---
title: "Real Numbers"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 10
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/10-real-numbers
---

# Chapter 10 — Real Numbers

All of analysis rests on one property that distinguishes $\R$ from $\Q$: every nonempty [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) bounded from above has a *least* [upper bound](#def-b1-reals-bounds). This chapter states it precisely, derives its first consequences — the [Archimedean property](#thm-b1-reals-archimedes), the [floor function](#thm-b1-reals-floor), the density of the rationals and of the irrationals — and [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) up the vocabulary (sup, inf, max, min) used constantly from [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) onward.

## 10.1 The upper bound property

**Definition 10.1 (Bounds, sup and inf).**

Let $A \subseteq \R$ be nonempty. A real $M$ is an *upper bound* of $A$ when $a \leq M$ for all $a \in A$; $A$ is *bounded above* when it has an upper bound (similarly below, with lower bounds; *bounded* means both). A *maximum* of $A$ is an upper bound belonging to $A$.

The *supremum* $\sup A$ is the least upper bound of $A$, when it exists; the *infimum* $\inf A$ is the greatest lower bound.

**Theorem 10.2 (Completeness axiom of R\RR).**

$\R$ is an ordered [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) containing $\Q$ in which *every nonempty subset bounded above has a [supremum](#def-b1-reals-bounds)*.

**Proof.** *Admitted at this level.* ∎

**Remark 10.3.**

We take this as the defining axiom of $\R$; constructing a model (by Dedekind cuts or by Cauchy sequences of rationals) and proving its uniqueness is honest but long, and is left for further study. Note that $\Q$ fails the property: $\{x \in \Q : x^2 < 2\}$ is bounded above but has no least [upper bound](#def-b1-reals-bounds) *in $\Q$* — its candidate, $\sqrt 2$, is missing ([Example 1.11](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#ex-b1-logic-sqrt2)). By passing to opposites ($\sup(-A) = -\inf A$), every nonempty [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) bounded below has an [infimum](#def-b1-reals-bounds).

**Proposition 10.4 (The ε\varepsilonε-characterization).**

Let $A \neq \emptyset$ be bounded above and $s \in \R$. Then $s =
\sup A$ if and only if

1. $s$ is an [upper bound](#def-b1-reals-bounds) : $\forall a \in A$ , $a \leq s$ ; and
2. nothing smaller is: $\forall \varepsilon > 0$ , $\exists a \in  A$ , $a > s - \varepsilon$ .

**Proof.** If $s = \sup A$: (1) holds by definition, and for (2), $s -
\varepsilon < s$ is not an [upper bound](#def-b1-reals-bounds), which is exactly the existence of $a > s - \varepsilon$. Conversely, (1) says $s$ is an [upper bound](#def-b1-reals-bounds); (2) says no $t < s$ is an [upper bound](#def-b1-reals-bounds) (take $\varepsilon
= s - t$): $s$ is the least one. ∎

**Example 10.5.**

$\sup \intoo{0}{1} = 1$, not attained (no maximum); $\sup \intcc{0}{1} = 1 = \max$. For $A = \{1 - \frac 1n : n \in
\N^*\}$: $\sup A = 1$, not attained; $\inf A = \min A = 0$. A maximum, when it exists, is the [supremum](#def-b1-reals-bounds); the whole point of $\sup$ is to have a substitute when the maximum does not exist.

![The set A = \1 - 1n : n ∈ ℕ*\ on the number line: its points pile up toward 1 without reaching it. Every number ≥ 1 is an upper bound (the ray), and nothing smaller is, because an element of A enters each interval (1 - , 1): the two clauses of in one picture. The supremum is the left endpoint of the ray of upper bounds — and the completeness axiom is precisely the guarantee that this ray always has a left endpoint.](https://one-course.com/images/onecourse/chapters/math-3/b1-reals/fig-44e38c4c9fa3.svg)

*The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $A = \{1 - \frac1n : n \in \N^*\}$ on the number line: its points pile up toward $1$ without reaching it. Every number $\geq 1$ is an [upper bound](#def-b1-reals-bounds) (the ray), and nothing smaller is, because an element of $A$ enters each [interval](#prop-b1-reals-intervals) $\intoo{1 - \varepsilon}{1}$: the two clauses of [Proposition 10.4](#prop-b1-reals-epsilon) in one picture. The [supremum](#def-b1-reals-bounds) is the left endpoint of the ray of [upper bounds](#def-b1-reals-bounds) — and the completeness axiom is precisely the guarantee that this ray always has a left endpoint.*

**Example 10.6 (Computing suprema in practice).**

Two full workouts of [Proposition 10.4](#prop-b1-reals-epsilon).

*The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $A = \{x + \frac1x : x > 0\}$.* For every $x > 0$, $x + \frac1x - 2 = \frac{(\,\sqrt x - 1/\sqrt x\,)^2}{1} \geq 0$, so $2$ is a lower bound; and $2 = 1 + \frac11 \in A$: therefore $\inf A = \min A = 2$, attained at $x = 1$. Above, $A$ is unbounded ($x + \frac1x > x$ can exceed any $M$ by [Theorem 10.10](#thm-b1-reals-archimedes)): $\sup A$ does not exist in $\R$ (it is $+\infty$ in $\overline\R$).

*The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $B = \bigl\{\frac{m}{m + n} : m, n \in
\N^*\bigr\}$.* Every element lies in $\intoo{0}{1}$, so $0$ and $1$ are bounds. Neither is attained: $\frac{m}{m+n} = 1$ would force $n = 0$. For the [supremum](#def-b1-reals-bounds), freeze $n = 1$ and let $m$ grow: $\frac{m}{m+1} = 1 - \frac{1}{m+1} > 1 - \varepsilon$ as soon as $m + 1 > \frac1\varepsilon$ (Archimedes): $\sup B = 1$. Symmetrically ($m = 1$, $n$ large), $\inf B = 0$. The closing insight: to pin a [supremum](#def-b1-reals-bounds), one well-chosen *one-parameter path* inside the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) suffices — here the path $n = 1$ — and the $\varepsilon$-characterization asks for nothing more.

**Example 10.7 (The infimum mirror).**

The [infimum](#def-b1-reals-bounds) has its own $\varepsilon$-characterization, obtained from [Proposition 10.4](#prop-b1-reals-epsilon) through $\inf A = -\sup(-A)$: $i = \inf A$ iff $i$ bounds $A$ below and, for every $\varepsilon > 0$, some $a \in A$ has $a < i + \varepsilon$. A workout with both bounds at once: let

$$
A = \Bigl\{(-1)^n + \frac1n : n \in \N^*\Bigr\}
= \Bigl\{0,\ \tfrac32,\ -\tfrac23,\ \tfrac54,\ -\tfrac45,\
\dots\Bigr\} .
$$

Even indices give $1 + \frac1n \leq \frac32$, with equality at $n = 2$: since also the odd-index values are $\leq 0 <
\frac32$, we get $\sup A = \max A = \frac32$. Odd indices give $-1 + \frac1n > -1$, decreasing toward $-1$: every element of $A$ is $> -1$, and $-1 + \varepsilon$ is beaten by $-1 +
\frac1n$ for odd $n > \frac1\varepsilon$: $\inf A = -1$, not attained. One [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets), all four behaviors on display: a [supremum](#def-b1-reals-bounds) that is a maximum, an [infimum](#def-b1-reals-bounds) that is not a minimum.

**Remark 10.8 (Common pitfalls with sup and inf).**

Four errors account for most lost points. (i) *Confusing $\sup$ and $\max$*: $\sup A$ need not belong to $A$; write $\max$ only after exhibiting an element of $A$ that is an [upper bound](#def-b1-reals-bounds). (ii) *Passing strict inequalities to the [supremum](#def-b1-reals-bounds)*: if $a <
b$ for all $a \in A$, one may only conclude $\sup A \leq b$ — witness $A = \intoo{0}{1}$, $b = 1$. (iii) *Writing $\sup A$ before checking legality*: the symbol requires $A$ nonempty and bounded above ([Method 10.18](#met-b1-reals-supproofs)); $\sup \emptyset$ and $\sup \N$ are undefined in $\R$ (the conventions of $\overline\R$ are a separate, explicit act). (iv) *Set operations*: $\sup(A \cup B) = \max(\sup A, \sup B)$ always, but nothing general holds for $A \cap B$ — it may be empty, and even when it is not, $\sup(A \cap B)$ can be far below $\min(\sup A, \sup B)$: take $A = \{0, 2\}$ and $B = \{0, 3\}$, where $\sup(A \cap B) = 0$.

**Example 10.9 (Finite sets have maxima — a lemma used silently).**

Every finite nonempty $F \subseteq \R$ has a maximum (and a minimum). Induction on the number of elements: a singleton $\{a\}$ has $\max = a$; if the claim holds for $n$-element [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) and $F$ has $n + 1$ elements, pick any $a \in F$: the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $F
\setminus \{a\}$ has a maximum $m$, and $\max F$ is $m$ if $a
\leq m$, else $a$. No completeness is involved — this is pure order plus induction, valid already in $\Q$ — yet the lemma deserves one honest [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) because the coming proofs invoke it silently: the [floor](#thm-b1-reals-floor) construction below (“a [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of integers trapped in a finite range has a greatest element”), every $\max(\abs{u_0}, \dots, \abs{u_{N-1}}, \dots)$ bound in [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq), every “take the largest of the finitely many $\delta$s” in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity). Infinite [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) are where maxima die and suprema take over: this chapter exists for the infinite case.

**Theorem 10.10 (Archimedean property).**

For every $x \in \R$ there is $n \in \N$ with $n > x$. Equivalently: for all $\varepsilon > 0$ and $y > 0$, some multiple $n\varepsilon$ exceeds $y$.

**Proof.** Suppose not: some $x$ is an [upper bound](#def-b1-reals-bounds) of $\N$. Then $s = \sup \N$ exists ([Theorem 10.2](#thm-b1-reals-sup)). By [Proposition 10.4](#prop-b1-reals-epsilon) (2) with $\varepsilon = 1$, there is $n
\in \N$ with $n > s - 1$; but then $n + 1 \in \N$ and $n + 1 > s$, contradicting that $s$ is an [upper bound](#def-b1-reals-bounds). For the second form, let $\varepsilon > 0$ and $y > 0$: the first form applied to $x
= \frac{y}{\varepsilon}$ produces $n \in \N$ with $n >
\frac{y}{\varepsilon}$, and multiplying by $\varepsilon > 0$ (which preserves strict inequalities) gives $n\varepsilon > y$. Conversely, the second form with $\varepsilon = 1$ and $y = x$ recovers the first for $x > 0$, and $n = 1$ handles $x \leq 0$: the two [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) are strictly equivalent. ∎

**Example 10.11 (Archimedes at work).**

Three immediate uses, constantly needed later. (i) *No positive real is below every $\frac1n$*: if $0 < \varepsilon$, pick $n > \frac1\varepsilon$; then $\frac1n < \varepsilon$. In other words, $\R$ contains no infinitesimals — the informal “$\frac1n$ becomes arbitrarily small” is exactly this theorem. (ii) *Explicit thresholds*: how large must $n$ be for $\frac{1}{n^2} \leq 10^{-6}$? It suffices that $n \geq 10^3$ — Archimedes guarantees such $n$ exist, and the algebra locates them. (iii) *Powers beat any bound*: $2^n \geq n +
1$ (induction), so for every $M$ some power of $2$ exceeds $M$: the geometric growth used for the dyadics in [Exercise 10.8](#exo-b1-reals-8). The closing insight: the [Archimedean property](#thm-b1-reals-archimedes) is the licence behind every phrase of the form “take $n$ large enough” — from now on we use that phrase freely, and this example is its one-time justification.

**Theorem 10.12 (Floor function).**

For every $x \in \R$ there is exactly one integer, the *floor* $\lfloor x \rfloor$, with

$$
\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1 .
$$

**Proof.** *Existence.* The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $E = \{k \in \Z : k \leq x\}$ is nonempty: by [Theorem 10.10](#thm-b1-reals-archimedes) there is $m \in \N$ with $m > -x$, and then $-m < x$, so $-m \in E$. It is bounded above (by any integer $n > x$, which exists for the same reason), so, being a [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of integers trapped in the finite range $\intint{-m}{n}$, it has a greatest element $k = \max E$. Then $k \leq x$, and $k + 1 \notin E$ means $x < k + 1$.

*Uniqueness.* If $k$ and $k'$ both satisfy the inequalities, then $k \leq x < k' + 1$ gives $k \leq k'$, and symmetrically $k'
\leq k$. ∎

**Example 10.13 (Floors in practice).**

$\lfloor 3.7 \rfloor = 3$, $\lfloor 5 \rfloor = 5$, and $\lfloor
-3.7 \rfloor = -4$: the [floor](#thm-b1-reals-floor) goes *down*, not toward $0$. Two consequences of uniqueness in [Theorem 10.12](#thm-b1-reals-floor) that we will use silently. First, for $n \in \Z$,

$$
\lfloor x + n \rfloor = \lfloor x \rfloor + n ,
$$

because $\lfloor x \rfloor + n$ is an integer satisfying the two defining inequalities for $x + n$ — and only one integer does. Second, $\lfloor \, \cdot \, \rfloor$ is nondecreasing: if $x \leq
y$ then $\lfloor x \rfloor \leq x \leq y < \lfloor y \rfloor + 1$, and an integer $< \lfloor y \rfloor + 1$ is $\leq \lfloor y
\rfloor$. Beware, however, that $\lfloor 2x \rfloor \neq 2\lfloor
x \rfloor$ in general: $x = 0.6$ gives $\lfloor 1.2 \rfloor = 1
\neq 0 = 2\lfloor 0.6 \rfloor$.

What *is* true is a worked identity worth keeping (Hermite’s, in its simplest case): for every real $x$,

$$
\lfloor x \rfloor + \Bigl\lfloor x + \frac12 \Bigr\rfloor
= \lfloor 2x \rfloor .
$$

Write $x = \lfloor x\rfloor + u$ with $u \in \intco{0}{1}$ and separate two cases. If $u < \frac12$: the left side is $\lfloor
x\rfloor + \lfloor x\rfloor = 2\lfloor x\rfloor$, and $2x =
2\lfloor x\rfloor + 2u$ with $2u \in \intco{0}{1}$, so the right side is $2\lfloor x\rfloor$ too. If $u \geq \frac12$: the left side is $\lfloor x\rfloor + (\lfloor x\rfloor + 1)$, and $2u \in
\intco{1}{2}$ makes the right side $2\lfloor x\rfloor + 1$. The closing insight: $\lfloor x + \frac12\rfloor$ is the *rounding* of $x$ to the nearest integer, so the identity says that [floor](#thm-b1-reals-floor) plus rounding equals [floor](#thm-b1-reals-floor) of the double — and the case split on the fractional part $u$ is the standard technique behind every [floor](#thm-b1-reals-floor) identity (Exercises [10.2](#exo-b1-reals-2) and [10.3](#exo-b1-reals-3) run on it too).

**Theorem 10.14 (Density of Q\QQ and of R∖Q\R \setminus \QR∖Q).**

Between any two reals $x < y$ there lie a rational and an irrational.

**Proof.** *A rational.* By [Theorem 10.10](#thm-b1-reals-archimedes), pick $n \in
\N^*$ with $n > \frac{1}{y - x}$, so $ny - nx > 1$. Let $m = \lfloor
nx \rfloor + 1$. On one side, $nx < \lfloor nx \rfloor + 1 = m$ ([Theorem 10.12](#thm-b1-reals-floor)); on the other, $m = \lfloor nx \rfloor +
1 \leq nx + 1 < ny$. Dividing by $n$: $x < \frac mn < y$.

*An irrational.* Apply the previous point to the pair $x -
\sqrt 2 < y - \sqrt 2$: some rational $q$ lies between them, and then $q + \sqrt 2 \in \intoo{x}{y}$ is irrational (if $q + \sqrt 2$ were rational, so would be $\sqrt 2$). ∎

**Example 10.15 (Running the proof of density).**

The proof is an algorithm; let us execute it on $x = 1.414$ and $y = \sqrt 2$. Since $1.4142^2 = 1.99996164 < 2$, we have $\sqrt 2
> 1.4142$, so $y - x > 0.0002$ and $\frac{1}{y - x} < 5000$: the choice $n = 5000$ is legitimate. Then $nx = 7070$, so $m = \lfloor
7070 \rfloor + 1 = 7071$, and the rational produced is

$$
\frac{m}{n} = \frac{7071}{5000} = 1.4142,
\qquad
1.414 < 1.4142 < \sqrt 2 .
$$

The closing insight: the proof needs $n$ only slightly larger than $\frac{1}{y-x}$, and returns the first multiple of $\frac 1n$ beyond $x$. Density is not an abstract miracle — it is long division in disguise, a theme developed at length in the weekend problem ([Problem 10.1](#pb-b1-reals-1)).

**Remark 10.16 (Where completeness is used next).**

[Theorem 10.2](#thm-b1-reals-sup) is the single non-algebraic axiom of this book, and every existence theorem of analysis is that axiom wearing different clothes: the monotone convergence theorem ([Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq)), the Bolzano–Weierstrass theorem ([Chapter 12](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#ch-b1-topology)), the intermediate value and extreme value theorems ([Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity)), and the very definition of the integral as a [supremum](#def-b1-reals-bounds) of lower sums ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)). The Year 3 volume builds measure theory and Hilbert spaces on the same single axiom. When a proof in the coming chapters produces a real number out of thin air, look for the hidden [supremum](#def-b1-reals-bounds).

**Remark 10.17 (Between discreteness and density).**

$\Z$ and $\Q$ sit at opposite extremes inside $\R$: around each integer there is a gap of length $1$ containing no other (discreteness — this is what makes the [floor](#thm-b1-reals-floor) well defined), while between any two reals there are infinitely many rationals (density). Remarkably, for *additive [subgroups](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup)* of $\R$ there is nothing in between: [Exercise 10.9](#exo-b1-reals-9) proves that such a [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) is either of the form $\alpha\Z$ (discrete) or dense — a dichotomy that powers the density of $\{\sin n\}$ in [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) and the constructive monster of [Problem 13.1](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#pb-b1-continuity-1). General [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets), of course, mix the behaviors freely: $\Z \cup \Q\cap\intcc{0}{1}$ is discrete far away and dense in the middle.

**Method 10.18 (Proving equalities with sup and inf).**

To prove $\sup A = s$: check that $s$ bounds $A$ above, then produce, for each $\varepsilon > 0$ (or for a sequence $\varepsilon = \frac
1n$), an element of $A$ above $s - \varepsilon$. To compare suprema, use: $A \subseteq B \implies \sup A \leq \sup B$; and for all $a, b$: $\sup(A + B) = \sup A + \sup B$, where $A + B = \{a + b\}$ ([Exercise 10.5](#exo-b1-reals-5)). Never write $\sup A$ before knowing $A$ is nonempty and bounded above.

## 10.2 Intervals

**Proposition 10.19 (Characterization of intervals).**

A subset $I \subseteq \R$ is an *interval* (one of the familiar types $\intoo{a}{b}$, $\intcc{a}{b}$, $\intco{a}{b}$, $\intoc{a}{b}$, half-lines, $\R$, $\emptyset$, singletons) if and only if it is *convex*:

$$
\forall x, y \in I,\ \forall z \in \R, \quad
x \leq z \leq y \implies z \in I .
$$

**Proof.** Every listed type is clearly convex. Conversely, let $I$ be convex and nonempty. Set $a = \inf I$ if $I$ is bounded below, else $a =
-\infty$; likewise $b = \sup I$ or $+\infty$. We claim $\intoo{a}{b}
\subseteq I \subseteq \intcc{a}{b}$ (with obvious conventions at $\pm\infty$). The second inclusion is the definition of bounds. For the first, let $z \in \intoo{a}{b}$: since $z > a$, $z$ is not a lower bound (or $a = -\infty$), so some $x \in I$ has $x < z$; similarly some $y \in I$ has $y > z$; convexity puts $z \in I$.

It remains to read off the type from the double inclusion $\intoo{a}{b} \subseteq I \subseteq \intcc{a}{b}$: the [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) squeezed between an open [interval](#prop-b1-reals-intervals) and its closure differ from $\intoo{a}{b}$ only by the presence or absence of the (finite) endpoints. Explicitly: if $a, b \in \R$, the four possibilities for $(a \in I,\ b \in I)$ give $\intoo{a}{b}$, $\intco{a}{b}$, $\intoc{a}{b}$, $\intcc{a}{b}$ (including the degenerate cases $a = b$: singleton if $a \in I$); if $a = -\infty$ and $b \in
\R$, one gets $\intoo{-\infty}{b}$ or $\intoc{-\infty}{b}$; symmetrically for $a \in \R$, $b = +\infty$; and $a = -\infty$, $b = +\infty$ gives $I = \R$. Every case is on the list: done. ∎

**Remark 10.20 (Why convexity is the right test).**

The proposition converts a *geometric* definition (a list of ten shapes) into a *one-line logical* test, and the test is what one actually uses: to prove that a [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) is an [interval](#prop-b1-reals-intervals), never chase which of the ten shapes it is — verify convexity and let the proposition sort out the type. The intermediate value theorem of [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity) will be stated exactly this way (“the continuous image of an [interval](#prop-b1-reals-intervals) is an [interval](#prop-b1-reals-intervals)”), and its proof produces the convexity, not the shape.

**Remark 10.21 (Extended real line).**

It is convenient to adjoin two symbols and work in $\overline\R = \R
\cup \{-\infty, +\infty\}$, with the conventions $\sup A = +\infty$ when $A$ is not bounded above and $\sup \emptyset = -\infty$. Then *every* subset of $\R$ has a [supremum](#def-b1-reals-bounds) in $\overline\R$ — a notational comfort used freely for limits in [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq).

**Example 10.22 (Computing in R‾\overline\RR).**

With the conventions in force: $\sup \Z = +\infty$, $\inf \Z =
-\infty$; for $A = \{n + (-1)^n n : n \in \N\} = \{0, 4, 0, 8,
\dots\} \cup \{0\}$, $\sup A = +\infty$ (the even terms $2n$ are unbounded) and $\inf A = \min A = 0$; and $\sup\emptyset =
-\infty \leq \inf\emptyset = +\infty$ — the one [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) whose [supremum](#def-b1-reals-bounds) is *smaller* than its [infimum](#def-b1-reals-bounds), a reminder that the conventions are chosen to make $\sup$ increasing and $\inf$ decreasing with respect to inclusion:

$$
A \subseteq B \implies \sup A \leq \sup B
\quad\text{and}\quad \inf A \geq \inf B ,
$$

now valid with no nonemptiness caveat. What the conventions do *not* provide is arithmetic: $+\infty + (-\infty)$ and $0
\times (+\infty)$ stay undefined, and every algebraic manipulation of suprema must first check it never forms these. The extended line is bookkeeping, not a number system.

**Example 10.23 (The supremum that escaped Q\QQ).**

Return to the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of the opening remark, $A = \{x \in \Q : x^2 <
2\}$, and compute its [supremum](#def-b1-reals-bounds) *in $\R$*. It is nonempty ($1 \in A$) and bounded above by $1.5$ (if $x > 1.5$ then $x^2 >
2.25 > 2$), so $s = \sup A$ exists. We claim $s = \sqrt 2$ (the real number built in [Exercise 10.12](#exo-b1-reals-12)). [Upper bound](#def-b1-reals-bounds): every $a \in A$ satisfies $a < \sqrt2$ — for $a \leq 0$ this is clear, and for $a > 0$, $a \geq \sqrt2$ would give $a^2 \geq 2$. Nothing smaller works: given $t < \sqrt2$, density ([Theorem 10.14](#thm-b1-reals-density)) provides a rational $q$ with $\max(1, t) < q < \sqrt 2$, and then $q^2 < 2$, so $q \in A$ exceeds $t$. By [Proposition 10.4](#prop-b1-reals-epsilon), $s = \sqrt2
\notin \Q$. The closing insight: the [supremum](#def-b1-reals-bounds) of a [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of rationals need not be rational — completeness is precisely the promise that $\R$, unlike $\Q$, never lets a [supremum](#def-b1-reals-bounds) escape; this example is the opening remark of the chapter, now proved rather than pointed at.

**Remark 10.24 (Perspectives inside this volume).**

The chapter’s three tools have distinct careers ahead. The [supremum](#def-b1-reals-bounds) runs the analysis half: monotone limits ([Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq)), the integral’s very definition ([Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration)), and, in the geometry of [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid), the distance from a point to a subspace — an [infimum](#def-b1-reals-bounds) that orthogonal projection turns into a minimum. The [floor function](#thm-b1-reals-floor) returns wherever the discrete meets the continuous: digit expansions (this chapter’s weekend problem), Dirichlet’s pigeonhole approximation ([Problem 14.1](https://one-course.com/books/math/3/en/chapter/14-differentiation#pb-b1-derivative-1)), integral comparisons of sums ([Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series)). Density arguments upgrade into a method in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity): an identity of continuous functions needs checking only on $\Q$ — half of Cauchy’s functional equation ([Problem 13.1](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#pb-b1-continuity-1)) is exactly that move. When in doubt about where a proof in this volume gets its existence [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement), the answer is nearly always: this chapter.

## 10.3 Exercises

**Exercise 10.1 ★.**

Determine (with proofs) sup, inf, max, min — when they exist — of:

$$
A = \Bigl\{\frac{1}{n} : n \in \N^*\Bigr\},
\qquad
B = \Bigl\{\frac{(-1)^n n}{n+1} : n \in \N\Bigr\},
\qquad
C = \{x \in \R : x^2 < 3\}.
$$

**Solution of Exercise 10.1.**

$A$: every element is $\leq 1$ and $1 \in A$: $\sup A = \max A = 1$. Lower bounds: $0$ bounds below; for $\varepsilon > 0$, Archimedes provides $n$ with $\frac 1n < \varepsilon$, so no positive number bounds $A$ below: $\inf A = 0$, not attained (no min).

$B$: terms $0, -\frac12, \frac23, -\frac34, \frac45, \dots$ The even terms $\frac{n}{n+1}$ ($n$ even) increase to $1$ without reaching it; the odd terms $-\frac{n}{n+1}$ decrease to $-1$. So $\sup B = 1$ and $\inf B = -1$, neither attained: no max, no min. (Bounds: $\abs{b}
< 1$ for all $b \in B$; and $\frac{n}{n+1} = 1 - \frac{1}{n+1} > 1 -
\varepsilon$ for $n$ large, similarly below.)

$C = \intoo{-\sqrt 3}{\sqrt 3}$: $\sup C = \sqrt 3$, $\inf C =
-\sqrt 3$, neither attained.

**Exercise 10.2 ★.**

Prove that for all $x, y \in \R$: $\lfloor x \rfloor + \lfloor y
\rfloor \leq \lfloor x + y \rfloor \leq \lfloor x \rfloor + \lfloor y
\rfloor + 1$, and that both bounds are attained.

**Solution of Exercise 10.2.**

Write $x = \lfloor x \rfloor + u$, $y = \lfloor y \rfloor + v$ with $u, v \in \intco{0}{1}$. Then $x + y = \lfloor x \rfloor + \lfloor y
\rfloor + (u + v)$ with $u + v \in \intco{0}{2}$. If $u + v < 1$, $\lfloor x + y\rfloor = \lfloor x\rfloor + \lfloor y \rfloor$; if $1 \leq u + v < 2$, $\lfloor x+y \rfloor = \lfloor x \rfloor +
\lfloor y \rfloor + 1$. Both cases occur: $(x, y) = (0.2,\, 0.3)$ gives the left equality, $(0.7,\, 0.8)$ the right one.

**Exercise 10.3 ★.**

Prove that for every $x \in \R$ and $n \in \N^*$: $\Bigl\lfloor \frac{\lfloor nx \rfloor}{n} \Bigr\rfloor = \lfloor x
\rfloor$.

**Solution of Exercise 10.3.**

Let $k = \lfloor x \rfloor$, so $k \leq x < k + 1$. Multiplying by $n$: $nk \leq nx < nk + n$, and taking [floors](#thm-b1-reals-floor) (an increasing operation on integers’ side): $nk \leq \lfloor nx \rfloor \leq nk + n
- 1$. Dividing by $n$: $k \leq \frac{\lfloor nx \rfloor}{n} < k + 1$, so the outer [floor](#thm-b1-reals-floor) is $k$.

**Exercise 10.4 ★.**

Let $A \subseteq B$ be nonempty subsets of $\R$, $B$ bounded. Prove $\inf B \leq \inf A \leq \sup A \leq \sup B$.

**Solution of Exercise 10.4.**

Every element of $A$ is in $B$, so $\sup B$ bounds $A$ above: therefore $\sup A \leq \sup B$ ($\sup A$ is the *least* [upper bound](#def-b1-reals-bounds)). Symmetrically $\inf B \leq \inf A$. Finally $\inf A \leq
\sup A$ because $A$ is nonempty: any $a \in A$ sits between them.

**Exercise 10.5 ★★.**

For nonempty bounded $A, B \subseteq \R$, define $A + B = \{a + b : a
\in A,\ b \in B\}$ and $-A = \{-a : a \in A\}$. Prove:

$$
\sup(A + B) = \sup A + \sup B,
\qquad
\sup(-A) = -\inf A .
$$

**Solution of Exercise 10.5.**

Let $s = \sup A$, $t = \sup B$. Every $a + b \leq s + t$: [upper bound](#def-b1-reals-bounds). For $\varepsilon > 0$, choose $a > s - \frac\varepsilon2$ and $b > t - \frac\varepsilon2$ ([Proposition 10.4](#prop-b1-reals-epsilon)): then $a +
b > s + t - \varepsilon$. By the $\varepsilon$-characterization, $\sup(A+B) = s + t$.

For $-A$: $m$ bounds $-A$ above $\iff$ $-m$ bounds $A$ below; the least [upper bound](#def-b1-reals-bounds) of $-A$ therefore corresponds to the greatest lower bound of $A$: $\sup(-A) = -\inf A$.

**Exercise 10.6 ★★.**

Let $f, g \colon E \to \R$ be bounded functions. Prove

$$
\sup_{x \in E}\, \bigl(f(x) + g(x)\bigr) \leq \sup_{x \in E} f(x) +
\sup_{x \in E} g(x),
$$

and give an example where the inequality is strict. Why does this not contradict [Exercise 10.5](#exo-b1-reals-5)?

**Solution of Exercise 10.6.**

For every $x$: $f(x) + g(x) \leq \sup f + \sup g$; taking the sup of the left side gives the inequality. Strict example: $E = \{0, 1\}$, $f = \mathbf{1}_{\{0\}}$ (value $1$ at $0$, else $0$), $g =
\mathbf{1}_{\{1\}}$: $\sup(f + g) = 1 < 2 = \sup f + \sup g$.

No contradiction with [Exercise 10.5](#exo-b1-reals-5): there, $a \in A$ and $b
\in B$ vary *independently*; here the same $x$ feeds both $f$ and $g$ — the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\{f(x) + g(x) : x \in E\}$ is smaller than the set $\{f(x) + g(y) : x, y \in E\}$.

**Exercise 10.7 ★★.**

Prove that $\sqrt 2 + \sqrt 3$ is irrational. *(Square it and use the irrationality of $\sqrt 6$, to be proved via [Exercise 6.7](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#exo-b1-arith-7).)*

**Solution of Exercise 10.7.**

$\sqrt 6$ is irrational: $6 = 2 \times 3$ is not a perfect square, and $v_2(6q^2) = 1 + 2v_2(q)$ odd prevents $6q^2 = r^2$ (as in [Exercise 6.7](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#exo-b1-arith-7)). Now suppose $x = \sqrt 2 + \sqrt 3 \in \Q$. Then $x^2 = 5 + 2\sqrt 6 \in \Q$, so $\sqrt 6 = \frac{x^2 - 5}{2} \in
\Q$: contradiction. Hence $\sqrt 2 + \sqrt 3 \notin \Q$.

**Exercise 10.8 ★★.**

Prove that the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $D = \bigl\{\frac{m}{2^n} : m \in \Z,\ n \in
\N\bigr\}$ of dyadic rationals is dense in $\R$: between any two reals lies a dyadic rational.

**Solution of Exercise 10.8.**

Let $x < y$. Pick $n \in \N$ with $2^n > \frac{1}{y - x}$ (Archimedes: $2^n \geq n + 1$ by an easy induction, so some power of $2$ exceeds any real). Then, as in the proof of [Theorem 10.14](#thm-b1-reals-density) with $2^n$ in place of $n$: $m = \lfloor 2^n x \rfloor + 1$ satisfies $x <
\frac{m}{2^n} < y$. So $D$ is dense.

**Exercise 10.9 ★★★.**

Let $G$ be a [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) of $(\R, +)$ with $G \neq \{0\}$. Set $\alpha =
\inf\,(G \cap \intoo{0}{+\infty})$. Prove:

1. if $\alpha > 0$ , then $G = \alpha\Z$ ;
2. if $\alpha = 0$ , then $G$ is dense in $\R$ .

Deduce that $\Z + \sqrt 2\,\Z$ is dense in $\R$.

**Solution of Exercise 10.9.**

1. Suppose $\alpha > 0$ . First, $\alpha \in G$ . Suppose not: by the $\varepsilon$ -characterization of the [infimum](#def-b1-reals-bounds) with $\varepsilon = \alpha$ , there is $g \in G$ with $\alpha < g <  2\alpha$ (strict on the left since $\alpha \notin G$ ); then, with $\varepsilon = g - \alpha$ , there is $h \in G$ with $\alpha < h < g$ . Now $g - h \in G$ and $0 < g - h < g -  \alpha < \alpha$ : an element of $G \cap \intoo{0}{+\infty}$ below its [infimum](#def-b1-reals-bounds) , absurd. So $\alpha \in G$ , and $\alpha\Z  \subseteq G$ ( $G$ is a [group](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-group) ). Conversely, for $x \in G$ , let $k = \lfloor x/\alpha \rfloor$ : then $x - k\alpha \in G$ and $0 \leq x - k\alpha < \alpha$ , and the definition of $\alpha$ forces $x - k\alpha = 0$ . Hence $G = \alpha\Z$ .
2. Suppose $\alpha = 0$ , and let $x < y$ . There is $g \in G$ with $0 < g < y - x$ . The multiple $kg$ with $k = \lfloor x/g  \rfloor + 1$ satisfies $x < kg \leq x + g < y$ , and $kg \in  G$ : density.

$G = \Z + \sqrt 2\,\Z$ is a [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) of $(\R, +)$. It is not of the form $\alpha\Z$: otherwise $1 = p\alpha$ and $\sqrt 2 = q\alpha$ ($p, q \in \Z$) would give $\sqrt 2 = \frac qp \in \Q$, contradiction. By the dichotomy, $G$ is dense in $\R$.

**Exercise 10.10 ★★★.**

For $A, B$ nonempty [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of positive reals, let $AB = \{ab : a \in
A, b \in B\}$. Prove $\sup(AB) = \sup A \cdot \sup B$ (bounded case), and show by example that positivity is essential.

**Solution of Exercise 10.10.**

Let $s = \sup A > 0$, $t = \sup B > 0$. For $a \in A$, $b \in B$: $ab
\leq st$ (multiplying inequalities between *positive* numbers). For $0 < \varepsilon < \min(s, t)$: choose $a > s - \varepsilon$ and $b > t - \varepsilon$; then

$$
ab > (s - \varepsilon)(t - \varepsilon)
= st - \varepsilon(s + t) + \varepsilon^2
> st - \varepsilon (s + t),
$$

and $\varepsilon(s+t)$ can be made arbitrarily small: by the $\varepsilon$-characterization (in the form: no number $< st$ bounds $AB$ above), $\sup AB = st$.

Positivity is essential: $A = B = \{-1, 0\}$ gives $AB = \{0, 1\}$, $\sup AB = 1$, while $\sup A \cdot \sup B = 0 \times 0 = 0$.

**Exercise 10.11 ★★.**

For a nonempty bounded $A \subseteq \R$, define the *diameter*

$$
\operatorname{diam} A = \sup\,\{\abs{a - a'} : a, a' \in A\} .
$$

Prove that $\operatorname{diam} A = \sup A - \inf A$, and that $\intcc{\inf A}{\sup A}$ is the smallest closed [interval](#prop-b1-reals-intervals) containing $A$.

**Solution of Exercise 10.11.**

Write $s = \sup A$, $i = \inf A$. For $a, a' \in A$: $a \leq s$ and $a' \geq i$ give $a - a' \leq s - i$; by symmetry $\abs{a -
a'} \leq s - i$, so $s - i$ bounds the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of gaps above. For $\varepsilon > 0$, choose $a > s - \frac\varepsilon2$ and $a' < i
+ \frac\varepsilon2$ ([Proposition 10.4](#prop-b1-reals-epsilon) and its mirror for the [infimum](#def-b1-reals-bounds)): then $\abs{a - a'} \geq a - a' > s - i -
\varepsilon$. By the $\varepsilon$-characterization, $\operatorname{diam} A = s - i$.

Every $a \in A$ satisfies $i \leq a \leq s$, so $A \subseteq
\intcc{i}{s}$, a closed [interval](#prop-b1-reals-intervals) of length $\operatorname{diam}
A$. If a closed [interval](#prop-b1-reals-intervals) $\intcc{u}{v}$ contains $A$, then $v$ is an [upper bound](#def-b1-reals-bounds) and $u$ a lower bound of $A$, so $u \leq i$ and $v
\geq s$: $\intcc{i}{s} \subseteq \intcc{u}{v}$. Hence $\intcc{i}{s}$ is the smallest one.

**Exercise 10.12 ★★★.**

Let $y > 0$ and $E = \{x \geq 0 : x^2 \leq y\}$. Prove that $E$ is nonempty and bounded above, and that $s = \sup E$ satisfies $s^2 =
y$ *(rule out $s^2 < y$ and $s^2 > y$ by exhibiting, in each case, a small $h > 0$ contradicting the definition of the [supremum](#def-b1-reals-bounds))*. Deduce that every $y > 0$ has a unique square root $\sqrt y > 0$ and that $y \mapsto \sqrt y$ is increasing on $\intoo{0}{+\infty}$.

**Solution of Exercise 10.12.**

$0 \in E$, so $E \neq \emptyset$. If $x > \max(1, y)$ then $x^2 >
x > y$, so $E$ is bounded above by $\max(1, y)$: $s = \sup E$ exists ([Theorem 10.2](#thm-b1-reals-sup)), and $s \geq \min(1, y) > 0$ because $\min(1, y) \in E$: indeed if $y \geq 1$ then $1^2 = 1
\leq y$, and if $y < 1$ then $y^2 < y$.

*$s^2 < y$ is impossible.* Choose $0 < h < 1$ with $h <
\frac{y - s^2}{2s + 1}$. Then

$$
(s + h)^2 = s^2 + 2sh + h^2 \leq s^2 + (2s + 1)h < y ,
$$

so $s + h \in E$, contradicting that $s$ bounds $E$ above.

*$s^2 > y$ is impossible.* Choose $0 < h < s$ with $h <
\frac{s^2 - y}{2s}$. Then $(s - h)^2 = s^2 - 2sh + h^2 > s^2 -
2sh > y$; every $x \in E$ satisfies $x^2 \leq y < (s - h)^2$, hence $x < s - h$ (both are $\geq 0$): $s - h$ is an [upper bound](#def-b1-reals-bounds) of $E$ smaller than $s$, contradicting leastness.

Therefore $s^2 = y$. Uniqueness: if $0 < s < s'$ then $s^2 <
s'^2$, so two distinct positive roots cannot both square to $y$. Monotonicity: if $0 < y < y'$, then $\sqrt y \neq \sqrt{y'}$, and $\sqrt y > \sqrt{y'}$ would give $y > y'$ by squaring: so $\sqrt
y < \sqrt{y'}$.

## 10.4 Problem: Digit expansions and the rhythm of the rationals

**Problem 10.1.**

Weekend problem — $b$-adic expansions: existence, uniqueness, and periodicity characterizes $\Q$

Every real in $\intco{0}{1}$ has a digit expansion in every base $b \geq 2$; the expansion is unique once trailing strings of the digit $b - 1$ are outlawed; and it is eventually periodic exactly when the number is rational. This problem proves all three facts from the completeness axiom alone — no sequences, no series: only the [supremum](#def-b1-reals-bounds), the [Archimedean property](#thm-b1-reals-archimedes) and the [floor function](#thm-b1-reals-floor) — and closes with Cantor’s diagonal argument in digit form. Throughout, $b \geq 2$ is a fixed integer (the *base*), a *digit* is an element of $\intint{0}{b-1}$, and a digit string $(d_n)_{n \geq 1}$ is *proper* when it is not eventually equal to $b - 1$ (that is: for every $N$ there is $n >
N$ with $d_n \leq b - 2$).

**Part I — Digits by hand.** Long division of $p$ by $q$ in base $b$: multiply the current remainder by $b$, divide by $q$, record the quotient as the next digit, keep the remainder.

1. In base $10$ , run the algorithm on $\frac 18$ and on $\frac 17$ , recording at each step the digit *and* the remainder. Check that the remainders for $\frac 17$ cycle through $1, 3, 2, 6, 4, 5$ and that the digits $142857$ then repeat forever.
2. Compute the base- $2$ expansions of $\frac 13$ and of $\frac{5}{16}$ , and the base- $3$ expansion of $\frac 12$ . Observe: one number terminates, the two others repeat — and $\frac 12$ , so tame in base $10$ , repeats forever in base $3$ .
3. For $x = \frac pq \in \intco{0}{1}$ in lowest terms, show that the digits produced by the algorithm are eventually all $0$ if and only if the remainder $b^N p \bmod q$ vanishes for some $N$ , if and only if $q$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) some power $b^N$ , if and only if every [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) factor of $q$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $b$ . Check: $\frac{1}{20}$ terminates in base $10$ , not in base $3$ .
4. Define the *truncation* $s_n = \lfloor b^n x \rfloor  / b^n$ . For $x = \sqrt 2$ and $b = 10$ , compute $s_0,  \dots, s_4$ by verifying at each step that two consecutive squares straddle $2$ (for instance $1.4142^2  = 1.99996164 < 2 < 2.00024449 = 1.4143^2$ ), and check $s_n \leq \sqrt 2 < s_n + 10^{-n}$ each time.

**Part II — Existence, from the [supremum](#def-b1-reals-bounds).** Fix $x \in \intco{0}{1}$ and set $A_n = \lfloor b^n x \rfloor$ and $d_n = A_n - b\,A_{n-1}$ for $n \geq 1$.

5. Show $A_0 = 0$ and $b\,A_{n-1} \leq A_n \leq b\,A_{n-1} +  b - 1$ ; conclude that each $d_n$ is a digit.
6. Show that $s_n := A_n b^{-n}$ satisfies $$s_n = \sum_{k=1}^{n} d_k\,b^{-k}  \qquad\text{and}\qquad  s_n \leq x < s_n + b^{-n} .$$
7. Prove $b^n \geq n + 1$ by induction, then show that $(s_n)$ is nondecreasing and that $x = \sup_n s_n$ *(use [Proposition 10.4](#prop-b1-reals-epsilon) and [Theorem 10.10](#thm-b1-reals-archimedes))* .
8. Show that the string $(d_n)$ is proper: if $d_k = b - 1$ for all $k > N$ , compute $s_n$ for $n > N$ by a finite geometric sum and contradict question 6.
9. Conversely, let $(e_n)_{n \geq 1}$ be any proper digit string and $t_n = \sum_{k=1}^n e_k b^{-k}$ . Show that $y  = \sup_n t_n$ exists, lies in $\intco{0}{1}$ , and satisfies $t_n \leq y < t_n + b^{-n}$ for every $n$ *(for the strict inequality, use a digit $e_m \leq b  - 2$ with $m > n$)* . Deduce $\lfloor b^n y \rfloor = b^n  t_n$ , then that the digits of $y$ , in the sense of question 5, are exactly the $e_n$ .

**Part III — Uniqueness, order, shift.**

10. Assemble questions 5–9 into the *$b$-adic expansion theorem* : the [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $x \mapsto (d_n)$ and $(e_n) \mapsto  \sup_n t_n$ are mutually inverse bijections between $\intco{0}{1}$ and the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of proper digit strings. In particular no two distinct proper strings have the same value.
11. Now allow improper strings. Show that a string with $e_n  = b - 1$ for all $n > M$ (with $M \geq 0$ minimal) has value $t_M + b^{-M}$ ; conclude that $0.999\dots = 1$ in base $10$ , and that the reals with two digit representations are exactly the $b$ -adic fractions $m/b^N \in \intoo{0}{1}$ — every other real has just one, even among improper strings.
12. Prove that the bijection of question 10 is order-preserving for the lexicographic order: if the proper strings of $x$ and $y$ first differ at index $m$ , then $x < y$ if and only if $d_m < e_m$ .
13. (Shift lemma) Let $x \in \intco{0}{1}$ have digits $(d_n)$ . Show that the fractional part of $bx$ has digits $(d_{n+1})_{n \geq 1}$ *(compute $\lfloor b^n(bx -  A_1)\rfloor$ using $\lfloor u - K \rfloor = \lfloor u  \rfloor - K$ for integer $K$)* , and deduce by induction that the fractional part of $b^m x$ has digits $(d_{n+m})_{n \geq 1}$ .

**Part IV — Rationality is periodicity.** Let $x
= \frac pq \in \intco{0}{1}$ be in lowest terms and $r_n = b^n p
\bmod q$ the remainder of the Euclidean division of $b^n p$ by $q$.

14. Show $A_n = \dfrac{b^n p - r_n}{q}$ and $r_n =  (b\,r_{n-1}) \bmod q$ .
15. Show $d_n = \Bigl\lfloor \dfrac{b\,r_{n-1}}{q}  \Bigr\rfloor$ : each digit is a function of the previous remainder alone. This is exactly the long division of Part I.
16. Apply the pigeonhole principle ( [Corollary 2.3](https://one-course.com/books/math/3/en/chapter/2-counting#cor-b1-counting-pigeonhole) ) to $r_0, \dots, r_q$ and conclude: the expansion of every rational is eventually periodic, with preperiod and period at most $q$ .
17. Conversely, suppose the digits of $y \in \intco{0}{1}$ are *purely* periodic: $d_{n+T} = d_n$ for all $n  \geq 1$ . Using the shift lemma and the uniqueness of question 10, show that the fractional part of $b^T y$ equals $y$ , and deduce $(b^T - 1)\,y \in \N$ : thus $y$ is rational with denominator dividing $b^T - 1$ . Verify the mechanism on $0.(142857)$ : $142857 \times 7 = 999999$ .
18. Treat the eventually periodic case by shifting, and state the *periodicity criterion* : $x \in \intco{0}{1}$ is rational if and only if its proper $b$ -adic expansion is eventually periodic — in one base if and only if in all.
19. For $x = \frac 1q$ with $\gcd(q, b) = 1$ , show that the expansion is purely periodic and that its least period is the least $T \geq 1$ with $b^T \equiv 1 \pmod q$ (the multiplicative order of $b$ modulo $q$ ). Check that for $q = 7$ , $b = 10$ the powers of $10$ modulo $7$ run through $3, 2, 6, 4, 5, 1$ : order $6$ , matching question 1.

**Part V — Dividends and the diagonal.**

20. Let $x^*$ be the real of $\intco{0}{1}$ whose base- $10$ digits are $1$ at the triangular positions $\frac{j(j +  1)}{2}$ ( $j \geq 1$ ) and $0$ elsewhere: $x^* =  0.101001000100001\dots$ Show that its digit string is proper but not eventually periodic *(a period $T$ would force ones at gaps at most $T$, but the gaps grow)* , and conclude that $x^*$ is irrational: a number proved irrational by pure rhythm.
21. Show that for every base $b$ the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\{m/b^n : m \in \Z,  n \in \N\}$ is dense in $\R$ (generalizing [Exercise 10.8](#exo-b1-reals-8) ), and that every rational $\frac pq  \in \intoo{0}{1}$ has a *terminating* expansion in base $q$ . Moral: terminating is a property of the pair (number, base); periodicity — rationality — is intrinsic.
22. (Cantor’s diagonal) Let $k \mapsto x_k$ be any [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) from $\N^*$ to $\intco{0}{1}$ . Define the digit string $e_k =  1$ if the $k$ -th digit of $x_k$ differs from $1$ , and $e_k = 2$ otherwise. Show that $(e_k)$ is proper, that its value $y$ lies in $\intco{0}{1}$ , and that $y \neq  x_k$ for every $k$ . Conclude: no [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\N^* \to  \intco{0}{1}$ is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) . (The vocabulary of countability, and this theorem’s proper home, is [Chapter 12](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#ch-b1-topology) .)
23. Show that if the proper expansions of $x$ and $y$ agree up to index $n$ then $\abs{x - y} < b^{-n}$ , and disprove the converse with $x = 0.1$ , $y = 0.0999$ in base $10$ : closeness of numbers does not force agreement of digits. Which reals are to blame?
24. Run Part IV on $x = \frac{1}{10}$ in base $b = 2$ : compute remainders and digits until they cycle, and conclude $\frac{1}{10} = (0.0\overline{0011})_2$ , with preperiod $1$ and period $4$ . Explain via question 3 why no finite binary string will ever equal $\frac{1}{10}$ — the reason a computer’s floating-point $0.1 + 0.2$ is not exactly $0.3$ .
25. Synthesis. In one sentence each: where did the proof use (i) completeness, (ii) the [Archimedean property](#thm-b1-reals-archimedes) , (iii) the uniqueness clause of the [floor](#thm-b1-reals-floor) , (iv) the pigeonhole principle? And the moral: $\intco{0}{1}$ is faithfully coded by proper digit strings, rationality reads off as periodicity — yet analysis prefers the [supremum](#def-b1-reals-bounds) to the digits. Why? (Think about adding two digit strings.)

**Solution of Problem 10.1.**

**1.** For $\frac 18$: $10 = 8 \cdot 1 + 2$, $20 = 8 \cdot 2
+ 4$, $40 = 8 \cdot 5 + 0$; digits $1, 2, 5$, remainder $0$, then only zeros: $\frac 18 = 0.125$. For $\frac 17$: $10 = 7 \cdot 1 +
3$, $30 = 7 \cdot 4 + 2$, $20 = 7 \cdot 2 + 6$, $60 = 7 \cdot 8 +
4$, $40 = 7 \cdot 5 + 5$, $50 = 7 \cdot 7 + 1$: digits $1, 4, 2,
8, 5, 7$, remainders $3, 2, 6, 4, 5, 1$. The remainder has returned to $r = 1$, so the six steps repeat verbatim forever: $\frac 17 = 0.(142857)$, remainders cycling through $1, 3, 2, 6,
4, 5$.

**2.** $\frac 13$ in base $2$ ($r_0 = 1$): $2 = 3 \cdot 0 +
2$, $4 = 3 \cdot 1 + 1$, and $r = 1$ recurs: $\frac 13 =
(0.\overline{01})_2$. $\frac{5}{16}$ in base $2$: $10 = 16 \cdot
0 + 10$, $20 = 16 \cdot 1 + 4$, $8 = 16 \cdot 0 + 8$, $16 = 16
\cdot 1 + 0$: $\frac{5}{16} = (0.0101)_2$, terminating. $\frac 12$ in base $3$: $3 = 2 \cdot 1 + 1$, and $r = 1$ recurs at once: $\frac 12 = (0.\overline{1})_3$.

**3.** The algorithm’s remainder after $N$ steps is $r_N =
b^N p \bmod q$ (proved formally in question 14; here it is the observation that each step multiplies the remainder by $b$ and reduces mod $q$). All later digits are $0$ if and only if some $r_N = 0$, i.e. $q \mid b^N p$; since $\gcd(p, q) = 1$, Gauss’s lemma gives $q \mid b^N$. If $q \mid b^N$, every [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) factor of $q$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $b^N$, hence [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $b$ (primality). Conversely, if every [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) of $q = p_1^{a_1} \cdots p_r^{a_r}$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $b$, then with $A = \max_i a_i$ each $p_i^{a_i}$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $b^A$, and the $p_i^{a_i}$ are pairwise [coprime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#cor-b1-arith-bezout), so $q \mid b^A$. For $q = 20 =
2^2 \cdot 5$: both [primes](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) divide $10$ ($\frac{1}{20} = 0.05$), but $2 \nmid 3$, so $\frac{1}{20}$ repeats forever in base $3$.

**4.** $1^2 = 1 < 2 < 4 = 2^2$ gives $s_0 = 1$. Then $1.4^2
= 1.96 < 2 < 2.25 = 1.5^2$: $\lfloor 10\sqrt 2 \rfloor = 14$, $s_1 = 1.4$. Next $1.41^2 = 1.9881 < 2 < 2.0164 = 1.42^2$: $s_2 =
1.41$; $1.414^2 = 1.999396 < 2 < 2.002225 = 1.415^2$: $s_3 =
1.414$; $1.4142^2 = 1.99996164 < 2 < 2.00024449 = 1.4143^2$: $s_4
= 1.4142$. In each case the displayed inequalities say exactly $s_n \leq \sqrt 2 < s_n + 10^{-n}$, which is the definition of the [floor](#thm-b1-reals-floor) of $10^n \sqrt 2$.

**5.** $A_0 = \lfloor x \rfloor = 0$ since $0 \leq x < 1$. From $A_{n-1} \leq b^{n-1} x < A_{n-1} + 1$, multiply by $b$:

$$
b\,A_{n-1} \leq b^n x < b\,A_{n-1} + b .
$$

The integer $b\,A_{n-1}$ is $\leq b^n x$, so $b\,A_{n-1} \leq
A_n$; and $b^n x < b\,A_{n-1} + b$ with $b\,A_{n-1} + b$ an integer forces $A_n \leq b\,A_{n-1} + b - 1$. Hence $0 \leq d_n =
A_n - b\,A_{n-1} \leq b - 1$: a digit.

**6.** Telescoping: $d_k b^{-k} = A_k b^{-k} - A_{k-1}
b^{-(k-1)}$, so

$$
\sum_{k=1}^n d_k b^{-k} = A_n b^{-n} - A_0 = s_n .
$$

Dividing $A_n \leq b^n x < A_n + 1$ by $b^n$ gives $s_n \leq x <
s_n + b^{-n}$.

**7.** Induction: $b^0 = 1 \geq 1$, and $b^{n+1} = b \cdot
b^n \geq 2(n + 1) \geq n + 2$. Monotonicity: $s_n - s_{n-1} = d_n
b^{-n} \geq 0$. Each $s_n \leq x$ (question 6): $x$ is an [upper bound](#def-b1-reals-bounds) of $\{s_n\}$. For $\varepsilon > 0$, the [Archimedean property](#thm-b1-reals-archimedes) provides $n$ with $n + 1 > \frac1\varepsilon$, hence $b^{-n} < \varepsilon$, and then $s_n > x - b^{-n} > x -
\varepsilon$ by question 6. By [Proposition 10.4](#prop-b1-reals-epsilon), $x =
\sup_n s_n$.

**8.** Suppose $d_k = b - 1$ for all $k > N$. For $n > N$, the finite geometric sum gives

$$
s_n = s_N + (b - 1)\sum_{k=N+1}^{n} b^{-k}
= s_N + b^{-N} - b^{-n} .
$$

So $x \geq s_n = s_N + b^{-N} - b^{-n}$ for every $n$; letting the last term shrink below any $\varepsilon$ (question 7), $x
\geq s_N + b^{-N}$. But question 6 at rank $N$ says $x < s_N +
b^{-N}$: contradiction. The string $(d_n)$ is proper.

**9.** Bounded: $t_n \leq (b-1)\sum_{k=1}^n b^{-k} = 1 -
b^{-n} < 1$, and $(t_n)$ is nondecreasing, so $y = \sup t_n$ exists with $0 \leq y \leq 1$. Fix $n$. For the two-sided estimate: $t_n \leq y$ is clear. By properness pick $m > n$ with $e_m \leq b - 2$. For $p \geq m$:

$$
t_p - t_n = \sum_{k=n+1}^{p} e_k b^{-k}
\leq (b^{-n} - b^{-p}) - b^{-m} < b^{-n} - b^{-m},
$$

the middle sum losing at least $b^{-m}$ against the all-$(b-1)$ maximum; for $p \leq m$, $t_p \leq t_m \leq t_n + b^{-n} -
b^{-m}$ as well (monotonicity plus the case $p = m$). Hence every $t_p \leq t_n + b^{-n} - b^{-m}$, so $y \leq t_n + b^{-n} -
b^{-m} < t_n + b^{-n}$. (With $n = 0$: $y < 1$, so $y \in
\intco{0}{1}$.) Now $b^n t_n = \sum_{k \leq n} e_k b^{n-k}$ is an integer, and $b^n t_n \leq b^n y < b^n t_n + 1$: so $\lfloor b^n
y \rfloor = b^n t_n$. Finally the digits of $y$: $d_n(y) = b^n
t_n - b \cdot b^{n-1} t_{n-1} = b^n(t_n - t_{n-1}) = e_n$.

**10.** Question 9 says: (value of string) has digits (the string); questions 5–8 say: (digits of $x$) form a proper string whose truncations have [supremum](#def-b1-reals-bounds) $x$ (question 7). So the two [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) compose to the identity in both orders: they are mutually inverse bijections between $\intco{0}{1}$ and the proper strings. If two proper strings had equal value, applying the digit [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) would make them equal: uniqueness. This is the *$b$-adic expansion theorem*.

**11.** Let $e_n = b - 1$ for $n > M$, $M \geq 0$ minimal. As in question 8, $t_n = t_M + b^{-M} - b^{-n}$ for $n \geq M$, so the value is $\sup t_n = t_M + b^{-M}$. If $M = 0$ the value is $0 + 1 = 1$: in base $10$, $0.999\dots = 1$ exactly — not approximately. If $M \geq 1$, minimality gives $e_M \leq b - 2$, and the value is

$$
t_M + b^{-M} = \frac{b^M t_M + 1}{b^M} \in \intoo{0}{1},
$$

a $b$-adic fraction, whose *proper* expansion is $e_1 \dots
e_{M-1}\,(e_M + 1)\,000\dots$ (a terminating string is proper, and its value is the same number). Conversely a real with two representations must have one improper (properness pins down the representation, question 10), hence be of this form. And each $m/b^N \in \intoo{0}{1}$, written with last nonzero digit $d_N$, does have the improper twin $d_1 \dots d_{N-1}(d_N -
1)(b-1)(b-1)\dots$: exactly the $b$-adic fractions carry two names, all other reals one.

**12.** Say the strings agree up to $m - 1$, with common truncation $P = s_{m-1}$, and $d_m < e_m$. By question 9 (strict upper estimate at rank $m$), $x < P + d_m b^{-m} + b^{-m} = P +
(d_m + 1)b^{-m} \leq P + e_m b^{-m} \leq y$, the last step because $P + e_m b^{-m}$ is $y$’s truncation $t_m \leq y$. So $d_m < e_m \implies x < y$; exchanging roles, $e_m < d_m \implies
y < x$; and since the strings differ at $m$, one of the two holds. Both directions follow.

**13.** Let $z = bx - A_1 \in \intco{0}{1}$ (indeed $A_1
\leq bx < A_1 + 1$). For $n \geq 0$: $b^n z = b^{n+1} x - b^n
A_1$ with $b^n A_1 \in \Z$, so by $\lfloor u - K \rfloor =
\lfloor u \rfloor - K$ ($K$ integer),

$$
A_n(z) = A_{n+1}(x) - b^n A_1(x) .
$$

Hence $d_n(z) = A_n(z) - b\,A_{n-1}(z) = A_{n+1} - b^n A_1 -
b\,A_n + b^n A_1 = d_{n+1}(x)$. So the fractional part of $bx$ carries the shifted digits; iterating $m$ times, the fractional part of $b^m x$ has digits $(d_{n+m})_{n \geq 1}$.

**14.** Euclidean division: $b^n p = q\,Q_n + r_n$ with $0
\leq r_n < q$. Divide by $q$: $b^n x = Q_n + \frac{r_n}{q}$ with $0 \leq \frac{r_n}{q} < 1$, so $Q_n = \lfloor b^n x \rfloor =
A_n$, giving $A_n = \frac{b^n p - r_n}{q}$. For the recurrence: $b^n p = b(q\,A_{n-1} + r_{n-1}) = q\,(b\,A_{n-1}) +
b\,r_{n-1}$, so $b^n p$ and $b\,r_{n-1}$ differ by a multiple of $q$: $r_n = (b\,r_{n-1}) \bmod q$.

**15.** Divide $b\,r_{n-1}$ by $q$: $b\,r_{n-1} = q\,c +
r_n$ with $c = \lfloor b\,r_{n-1}/q \rfloor$. Substituting into the display of question 14: $b^n p = q(b\,A_{n-1} + c) + r_n$, and uniqueness of Euclidean division identifies $A_n = b\,A_{n-1}
+ c$, that is $d_n = c = \lfloor b\,r_{n-1}/q \rfloor$. Digit $n$ depends only on $r_{n-1}$ — the long division loop of Part I, now certified.

**16.** The $q + 1$ remainders $r_0, \dots, r_q$ take values in the $q$-element [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\intint{0}{q-1}$: by the pigeonhole principle ([Corollary 2.3](https://one-course.com/books/math/3/en/chapter/2-counting#cor-b1-counting-pigeonhole)) two coincide, say $r_N = r_{N+T}$ with $0 \leq N < N + T \leq q$. Since $r_n$ determines $r_{n+1}$ (question 14), induction gives $r_{n+T} =
r_n$ for all $n \geq N$; since $r_{n-1}$ determines $d_n$ (question 15), $d_{n+T} = d_n$ for all $n \geq N + 1$. Every rational’s expansion is eventually periodic, with preperiod $\leq
q$ and period $\leq q$.

**17.** The digits of the fractional part of $b^T y$ are $(d_{n+T}) = (d_n)$ (shift lemma, then pure periodicity): the same proper string as $y$. By question 10 the values are equal: $b^T y - \lfloor b^T y \rfloor = y$, so $(b^T - 1)\,y = \lfloor
b^T y \rfloor = A_T \in \N$ and

$$
y = \frac{A_T}{b^T - 1} ,
$$

rational with denominator dividing $b^T - 1$; the numerator $A_T$ is the integer whose base-$b$ digits are $d_1 \dots d_T$. Check: $0.(142857) = \frac{142857}{999999}$, and $142857 \times 7 =
999999$, so this is $\frac 17$.

**18.** If $d_{n+T} = d_n$ for $n > N$, the fractional part $z$ of $b^N x$ has digits $(d_{N+n})_{n\geq1}$ (shift lemma), which are purely periodic; by question 17, $z \in \Q$. Then $b^N
x = A_N + z$ gives $x = (A_N + z)/b^N \in \Q$. With question 16: $x$ rational $\iff$ expansion eventually periodic. The right side mentions the base, the left does not: periodicity in one base is equivalent to rationality, hence to periodicity in every base.

**19.** For $x = \frac 1q$, $r_n = b^n \bmod q$. If $\gcd(b, q) = 1$, then $r_T = r_0 = 1$ if and only if $b^T \equiv
1 \pmod q$; such $T$ exists (pigeonhole gives $b^i \equiv b^j$, $i < j$, and $b$ is invertible modulo $q$, so $b^{j-i} \equiv
1$), and the least one — the multiplicative order — makes the remainders, hence the digits, purely periodic of period $T$. No smaller period is possible: a period $T'$ would give $(b^{T'}
- 1)\frac1q \in \N$ (question 17), i.e. $q \mid b^{T'} - 1$. For $q = 7$, $b = 10$: $10 \equiv 3$, $10^2 \equiv 2$, $10^3
\equiv 6$, $10^4 \equiv 4$, $10^5 \equiv 5$, $10^6 \equiv 1
\pmod 7$: order $6$, and indeed $\frac 17$ has period six.

**20.** The string has infinitely many $0$s ($x^*$’s digits are mostly zero), so it is proper, and $x^*$ is well defined (question 9). Suppose the digits eventually periodic with period $T$ beyond $N$. Infinitely many digits equal $1$ (one per triangular number), so some $1$ sits at a position $j > N$; then periodicity puts a $1$ at every position $j + kT$: from $j$ onward, gaps between consecutive $1$s are at most $T$. But the $1$s sit exactly at the triangular numbers, whose consecutive gaps $\frac{(j+1)(j+2)}{2} - \frac{j(j+1)}{2} = j + 1$ exceed $T$ eventually: contradiction. Not eventually periodic, so by question 18, $x^* \notin \Q$ — irrationality read off the rhythm of the digits alone.

**21.** Given $x < y$, question 7 provides $n$ with $b^{-n}
< y - x$; set $m = \lfloor b^n x \rfloor + 1$. Then $b^n x < m
\leq b^n x + 1 < b^n y$, so $x < \frac{m}{b^n} < y$: density, for every base at once ($b = 2$ recovers [Exercise 10.8](#exo-b1-reals-8)). For $\frac pq \in \intoo{0}{1}$ in base $b = q$: the first digit is $\lfloor q \cdot \frac pq \rfloor =
p$ and the fractional part of $q \cdot \frac pq = p$ is $0$: all later digits vanish, a terminating expansion $\frac pq =
(0.p)_q$. Terminating depends on the base; periodicity — rationality — does not (question 18).

**22.** Each $e_k \in \{1, 2\}$ is a digit of base $10$, and the string never ends in all $9$s: proper. Its value $y$ lies in $\intco{0}{1}$ and has digits exactly $(e_k)$ (question 9). Fix $k$: the $k$-th digit of $y$ is $e_k$, chosen $\neq$ the $k$-th digit of $x_k$, so the proper strings of $y$ and $x_k$ differ, so $y \neq x_k$ (question 10: the coding is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj)). Thus $y$ is in no list: no [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\N^* \to \intco{0}{1}$ is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj). The reals, unlike the rationals, cannot be enumerated — uncountability, whose theory [Chapter 12](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#ch-b1-topology) develops.

**23.** If the expansions agree up to $n$, then $x$ and $y$ have the same truncation $s_n$, and question 6 puts both in $\intco{s_n}{s_n + b^{-n}}$, an [interval](#prop-b1-reals-intervals) of length $b^{-n}$: $\abs{x - y} < b^{-n}$. Converse: $x = 0.1$ and $y = 0.0999$ (terminating, hence proper) satisfy $\abs{x - y} = 10^{-4} <
10^{-3}$, yet their expansions differ at the very first digit. The culprits are the $b$-adic fractions of question 11: near them, a tiny move flips every displayed digit ($0.0999 \to
0.1000$), because they are precisely the reals where the improper twin lurks.

**24.** $p = 1$, $q = 10$, $b = 2$, $r_0 = 1$: $2 = 10 \cdot
0 + 2$, $4 = 10 \cdot 0 + 4$, $8 = 10 \cdot 0 + 8$, $16 = 10
\cdot 1 + 6$, $12 = 10 \cdot 1 + 2$ — and $r_5 = 2 = r_1$: the remainders cycle $(2, 4, 8, 6)$ from index $1$. Digits: $d_1 =
0$, then the repeating block $d_2 d_3 d_4 d_5 = 0, 0, 1, 1$:

$$
\tfrac{1}{10} = (0.0\overline{0011})_2 ,
$$

preperiod $1$, period $4$. By question 3, a terminating base-$2$ expansion would need every [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) of $10$ to divide $2$; the [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) $5$ refuses. So $0.1$ is *not* representable by any finite binary string — a computer storing finitely many bits keeps only a truncation, and the accumulated truncation errors are why floating-point $0.1 + 0.2$ differs from $0.3$ in the last bits.

**25.** (i) Completeness produced the values: $x = \sup s_n$ and $y = \sup t_n$ (questions 7 and 9) — over $\Q$ alone, the proper string of $\sqrt 2$ would name nothing. (ii) The [Archimedean property](#thm-b1-reals-archimedes) made $b^{-n}$ eventually smaller than any $\varepsilon$, forcing the truncations to close in on their [supremum](#def-b1-reals-bounds) (questions 7, 21). (iii) The uniqueness clause of the [floor](#thm-b1-reals-floor) identified $Q_n = A_n$ in question 14 and legitimized every digit extraction $\lfloor u - K \rfloor = \lfloor u \rfloor - K$ (question 13). (iv) The pigeonhole principle, applied to finitely many remainders, is the sole engine of periodicity (question 16). Moral: proper strings code $\intco{0}{1}$ faithfully and turn rationality into a visible rhythm; but addition of digit strings requires carries propagating from infinitely far right, so no finite-stage rule computes even the first digit of a sum — whereas the [supremum](#def-b1-reals-bounds) interface of [Theorem 10.2](#thm-b1-reals-sup) handles all of analysis with one axiom. Digits are a magnificent *picture* of $\R$; the [supremum](#def-b1-reals-bounds) is its *engine*.