§3 The Set ℝ of Real Numbers

You are probably comfortable with $\mathbb{Q}$, even though a careful construction of it would need equivalence classes (for example, $\frac24$ and $\frac36$ are the same element of $\mathbb{Q}$). Rather than build $\mathbb{Q}$, we list the properties of it that we actually use.

Why this matters

Listing the axioms tells us exactly what we are allowed to use. From now on, “obvious” facts such as $0 \cdot a = 0$ or $0 < 1$ must be derived from the list, and Theorems 3.1 and 3.2 show how. Keep in mind that everything in this section is equally true in $\mathbb{Q}$. Nothing here can tell $\mathbb{Q}$ and $\mathbb{R}$ apart; the one new ingredient, completeness, arrives in §4.

Field and order axioms

Given $a, b \in \mathbb{Q}$, the sum $a + b$ and the product $ab$ are again rational, and for all $a, b, c$:

1. A1. $a + (b + c) = (a + b) + c$
2. M1. $a(bc) = (ab)c$
3. A2. $a + b = b + a$
4. M2. $ab = ba$
5. A3. $a + 0 = a$
6. M3. $a \cdot 1 = a$
7. A4. For each $a$ there is $-a$ with $a + (-a) = 0$.
8. M4. For each $a \ne 0$ there is $a^{-1}$ with $a a^{-1} = 1$.

and the distributive law

• DL. $a(b + c) = ab + ac$.

A1 and M1 are the associative laws, A2 and M2 the commutative laws. DL is the least obvious one; it is what justifies factoring and multiplying out. A system with more than one element satisfying these nine properties is called a field.

$\mathbb{Q}$ also has an order $\le$ satisfying:

• O1. Given $a$ and $b$, either $a \le b$ or $b \le a$.
• O2. If $a \le b$ and $b \le a$, then $a = b$.
• O3. If $a \le b$ and $b \le c$, then $a \le c$. (the transitive law)
• O4. If $a \le b$, then $a + c \le b + c$.
• O5. If $a \le b$ and $0 \le c$, then $ac \le bc$.

A field with an ordering satisfying O1–O5 is an ordered field.

The real numbers $\mathbb{R}$ will contain all rationals, all algebraic numbers, $\pi$, $e$ and more, and can be pictured as the whole number line with no gaps. We accept $\mathbb{R}$ as an ordered field: it satisfies A1–A4, M1–M4, DL and O1–O5. (Enrichment §6 sketches one way to construct $\mathbb{R}$ from $\mathbb{Q}$.) What distinguishes $\mathbb{R}$ from $\mathbb{Q}$ is the completeness axiom of §4. Everything in this section holds in any ordered field, $\mathbb{Q}$ included.

3.1 Theorem.

The following are consequences of the field properties, for $a, b, c \in \mathbb{R}$:

1. $a + c = b + c$ implies $a = b$;
2. $a \cdot 0 = 0$ for all $a$;
3. $(-a)b = -ab$ for all $a, b$;
4. $(-a)(-b) = ab$ for all $a, b$;
5. $ac = bc$ and $c \ne 0$ imply $a = b$;
6. $ab = 0$ implies either $a = 0$ or $b = 0$.

Used in Exercise 3.2, Exercise 3.3, Exercise 3.4

Proof

(i) Add $-c$ to both sides: $(a + c) + (-c) = (b + c) + (-c)$. By A1, $a + [c + (-c)] = b + [c + (-c)]$. By A4 this is $a + 0 = b + 0$, so $a = b$ by A3.

(ii) By A3 and DL, $a \cdot 0 = a(0 + 0) = a\cdot0 + a\cdot0$, so $0 + a\cdot0 = a\cdot0 + a\cdot0$. By (i), $0 = a \cdot 0$.

(iii) $ab + (-a)b = [a + (-a)]b = 0 \cdot b = 0 = ab + (-(ab))$. By (i), $(-a)b = -(ab)$.

(iv) and (v) are Exercise 3.3.

(vi) If $ab = 0$ and $b \ne 0$, then $0 = b^{-1} \cdot 0 = (ab) b^{-1} = a(bb^{-1}) = a \cdot 1 = a$.

∎

3.2 Theorem.

The following are consequences of the properties of an ordered field, for $a, b, c \in \mathbb{R}$:

1. If $a \le b$, then $-b \le -a$;
2. If $a \le b$ and $c \le 0$, then $bc \le ac$;
3. If $0 \le a$ and $0 \le b$, then $0 \le ab$;
4. $0 \le a^2$ for all $a$;
5. $0 < 1$;
6. If $0 < a$, then $0 < a^{-1}$;
7. If $0 < a < b$, then $0 < b^{-1} < a^{-1}$.

Here $a < b$ means $a \le b$ and $a \ne b$.

Used in Exercise 3.4, Exercise 3.5

Proof

(i) Apply O4 with $c = (-a) + (-b)$: $a + [(-a) + (-b)] \le b + [(-a) + (-b)]$, i.e. $-b \le -a$.

(ii) If $c \le 0$ then $0 \le -c$ by (i). By O5, $a(-c) \le b(-c)$, i.e. $-ac \le -bc$. By (i) again, $bc \le ac$.

(iii) Put $a = 0$ in O5: $0 \le b$ and $0 \le c$ imply $0 \le bc$.

(iv) By O1 either $a \ge 0$ or $a \le 0$. If $a \ge 0$, then $a^2 \ge 0$ by (iii). If $a \le 0$, then $-a \ge 0$, so $(-a)^2 \ge 0$, i.e. $a^2 \ge 0$.

(v) is Exercise 3.4.

(vi) Suppose $0 < a$ but $a^{-1} \le 0$. Then $0 \le -a^{-1}$, and by (iii) $0 \le a(-a^{-1}) = -1$, so $1 \le 0$, contradicting (v).

(vii) is Exercise 3.4.

∎

Absolute value and distance

3.3 Definition.

$|a| = a$ if $a \ge 0$, and $|a| = -a$ if $a \le 0$. $|a|$ is the absolute value of $a$.

3.4 Definition.

For numbers $a$ and $b$, $\operatorname{dist}(a, b) = |a - b|$ is the distance between $a$ and $b$.

So “distance” is defined in terms of absolute value, which is defined in terms of the ordering.

Intuitionreading absolute values as distances

Most inequalities in analysis are statements about distance, and it pays to translate them on sight:

$$|x - a| < r \iff a - r < x < a + r \iff x \text{ lies within } r \text{ of } a.$$

So “$|s_n - s| < \varepsilon$” in §7 says that $s_n$ sits in the interval $(s - \varepsilon, s + \varepsilon)$. In the same way, $|x| \le r$ means $-r \le x \le r$ (Exercise 3.5(a)).

3.5 Theorem.

1. $|a| \ge 0$ for all $a \in \mathbb{R}$.
2. $|ab| = |a| \cdot |b|$ for all $a, b \in \mathbb{R}$.
3. $|a + b| \le |a| + |b|$ for all $a, b \in \mathbb{R}$.

Used in Corollary 3.6

Proof

(i) If $a \ge 0$ then $|a| = a \ge 0$; if $a < 0$ then $|a| = -a > 0$.

(ii) There are four easy cases. If $a, b \ge 0$, then $ab \ge 0$ and $|a||b| = ab = |ab|$. If $a, b \le 0$, then $(-a)(-b) = ab \ge 0$ and $|a||b| = ab = |ab|$. If $a \ge 0 \ge b$, then $a(-b) \ge 0$ and $|a||b| = -(ab) = |ab|$. The case $a \le 0 \le b$ is the same.

(iii) Clearly $-|a| \le a \le |a|$ and $-|b| \le b \le |b|$. Adding (four applications of O4),

$$-(|a| + |b|) \le a + b \le |a| + |b|.$$

So both $a + b$ and $-(a + b)$ are at most $|a| + |b|$. Since $|a + b|$ is one of these, $|a + b| \le |a| + |b|$.

∎

3.6 Corollary.

$\operatorname{dist}(a, c) \le \operatorname{dist}(a, b) + \operatorname{dist}(b, c)$ for all $a, b, c \in \mathbb{R}$.

Used in §13

Proof

Apply Theorem 3.5(iii) to $a - b$ and $b - c$: $|a - c| = |(a - b) + (b - c)| \le |a - b| + |b - c|$.

∎

In the plane, the analogous inequality says that one side of a triangle is no longer than the sum of the other two sides. That is why Corollary 3.6 and Theorem 3.5(iii) are both called the Triangle Inequality.

3.7 Triangle Inequality.

$|a + b| \le |a| + |b|$ for all $a, b$.

A useful variant is given in Exercise 3.5(b).

Intuitionthe “for every ε” principle

Here is a small fact that analysis uses constantly:

$$\text{if } a \le b + \varepsilon \text{ for every } \varepsilon > 0, \text{ then } a \le b.$$

Indeed, if $a > b$, then $\varepsilon = \frac{a - b}{2}$ is positive and $b + \varepsilon < a$, contradicting the hypothesis. Applying it twice gives the version for equality: if $|a - b| < \varepsilon$ for every $\varepsilon > 0$, then $a = b$. This is how we prove two numbers are equal when we cannot compute either one exactly: we show they are closer than any positive tolerance. The proof that limits are unique (§7) ends exactly this way.

Common pitfall

• The Triangle Inequality gives an upper bound only. $|a + b|$ can be much smaller than $|a| + |b|$ because of cancellation: $|5 + (-5)| = 0$. For a lower bound, use $|a + b| \ge |a| - |b|$, which follows from Exercise 3.5(b).
• Inequalities can be added but not subtracted or divided. From $1 < 2$ and $0 < 5$ you cannot conclude $1 - 0 < 2 - 5$.
• Multiplying both sides by a negative number reverses the inequality (Theorem 3.2(ii)). When the sign of a quantity is unknown, split into cases or take absolute values first.

Try it: walk along the line

Think of $a + b$ as a walk: first $a$ units from $0$, then $b$ more. The walk covers $|a| + |b|$ units but ends only $|a + b|$ from where it started. Drag $a$ and $b$ to see when the two agree.

Keyboard: Tab to a marker on the number line and use the left and right arrow keys to move it in steps of 0.1. Hold Shift for bigger steps.

drag a: walk |a| = 2.0drag b: then walk |b| = 3.0a + b = −1.0, net distance |a + b| = 1.0

|a + b| = 1.0 < 5.0 = |a| + |b|. Strict: a and b have opposite signs, so the second leg backtracks 4.0 units of the first.

Reading alongside Abbott

Abbott covers absolute value and the Triangle Inequality in Section 1.2, together with the “for every $\varepsilon$” principle above, stated there as a theorem about equality. Abbott simply assumes $\mathbb{R}$ is an ordered field (Section 1.3) rather than listing the axioms one by one, and postpones a precise definition of “ordered field” to Section 8.6.

Check your understanding

Take the quiz (3 questions)

Which inequality holds for all real $a$ and $b$?

$|a + b| \le |a| + |b|$$|a + b| = |a| + |b|$$|a - b| \le |a| - |b|$

For $\varepsilon > 0$, what does $\operatorname{dist}(a, b) < \varepsilon$ mean?

$a - \varepsilon < b < a + \varepsilon$$b < a + \varepsilon$$|a| - |b| < \varepsilon$

Which property does $\mathbb{R}$ have that $\mathbb{Q}$ lacks?

The field properties (A1–A4, M1–M4, DL)The order properties (O1–O5)The Completeness Axiom

Exercises

Exercise 3.1

1. Which of the properties A1–A4, M1–M4, DL, O1–O5 fail for $\mathbb{N}$?
2. Which of these properties fail for $\mathbb{Z}$?

Solution

(a) A3 and A4 fail ($0 \notin \mathbb{N}$, no negatives), and M4 fails. The others hold. (b) Only M4 fails: for example, $2$ has no multiplicative inverse in $\mathbb{Z}$.

Tags: Warm-upFrom the definitionsExplain why

Exercise 3.2

1. The commutative law A2 was used in the proof of (ii) in Theorem 3.1. Where?
2. The commutative law A2 was also used in the proof of (iii) in Theorem 3.1. Where?

Solution

(a) In passing from $a\cdot0 = a\cdot0 + a\cdot0$ to $0 + a\cdot0 = a\cdot0 + a\cdot0$: A3 gives $a\cdot0 + 0 = a\cdot0$, and A2 is needed to write it as $0 + a\cdot0$. Also (i) is stated as $a + c = b + c$, so the common term must be moved to the right.

(b) The last step compares $ab + (-a)b$ with $ab + (-(ab))$. To apply (i), which cancels a term on the right, we need A2 to put $ab$ on the right of each sum.

Tags: Warm-upFrom the definitions

Exercise 3.3

Prove (iv) and (v) of Theorem 3.1.

Hint

For (iv), apply (iii) twice: $(-a)(-b) = -(a(-b)) = -(-(ab))$, and show $-(-x) = x$ using (i). For (v), multiply by $c^{-1}$.

Solution

First, additive inverses are unique: if $y + x = 0 = (-x) + x$ then $y = -x$ by Theorem 3.1(i). Since $(-x) + x = 0$, this shows $-(-x) = x$.

(iv) By (iii), $(-a)(-b) = -(a(-b))$. Also $a(-b) = (-b)a = -(ba) = -(ab)$ by M2 and (iii). Hence $(-a)(-b) = -(-(ab)) = ab$.

(v) Suppose $ac = bc$ with $c \ne 0$. Multiply by $c^{-1}$: $a = a\cdot1 = a(cc^{-1}) = (ac)c^{-1} = (bc)c^{-1} = b(cc^{-1}) = b$.

Tags: StandardFrom the definitions

Used in Theorem 3.1

Exercise 3.4

Prove (v) and (vii) of Theorem 3.2.

Hint

For (v): by (iv), $1 = 1^2 \ge 0$, and $1 \ne 0$ since a field has more than one element. For (vii): $a^{-1}$ and $b^{-1}$ are positive by (vi). Multiply $a < b$ by $a^{-1}b^{-1} > 0$.

Solution

(v) By (iv), $1 = 1^2 \ge 0$. Also $1 \ne 0$: otherwise every $a$ would satisfy $a = a\cdot1 = a\cdot0 = 0$ (Theorem 3.1(ii)), but a field has more than one element. So $0 < 1$.

(vii) Suppose $0 < a < b$. Then $b > 0$ too, so $a^{-1} > 0$ and $b^{-1} > 0$ by (vi), and $a^{-1}b^{-1} \ge 0$ by (iii). Multiplying $a \le b$ by $a^{-1}b^{-1}$ (O5) gives $b^{-1} \le a^{-1}$. Equality is impossible: if $b^{-1} = a^{-1}$, multiplying by $ab$ would give $a = b$. So $0 < b^{-1} < a^{-1}$.

Tags: StandardFrom the definitions

Used in Theorem 3.2

Exercise 3.5

1. Show $|b| \le a$ if and only if $-a \le b \le a$.
2. Prove $\bigl||a| - |b|\bigr| \le |a - b|$ for all $a, b \in \mathbb{R}$.

Hint

For (b), $|a| = |(a - b) + b| \le |a - b| + |b|$, so $|a| - |b| \le |a - b|$. Swap $a$ and $b$ and use (a).

Solution

1. ($\Rightarrow$) Suppose $|b| \le a$; then $a \ge 0$. If $b \ge 0$ then $b = |b| \le a$ and $-a \le 0 \le b$. If $b \le 0$ then $-b = |b| \le a$, so $-a \le b$ by Theorem 3.2(i), and $b \le 0 \le a$. ($\Leftarrow$) Suppose $-a \le b \le a$. If $b \ge 0$ then $|b| = b \le a$. If $b \le 0$ then $|b| = -b \le a$ because $-a \le b$.
2. By the triangle inequality, $|a| = |(a - b) + b| \le |a - b| + |b|$, so $|a| - |b| \le |a - b|$. Exchanging $a$ and $b$ gives $|b| - |a| \le |b - a| = |a - b|$. Thus $-|a - b| \le |a| - |b| \le |a - b|$, and (a) gives $\bigl||a| - |b|\bigr| \le |a - b|$.

Tags: StandardInequalities

Used in Exercise 3.7, Theorem 17.3

Exercise 3.6

1. Prove $|a + b + c| \le |a| + |b| + |c|$ for all $a, b, c \in \mathbb{R}$. (Apply the triangle inequality twice. Do not consider eight cases.)
2. Use induction to prove $|a_1 + a_2 + \cdots + a_n| \le |a_1| + |a_2| + \cdots + |a_n|$.

Solution

(a) $|(a + b) + c| \le |a + b| + |c| \le |a| + |b| + |c|$. (b) The step is $|(a_1 + \cdots + a_n) + a_{n+1}| \le |a_1 + \cdots + a_n| + |a_{n+1}|$, followed by the induction hypothesis.

Tags: StandardInequalitiesInduction

Used in Theorem 14.6

Exercise 3.7

1. Show $|b| < a$ if and only if $-a < b < a$.
2. Show $|a - b| < c$ if and only if $b - c < a < b + c$.
3. Show $|a - b| \le c$ if and only if $b - c \le a \le b + c$.

Hint

For (b) and (c), apply (a) and Exercise 3.5(a) to the number $a - b$.

Solution

1. ($\Rightarrow$) Suppose $|b| < a$; then $a > 0$. If $b \ge 0$, then $-a < 0 \le b = |b| < a$. If $b \le 0$, then $b \le 0 < a$ and $-b = |b| < a$, so $-a < b$. ($\Leftarrow$) If $-a < b < a$: when $b \ge 0$, $|b| = b < a$; when $b \le 0$, $|b| = -b < a$ because $-a < b$.
2. Apply (a) to the number $a - b$ and bound $c$: $|a - b| < c$ iff $-c < a - b < c$ iff $b - c < a < b + c$ (add $b$ throughout; adding preserves strict inequalities).
3. Same argument as (b), using Exercise 3.5(a): $|a - b| \le c$ iff $-c \le a - b \le c$ iff $b - c \le a \le b + c$.

Tags: Warm-upInequalities

Used in Exercise 4.13

Exercise 3.8

Let $a, b \in \mathbb{R}$. Show if $a \le b_1$ for every $b_1 > b$, then $a \le b$.

Hint

Suppose $a > b$ and take $b_1 = \frac{a + b}{2}$.

Solution

If $a > b$, then $b_1 = \frac{a + b}2$ satisfies $b < b_1 < a$, contradicting the hypothesis $a \le b_1$.

Tags: StandardInequalitiesContradiction

Used in Exercise 4.15