---
title: "Numbers and Sets of Numbers"
book: "High School Mathematics"
subject: math
language: en
chapter: 1
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers
---

# Chapter 1 — Numbers and Sets of Numbers

Mathematics begins with numbers, and not all numbers are of the same kind: counting numbers, negative numbers, fractions, and numbers like $\sqrt 2$ or $\pi$ that no fraction can express. This chapter organizes them into nested families, introduces [intervals](#def-g10-numbers-interval) to describe portions of the number line, and uses the [absolute value](#def-g10-numbers-abs) to measure distances between numbers.

## 1.1 The families of numbers

**Definition 1.1 (Sets of numbers).**

- $\N$ is the set of *natural numbers* : $0, 1, 2, 3, \dots$
- $\Z$ is the set of *integers* : $\dots, -2, -1,  0, 1, 2, \dots$
- $\Q$ is the set of *rational numbers* : all quotients $\frac{p}{q}$ with $p \in \Z$ , $q \in \N$ and $q \neq 0$ .
- $\R$ is the set of *real numbers* : all the numbers that can be placed on the number line.

**Notation 1.2.**

The symbol $\in$ reads “belongs to”: $3 \in \N$, $-5 \in \Z$, $\frac23 \in \Q$. The symbol $\subset$ reads “is included in”: every [natural number](#def-g10-numbers-sets) is an [integer](#def-g10-numbers-sets), every [integer](#def-g10-numbers-sets) is a [rational number](#def-g10-numbers-sets) (e.g. $-5 = \frac{-5}{1}$), and every [rational number](#def-g10-numbers-sets) is real, so

$$
\N \subset \Z \subset \Q \subset \R .
$$

A slash negates a symbol: $\frac12 \notin \Z$.

![The nested families of numbers: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ. Each ring contains numbers that do not belong to the smaller ones.](https://one-course.com/images/onecourse/chapters/math-2/g10-numbers/fig-6517b7f5c6d7.svg)

*The nested families of numbers: $\N \subset \Z \subset \Q \subset
\R$. Each ring contains numbers that do not belong to the smaller ones.*

**Example 1.3.**

Let us place a few numbers in the smallest family that contains them.

- $\frac{28}{4} = 7$ , so $\frac{28}{4} \in \N$ even though it is written as a fraction: always simplify first.
- $-3.5 = -\frac{35}{10} = -\frac72$ is rational but not an [integer](#def-g10-numbers-sets) .
- $0.333\dots$ (the digit $3$ repeating forever) equals $\frac13$ , a [rational number](#def-g10-numbers-sets) .
- $\sqrt 2$ and $\pi$ are real but not rational, as we are about to see for $\sqrt 2$ ; such numbers are called *irrational* .

**Theorem 1.4 (Irrationality of 2\sqrt 22​).**

The number $\sqrt 2$ is not rational: no fraction of [integers](#def-g10-numbers-sets) has square $2$.

**Proof.** We reason by contradiction, in small steps.

1. Suppose $\sqrt 2 = \frac{p}{q}$ where $p$ and $q$ are positive [integers](#def-g10-numbers-sets) , and the fraction is fully simplified, so $p$ and $q$ are not both even.
2. Squaring both sides gives $2 = \frac{p^2}{q^2}$ , that is $p^2 = 2q^2$ . So $p^2$ is even.
3. If $p$ were odd, say $p = 2k+1$ , then $p^2 = 4k^2 + 4k + 1$ would be odd. Since $p^2$ is even, $p$ must be even: $p = 2k$ for some [integer](#def-g10-numbers-sets) $k$ .
4. Substituting: $(2k)^2 = 2q^2$ , so $4k^2 = 2q^2$ , so $q^2 = 2k^2$ . By the same argument as in step 3, $q$ is even.
5. Now $p$ and $q$ are both even, contradicting step 1. The assumption was impossible: $\sqrt 2$ is [irrational](#ex-g10-numbers-classify) .

∎

**Remark 1.5.**

[Rational numbers](#def-g10-numbers-sets) are exactly the numbers whose decimal expansion either stops (like $\frac72 = 3.5$) or eventually repeats the same block forever (like $\frac13 = 0.333\dots$ or $\frac{1}{7} = 0.142857\,142857\dots$). [Irrational numbers](#ex-g10-numbers-classify) like $\sqrt2 = 1.41421356\dots$ have decimal digits that never fall into a repeating pattern. We admit this characterization at this level.

## 1.2 The number line and intervals

[Real numbers](#def-g10-numbers-sets) fill a line: choosing an origin $0$ and a unit of length, every [real number](#def-g10-numbers-sets) corresponds to exactly one point.

**Definition 1.6 (Interval).**

Let $a$ and $b$ be [real numbers](#def-g10-numbers-sets) with $a < b$. An *interval* is the set of all [real numbers](#def-g10-numbers-sets) between two bounds. A square bracket means the bound is included, a parenthesis means it is excluded:

| notation | description |
| --- | --- |
| $\intcc{a}{b}$ | $a \leq x \leq b$ (both ends included) |
| $\intoo{a}{b}$ | $a < x < b$ (both ends excluded) |
| $\intco{a}{b}$ | $a \leq x < b$ |
| $\intoc{a}{b}$ | $a < x \leq b$ |
| $\intco{a}{+\infty}$ | $x \geq a$ |
| $\intoo{-\infty}{b}$ | $x < b$ |

The symbols $-\infty$ and $+\infty$ (“infinity”) are not numbers, only a way of saying that the interval continues forever; the bracket next to them is always a parenthesis. The whole line $\R$ is the interval $\intoo{-\infty}{+\infty}$.

![Intervals on the number line: a filled dot for an included bound, a hollow dot for an excluded one.](https://one-course.com/images/onecourse/chapters/math-2/g10-numbers/fig-cfc1de785072.svg)

*[Intervals](#def-g10-numbers-interval) on the number line: a filled dot for an included bound, a hollow dot for an excluded one.*

**Definition 1.7 (Intersection and union).**

Let $I$ and $J$ be two sets of [real numbers](#def-g10-numbers-sets). The *intersection* $I \cap J$ (“$I$ and $J$”) is the set of numbers belonging to both; the *union* $I \cup J$ (“$I$ or $J$”) is the set of numbers belonging to at least one of them.

**Example 1.8.**

Take $I = \intcc{-1}{3}$ and $J = \intoo{1}{5}$. Draw both on the same line: they overlap between $1$ and $3$. Therefore

$$
I \cap J = \intoc{1}{3},
\qquad
I \cup J = \intco{-1}{5}.
$$

Note the brackets: $1 \notin J$ so $1 \notin I \cap J$, but $3 \in I$ and $3 \in J$, so $3 \in I \cap J$.

**Method 1.9 (Working with intervals).**

To find the [intersection](#def-g10-numbers-interunion) or [union](#def-g10-numbers-interunion) of two [intervals](#def-g10-numbers-interval):

1. draw the number line and mark the four bounds;
2. shade the first [interval](#def-g10-numbers-interval) above the line and the second below it;
3. the [intersection](#def-g10-numbers-interunion) is where the shadings overlap, the [union](#def-g10-numbers-interunion) is where at least one shading is present;
4. decide each bracket by testing whether the bound itself belongs to both sets ( [intersection](#def-g10-numbers-interunion) ) or to at least one ( [union](#def-g10-numbers-interunion) ).

## 1.3 Absolute value and distance

**Definition 1.10 (Absolute value).**

The *absolute value* of a [real number](#def-g10-numbers-sets) $x$ is

$$
\abs{x} =
\begin{cases}
x & \text{if } x \geq 0, \\
-x & \text{if } x < 0.
\end{cases}
$$

For instance $\abs{7} = 7$ and $\abs{-4} = 4$. In every case $\abs{x} \geq 0$.

**Proposition 1.11 (Distance on the line).**

For all [real numbers](#def-g10-numbers-sets) $a$ and $b$, the distance between the points $a$ and $b$ on the number line is $\abs{b - a}$. In particular $\abs{x}$ is the distance from $x$ to $0$.

**Proof.** If $b \geq a$, the distance from $a$ to $b$ is $b - a \geq 0$, which equals $\abs{b-a}$. If $b < a$, the distance is $a - b = -(b - a) > 0$, which is again $\abs{b - a}$ by definition of the [absolute value](#def-g10-numbers-abs). ∎

**Proposition 1.12 (Absolute value and intervals).**

Let $a$ be a [real number](#def-g10-numbers-sets) and $r > 0$. Then

$$
\abs{x - a} \leq r
\quad\text{exactly when}\quad
x \in \intcc{a - r}{a + r}.
$$

**Proof.** $\abs{x-a} \leq r$ says that the distance from $x$ to $a$ is at most $r$, i.e. that $x$ lies no further than $r$ from $a$ on either side. The numbers satisfying this are exactly those between $a - r$ and $a + r$, bounds included. ∎

![The inequality |x - a| ≤ r describes the interval of numbers at distance at most r from a.](https://one-course.com/images/onecourse/chapters/math-2/g10-numbers/fig-fd741c75d5ad.svg)

*The inequality $\abs{x - a} \leq r$ describes the [interval](#def-g10-numbers-interval) of numbers at distance at most $r$ from $a$.*

**Example 1.13.**

Solve $\abs{x - 3} \leq 2$. The solutions are the numbers at distance at most $2$ from $3$: the [interval](#def-g10-numbers-interval) $\intcc{1}{5}$. Conversely, the [interval](#def-g10-numbers-interval) $\intcc{-1}{7}$ has center $\frac{-1+7}{2} = 3$ and radius $\frac{7-(-1)}{2} = 4$, so it is described by $\abs{x - 3} \leq 4$.

## 1.4 Approximations

[Irrational numbers](#ex-g10-numbers-classify), and even most fractions, cannot be written exactly with finitely many decimal digits, so in practice we approximate them.

**Definition 1.14 (Approximation to a given accuracy).**

A number $d$ is an *approximation* of $x$ to within $10^{-n}$ when $\abs{x - d} \leq 10^{-n}$. Truncating or rounding the decimal expansion after the $n$-th digit both give such approximations.

**Example 1.15.**

From $\pi = 3.14159\,26\dots$: the truncation $3.141$ and the rounding $3.142$ are both [approximations](#def-g10-numbers-approx) of $\pi$ to within $10^{-3}$. The rounding is at distance less than $\frac12 \times 10^{-3}$, the truncation only guarantees $10^{-3}$. Writing $3.141 \leq \pi \leq 3.142$ *frames* $\pi$ between two decimal bounds.

## 1.5 Exercises

**Exercise 1.1 ★.**

For each number, give the smallest of the sets $\N$, $\Z$, $\Q$, $\R$ to which it belongs:

$$
\frac{15}{3}, \qquad -7, \qquad \frac{22}{7}, \qquad \sqrt{9}, \qquad
\sqrt{10}, \qquad -2.4, \qquad 0 .
$$

**Solution of Exercise 1.1.**

$\frac{15}{3} = 5 \in \N$. $-7 \in \Z$. $\frac{22}{7} \in \Q$ (it is not an [integer](#def-g10-numbers-sets): $22 = 7 \times 3 + 1$). $\sqrt 9 = 3 \in \N$. $\sqrt{10} \in \R$ ([irrational](#ex-g10-numbers-classify), since $10$ is not the square of a rational — admitted here, in the spirit of [Theorem 1.4](#thm-g10-numbers-sqrt2)). $-2.4 = -\frac{24}{10} \in \Q$. $0 \in \N$.

**Exercise 1.2 ★.**

Write each statement with [interval](#def-g10-numbers-interval) notation, then draw it on a number line: (a) $-2 \leq x < 5$; (b) $x > 3$; (c) $x \leq -1$; (d) the distance from $x$ to $2$ is at most $3$.

**Solution of Exercise 1.2.**

(a) $\intco{-2}{5}$: filled dot at $-2$, hollow dot at $5$. (b) $\intoo{3}{+\infty}$: hollow dot at $3$, shading to the right. (c) $\intoc{-\infty}{-1}$: shading from the left up to a filled dot at $-1$. (d) “distance from $x$ to $2$ at most $3$” means $\abs{x - 2} \leq 3$, i.e. $x \in \intcc{-1}{5}$: filled dots at $-1$ and $5$.

**Exercise 1.3 ★.**

Compute $I \cap J$ and $I \cup J$ for

$$
\text{(a) } I = \intcc{-3}{2},\ J = \intco{0}{4};
\qquad
\text{(b) } I = \intoo{-\infty}{1},\ J = \intco{-2}{+\infty}.
$$

**Solution of Exercise 1.3.**

(a) The two [intervals](#def-g10-numbers-interval) overlap between $0$ and $2$: $I \cap J = \intcc{0}{2}$ ($0 \in J$ and $2 \in I$, and both belong to the other set as well), and $I \cup J = \intco{-3}{4}$.

(b) $I \cap J$ is the set of $x$ with $-2 \leq x$ and $x < 1$: $\intco{-2}{1}$. The [union](#def-g10-numbers-interunion) covers everything: $I \cup J = \R$.

**Exercise 1.4 ★.**

Compute without a calculator:

$$
\abs{-6}, \qquad \abs{4 - 9}, \qquad \abs{-3 - 5}, \qquad
\abs{\sqrt 2 - 1}, \qquad \abs{1 - \sqrt 2}.
$$

**Solution of Exercise 1.4.**

$\abs{-6} = 6$; $\abs{4 - 9} = \abs{-5} = 5$; $\abs{-3 - 5} = \abs{-8} = 8$; $\sqrt 2 > 1$ so $\abs{\sqrt2 - 1} = \sqrt2 - 1$; $1 - \sqrt2 < 0$ so $\abs{1 - \sqrt2} = \sqrt2 - 1$ as well (a number and its opposite have the same [absolute value](#def-g10-numbers-abs)).

**Exercise 1.5 ★.**

Solve the equations and inequalities and give the solution sets:

$$
\abs{x} = 5, \qquad \abs{x - 1} = 3, \qquad \abs{x - 4} \leq 1, \qquad
\abs{x + 2} < 3 .
$$

(Note that $\abs{x+2} = \abs{x - (-2)}$ is a distance to $-2$.)

**Solution of Exercise 1.5.**

$\abs{x} = 5$: distance $5$ from $0$, so $x = 5$ or $x = -5$.

$\abs{x - 1} = 3$: distance $3$ from $1$, so $x = 4$ or $x = -2$.

$\abs{x - 4} \leq 1$: distance at most $1$ from $4$, so $x \in \intcc{3}{5}$.

$\abs{x + 2} < 3$: distance strictly less than $3$ from $-2$, so $x \in \intoo{-5}{1}$.

**Exercise 1.6 ★★.**

Describe each [interval](#def-g10-numbers-interval) by an inequality of the form $\abs{x - a} \leq r$ or $\abs{x - a} < r$:

$$
\intcc{2}{8}, \qquad \intoo{-5}{1}, \qquad \intcc{-7}{-3}.
$$

**Solution of Exercise 1.6.**

Each [interval](#def-g10-numbers-interval) is described by its center $a$ (midpoint of the bounds) and radius $r$ (half the length).

$\intcc{2}{8}$: $a = 5$, $r = 3$, so $\abs{x - 5} \leq 3$.

$\intoo{-5}{1}$: $a = -2$, $r = 3$, open bounds, so $\abs{x + 2} < 3$.

$\intcc{-7}{-3}$: $a = -5$, $r = 2$, so $\abs{x + 5} \leq 2$.

**Exercise 1.7 ★★.**

Show that $0.272727\dots$ (the block $27$ repeating forever) is rational. (Hint: call it $x$ and compute $100x - x$.)

**Solution of Exercise 1.7.**

Let $x = 0.272727\dots$ Then $100x = 27.2727\dots$, and subtracting:

$$
100x - x = 27.2727\dots - 0.2727\dots = 27,
$$

so $99x = 27$ and $x = \frac{27}{99} = \frac{3}{11}$, a quotient of [integers](#def-g10-numbers-sets): $x$ is rational.

**Exercise 1.8 ★★.**

True or false? Justify each answer with an argument or a counterexample.

1. The sum of two [integers](#def-g10-numbers-sets) is an [integer](#def-g10-numbers-sets) .
2. The quotient of two [integers](#def-g10-numbers-sets) is an [integer](#def-g10-numbers-sets) .
3. The sum of two [rational numbers](#def-g10-numbers-sets) is rational.
4. The sum of a [rational number](#def-g10-numbers-sets) and an [irrational number](#ex-g10-numbers-classify) is [irrational](#ex-g10-numbers-classify) .

**Solution of Exercise 1.8.**

*1. True:* adding [integers](#def-g10-numbers-sets) (positive or negative whole numbers) always produces an [integer](#def-g10-numbers-sets).

*2. False:* $\frac{1}{2}$ is a quotient of the [integers](#def-g10-numbers-sets) $1$ and $2$ and is not an [integer](#def-g10-numbers-sets).

*3. True:* $\frac pq + \frac{p'}{q'} = \frac{pq' + p'q}{qq'}$ is again a quotient of [integers](#def-g10-numbers-sets) (with nonzero denominator).

*4. True:* suppose $r$ is rational, $t$ [irrational](#ex-g10-numbers-classify), and $r + t = s$ were rational. Then $t = s - r$ would be a difference of two rationals, hence rational (by 3, applied with $-r$) — contradiction. So $r + t$ is [irrational](#ex-g10-numbers-classify).

**Exercise 1.9 ★★.**

Using $1.414 \leq \sqrt 2 \leq 1.415$, frame the numbers $2\sqrt2$, $\sqrt2 + 3$ and $-\sqrt2$ between two decimal bounds.

**Solution of Exercise 1.9.**

Multiplying $1.414 \leq \sqrt2 \leq 1.415$ by $2 > 0$: $2.828 \leq 2\sqrt2 \leq 2.830$.

Adding $3$: $4.414 \leq \sqrt2 + 3 \leq 4.415$.

Multiplying by $-1 < 0$ reverses the inequalities: $-1.415 \leq -\sqrt2 \leq -1.414$.

**Exercise 1.10 ★★★.**

Adapt the proof of [Theorem 1.4](#thm-g10-numbers-sqrt2) to show that $\sqrt 3$ is [irrational](#ex-g10-numbers-classify). (Replace “even” by “multiple of $3$”: first check that if $p^2$ is a multiple of $3$, then so is $p$, by examining the remainders $0$, $1$, $2$ of $p$ upon division by $3$.)

**Solution of Exercise 1.10.**

First the auxiliary fact. Divide $p$ by $3$: the remainder is $0$, $1$ or $2$, i.e. $p = 3k$, $p = 3k+1$ or $p = 3k+2$. Squaring:

$$
(3k)^2 = 3(3k^2), \quad
(3k+1)^2 = 3(3k^2 + 2k) + 1, \quad
(3k+2)^2 = 3(3k^2 + 4k + 1) + 1 .
$$

Only the first is a multiple of $3$: if $p^2$ is a multiple of $3$, then $p$ is too.

Now suppose $\sqrt3 = \frac pq$ fully simplified. Squaring gives $p^2 = 3q^2$, so $p^2$ is a multiple of $3$, so $p = 3k$. Then $9k^2 = 3q^2$, so $q^2 = 3k^2$ and $q$ is also a multiple of $3$ — but then the fraction $\frac pq$ was not fully simplified: contradiction. So $\sqrt3$ is [irrational](#ex-g10-numbers-classify).

## 1.6 Problem: Between any two numbers

**Problem 1.1.**

Weekend problem — rationals and irrationals interlace: every interval, however small, contains infinitely many of each, and no measurement can ever tell them apart

The rationals look like a crowd (all the fractions!) and the [irrationals](#ex-g10-numbers-classify) like exotic exceptions ($\sqrt2$, $\pi$). This problem reveals the true picture: the two families *interlace* so finely that every [interval](#def-g10-numbers-interval) of the number line, however microscopic, contains infinitely many of each — with strange consequences, such as this: no physical measurement, however precise, can ever decide whether a length is rational.

**Part I — The four kingdoms.**

1. For each number, name the smallest of the sets $\N$ , $\Z$ , $\Q$ , $\R$ containing it ( [Definition 1.1](#def-g10-numbers-sets) ): $-7$ ; $\frac{13}{4}$ ; $\sqrt{16}$ ; $0.121212\ldots$ (recall the repeating-decimals weekend problem of the Middle School volume); $\sqrt8$ ; $\pi$ (admit its irrationality — proved only in the university volumes).
2. Prove that $\Q$ is stable under addition and multiplication: if $x = \frac pq$ and $y = \frac rs$ are rational, write $x + y$ and $xy$ as single fractions.
3. Deduce by contradiction: (a) the sum of a rational and an [irrational](#ex-g10-numbers-classify) is [irrational](#ex-g10-numbers-classify) ; (b) the product of a *nonzero* rational and an [irrational](#ex-g10-numbers-classify) is [irrational](#ex-g10-numbers-classify) .
4. Show that the set of [irrationals](#ex-g10-numbers-classify) is stable under *neither* operation: exhibit two [irrationals](#ex-g10-numbers-classify) whose sum is rational, and two whose product is rational.
5. Locate $\sqrt2 + \sqrt8$ and $\sqrt2 \times \sqrt8$ in the four kingdoms (simplify $\sqrt8$ first, the Middle School volume’s method).

**Part II — The rationals are dense.**

6. Find a rational strictly between $3.47$ and $3.48$ ; then a second one; then describe how to produce as many as desired (the zoom of the no-next-number weekend problem of the Middle School volume, now with a proof in sight).
7. The general theorem. Let $a < b$ be any two reals, with gap $g = b - a > 0$ . Choose $n$ with $10^{-n} < g$ , and consider the multiples of $10^{-n}$ (the decimal grid of step $10^{-n}$ ). Explain why at least one grid point falls strictly between $a$ and $b$ , and conclude: *every [interval](#def-g10-numbers-interval) of positive length contains a [rational number](#def-g10-numbers-sets)* .
8. Upgrade the conclusion: every such [interval](#def-g10-numbers-interval) contains *infinitely many* rationals. (Apply question 7 again, inside a smaller [interval](#def-g10-numbers-interval) .)
9. Now the [irrationals](#ex-g10-numbers-classify) : given $a < b$ , take a rational $r$ strictly inside (question 7) and consider the numbers $r + \frac{\sqrt2}{10^k}$ . Using question 3, show they are [irrational](#ex-g10-numbers-classify) , and that for $k$ large enough they still lie in the [interval](#def-g10-numbers-interval) : *every [interval](#def-g10-numbers-interval) also contains infinitely many [irrationals](#ex-g10-numbers-classify)* .
10. Two classic riddles settled: is there a smallest positive [real number](#def-g10-numbers-sets) ? A [real number](#def-g10-numbers-sets) “just after $3$ ”? Answer both with question 7’s theorem, and salute the childhood version (the no-next-number weekend problem of the Middle School volume).

**Part III — [Absolute value](#def-g10-numbers-abs), the geometry of $\R$.**

11. Solve, and express the solutions as [intervals](#def-g10-numbers-interval) or [unions](#def-g10-numbers-interunion) ( [Proposition 1.12](#prop-g10-numbers-absinterval) ): $\abs{x - 5} = 2$ ; $\abs{x - 5} < 2$ ; $\abs{x + 1} \geq 3$ .
12. A piston must be manufactured to $80$ mm within a tolerance of $0.05$ mm. Write the requirement with an [absolute value](#def-g10-numbers-abs) , then as an [interval](#def-g10-numbers-interval) . Two pistons measure $79.97$ and $80.06$ mm: verdicts?
13. The triangle inequality on the line: $\abs{a + b} \leq \abs a + \abs b$ . Check it on $(a, b) = (3, -5)$ and $(-2, -7)$ , prove it when $a$ and $b$ have the same sign and when they have opposite signs, and state exactly when equality holds.
14. Solve $\abs{x - 2} = \abs{x + 4}$ by reading it as an equality of distances on the line. Which point of the two-mirrors weekend problem of the Middle School volume have you just computed, one dimension down?
15. Simplify $\sqrt{x^2}$ — carefully. Test your formula on $x = 3$ and $x = -3$ , then use it to solve $x^2 < 9$ with an [absolute value](#def-g10-numbers-abs) .

**Part IV — What no measurement can decide.**

16. A physicist measures a rod: $1.41421 \pm 0.00001$ m. Can any such measurement — this one or a future, more precise one — ever *prove* that the rod’s length is [irrational](#ex-g10-numbers-classify) ? Or that it is rational? Use questions 7 and 9 on the tolerance [interval](#def-g10-numbers-interval) , and explain why only *proof* (as for the diagonal, the irrationality weekend problem of the Middle School volume) can settle irrationality.
17. The truncations $1.4$ , $1.41$ , $1.414$ , $1.4142,\dots$ of $\sqrt2$ are all rational. What do they show about how close $\Q$ presses against every [irrational](#ex-g10-numbers-classify) ? Formulate the general fact.
18. [Interval](#def-g10-numbers-interval) arithmetic: measured values $a = 2.5 \pm 0.1$ and $b = 1.2 \pm 0.1$ . Bracket $a + b$ and $a \times b$ between certain bounds. Which operation degrades the precision more?
19. An engineer’s rule says errors of independent measurements “add” for sums: express with the triangle inequality (question 13) why the error of $a + b$ is at most the sum of the errors — the inequality is the engineering rule.
20. Finale, in a short paragraph: assemble the picture of the real line established by this problem and its ancestors — no next number (the no-next-number weekend problem of the Middle School volume), rationals $=$ repeating decimals (the repeating-decimals weekend problem of the Middle School volume), both families dense (questions 8 and 9), measurement forever undecided (question 16). End with the teaser this chapter cannot yet prove: in a precise sense there are *vastly more* [irrationals](#ex-g10-numbers-classify) than rationals — the university volumes count the infinities.

**Solution of Problem 1.1.**

**1.** $-7 \in \Z$; $\frac{13}{4} \in \Q$; $\sqrt{16} = 4 \in \N$; $0.121212\ldots = \frac{12}{99} =
\frac{4}{33} \in \Q$ (the repeating-decimals weekend problem of the Middle School volume); $\sqrt8 = 2\sqrt2$ is [irrational](#ex-g10-numbers-classify): smallest set $\R$; $\pi$: $\R$.

**2.** $\frac pq + \frac rs = \frac{ps + qr}{qs}$ and $\frac pq \times \frac rs = \frac{pr}{qs}$: quotients of [integers](#def-g10-numbers-sets) with nonzero denominators — rational.

**3.** (a) If $r$ is rational, $x$ [irrational](#ex-g10-numbers-classify) and $r + x = q$ were rational, then $x = q - r$ would be a difference of rationals, hence rational (question 2): contradiction. (b) If $r \neq 0$ and $r x = q$ rational, then $x = \frac qr$ is rational: contradiction.

**4.** $\sqrt2$ and $-\sqrt2$ are [irrational](#ex-g10-numbers-classify) with sum $0$; $\sqrt2$ and $\sqrt2$ have product $2$. Rational results from [irrational](#ex-g10-numbers-classify) ingredients: the [irrationals](#ex-g10-numbers-classify) form no stable kingdom.

**5.** $\sqrt8 = 2\sqrt2$, so $\sqrt2 + \sqrt8 = 3\sqrt2$: [irrational](#ex-g10-numbers-classify) (question 3b, multiplier $3$). And $\sqrt2 \times \sqrt8 = \sqrt{16} = 4 \in \N$: the product of two [irrationals](#ex-g10-numbers-classify) lands in the smallest kingdom of all.

**6.** $3.475$, then $3.471$ (or $3.4701$, $3.47001$, …): appending decimal digits produces fresh rationals between the two forever.

**7.** The multiples of $10^{-n}$ march along the line in steps of $10^{-n} < g$. The first multiple strictly greater than $a$ — it exists, since the multiples eventually exceed $a$ — lies at most one step beyond $a$, hence before $a + g = b$: strictly between $a$ and $b$. A grid point is a decimal, hence a rational: every [interval](#def-g10-numbers-interval) of positive length contains one.

**8.** Between $a$ and the rational $r_1$ found in question 7 there is (question 7 again) a rational $r_2$; between $a$ and $r_2$ a rational $r_3$; and so on: infinitely many, all distinct, all in the original [interval](#def-g10-numbers-interval).

**9.** $r + \frac{\sqrt2}{10^k}$ is the sum of a rational and an [irrational](#ex-g10-numbers-classify) ($\frac{\sqrt2}{10^k}$ is [irrational](#ex-g10-numbers-classify) by question 3b), hence [irrational](#ex-g10-numbers-classify). Since $\frac{\sqrt2}{10^k} < \frac{2}{10^k}$ shrinks below any bound, for $k$ large enough $r + \frac{\sqrt2}{10^k}$ still lies before $b$: an [irrational](#ex-g10-numbers-classify) inside the [interval](#def-g10-numbers-interval) — and varying $k$ gives infinitely many.

**10.** No smallest positive real: if $s > 0$ were it, the [interval](#def-g10-numbers-interval) $(0, s)$ would still contain a rational (question 7), positive and smaller than $s$. No number just after $3$: any candidate $c > 3$ leaves the [interval](#def-g10-numbers-interval) $(3, c)$, which is not empty — the no-next-number weekend problem of the Middle School volume’s game, now a theorem about $\R$.

**11.** $\abs{x - 5} = 2$: $x = 3$ or $x = 7$. $\abs{x - 5} < 2$: $x \in \intoo{3}{7}$. $\abs{x + 1} \geq 3$: distance to $-1$ at least $3$: $x \in \intoc{-\infty}{-4} \cup \intco{2}{+\infty}$.

**12.** $\abs{d - 80} \leq 0.05$, i.e. $d \in \intcc{79.95}{80.05}$. The piston at $79.97$ passes; the one at $80.06$ fails by a hundredth of a millimeter.

**13.** $(3, -5)$: $\abs{-2} = 2 \leq 8$. $(-2, -7)$: $\abs{-9} = 9 = 2 + 7$: equality. Same signs: $\abs{a + b}$ is the sum of the distances, equal to $\abs a + \abs b$. Opposite signs: the walk doubles back, $\abs{a + b}$ is the *difference* of the distances, strictly less than their sum (unless one of them is $0$). Equality exactly when $a$ and $b$ have the same sign or one is zero.

**14.** The solutions are the points equidistant from $2$ and $-4$: the midpoint, $x = -1$. It is the perpendicular bisector of the two-mirrors weekend problem of the Middle School volume, collapsed to one dimension: a single point.

**15.** $\sqrt{x^2} = \abs{x}$, not $x$: for $x = -3$, $\sqrt{9} = 3 = \abs{-3} \neq -3$. Then $x^2 < 9$ reads $\sqrt{x^2} < 3$, i.e. $\abs x < 3$: $x \in \intoo{-3}{3}$.

**16.** The measurement asserts only that the length lies in the [interval](#def-g10-numbers-interval) $\intcc{1.41420}{1.41422}$ — and by questions 7 and 9 *that [interval](#def-g10-numbers-interval) contains infinitely many rationals and infinitely many [irrationals](#ex-g10-numbers-classify)*. The same holds for every future tolerance, however small. No measurement can ever separate the two families; only a proof about the exact length — like the one for the diagonal of the unit square (the irrationality weekend problem of the Middle School volume) — can.

**17.** They approach $\sqrt2$ with errors below $10^{-1}, 10^{-2}, 10^{-3}, \dots$: rationals press against $\sqrt2$ at every scale. The general fact: every [real number](#def-g10-numbers-sets) is approached as closely as desired by rationals (its decimal truncations) — density once more, seen from the target’s side.

**18.** $a + b \in \intcc{2.4 + 1.1}{2.6 + 1.3} =
\intcc{3.5}{3.9}$: uncertainty $\pm 0.2$ (errors added). $a \times b \in \intcc{2.4 \times 1.1}{2.6 \times 1.3} =
\intcc{2.64}{3.38}$: an uncertainty of about $\pm 0.37$ around $3$ — relative errors add for products, so precision degrades faster.

**19.** Write the true values $a = 2.5 + e_1$, $b = 1.2 + e_2$ with $\abs{e_1}, \abs{e_2} \leq 0.1$. Then the error of the sum is $\abs{e_1 + e_2} \leq \abs{e_1} +
\abs{e_2} \leq 0.2$: the engineer’s rule *is* the triangle inequality of question 13.

**20.** The real line has no gaps and no neighbours: after any number, no “next” one (question 10). Its points split into the rationals — exactly the stopping or repeating decimals (the repeating-decimals weekend problem of the Middle School volume) — and the [irrationals](#ex-g10-numbers-classify), and the two families are interwoven so tightly that every [interval](#def-g10-numbers-interval) holds infinitely many of each (questions 8 and 9), which is why no measurement, only proof, can tell them apart (question 16). And the final surprise, left as a promise: the [irrationals](#ex-g10-numbers-classify) outnumber the rationals — not by counting one by one, but in the precise sense of comparing infinities, a theory built in the university volumes.