What Is a Number? From the Naturals to the Reals, and Why 1 = 0.999… Is True

0. Key points

• The chain of extensions $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R} \to \mathbb{C}$ is driven by a single consistent motive: making solvable the equations that the current system can state but cannot solve.
• Between any two rationals there is another rational (density), and yet no rational satisfies $x^2 = 2$. This “gap” is stated precisely as the failure of a supremum to exist.
• The assertion that there are no such gaps is the completeness axiom for the reals (the least upper bound property). Dedekind cuts and Cauchy sequences of rationals are two constructions realizing this same property in different language.
• $1 = 0.999\cdots$ is neither an approximation nor a convention: it is an equality that follows from the definition of an infinite decimal (the supremum of the partial sums). We prove it along three routes — the geometric series, three times $1/3$, and the fact that nothing fits between the two numbers.
• Division by zero is forbidden not because someone made a rule. Once $b/a$ is defined as “the unique solution of $ax = b$”, the case $a = 0$ has either no solution or no unique one.

1. Motivation: where do numbers come from?

Numbers did not arrive in their present shape. The Pythagoreans held that “all is number”, meaning ratios of integers, until the discovery that the side and the diagonal of a square admit no common measure — that is, that their ratio is not a ratio of integers — wrecked that worldview. Negative numbers were doubted far longer; in the seventeenth century Descartes still called negative solutions of an equation “false roots”. That $\sqrt{-1}$ was named an imaginary number likewise records that the mathematicians of the day did not regard it as a legitimate number.

The decisive turn came in the nineteenth century. Fourier series, and functions that are continuous yet nowhere differentiable, made it plain that treating the real numbers on geometric intuition alone is unsafe. In 1872 Dedekind published Continuity and Irrational Numbers, defining real numbers by cuts of the rationals; in the same year Cantor gave a construction by Cauchy sequences of rationals. The real numbers, in other words, received a rigorous definition only some two hundred years after the invention of calculus.

We take three naive questions as our starting point.

1. Is $1 = 0.999\cdots$ really an equality, or are the two sides merely “arbitrarily close”?
2. Does the number $\sqrt{2}$ exist, or is it just a convenient symbol?
3. Why may we not divide by $0$? Who decided that?

None of these can be answered while “what a number is” remains undecided. Conversely, once the number system is written down as axioms, all three are settled by short proofs. For the language of mathematics (sets, logic, quantifiers) we presuppose The Grammar of Mathematics — Sets and Logic, in particular the universal and existential quantifiers（Definition 5.2）［The Grammar of Mathematics］.

2. Preliminaries: the framework of an ordered field

When mathematics answers “what is a number”, it does not pry into the inner nature of numbers. Instead it lists the rules a number must obey, and calls anything obeying those rules a number. Both $\mathbb{Q}$ and $\mathbb{R}$ belong to the same framework in the following sense.

Definition 2.1（Ordered field）

Let a set $K$ carry two operations $+ ,\ \cdot$ and a relation $<$ satisfying the following. Then $(K, +, \cdot, <)$ is called an ordered field.

Field axioms: for all $a, b, c \in K$,

1. $a + b = b + a$ and $(a+b)+c = a+(b+c)$.
2. There is an additive identity $0 \in K$ with $a + 0 = a$, and for each $a$ there is $-a \in K$ with $a + (-a) = 0$.
3. $ab = ba$ and $(ab)c = a(bc)$.
4. There is a multiplicative identity $1 \in K$ with $1 \ne 0$ and $a \cdot 1 = a$, and for each $a \ne 0$ there is $a^{-1} \in K$ with $a a^{-1} = 1$.
5. $a(b+c) = ab + ac$.

Order axioms:

6. For all $a, b$, exactly one of $a < b$, $a = b$, $b < a$ holds.
7. If $a < b$ and $b < c$, then $a < c$.
8. If $a < b$, then $a + c < b + c$.
9. If $a < b$ and $0 < c$, then $ac < bc$.

Subtraction and division are notations derived from these axioms: we set $a - b := a + (-b)$, and, for $b \ne 0$, $a / b := a \cdot b^{-1}$.

Remark 2.2

Note the proviso “for $b \ne 0$” in the definition of division. The problem of division by zero has already surfaced at this point; we take it up in Section 7.

Axioms 6–9 also yield $a \ne 0 \Rightarrow 0 < a^2$. Indeed, if $0 < a$, apply axiom 9 to the inequality $0 < a$ with $c = a$ to get $0 \cdot a < a \cdot a$, that is, $0 < a^2$ (the identity $0 \cdot a = 0$ is proved in Proposition 7.1 (1)). If $a < 0$, then axiom 8 gives $0 < -a$, and $a^2 = (-a)(-a) > 0$.

Both $\mathbb{Q}$ and $\mathbb{R}$ are ordered fields. To tell them apart we therefore need a property beyond the ordered field axioms; that property is the completeness axiom introduced in Section 5. Throughout, the natural numbers are $\mathbb{N} = \{1, 2, 3, \ldots\}$ (not containing $0$).

3. Unsolvable equations drive the extensions

Each extension of the number system has a definite reason behind it. When an equation appears that the system can write down but cannot solve, we enlarge the system so that a solution exists.

flowchart LR
N["ℕ naturals"] -->|"x + 3 = 1"| Z["ℤ integers"]
Z -->|"3x = 1"| Q["ℚ rationals"]
Q -->|"x² = 2"| R["ℝ reals"]
R -->|"x² = −1"| C["ℂ complex numbers"]

Number systems, and the equation that first becomes solvable in each

Example 3.1（Checking the unsolvable equation at each stage）

(1) $x + 3 = 1$ has no solution in $\mathbb{N}$. If $x \in \mathbb{N}$ then $x \ge 1$, so $x + 3 \ge 4$, and $x + 3 = 1$ fails. Making the inverse operation of addition (subtraction) freely available produces $\mathbb{Z}$.

(2) $3x = 1$ has no solution in $\mathbb{Z}$. If $x \le 0$ then $3x \le 0 < 1$; if $x \ge 1$ then $3x \ge 3 > 1$. As $x$ is an integer there is no other case, so no integer satisfies $3x = 1$. Making the inverse operation of multiplication (division, excluding $0$) freely available produces $\mathbb{Q}$.

(3) $x^2 = 2$ has no solution in $\mathbb{Q}$. This is proved in Proposition 4.2. Filling this lacuna produces $\mathbb{R}$.

(4) $x^2 = -1$ has no solution in $\mathbb{R}$. As seen in Remark 2.2, in an ordered field $x^2 > 0$ when $x \ne 0$ and $x^2 = 0$ when $x = 0$, so always $x^2 \ge 0 > -1$. Filling this lacuna produces $\mathbb{C}$ — but the same computation also shows that $\mathbb{C}$ cannot be made into an ordered field. Enlarging the number system gains something and loses something.

3.1. The extensions are built from equivalence classes

$\mathbb{Z}$ and $\mathbb{Q}$ do not fall from the sky; they can be constructed from the system below. $\mathbb{Z}$ is defined as the set of equivalence classes on $\mathbb{N} \times \mathbb{N}$ under

$$(a, b) \sim (c, d) \iff a + d = b + c ,$$

the pair $(a,b)$ being “intended as $a - b$”. Addition is defined by $[(a,b)] + [(c,d)] := [(a+c,\ b+d)]$, and here something must be checked. The left-hand side depends only on the equivalence classes, whereas the right-hand side appears to depend on the choice of representatives $(a,b)$, $(c,d)$. In fact it does not. Suppose $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$, that is, $a + b' = b + a'$ and $c + d' = d + c'$. Adding the two equations gives

$$(a + c) + (b' + d') = (b + d) + (a' + c') ,$$

which says precisely that $(a+c,\ b+d) \sim (a'+c',\ b'+d')$. Hence the sum is independent of the representatives chosen. This verification is expressed by saying that the operation is well-defined.

Similarly $\mathbb{Q}$ is the set of equivalence classes on $\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})$ under $(a,b) \sim (c,d) \iff ad = bc$. That $\tfrac{1}{2}$ and $\tfrac{2}{4}$ are the same number is exactly this equivalence relation (the equivalence relation on fractions（Proposition 6.1）［関係と同値関係］). For the general theory of equivalence relations and quotient sets see Relations and Equivalence — What Does “the Same” Mean?, in particular equivalence classes, quotient sets, and the canonical projection（Definition 4.1）［関係と同値関係］.

4. Density of the rationals, and the gaps that remain

4.1. The rationals are packed tight

For rationals $p < q$ consider $r = \dfrac{p+q}{2}$. Then $r$ is rational, and $p < r$ is equivalent to $2p < p + q$, that is, to $p < q$, while $r < q$ is equivalent to $p + q < 2q$, again to $p < q$; so both hold. Thus between any two rationals there is another rational. Iterating, there are infinitely many rationals in between. The number line looks completely filled by the rationals.

It is not.

4.2. $\sqrt{2}$ is not rational

Lemma 4.1

For an integer $n$, if $n^2$ is even then $n$ is even.

Proof（Lemma 4.1）

We prove the contrapositive: if $n$ is odd then $n^2$ is odd. If $n$ is odd, then $n = 2k+1$ for some integer $k$, and

$$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 .$$

Since $2k^2 + 2k$ is an integer, $n^2$ is odd. The contrapositive being true, so is the original statement. For why proof by contraposition is legitimate (equivalence of a statement and its contrapositive（Proposition 6.1）［Techniques of Proof］), see Techniques of Proof — Induction and Contradiction.

∎

Proposition 4.2（Irrationality of √2）

There is no rational number $x$ with $x^2 = 2$.

Proof（Proposition 4.2）

By contradiction. Suppose $x \in \mathbb{Q}$ satisfies $x^2 = 2$. Being rational, $x$ is a ratio of integers; choosing among such representations one with least denominator (by the well-ordering of the naturals（Axiom 3.1）［Techniques of Proof］), we may write $x = m/n$ with $m, n$ integers, $n \ge 1$, and $\gcd(m, n) = 1$.

Squaring gives $m^2 / n^2 = 2$, that is,

$$m^2 = 2n^2 .$$

The right-hand side is even, so $m^2$ is even, and by Lemma 4.1 $m$ is even. Writing $m = 2k$ with $k$ an integer gives $4k^2 = 2n^2$, and dividing by $2$,

$$n^2 = 2k^2 .$$

By the same reasoning $n^2$ is even, and again by Lemma 4.1 so is $n$. Then $m$ and $n$ are both divisible by $2$, contradicting $\gcd(m,n) = 1$. Hence no rational satisfies $x^2 = 2$.

∎

Remark 4.3

The proposition does not say that $\sqrt{2}$ does not exist. It says only that it is not in $\mathbb{Q}$. The diagonal of a square of side $1$ certainly has a length. It exists, and it is not in $\mathbb{Q}$; therefore $\mathbb{Q}$ must be enlarged. That is the direction of the argument.

4.3. Saying “gap” precisely, in the language of suprema

We need to state “there is a gap” without appealing to pictures. The key notion is the supremum.

Definition 4.4（Upper bound and supremum）

Let $K$ be an ordered field and $A \subset K$ a nonempty subset.

• An element $b \in K$ is an upper bound of $A$ if $a \le b$ for every $a \in A$. If at least one upper bound exists, $A$ is said to be bounded above.
• An element $s \in K$ is a supremum (least upper bound) of $A$ if (i) $s$ is an upper bound of $A$, and (ii) $s \le b$ for every upper bound $b$ of $A$.

A supremum, if it exists, is unique: if $s$ and $s'$ are both suprema, applying (i) and (ii) to each in turn gives $s \le s'$ and $s' \le s$, hence $s = s'$. This unique element is written $\sup A$.

Theorem 4.5（The rational field fails the least upper bound property）

Let $A = \{x \in \mathbb{Q} \mid x > 0,\ x^2 < 2\}$. Then $A$ is nonempty and bounded above in $\mathbb{Q}$, but has no supremum in $\mathbb{Q}$.

Proof（Theorem 4.5）

(a) $A \ne \varnothing$. Since $1 > 0$ and $1^2 = 1 < 2$, we have $1 \in A$.

(b) $2$ is an upper bound of $A$. Suppose $x \in A$ and $x > 2$. Applying axiom 9 of an ordered field twice to $0 < 2 < x$ gives $x^2 > 2x > 4 > 2$, contradicting $x^2 < 2$. Hence $x \le 2$ for every $x \in A$.

(c) The key transformation. For a positive rational $p$ set

$$q := p - \frac{p^2 - 2}{p + 2} = \frac{2p + 2}{p + 2} .$$

Since $p > 0$, $q$ is a positive rational. We record two identities:

$$q - p = -\,\frac{p^2 - 2}{p + 2}, \qquad q^2 - 2 = \frac{(2p+2)^2 - 2(p+2)^2}{(p+2)^2} = \frac{2p^2 - 4}{(p+2)^2} = \frac{2(p^2 - 2)}{(p+2)^2}.$$

(The numerator in the second identity is $4p^2 + 8p + 4 - 2p^2 - 8p - 8 = 2p^2 - 4$.) As the denominators $(p+2)$ and $(p+2)^2$ are positive, the signs of $q - p$ and $q^2 - 2$ are as follows.

• If $p^2 < 2$: $q > p$ and $q^2 < 2$.
• If $p^2 > 2$: $q < p$ and $q^2 > 2$.

(d) No supremum in $\mathbb{Q}$. Suppose $s \in \mathbb{Q}$ were a supremum of $A$. From $1 \in A$ we get $s \ge 1 > 0$. By Proposition 4.2, $s^2 = 2$ is impossible, so either $s^2 < 2$ or $s^2 > 2$.

Case $s^2 < 2$. Together with $s > 0$ this gives $s \in A$. By (c), $q > s$, $q^2 < 2$ and $q > 0$, so $q \in A$. This contradicts $s$ being an upper bound of $A$.

Case $s^2 > 2$. We construct an upper bound smaller than $s$. The element $q$ of (c) satisfies $q < s$, $q^2 > 2$ and $q > 0$. That $q$ is an upper bound of $A$ follows by the argument of (b): if $x \in A$ with $x > q > 0$, then $x^2 > q^2 > 2$, contradicting $x \in A$; hence $x \le q$ for every $x \in A$. So $q$ is an upper bound smaller than $s$, contradicting condition (ii) in the definition of supremum ($s$ is at most every upper bound).

Both cases lead to a contradiction, so $A$ has no supremum in $\mathbb{Q}$.

∎

As seen in Section 4.1, the rationals are dense. Even so, no rational sits at the “right edge” of this set $A$. The lesson here is that being dense and having no gaps are two different things.

A cut of the rational line. A and B together exhaust the rationals, yet no rational lies at the boundary

5. The reals: filling the gaps by the completeness axiom

Axiom 5.1（Completeness axiom for the reals (least upper bound property)）

The field of real numbers $\mathbb{R}$ is an ordered field satisfying the following.

Every nonempty subset $A \subset \mathbb{R}$ that is bounded above has a supremum $\sup A$ in $\mathbb{R}$.

Such an ordered field exists, and is unique up to isomorphism.

Theorem 4.5 says that $\mathbb{Q}$ fails this axiom. The completeness axiom is thus the single point separating $\mathbb{R}$ from $\mathbb{Q}$.

Remark 5.2

The clause “such a field exists” is guaranteed by actually building one. There are two standard constructions.

Dedekind cuts. Split the rationals into a nonempty downward-closed set $A$ (if $x \in A$ and $y < x$ then $y \in A$) that is bounded above and has no greatest element, together with its complement. Define $A$ itself to be a real number. To a rational $r$ corresponds $A_r = \{x \in \mathbb{Q} \mid x < r\}$; when $A$ corresponds to no rational, the cut defines an irrational number. Suprema are obtained as unions of cuts, so the completeness axiom holds almost automatically.

Cauchy sequences. On the set of all Cauchy sequences of rationals, impose the equivalence relation “the difference tends to $0$”, and define the real numbers to be the quotient set. Here $\sqrt{2}$ is the equivalence class of the sequence $1,\ 1.4,\ 1.41,\ 1.414, \ldots$. Again the crux is checking that sums and products do not depend on the representatives chosen (well-definedness). See Relations and Equivalence — What Does “the Same” Mean?; an outline is collected in the Appendix.

The two constructions yield isomorphic ordered fields. Whichever route one takes, everything that follows depends on the completeness axiom alone.

Example 5.3（The cut that defines √2）

Let $A = \{x \in \mathbb{Q} \mid x \le 0 \ \text{or}\ x^2 < 2\}$. It is nonempty ($0 \in A$) and bounded above (by (b) of Theorem 4.5, $2$ is an upper bound).

We check that $A$ has no greatest element. Take any $p \in A$. If $p \le 0$, then $1 \in A$ and $p < 1$. If $p > 0$, then $p^2 < 2$, and the element $q = (2p+2)/(p+2)$ from (c) of Theorem 4.5 satisfies $q > p$, $q^2 < 2$ and $q > 0$, so $q \in A$ with $q > p$. In either case $A$ contains an element larger than $p$, so there is no greatest element.

Likewise the complement $B = \mathbb{Q} \setminus A$ (the set of rationals with $x > 0$ and $x^2 > 2$; the case $x^2 = 2$ does not occur, by Proposition 4.2) has no least element: for $p \in B$ the same $q$ satisfies $q < p$, $q^2 > 2$ and $q > 0$.

Thus $A$ and $B$ exhaust $\mathbb{Q}$, and nothing sits at the boundary. Dedekind’s idea is to regard the cut itself as a number.

5.1. Two tools that come out of the completeness axiom

Proposition 5.4（Archimedean property）

(1) For every real number $x$ there is a natural number $n$ with $n > x$. (2) For every real $\varepsilon > 0$ there is a natural number $n$ with $\dfrac{1}{n} < \varepsilon$.

Proof（Proposition 5.4）

(1) By contradiction. Suppose that for some real $x$ every $n \in \mathbb{N}$ satisfies $n \le x$. Then $\mathbb{N}$ is nonempty and bounded above, so by Axiom 5.1 the supremum $s = \sup \mathbb{N}$ exists. Since $s - 1 < s$ and the supremum is the least upper bound, $s - 1$ is not an upper bound; that is, some $n_0 \in \mathbb{N}$ satisfies $n_0 > s - 1$. Then $n_0 + 1 > s$, yet $n_0 + 1$ is also a natural number and $s$ is an upper bound of $\mathbb{N}$, so $n_0 + 1 \le s$. Contradiction.

(2) Apply (1) with $x = 1/\varepsilon$ to obtain $n > 1/\varepsilon$. Multiplying both sides by $\varepsilon / n > 0$ (using $n > 0$, $\varepsilon > 0$ and axiom 9) gives $\varepsilon > 1/n$.

∎

Proposition 5.5（Density of the rationals）

If real numbers $a, b$ satisfy $a < b$, then there is a rational $r$ with $a < r < b$.

Proof（Proposition 5.5）

Since $b - a > 0$, part (2) of Proposition 5.4 yields a natural number $n$ with $\dfrac{1}{n} < b - a$, that is, $1 < n(b-a)$.

Next we find the least integer $m$ with $m > na$. By part (1) of Proposition 5.4 there is a natural number $N$ with $N > |na|$, so the set $S = \{k \in \mathbb{Z} \mid k > na\}$ is nonempty ($N \in S$) and bounded below by $-N$. Hence $S$ has a least element $m$ (well-ordering of the integers). By minimality $m - 1 \notin S$, that is, $m - 1 \le na$. Altogether

$$m - 1 \le na < m .$$

The left inequality gives $m \le na + 1$, and using $1 < n(b-a)$,

$$m \le na + 1 < na + n(b - a) = nb .$$

Therefore $na < m < nb$. Since $n > 0$, dividing through by $n$ gives $a < \dfrac{m}{n} < b$. As $r = m/n$ is rational, this is the required number.

∎

Theorem 5.6（Existence of √2）

There is exactly one positive real number $s$ with $s^2 = 2$.

Proof（Theorem 5.6）

Existence. Let $A = \{x \in \mathbb{R} \mid x > 0,\ x^2 < 2\}$. It is nonempty since $1 \in A$, and $2$ is an upper bound by the same argument as in (b) of Theorem 4.5 (which nowhere required the numbers to be rational). Hence by Axiom 5.1 the supremum $s = \sup A$ exists, and $1 \in A$ gives $s \ge 1 > 0$.

Suppose $s^2 < 2$. Applying the transformation of (c) in Theorem 4.5 to the real number $p = s$ (that computation uses only the ordered field operations, so it remains valid over the reals), the element $q = (2s+2)/(s+2)$ satisfies $q > s$, $q^2 < 2$ and $q > 0$. Thus $q \in A$ with $q > s$, contradicting that $s$ is an upper bound.

Suppose $s^2 > 2$. The same transformation gives $q < s$, $q^2 > 2$ and $q > 0$, and by the argument of (d) in Theorem 4.5, $q$ is an upper bound of $A$. An upper bound smaller than $s$ contradicts $s$ being the least upper bound.

By order axiom 6 the only remaining possibility is $s^2 = 2$.

Uniqueness. Suppose $s, t > 0$ both satisfy $s^2 = t^2 = 2$. If $s \ne t$, say $s < t$, then axiom 9 gives $s^2 < st < t^2$, that is, $2 < 2$, a contradiction. Hence $s = t$.

∎

The argument that failed over $\mathbb{Q}$ went through over $\mathbb{R}$ because we invoked Axiom 5.1, which guarantees the existence of the supremum, exactly once. This contrast is, I think, the clearest way to see where the completeness axiom does its work.

6. What $1 = 0.999\cdots$ means, and three proofs

6.1. First, fix the meaning of “infinite decimal”

Most disputes about $1 = 0.999\cdots$ get tangled because they begin without deciding what the symbol $0.999\cdots$ denotes. Let us start from the definition.

Definition 6.1（Value of an infinite decimal）

Let $a_0 \in \mathbb{Z}$ and, for each $k \ge 1$, $a_k \in \{0, 1, \ldots, 9\}$ be given. The symbol $a_0.a_1a_2a_3\cdots$ denotes the supremum of the set $\{s_n \mid n \in \mathbb{N}\}$ of partial sums

$$s_n = a_0 + \sum_{k=1}^{n} \frac{a_k}{10^k} \qquad (n = 1, 2, \ldots).$$

We check that this definition makes sense, that is, that the supremum exists. First, $s_{n+1} - s_n = a_{n+1}/10^{n+1} \ge 0$, so $(s_n)$ is nondecreasing. Next we show it is bounded above. Putting $T_n = \sum_{k=1}^n 10^{-k}$,

$$10 T_n - T_n = \left(1 + 10^{-1} + \cdots + 10^{-(n-1)}\right) - \left(10^{-1} + \cdots + 10^{-n}\right) = 1 - 10^{-n} ,$$

so $9T_n = 1 - 10^{-n}$, that is, $\sum_{k=1}^{n} 9 \cdot 10^{-k} = 1 - 10^{-n}$. Since $a_k \le 9$,

$$s_n \le a_0 + \sum_{k=1}^n \frac{9}{10^k} = a_0 + 1 - \frac{1}{10^n} < a_0 + 1 ,$$

so $a_0 + 1$ is an upper bound. Hence by Axiom 5.1 the supremum $s = \sup\{s_n\}$ exists.

Moreover $s_n$ converges to $s$. Given any $\varepsilon > 0$, since $s - \varepsilon < s$ the number $s - \varepsilon$ is not an upper bound, so some $N$ satisfies $s_N > s - \varepsilon$. By monotonicity, for $n \ge N$ we have $s - \varepsilon < s_N \le s_n \le s$, hence $|s_n - s| < \varepsilon$. From now on we use “value of an infinite decimal” and “limit of the partial sums” interchangeably.

Note

This is where misunderstandings branch off. The symbol $0.999\cdots$ does not denote the process of writing more and more $9$s; it names the single number that this process determines, namely the supremum of all the partial sums. “It never quite reaches $1$” is a statement about the individual partial sums $s_n = 1 - 10^{-n}$, not about their supremum. The subject of the sentence has been switched.

6.2. A tool: the sum of a geometric series

Proposition 6.2（Geometric series）

Let $r$ be a real number with $0 \le r < 1$. For every integer $n \ge 0$,

$$\sum_{k=0}^{n} r^k = \frac{1 - r^{n+1}}{1 - r} ,$$

and moreover $\displaystyle\sum_{k=0}^{n} r^k \to \frac{1}{1-r}$ as $n \to \infty$.

Proof（Proposition 6.2）

The finite sum. Putting $S_n = \sum_{k=0}^n r^k$,

$$(1-r)S_n = \sum_{k=0}^{n} r^k - \sum_{k=0}^{n} r^{k+1} = \sum_{k=0}^{n} r^k - \sum_{k=1}^{n+1} r^{k} = r^0 - r^{n+1} = 1 - r^{n+1}$$

(the intermediate terms $r^1, \ldots, r^n$ cancel). Since $r < 1$ we have $1 - r \ne 0$, so dividing both sides by $1-r$ gives the formula.

$r^n \to 0$. For $r = 0$ this is clear, as $r^n = 0$ for $n \ge 1$. Let $0 < r < 1$. Then $1/r > 1$, so we may write $1/r = 1 + h$ with $h > 0$. Bernoulli’s inequality $(1+h)^n \ge 1 + nh$ holds: for $n = 0$ both sides equal $1$, and assuming $(1+h)^n \ge 1+nh$,

$$(1+h)^{n+1} = (1+h)^n (1+h) \ge (1+nh)(1+h) = 1 + (n+1)h + nh^2 \ge 1 + (n+1)h ,$$

so by the principle of mathematical induction（Theorem 3.2）［Techniques of Proof］ it holds for all $n \ge 0$ (see Techniques of Proof — Induction and Contradiction). Consequently

$$0 < r^n = \frac{1}{(1+h)^n} \le \frac{1}{1 + nh} < \frac{1}{nh} \qquad (n \ge 1).$$

Given $\varepsilon > 0$, applying part (2) of Proposition 5.4 to $h\varepsilon > 0$ produces $n_0$ with $1/n_0 < h\varepsilon$, and for $n \ge n_0$ we get $r^n < 1/(nh) \le 1/(n_0 h) < \varepsilon$. Hence $r^n \to 0$.

The limit. Therefore $S_n = \dfrac{1 - r^{n+1}}{1-r} \to \dfrac{1 - 0}{1 - r} = \dfrac{1}{1-r}$.

∎

6.3. The theorem and three proofs

Theorem 6.3（1 = 0.999…）

The infinite decimal $0.999\cdots$ determined by $a_0 = 0$ and $a_k = 9\ (k \ge 1)$ has value $1$.

Proof（Theorem 6.3）

Put $s_n = \sum_{k=1}^{n} 9 \cdot 10^{-k} = 1 - 10^{-n}$ (this identity was verified in Section 6.1). By Definition 6.1, what has to be shown is $\sup\{s_n \mid n \in \mathbb{N}\} = 1$. We give three routes.

Proof 1 (direct computation as a geometric series).

$$\sum_{k=1}^{n} \frac{9}{10^k} = \frac{9}{10}\sum_{j=0}^{n-1} \left(\frac{1}{10}\right)^{j}$$

(substituting $k = j+1$). Applying Proposition 6.2 with $r = 1/10$, the sum on the right converges to $\dfrac{1}{1 - 1/10} = \dfrac{10}{9}$ as $n \to \infty$. Therefore

$$0.999\cdots = \frac{9}{10} \cdot \frac{10}{9} = 1 .$$

Proof 2 (as three times $1/3$). First we verify $0.333\cdots = 1/3$. The partial sums are $t_n = \sum_{k=1}^n 3 \cdot 10^{-k} = 3 T_n = \dfrac{1 - 10^{-n}}{3}$, and since $10^{-n} \to 0$ (shown in the course of proving Proposition 6.2) we get $t_n \to 1/3$. Next, for every $n$,

$$3 t_n = 1 - 10^{-n} = s_n .$$

The limit of the left-hand side is $3 \cdot \dfrac13 = 1$ and the limit of the right-hand side is $0.999\cdots$, so by uniqueness of limits $0.999\cdots = 1$.

The schoolroom manipulation “set $x = 0.999\cdots$, then $10x = 9.999\cdots$, and subtracting, $9x = 9$” also becomes a legitimate argument once written in terms of partial sums. With $s = \lim s_n$ we have $9 s_n = 9 - 9 \cdot 10^{-n} \to 9$, while on the other hand $9 s_n \to 9s$; hence $9s = 9$, that is, $s = 1$. The computation with partial sums takes over the role of the claim that multiplying an infinite decimal by $10$ amounts to shifting the decimal point one place to the right.

Proof 3 (nothing fits between the two numbers). Put $s = 0.999\cdots$. For every $n$ we have $s_n = 1 - 10^{-n} < 1$, so $1$ is an upper bound of $\{s_n\}$, and minimality of the supremum gives $s \le 1$.

Suppose $s < 1$. By Proposition 5.5 there is a rational $r$ with $s < r < 1$. Write $r = p/q$ with $p, q$ integers and $q \ge 1$. From $r < 1$ we get $p < q$, and as both are integers, $q - p \ge 1$. Hence

$$1 - r = \frac{q-p}{q} \ge \frac{1}{q} .$$

On the other hand, for every $n$ we have $s_n \le s < r$, that is, $1 - 10^{-n} < r$, so $1 - r < 10^{-n}$. Combining the two, for every $n$,

$$\frac{1}{q} \le 1 - r < \frac{1}{10^n}, \qquad \text{that is,} \qquad 10^n < q .$$

But by Bernoulli’s inequality above, $10^n = (1+9)^n \ge 1 + 9n > n$, so taking $n = q$ gives $10^q > q$, a contradiction. Hence $s < 1$ is impossible, and together with $s \le 1$ we conclude $s = 1$.

This proof simply writes out the reasoning: no rational number lies between $0.999\cdots$ and $1$; between any two distinct reals there must be a rational; hence the two are equal.

∎

Remark 6.4

The three proofs are not independent. Proof 2 justifies $1/3 = 0.333\cdots$ by the same geometric series computation as Proof 1. Proof 3, pushed far enough, also rests on $10^{-n} \to 0$, that is, on the Archimedean property. There is a single common root: Axiom 5.1. It is accurate to say that there are three ways of seeing the fact; it would be an overstatement to say that there are three independent grounds for it.

Conversely, in an ordered field without the Archimedean property (a system containing infinitesimals) one can build a world in which the quantity corresponding to $1 - 0.999\cdots$ is not $0$. The statement $1 = 0.999\cdots$ is a statement about the system $\mathbb{R}$; change the system and the statement changes.

6.4. Examples, and decimal expansions as “names”

Example 6.5（Converting repeating decimals to fractions）

(1) $0.272727\cdots$. The partial sums are $\sum_{k=1}^{n} 27 \cdot 100^{-k}$. Applying Proposition 6.2 with $r = 1/100$,

$$0.272727\cdots = \frac{27}{100} \sum_{j=0}^{\infty}\left(\frac{1}{100}\right)^j = \frac{27}{100} \cdot \frac{1}{1 - \frac{1}{100}} = \frac{27}{100}\cdot\frac{100}{99} = \frac{27}{99} = \frac{3}{11}.$$

Checking, $3 \div 11 = 0.2727\cdots$, as it should be.

(2) $0.4999\cdots$. We compute the limit of the partial sums as the definition prescribes.

$$0.4999\cdots = \frac{4}{10} + \sum_{k=2}^{\infty} \frac{9}{10^k} = \frac{4}{10} + \frac{9}{100}\cdot\frac{1}{1 - \frac{1}{10}} = \frac{4}{10} + \frac{9}{100}\cdot\frac{10}{9} = \frac{4}{10} + \frac{1}{10} = \frac{1}{2}.$$

So $0.5$ and $0.4999\cdots$ are two names for the same real number.

Remark 6.6

Which real numbers have two decimal expansions? The answer: exactly those expressible as $m/10^k$ with $m, k$ integers and $k \ge 0$ — that is, the terminating decimals. A terminating decimal $0.a_1\cdots a_j$ (with $a_j \ne 0$) always also has the expansion $0.a_1 \cdots (a_j - 1)999\cdots$, obtained by decreasing the last digit by one and appending infinitely many $9$s; every other real number has a unique decimal expansion.

The essential point is that a decimal expansion is not the real number itself but a name for it. Two names may denote the same object, and then they are equal. The relation between $1$ and $0.999\cdots$ is of the same kind as that between $\tfrac12$ and $\tfrac24$.

This uniqueness question matters again in the proof that the reals are uncountable (Cantor's diagonal argument（Theorem 6.3）［濃度と無限］). Avoiding expansions ending in infinitely many $9$s when constructing the diagonal is precisely a response to this double naming. See Cardinality and Infinity — Infinities Come in Sizes.

7. Why we cannot divide by zero

“You may not divide by $0$” is not a prohibition to be memorized. Returning to the definition of division makes clear why division by $0$ alone cannot be defined. In Definition 2.1 we set $a/b := a \cdot b^{-1}$; this is the operation of finding “the $x$ with $bx = a$”, the inverse of multiplication.

Proposition 7.1（Solvability of linear equations）

Let $K$ be a field and $a, b \in K$.

1. $0 \cdot x = 0$ for every $x \in K$.
2. If $a \ne 0$, the equation $ax = b$ has exactly one solution, $x = a^{-1}b$.
3. If $a = 0$ and $b \ne 0$, then $ax = b$ has no solution.
4. If $a = 0$ and $b = 0$, then every element of $K$ is a solution of $ax = b$.

Proof（Proposition 7.1）

(1) From $0 = 0 + 0$ and distributivity (axiom 5 of Definition 2.1),

$$0 \cdot x = (0 + 0)\cdot x = 0\cdot x + 0 \cdot x .$$

Adding $-(0 \cdot x)$ to both sides (existence of additive inverses, axiom 2) makes the left side $0$ and the right side $0 \cdot x$, giving $0 = 0 \cdot x$.

(2) Existence: $a(a^{-1}b) = (a a^{-1})b = 1 \cdot b = b$ (associativity of multiplication, inverses and identity; axioms 3 and 4). Uniqueness: if $ax = b$ and $ax' = b$, then $ax = ax'$, and multiplying both sides by $a^{-1}$ gives $x = (a^{-1}a)x = (a^{-1}a)x' = x'$. Here $a \ne 0$ was indispensable for the existence of $a^{-1}$.

(3) By (1), $0 \cdot x = 0 \ne b$ for every $x$, so there is no solution.

(4) By (1), $0 \cdot x = 0 = b$ for every $x$, so every element is a solution.

∎

This proposition is the answer itself. The symbol $b/a$ is meaningful as a name for “the unique solution of $ax = b$”. But when $a = 0$, either there is no solution, by (3) (the case $1/0$), or there are too many solutions to single one out, by (4) (the case $0/0$). Since no value can be specified, no meaning can be given to the symbol. It is not that we may not divide, but that the result of dividing cannot be defined.

Example 7.2（What breaks if we force a definition）

(1) The system collapses. Suppose we could adjoin an element $c$ with $0^{-1} = c$ while keeping all the field axioms. By the definition of an inverse, $0 \cdot c = 1$, whereas by (1) of Proposition 7.1, $0 \cdot c = 0$; hence $1 = 0$. Then for every $x$,

$$x = x \cdot 1 = x \cdot 0 = 0 ,$$

so the system has exactly one element. The axiom $1 \ne 0$ in axiom 4 is there to prevent this collapse.

(2) A famous fallacy. Start from $a = b$ with $a \ne 0$.

$$a^2 = ab,\quad a^2 - b^2 = ab - b^2,\quad (a+b)(a-b) = b(a-b) .$$

So far the manipulations are correct. Next “divide both sides by $a - b$” to get $a + b = b$, then use $a = b$ to get $2b = b$, and finally $2 = 1$. The error is the division by $a - b = 0$. The identity $(a+b)(a-b) = b(a-b)$ is the correct statement $0 = 0$, from which $a+b = b$ does not follow: as (4) of Proposition 7.1 says, $0 \cdot x = 0$ determines nothing about $x$.

Caution

The “indeterminate form $0/0$” of calculus is a different matter from division by zero. As the examples $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x\to 0}\frac{x^2}{x} = 0$ show, the value of a limit in which numerator and denominator both tend to $0$ depends on the case at hand. That is why it is called indeterminate; it does not mean that a number $0/0$ has several values. There is no such number in the first place.

There are, to be sure, systems in which one sets $1/0 = \infty$, such as the Riemann sphere, and algebraic systems in which division is made total. But these discard part of the field axioms; they do not amount to dividing by zero inside $\mathbb{R}$.

Remark 7.3

Everything above has taken the form “posit axioms, then deduce logically”. What, then, guarantees the axiom system itself? This question gave birth to twentieth-century foundations of mathematics and leads to Gödel’s incompleteness theorems (the first incompleteness theorem（Theorem 5.1）［ゲーデルの不完全性定理］). We treat it in An Invitation to Foundations — The Incompleteness Theorems.

8. Exercises

Exercise 8.1Standard

Prove that no rational number satisfies $x^2 = 3$. First show that for an integer $n$, if $n^2$ is a multiple of $3$ then so is $n$; then argue as in Proposition 4.2.

Solution

Proof of the lemma. Split into cases according to the remainder of $n$ on division by $3$.

• $n = 3k$: $n^2 = 9k^2 = 3(3k^2)$, a multiple of $3$.
• $n = 3k+1$: $n^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$, remainder $1$.
• $n = 3k+2$: $n^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1$, remainder $1$.

Hence $n^2$ is a multiple of $3$ only when $n$ is.

Main argument. Suppose a rational $x$ satisfies $x^2 = 3$, and write it in lowest terms as $x = m/n$ ($m, n$ integers, $n \ge 1$, $\gcd(m,n) = 1$). Squaring gives $m^2 = 3n^2$. The right-hand side is a multiple of $3$, so $m^2$ is, and by the lemma $m = 3k$. Substituting, $9k^2 = 3n^2$, and dividing by $3$, $n^2 = 3k^2$. Again by the lemma $n$ is a multiple of $3$. Then $m$ and $n$ are both divisible by $3$, contradicting $\gcd(m,n) = 1$.

Exercise 8.2Easy

Express the value of the infinite decimal $0.135135135\cdots$ (with $135$ repeating) as a fraction in lowest terms. Use Definition 6.1 and Proposition 6.2.

Solution

The repetition has period three digits, so this is a geometric series with ratio $1/1000$.

$$0.135135\cdots = \sum_{k=1}^{\infty}\frac{135}{1000^{k}} = \frac{135}{1000}\sum_{j=0}^{\infty}\left(\frac{1}{1000}\right)^{j} = \frac{135}{1000}\cdot\frac{1}{1 - \frac{1}{1000}} = \frac{135}{1000}\cdot\frac{1000}{999} = \frac{135}{999}.$$

Since $135 = 27 \cdot 5$ and $999 = 27 \cdot 37$, cancelling gives $\dfrac{5}{37}$. Checking, $5 \div 37 = 0.135135\cdots$.

Exercise 8.3Standard

Let $A = \left\{1 - \dfrac{1}{n} \ \middle|\ n \in \mathbb{N}\right\}$. Following Definition 4.4, show that $\sup A = 1$, and show further that $A$ has no greatest element.

Solution

$1$ is an upper bound. For every $n \in \mathbb{N}$ we have $1/n > 0$, hence $1 - 1/n < 1$.

$1$ is the least upper bound. Let $b$ be any upper bound of $A$ and suppose $b < 1$. Then $1 - b > 0$, so by part (2) of Proposition 5.4 there is a natural number $n_0$ with $\dfrac{1}{n_0} < 1 - b$. Rearranging, $b < 1 - \dfrac{1}{n_0}$, and the right-hand side belongs to $A$, contradicting that $b$ is an upper bound. Hence $b \ge 1$. So $1$ is an upper bound and is at most every upper bound, that is, $\sup A = 1$.

No greatest element. Take any element $1 - 1/n$ of $A$. Since $\dfrac{1}{n+1} < \dfrac{1}{n}$, we have $1 - \dfrac{1}{n} < 1 - \dfrac{1}{n+1} \in A$. Every element is exceeded by another element of $A$, so there is no greatest element. This is an example of a supremum that does not belong to the set.

Exercise 8.4Standard

Show that if a real number $x$ satisfies $0 \le x < \dfrac{1}{n}$ for every natural number $n$, then $x = 0$. Then use this fact to give an alternative proof of Theorem 6.3.

Solution

First part. Suppose $x > 0$. Applying part (2) of Proposition 5.4 with $\varepsilon = x$ gives a natural number $n_0$ with $\dfrac{1}{n_0} < x$. But the hypothesis also gives $x < \dfrac{1}{n_0}$, whence $x < x$, contradicting order axiom 6. So $x > 0$ fails, and together with the hypothesis $x \ge 0$ we get $x = 0$.

Second part. Put $s = 0.999\cdots$. As seen in the proof of Theorem 6.3, the partial sums are $s_n = 1 - 10^{-n}$ and $s$ is their supremum, so $s \le 1$ and hence $x := 1 - s \ge 0$. Also $s \ge s_n = 1 - 10^{-n}$ gives $x \le 10^{-n}$. By Bernoulli’s inequality $10^n \ge 1 + 9n > n$, so $10^{-n} < \dfrac{1}{n}$, and therefore $0 \le x < \dfrac{1}{n}$ for every $n$. By the first part $x = 0$, that is, $s = 1$.

References

• Teiji Takagi, Kaiseki Gairon (in Japanese), Iwanami Shoten — Chapter 1, “Basic concepts”. The classical treatment introducing the continuity of the reals as the least upper bound property.
• Mitsuo Sugiura, Kaiseki Nyūmon I (in Japanese), University of Tokyo Press, 1980 — Chapter I, “Real numbers and continuity”. A careful account of both the axiomatic treatment and the constructions of the reals.
• W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976 — Chapter 1, “The Real and Complex Number Systems”. The transformation $q = p - (p^2-2)/(p+2)$ used in Theorem 4.5 is based on the discussion at the opening of that chapter.
• R. Dedekind, Kazu ni tsuite: Renzokusei to Kazu no Honshitsu (in Japanese), trans. Isaburō Kōno, Iwanami Bunko — the original source for the definition of the reals by cuts (original edition 1872).
• T. Tao, Analysis I, 3rd ed., Hindustan Book Agency / Springer, 2016 — Chapter 4, “Integers and rationals”, and Chapter 5, “The real numbers”. A textbook in which the construction by Cauchy sequences can be followed without any of the well-definedness checks being skipped.

Appendix: Outline of the construction of the reals by Cauchy sequences

Let us look, step by step, at the other construction mentioned in Remark 5.2. The details of the proofs are left to Tao’s textbook above and similar sources, but one can see where and what has to be checked.

Step 1: collect the Cauchy sequences. A sequence $(x_n)$ of rationals is a rational Cauchy sequence if for every rational $\varepsilon > 0$ there is an $N$ such that $|x_m - x_n| < \varepsilon$ whenever $m, n \ge N$. The point is that this says only “the terms grow close to one another far out”, without using the word limit. The limit value may fail to exist in $\mathbb{Q}$, so it is not yet available.

Step 2: identify sequences. Declare two rational Cauchy sequences $(x_n)$, $(y_n)$ equivalent when $|x_n - y_n| \to 0$. One checks that this is an equivalence relation (reflexive, symmetric, transitive); transitivity uses the triangle inequality $|x_n - z_n| \le |x_n - y_n| + |y_n - z_n|$. A real number is by definition an equivalence class under this relation. The class of $1,\ 1.4,\ 1.41,\ 1.414,\ \ldots$ is $\sqrt{2}$.

Step 3: introduce the operations. Define $[(x_n)] + [(y_n)] := [(x_n + y_n)]$ and $[(x_n)]\cdot[(y_n)] := [(x_n y_n)]$. Two things must be checked. First, that $(x_n + y_n)$ and $(x_n y_n)$ are again Cauchy sequences. Second, independence of the representatives (well-definedness), the same kind of work as the check carried out for addition in $\mathbb{Z}$ in Section 3.1. For products one passes through the fact that a Cauchy sequence is bounded.

Step 4: embed $\mathbb{Q}$ and introduce the order. Assigning to a rational $r$ the class of the constant sequence $(r, r, r, \ldots)$ embeds $\mathbb{Q}$ as a subfield of the new field. The order is introduced by declaring an element positive when $(x_n)$ is eventually at least some fixed positive rational.

Step 5: verify the completeness axiom. Finally one proves that the ordered field so constructed satisfies Axiom 5.1. This establishes that $\mathbb{R}$ “exists”.

If Dedekind cuts embody the idea of calling the gap itself a number, Cauchy sequences embody the idea of calling an approximating sequence a number. The difference is whether $\sqrt{2}$ is named as “something whose square is $2$” or as “the procedure of computing $1.414\ldots$”. Either road leads to isomorphic fields.