This is Newton's method for the equation $x^2 - 2 = 0$ (Figure 4.1), and Newton iteration is still one of the standard ways software computes square roots and reciprocals.

Figure 4.1. Divide-and-average is Newton's method: follow the tangent to

y = x^2 - 2

down to the axis. From

x_0 = 2

the tangents land at

\tfrac32

, then

\tfrac{17}{12}

, then

\tfrac{577}{408}

, each rational and each closer to the crossing point, which is not rational.

The sequence $x_0, x_1, x_2, \dots$ is a sequence of rational numbers that "wants" to converge, but there is no rational number for it to converge to. The construction of the reals takes that literally: the real number $\sqrt 2$ will be (the class of) this sequence. Two things have to be made precise first: what it means for a sequence to want to converge without mentioning its limit, and when two such sequences want to converge to the same place.

Cauchy sequences

A sequence of rationals is a function $n \mapsto a_n$ from the natural numbers (or from the naturals at least some starting index) to $\mathbb{Q}$ . We write it $(a_n)_{n=0}^\infty$ , or just $(a_n)$ .

The obvious definition of convergence, " $a_n$ gets close to $L$ ", needs the limit $L$ . The trick, due to Cauchy, is to ask only that the terms get close to each other.

Definition 4.1 Cauchy sequence

A sequence of rationals $(a_n)$ is a Cauchy sequence if for every rational $\varepsilon > 0$ there is an $N$ such that $|a_n - a_m| \leq \varepsilon$ for all $n, m \geq N$ .

In words: however small a tolerance you name, from some point on, all the terms are within that tolerance of each other. Tao builds this up in two steps, calling a sequence $\varepsilon$ -steady if all its terms are $\varepsilon$ -close, and eventually $\varepsilon$ -steady if all terms from some point on are. A Cauchy sequence is then one that is eventually $\varepsilon$ -steady for every $\varepsilon > 0$ .

Example 4.2 Which sequences are Cauchy?

Decimal truncations. Let $a_n$ be $\sqrt 2$ truncated to $n$ decimal places: $1, 1.4, 1.41, 1.414, \dots$ . (These are rational numbers, computable without knowing $\sqrt 2$ exists: $a_n$ is the largest number with $n$ decimal places whose square is less than $2$ .) For $n, m \geq N$ the two truncations agree in the first $N$ decimals, so $|a_n - a_m| \leq 10^{-N}$ . Given $\varepsilon$ , choose $N$ with $10^{-N} \leq \varepsilon$ . Cauchy.
$a_n = 1/n$ (for $n \geq 1$ ). For $n, m \geq N$ , both terms lie in $(0, 1/N]$ , so they differ by at most $1/N$ . Cauchy.
$a_n = (-1)^n$ . Any two consecutive terms differ by $2$ , so the definition fails for $\varepsilon = 1$ . Not Cauchy.
$a_n = 1 + \tfrac12 + \dots + \tfrac1n$ . Consecutive terms differ by only $1/n$ , which tends to $0$ , so this sequence looks Cauchy. It isn't: $a_{2m} - a_m \geq m \cdot \tfrac{1}{2m} = \tfrac12$ for every $m$ . Small consecutive differences are not enough; all pairs beyond $N$ must be close. (2A.7 Series returns to this, the harmonic series.)

Figure 4.2. Two Cauchy sequences of rationals: the decimal truncations of

\sqrt 2

(from below) and the divide-and-average iterates (from above). Beyond each index

N

, all later terms fit in a band whose width shrinks to

0

. That is the Cauchy property. Neither sequence has a rational limit.

Newton's iterates are Cauchy as well, and the proof never mentions $\sqrt 2$ , which is the point.

Proposition 4.3 The divide-and-average sequence is Cauchy

Let $x_0 = 2$ and $x_{n+1} = \tfrac12(x_n + 2/x_n)$ . Then $(x_n)$ is a Cauchy sequence of rationals.

Proof. Write $e_n = x_n^2 - 2$ . A direct computation gives

x_{n+1}^2 - 2 = \frac14\Big(x_n^2 + 4 + \frac{4}{x_n^2}\Big) - 2 = \frac{(x_n^2 - 2)^2}{4x_n^2}, \qquad\text{so}\qquad e_{n+1} = \frac{e_n^2}{4x_n^2}.

Since $e_0 = 2 > 0$ , induction gives $e_n > 0$ for all $n$ , so every $x_n^2 > 2$ , and hence $x_n > 1$ . Then $e_{n+1} \leq e_n^2/4$ , and $x_{n+1} - x_n = (2/x_n - x_n)/2 = -e_n/(2x_n) < 0$ : the sequence is decreasing.

Now let $m \geq n$ . Since $x_m \leq x_n$ and $x_m^2 > 2$ , we have $x_m > 2/x_m \geq 2/x_n$ . So $x_m$ lies between $2/x_n$ and $x_n$ , and

|x_n - x_m| \leq x_n - \frac{2}{x_n} = \frac{e_n}{x_n} \leq e_n.

Finally $e_1 = \tfrac14$ and $e_{n+1} \leq e_n^2/4$ force $e_n \leq 4^{-n}$ for $n \geq 1$ (by induction: $e_{n+1} \leq 4^{-2n}/4 \leq 4^{-(n+1)}$ ). Given $\varepsilon > 0$ , choose $N \geq 1$ with $4^{-N} \leq \varepsilon$ ; then $|x_n - x_m| \leq \varepsilon$ for all $m \geq n \geq N$ .

One property of Cauchy sequences is needed constantly.

Lemma 4.4 Cauchy sequences are bounded

If $(a_n)$ is Cauchy, there is a rational $M$ with $|a_n| \leq M$ for all $n$ .

Proof. Take $\varepsilon = 1$ : there is $N$ with $|a_n - a_N| \leq 1$ for all $n \geq N$ , so $|a_n| \leq |a_N| + 1$ for those $n$ . The finitely many terms $a_0, \dots, a_{N-1}$ have a largest absolute value. Let $M$ be the larger of that and $|a_N| + 1$ .

When two sequences approximate the same number

The decimal truncations and the Newton iterates are different sequences, but they are approximating the same thing. The sequences $1, 1, 1, \dots$ and $0.9, 0.99, 0.999, \dots$ are different too, yet both "are" the number $1$ . We need an equivalence relation that identifies them.

Definition 4.5 Equivalent sequences

Two sequences of rationals $(a_n)$ and $(b_n)$ are equivalent if for every rational $\varepsilon > 0$ there is an $N$ such that $|a_n - b_n| \leq \varepsilon$ for all $n \geq N$ .

So $0.9, 0.99, 0.999, \dots$ is equivalent to the constant sequence $1, 1, 1, \dots$ , because the difference at stage $n$ is $10^{-n}$ . That is the precise content of the slogan $0.999\ldots = 1$ : the two decimal expansions are different sequences in the same equivalence class.

Lemma 4.6 This is an equivalence relation

Equivalence of sequences is reflexive, symmetric and transitive.

Proof. Reflexivity and symmetry are immediate. For transitivity, suppose $(a_n) \sim (b_n)$ and $(b_n) \sim (c_n)$ , and let $\varepsilon > 0$ . Apply the definitions with $\varepsilon/2$ : there are $N_1$ and $N_2$ with $|a_n - b_n| \leq \varepsilon/2$ for $n \geq N_1$ and $|b_n - c_n| \leq \varepsilon/2$ for $n \geq N_2$ . For $n$ at least the larger of $N_1$ and $N_2$ , the triangle inequality (2A.3 Integers and Rationals) gives $|a_n - c_n| \leq \varepsilon$ .

The " $\varepsilon/2$ trick" in this proof, splitting a tolerance between two steps and recombining with the triangle inequality, will be used hundreds of times in this guidebook. It is the rigorous version of the closeness computation in the rehearsal exercise of 2A.3 Integers and Rationals.

The real numbers

Definition 4.7 Real numbers

A real number is an expression $\operatorname{LIM}_{n\to\infty} a_n$ , where $(a_n)$ is a Cauchy sequence of rationals. Two real numbers $\operatorname{LIM} a_n$ and $\operatorname{LIM} b_n$ are equal exactly when $(a_n)$ and $(b_n)$ are equivalent. Formally, $\mathbb{R}$ is the set of Cauchy sequences of rationals modulo equivalence.

The notation $\operatorname{LIM}$ (a "formal limit") is Tao's. It is a reminder that at this stage nothing converges to anything: $\operatorname{LIM} a_n$ is just a name for the class of $(a_n)$ . In 2A.6 Sequences we will prove that $a_n$ really does converge to the real number $\operatorname{LIM} a_n$ , and then $\operatorname{LIM}$ can be replaced by the ordinary $\lim$ .

The rationals sit inside the reals. A rational $q$ is identified with the constant sequence, $\operatorname{LIM} q$ . Two rationals are equal as reals exactly when they are equal as rationals, so nothing is lost. So $\sqrt 2$ , if it exists, is $\operatorname{LIM} x_n$ for the Newton sequence, and also $\operatorname{LIM}$ of the decimal truncations, since those two sequences are equivalent (Exercise 4.19).

Arithmetic

Definition 4.8 Operations on reals

\operatorname{LIM} a_n + \operatorname{LIM} b_n := \operatorname{LIM}(a_n + b_n), \qquad \operatorname{LIM} a_n \times \operatorname{LIM} b_n := \operatorname{LIM}(a_n b_n).

Two things must be checked, as for every definition on a quotient (2A.2 Sets, Functions and Equivalence). First, the right-hand sides must be Cauchy sequences, so that they define reals at all. Second, the operations must be well defined: replacing $(a_n)$ by an equivalent sequence must give an equivalent result.

Proposition 4.9 Multiplication of reals is well defined

If $(a_n)$ and $(b_n)$ are Cauchy, then so is $(a_n b_n)$ . If moreover $(a_n) \sim (a'_n)$ , then $(a_n b_n) \sim (a'_n b_n)$ .

Proof. The key identity is $a_n b_n - a_m b_m = a_n(b_n - b_m) + b_m(a_n - a_m)$ . By Lemma 4.4 there is $M$ with $|a_n| \leq M$ and $|b_n| \leq M$ for all $n$ . Given $\varepsilon$ , choose $N$ so that $|a_n - a_m|$ and $|b_n - b_m|$ are at most $\varepsilon/(2M)$ for $n, m \geq N$ . Then $|a_n b_n - a_m b_m| \leq M\frac{\varepsilon}{2M} + M\frac{\varepsilon}{2M} = \varepsilon$ .

For well-definedness, $a_n b_n - a'_n b_n = (a_n - a'_n)b_n$ , and $|b_n| \leq M$ ; choose $N$ with $|a_n - a'_n| \leq \varepsilon/M$ for $n \geq N$ .

Notice why boundedness was needed: a product is close to another product only if the factors that are not being compared are under control. This "split the difference, bound the other factor" step is the template for every product estimate in analysis, including the estimates for Ricci flow, where curvature is multiplied by curvature (11A.2 How Curvature Evolves).

Negation is $-\operatorname{LIM} a_n = \operatorname{LIM}(-a_n)$ , and subtraction is $x - y = x + (-y)$ . The laws of arithmetic (commutativity, associativity, distributivity) hold because they hold term by term for rationals.

Reciprocals need one more idea. If $x = \operatorname{LIM} a_n$ is not zero, some terms $a_n$ may still be zero, so $\operatorname{LIM}(1/a_n)$ makes no sense as written. The way out is to choose a better representative.

Lemma 4.10 Nonzero reals are bounded away from zero

If $x \neq 0$ , then $x = \operatorname{LIM} a_n$ for some Cauchy sequence $(a_n)$ and some rational $c > 0$ with $|a_n| \geq c$ for every $n$ .

Proof. Take any representative $(b_n)$ . Since $x \neq 0$ , $(b_n)$ is not equivalent to the zero sequence, so there is some $\varepsilon_0 > 0$ such that $|b_n| > \varepsilon_0$ for infinitely many $n$ . Choose $N$ with $|b_n - b_m| \leq \varepsilon_0/2$ for $n, m \geq N$ , and then some $n_0 \geq N$ with $|b_{n_0}| > \varepsilon_0$ . For every $n \geq N$ , $|b_n| \geq |b_{n_0}| - \varepsilon_0/2 > \varepsilon_0/2$ . Now let $a_n = b_N$ for $n < N$ and $a_n = b_n$ for $n \geq N$ . This sequence is equivalent to $(b_n)$ (they agree from $N$ on), and $|a_n| \geq \varepsilon_0/2 =: c$ for every $n$ .

For such a representative, $(1/a_n)$ is Cauchy, because $|1/a_n - 1/a_m| = |a_m - a_n|/|a_n a_m| \leq |a_n - a_m|/c^2$ . Define $x^{-1} := \operatorname{LIM}(1/a_n)$ . It doesn't depend on the representative chosen (Exercise 4.20), and $x \cdot x^{-1} = 1$ .

Proposition 4.11

\mathbb{R}

is a field

With these operations, the real numbers satisfy all the laws of a field, and the inclusion $\mathbb{Q} \subseteq \mathbb{R}$ respects addition and multiplication.

Order

Definition 4.12 Positive reals; order

A real number $x$ is positive if $x = \operatorname{LIM} a_n$ for a Cauchy sequence with $a_n \geq c$ for every $n$ , for some rational $c > 0$ , and negative if $-x$ is positive. Write $x > y$ if $x - y$ is positive.

Proposition 4.13 Trichotomy and order

Every real number is exactly one of positive, zero or negative. With this order, $\mathbb{R}$ is an ordered field, and its order agrees with the order of $\mathbb{Q}$ on rationals.

The proof combines Lemma 4.10 with the Cauchy property: once a sequence is bounded away from zero, it eventually keeps one sign, because its terms are eventually within $c/2$ of each other (Exercise 4.21).

Absolute value and distance on $\mathbb{R}$ are defined as for $\mathbb{Q}$ , and the triangle inequality carries over. Two properties connect $\mathbb{R}$ to the rationals it was built from.

Proposition 4.14 Archimedean property; density of

\mathbb{Q}

\mathbb{R}

For every real $x$ there is a natural number $N > x$ . For every real $\varepsilon > 0$ there is a natural number $n$ with $1/n < \varepsilon$ .
Between any two reals $x < y$ there is a rational $q$ with $x < q < y$ .

Proof. (1) Write $x = \operatorname{LIM} a_n$ . The sequence is bounded by some rational $M$ (Lemma 4.4), so $x \leq M$ ; by 2A.3 Integers and Rationals there is a natural number $N > M$ . The second statement follows from the first applied to $1/\varepsilon$ .

(2) By (1) choose $n$ with $1/n < y - x$ . Then the rationals $k/n$ , for integers $k$ , are spaced less than $y - x$ apart. Let $k$ be the smallest integer with $k/n > x$ (it exists by (1) and well-ordering). Then $(k-1)/n \leq x$ , so $k/n \leq x + 1/n < y$ .

So every real number can be approximated by rationals to any accuracy. This is what makes the reals usable in practice: in any computation, a real can be replaced by a rational close enough to it.

The least upper bound property

The defining feature of $\mathbb{R}$ , the property $\mathbb{Q}$ lacks, can be stated without sequences at all.

Definition 4.15 Upper bounds and suprema

Let $E$ be a set of reals. A real $M$ is an upper bound for $E$ if $x \leq M$ for every $x \in E$ . A real $S$ is a least upper bound, or supremum, of $E$ , written $\sup E$ , if $S$ is an upper bound for $E$ and $S \leq M$ for every upper bound $M$ of $E$ .

A set has at most one supremum (if $S$ and $S'$ are both least upper bounds, then $S \leq S'$ and $S' \leq S$ ). The supremum need not belong to the set: $\sup\{x : x < 1\} = 1$ .

Theorem 4.16 The least upper bound property

Every nonempty set of real numbers that has an upper bound has a least upper bound.

Proof (By bisection). Let $E$ be nonempty, $x_0 \in E$ , and $M$ an upper bound. We build the supremum as the formal limit of a Cauchy sequence of rationals, by repeatedly halving an interval that contains it.

Fix a positive integer $n$ . Among the rationals $k/n$ with $k$ an integer, some are upper bounds for $E$ (those at least $M$ ) and some aren't (those below $x_0$ ). By well-ordering, there is a smallest integer $K_n$ such that $K_n/n$ is an upper bound. Let $s_n = K_n/n$ . So $s_n$ is an upper bound, but $s_n - 1/n$ is not.

Claim: $(s_n)$ is Cauchy. For $n, m \geq N$ : $s_n$ is an upper bound and $s_m - 1/m$ is not, so some element of $E$ exceeds $s_m - 1/m$ , and that element is at most $s_n$ . Hence $s_m - 1/m < s_n$ . By symmetry $s_n - 1/n < s_m$ . So $|s_n - s_m| < \max(1/n, 1/m) \leq 1/N$ .

Let $S = \operatorname{LIM} s_n$ . $S$ is an upper bound: if some $x \in E$ had $x > S$ , then since $s_n - S \to 0$ (Exercise 4.22) we would get $s_n < x$ for large $n$ , contradicting that $s_n$ is an upper bound. $S$ is the least: if $M'$ were an upper bound with $M' < S$ , then for large $n$ , $s_n - 1/n > M'$ (since $s_n - 1/n$ also converges to $S$ ), so $s_n - 1/n$ would be an upper bound, contradicting the choice of $K_n$ .

Figure 4.3. Finding the supremum: on grids of spacing

1/n

, take the smallest grid point that is still an upper bound. Each such point is within

1/n

of the boundary of

E

, so they form a Cauchy sequence, and its formal limit is

\sup E

Over the rationals the theorem fails. The set $E = \{q \in \mathbb{Q} : q^2 < 2\}$ has rational upper bounds ( $2$ , $\tfrac32$ , $\tfrac{17}{12}$ , …), but no least rational upper bound, since any rational upper bound $u$ has $u^2 > 2$ (it can't be $2$ by 2A.3 Integers and Rationals), and then the divide-and-average step produces a smaller one. The least upper bound property is precisely the statement that the reals have no gaps.

√2 exists

Theorem 4.17 Square roots exist

There is a unique positive real number $x$ with $x^2 = 2$ . More generally, for every real $y \geq 0$ and positive integer $n$ there is a unique real $x \geq 0$ with $x^n = y$ , written $y^{1/n}$ .

Proof (For √2). Let $E = \{x \in \mathbb{R} : x \geq 0 \text{ and } x^2 < 2\}$ . It contains $1$ and is bounded above by $2$ (if $x > 2$ then $x^2 > 4$ ). Let $s = \sup E$ ; then $1 \leq s \leq 2$ . We rule out $s^2 < 2$ and $s^2 > 2$ .

If $s^2 < 2$ : for $0 < \delta < 1$ , $(s + \delta)^2 = s^2 + 2s\delta + \delta^2 \leq s^2 + 5\delta$ , using $s \leq 2$ and $\delta^2 < \delta$ . Choosing $\delta$ with $5\delta < 2 - s^2$ gives $(s + \delta)^2 < 2$ , so $s + \delta \in E$ , contradicting that $s$ is an upper bound.

If $s^2 > 2$ : for $0 < \delta < s$ , $(s - \delta)^2 \geq s^2 - 2s\delta \geq s^2 - 4\delta$ . Choosing $\delta$ with $4\delta < s^2 - 2$ gives $(s - \delta)^2 > 2$ , so every $x \in E$ satisfies $x < s - \delta$ (since $x^2 < 2 < (s-\delta)^2$ ), making $s - \delta$ a smaller upper bound, a contradiction.

So $s^2 = 2$ . Uniqueness: if $0 \leq x < y$ then $x^2 < y^2$ , so two different non-negative reals can't have the same square.

With $n$ -th roots in hand, rational powers $y^{p/q} := (y^{1/q})^p$ are defined for $y > 0$ , and the usual laws hold (Exercise 4.24). Powers with irrational exponents, such as $2^{\sqrt 2}$ , need limits and are defined in 2A.6 Sequences (Tao's "real exponentiation, part II").

Completion

The idea Completion

The construction in this chapter has a shape that recurs again and again:

Start with a space that has a notion of distance but has holes: here $\mathbb{Q}$ with $|x - y|$ .
Take the sequences that "want to converge": the Cauchy sequences.
Identify two of them when their distance tends to zero.
The resulting quotient is a complete space, one in which every Cauchy sequence converges, and it contains the original space as a dense subset.

This is called the completion. The reals are the completion of the rationals.

That last property, completeness, is proved in 2A.6 Sequences: every Cauchy sequence of real numbers converges to a real number. Completing a second time produces nothing new. Completeness is what makes existence theorems possible, because to show something exists, it is enough to build a sequence of better and better approximations and check that it is Cauchy. The approximations needn't converge to anything you can write down.

Figure 4.4. The same construction twice. Top: the rationals completed to the reals (this chapter). Bottom, a preview: smooth functions, completed with respect to an energy distance, give a Sobolev space (4A.9 Sobolev Spaces), the setting in which the PDE of later courses are solved.

Where this goes Where completion returns

3A.3 The Lebesgue Integral: integrable functions. The Riemann-integrable functions, with the distance $\int |f - g|$ , have holes (the limit of a Cauchy sequence of them may not be Riemann integrable). Completing gives the Lebesgue space $L^1$ .
4A.1 Banach Spaces and Bounded Operators: Banach spaces are, by definition, complete normed spaces.
4A.9 Sobolev Spaces: Sobolev spaces, completions of smooth functions under norms that measure derivatives. Weak solutions of PDE live here.
2B.2 Completeness and Contraction: the contraction mapping principle, which proves existence by building a Cauchy sequence. It is used for differential equations (2B.10 Ordinary Differential Equations) and for the short-time existence of Ricci flow (11A.3 Short-Time Existence and Uniqueness).
9B.4 Convergence of Manifolds: limits of whole spaces. Sequences of Riemannian manifolds are compared with the Gromov–Hausdorff distance, and their limits are taken in a complete space of metric spaces. This is how Perelman's blow-up limits are formed (12B.3 The Canonical Neighbourhood Theorem).

Computers don't use real numbers

The real numbers are an idealisation: infinitely many digits, infinitely precise. Machines can't store them. What computers use instead is a finite set of rationals, the floating-point numbers, standardised as IEEE 754.1 First published in 1985 and revised since. Almost all hardware today implements its "binary64" (double-precision) format. A double-precision number has a 53-bit binary significand and an exponent: numbers of the form $\pm m \times 2^e$ with $m$ an integer below $2^{53}$ .

That finite set lacks almost every property of this chapter:

It isn't closed under arithmetic. The exact sum or product of two floating-point numbers is usually not a floating-point number, so the machine rounds.
Familiar decimals aren't representable. $\tfrac{1}{10}$ has the infinite binary expansion $0.0001100110011\ldots$ , so the number stored for 0.1 is $\tfrac{3602879701896397}{36028797018963968}$ , which exceeds $\tfrac{1}{10}$ by about $5.6 \times 10^{-18}$ . As a result 0.1 + 0.2 evaluates to 0.30000000000000004.
Addition isn't associative. In double precision, (0.1 + 0.2) + 0.3 gives 0.6000000000000001, but 0.1 + (0.2 + 0.3) gives 0.6.
It isn't complete, and has no least upper bounds in general. The numbers are spaced unevenly, closer near $0$ and further apart for large values (Figure 4.5).

Figure 4.5. A toy floating-point system (4-bit significand): the representable numbers are rationals, spaced uniformly within each power-of-two range and twice as far apart in the next. The decimal

0.1

falls between two representable neighbours,

\tfrac{12}{128} = 0.09375

and

\tfrac{13}{128} = 0.1015625

, so the machine must round. Real double precision has 53 bits, but the picture is the same.

Numerical analysts manage these errors carefully, and usually they are harmless. The classic case where they were not shows how a tiny representation error can grow with time.

In the world Data The Patriot missile at Dhahran, 1991

On 25 February 1991, during the Gulf War, a Patriot air-defence battery at Dhahran, Saudi Arabia, failed to track and intercept an incoming Scud missile. The Scud hit an Army barracks and killed 28 American soldiers. A report by the US General Accounting Office (GAO/IMTEC-92-26, February 1992) traced the failure to time-keeping arithmetic. The system's clock counted time in tenths of a second as an integer. To be used in tracking calculations, that count had to be converted to seconds, a real number, in a computer whose registers were only 24 bits long. The conversion lost precision, and the error was proportional to how long the system had been running. The battery had been operating continuously for about 100 hours, and by then the error had moved the "range gate", the region where the radar looked for the target, so far that the Scud was outside it.

The mechanism, analysed by Robert Skeel (SIAM News, 1992), is that $\tfrac{1}{10}$ has no finite binary expansion, so the constant $\tfrac{1}{10}$ held in the register was chopped. The stored value was short of $\tfrac{1}{10}$ by about $9.54 \times 10^{-8}$ . After $100$ hours, that is $3{,}600{,}000$ tenths of a second, the accumulated error is

3{,}600{,}000 \times 9.54 \times 10^{-8} \approx 0.343 \text{ seconds},

which matches the GAO's figure of $0.3433$ seconds. A Scud travels at roughly Mach 5, and the GAO computed a shift in the range gate of about 687 metres. Israeli users had noticed the drift after 8 hours of operation, and the Army had prepared corrected software. It reached Dhahran on 26 February, the day after the attack.

Figure 4.6. Timing error and range-gate shift against hours of continuous operation, from the table in Appendix II of the GAO report. The error grows linearly, by about

0.0034

seconds per hour, because the same chopped constant

\tfrac{1}{10}

is used at every tick. Beyond about 20 hours the target falls outside the range gate.

The lesson for this guidebook is not about computers. It is that a quantity known only to within a tolerance, multiplied by a large number, can produce a large error. Every estimate in analysis tracks exactly this: how errors in the inputs, multiplied by the sizes of other quantities, combine in the output. Proposition 4.9 did it in miniature, with the bound $M$ on the other factor.

History: two ways to fill the gaps

History Cuts and sequences

By the 1860s the rigorous foundations of calculus rested on the real numbers, which were still undefined. Within a few years several definitions appeared. Charles Méray (1869) and Georg Cantor (1872) defined reals through sequences of rationals that converge "among themselves", the construction in this chapter. Richard Dedekind's Stetigkeit und irrationale Zahlen ("Continuity and irrational numbers", 1872) took a different route: a real number is a cut, a way of splitting the rationals into a lower set and an upper set, with every element of the lower set less than every element of the upper set. The cut at $\sqrt 2$ puts every rational whose square is less than $2$ (or which is negative) below, and the rest above. Dedekind's approach makes the least upper bound property almost immediate, while Cantor's makes completeness and the link to approximation natural. Both give the same real numbers, in the sense of Remark 4.18.

Remark 4.18 The reals are unique

Any two complete ordered fields can be matched up by a bijection that respects addition, multiplication and order. So, as with the natural numbers (2A.1 The Natural Numbers), it makes sense to speak of the real numbers, however they are built. From now on we use only the properties: $\mathbb{R}$ is an ordered field with the least upper bound property, containing $\mathbb{Q}$ as a dense subset.

Recall Where we stand

The real numbers are the completion of the rationals: Cauchy sequences of rationals, identified when their difference tends to zero. They form an ordered field in which the rationals are dense and every bounded nonempty set has a least upper bound. Square roots and rational powers exist. The next chapter, 2A.5 Quantifiers and the Shape of a Proof, gives the quantifier sentences of this chapter ("for every $\varepsilon$ there is an $N$ …") a chapter of their own, because from now on every definition is written in them.

Exercises

Exercise 4.19 Two names for √2

Show that the decimal truncations of $\sqrt 2$ and the Newton sequence of Proposition 4.3 are equivalent Cauchy sequences. (Define the truncations as in Example 4.2, without assuming $\sqrt 2$ exists.)

Hint

Let $d_n$ be the truncation and $x_n$ the Newton iterate. Both satisfy a squeeze: $d_n^2 < 2 < (d_n + 10^{-n})^2$ and $x_n^2 > 2$ . Show $0 < x_n - d_n \leq (x_n - 2/x_n) + 10^{-n}$ , using that $2/x_n < $ any number whose square exceeds $2$ .

Exercise 4.20 Reciprocals are well defined

Suppose $(a_n)$ and $(a'_n)$ are equivalent Cauchy sequences, both bounded away from zero ( $|a_n|, |a'_n| \geq c > 0$ ). Show that $(1/a_n)$ and $(1/a'_n)$ are equivalent.

Solution

Exercise 4.21 Eventually one sign

Let $(a_n)$ be Cauchy with $|a_n| \geq c > 0$ for every $n$ . Show that from some point on, either all $a_n \geq c$ or all $a_n \leq -c$ . Use this to complete the proof of Proposition 4.13.

Exercise 4.22 Terms approach their formal limit

Let $(a_n)$ be a Cauchy sequence of rationals and $x = \operatorname{LIM} a_n$ . Show that for every rational $\varepsilon > 0$ there is $N$ with $|a_n - x| \leq \varepsilon$ for all $n \geq N$ , where $|a_n - x|$ is computed in $\mathbb{R}$ . This is the statement that the sequence converges to its own formal limit, used in the proof of Theorem 4.16.

Hint

$a_n - x$ is the real number $\operatorname{LIM}_{m \to \infty}(a_n - a_m)$ , with $n$ fixed. For $n \geq N$ , all its terms with $m \geq N$ lie in $[-\varepsilon, \varepsilon]$ .

Exercise 4.23 Infima

Define the infimum (greatest lower bound) of a set, and prove that every nonempty set of reals with a lower bound has one, by applying Theorem 4.16 to $-E = \{-x : x \in E\}$ .

Exercise 4.24 Rational powers

Show that if $y > 0$ and $p/q = p'/q'$ (positive integers $q, q'$ ), then $(y^{1/q})^p = (y^{1/q'})^{p'}$ , so $y^{p/q}$ is well defined. Then show $y^{r + s} = y^r y^s$ for rational $r, s$ .

Exercise 4.25 Floating point, by hand

In a toy decimal system that keeps only 3 significant digits and rounds to nearest, compute $(1000 + 0.4) + 0.4$ and $1000 + (0.4 + 0.4)$ . Explain why adding many small numbers to a large one, one at a time, can lose all of them, and why summing small numbers first helps. (2A.7 Series returns to this with Kahan's compensated summation.)

Solution

$1000 + 0.4 = 1000.4$ rounds to $1000$ , and then $1000 + 0.4$ rounds to $1000$ again. But $0.4 + 0.4 = 0.8$ , and $1000 + 0.8 = 1000.8$ rounds to $1001$ . Each small addend on its own is below half the spacing of representable numbers near $1000$ (which is $1$ here), so it is rounded away. Combined first, they cross the threshold.

Exercise 4.26 Rehearsal: an error that grows with time

The Patriot error was $\delta \approx 9.54 \times 10^{-8}$ seconds per tick, at $10$ ticks per second. (a) Find the time error after $T$ hours, and check it against the GAO's $0.0275$ seconds at 8 hours. (b) If the range gate tolerated a shift equal to 50 percent of its size, and the GAO estimated that this was reached at about 20 hours, how long a run would a register with twice as many bits (about $2^{-24}$ times the error per tick) have tolerated? (c) The same arithmetic, "small error per step × number of steps", governs the accuracy of numerical solutions of differential equations. 2B.10 Ordinary Differential Equations makes it precise with Gronwall's inequality, where the errors can also be amplified at every step.

Solution

(a) $36{,}000\,T \times 9.54 \times 10^{-8} \approx 0.00343\,T$ seconds; at $T = 8$ this is $0.0275$ . (b) The error per tick scales by $2^{-24}$ , so the tolerable run time scales by $2^{24} \approx 1.7 \times 10^7$ , from $20$ hours to roughly $3.4 \times 10^8$ hours, about $38{,}000$ years.

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4.1Approximations, ancient and modern

4.2Cauchy sequences

4.3When two sequences approximate the same number

4.4The real numbers

4.4.1Arithmetic

4.5Order

4.6The least upper bound property

4.6.1√2 exists

4.7Completion

4.8Computers don't use real numbers

4.9History: two ways to fill the gaps

4.10Exercises

Approximations, ancient and modern

Cauchy sequences

When two sequences approximate the same number

The real numbers

Arithmetic

Order

The least upper bound property

√2 exists

Completion

Computers don't use real numbers

History: two ways to fill the gaps

Exercises