Contents / Real Analysis / The Real Number System

Chapter 1

The Real Number System

The reals as a complete ordered field; why Q fails completeness and the same algebra produces sqrt(2) once a supremum exists; the Archimedean property proved from completeness; density of the rationals and irrationals; both the Cauchy and Dedekind constructions with the uniqueness theorem that makes them interchangeable; and Cantor's diagonal argument with the cardinality of the continuum.

Introduction

Calculus is built on limits, and limits are built on a property of the real numbers that is easy to use and surprisingly hard to state: that there are no gaps. Every theorem in analysis that guarantees something exists — a maximum, a limit, a root, an integral — traces back to that single property, and none of them hold over the rationals. This chapter is about what the property says, why it cannot be proved from the arithmetic, and what the real numbers actually are.

The order of business is deliberate. First the reals are described axiomatically, as a complete ordered field, because that is the only description one ever uses in practice: no argument in analysis ever unpacks a real number into whatever it was built from. Then completeness is isolated and its consequences drawn out — the Archimedean property, the density of the rationals, the nested interval theorem — each of which looks obvious and none of which is.

Then the awkward question the axioms leave open: does such a thing exist? Two constructions answer it, Cauchy sequences and Dedekind cuts, and they are genuinely different in character — one completes $Q$ by adding limits, the other by filling gaps in the order. That they produce isomorphic results is a theorem, and it is what licenses speaking of the real numbers.

The chapter closes with the fact that makes the reals strange rather than merely large. They are uncountable: no list can contain them all, and Cantor's diagonal argument proves it in half a page. So $R$ is not $Q$ with a few gaps patched — the gaps vastly outnumber the rationals, and almost every real number is one you can never name.

1.1What the real numbers are

The reals are not defined by a formula. They are characterised by the properties they satisfy, and the characterisation is complete in a precise sense made exact later in the chapter.

Definition 1.1 (The reals, axiomatically). $R$ is a complete ordered field: a set with two operations $+$ and $\cdot$ and a relation $<$ satisfying

the field axioms — $+$ and $\cdot$ are associative and commutative with identities $0 \neq = 1$ and inverses (multiplicative inverses for non-zero elements), and $\cdot$ distributes over $+$ ;
the order axioms — $<$ is a total order compatible with the operations: $a < b$ implies $a + c < b + c$ , and $a < b$ with $c > 0$ implies $a c < b c$ ;
completeness — every non-empty subset bounded above has a least upper bound in $R$ .

The first two groups are shared with $Q$ , which is also an ordered field. Everything that distinguishes the reals is in the third, and it is worth noticing how strange that axiom looks beside the others: the first two are statements about elements, and the third is a statement about arbitrary subsets. That difference in logical shape is why completeness cannot be derived from the rest, and why it is the axiom that does all the work.

Example 1.2 (Small consequences of the field axioms). Prove from the axioms alone that $0$ is unique, that $a \cdot 0 = 0$ for every $a$ , and that $ab = 0$ forces $a = 0$ or $b = 0$ .

Solution.

Uniqueness of $0$ . Suppose $0$ and $0^{'}$ are both additive identities. Then $0 = 0 + 0^{'} = 0^{'}$ , using the identity property of $0^{'}$ first and of $0$ second.
$a \cdot 0 = 0$ . Since $0 + 0 = 0$ , distributivity gives $a \cdot 0 = a (0 + 0) = a \cdot 0 + a \cdot 0$ . Adding $- (a \cdot 0)$ to both sides leaves $0 = a \cdot 0$ .
No zero divisors. Suppose $ab = 0$ and $a \neq = 0$ . Then $a^{- 1}$ exists, and $b = 1 \cdot b = (a^{- 1} a) b = a^{- 1} (ab) = a^{- 1} \cdot 0 = 0$ by part 2.
Sanity check on what was used: only the axioms, never any picture of a number line. These facts hold in every field — in $Q$ , in the two-element field, in the complex numbers — which is exactly why they cannot distinguish $R$ from anything else.

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Intuition. A useful way to hold the three groups: the field axioms say you can do arithmetic, the order axioms say the arithmetic respects a left-to-right arrangement, and completeness says the arrangement has no holes.

The rationals satisfy the first two perfectly well and fail the third, and the failure is visible at a single point: $2$ . Slice the rational line at that place and you get two halves with nothing at the join — no largest rational below, no smallest above. Completeness is precisely the assertion that such a slice always has something sitting at it.

Exercises for §1.1 →

1.2The reals as an ordered field

The order axioms have consequences used so constantly that they are rarely cited, and it is worth proving a few to see that they really do follow.

Proposition 1.3 (Basic order facts). In any ordered field: (a) $a < b$ implies $- b < - a$ ; (b) $a \neq = 0$ implies $a^{2} > 0$ , and hence $1 > 0$ ; (c) $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ .

Proof. (a) Add $- a - b$ to both sides of $a < b$ : the left becomes $- b$ and the right $- a$ , and adding a constant preserves order by the first order axiom.

(b) If $a > 0$ then $a \cdot a > 0 \cdot a = 0$ by the second order axiom. If $a < 0$ then $- a > 0$ by (a), so $(- a) (- a) > 0$ ; and $(- a) (- a) = a^{2}$ by the field axioms. Since $1 = 1^{2}$ and $1 \neq = 0$ , we get $1 > 0$ .

(c) First, $a > 0$ forces $a^{- 1} > 0$ : if $a^{- 1} < 0$ then $1 = a \cdot a^{- 1} < 0$ by the order axiom, contradicting (b). Now multiply $a < b$ by the positive quantity $a^{- 1} b^{- 1}$ , giving $b^{- 1} < a^{- 1}$ .∎

Definition 1.4 (Absolute value). $∣ a ∣ = a$ if $a \geq 0$ and $∣ a ∣ = - a$ if $a < 0$ .

Theorem 1.5 (Triangle inequality). For all real $a, b$ : $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ , and consequently $∣ a ∣ - ∣ b ∣ \leq ∣ a - b ∣$ .

Proof. From the definition, $- ∣ a ∣ \leq a \leq ∣ a ∣$ and $- ∣ b ∣ \leq b \leq ∣ b ∣$ . Adding the two chains,

- (∣ a ∣ + ∣ b ∣) \leq a + b \leq ∣ a ∣ + ∣ b ∣,

and a number squeezed between $- c$ and $c$ has absolute value at most $c$ . That gives the first claim.

For the second, write $∣ a ∣ = ∣ (a - b) + b ∣ \leq ∣ a - b ∣ + ∣ b ∣$ , so $∣ a ∣ - ∣ b ∣ \leq ∣ a - b ∣$ . Swapping the roles of $a$ and $b$ gives $∣ b ∣ - ∣ a ∣ \leq ∣ b - a ∣ = ∣ a - b ∣$ . Both $∣ a ∣ - ∣ b ∣$ and its negative are at most $∣ a - b ∣$ , which is the claim.∎

Example 1.6 (Solving an inequality with absolute values). Describe ${x \in R : ∣ x - 3∣ < 2∣ x ∣}$ explicitly.

Solution.

Both sides are non-negative, so the inequality is equivalent to the squared one: $(x - 3)^{2} < 4 x^{2}$ .
Expand and collect: $x^{2} - 6 x + 9 < 4 x^{2}$ , that is $3 x^{2} + 6 x - 9 > 0$ , or $x^{2} + 2 x - 3 > 0$ after dividing by $3 > 0$ .
Factor: $(x + 3) (x - 1) > 0$ , which holds when both factors are positive or both negative — that is, $x > 1$ or $x < - 3$ .
The set is $(- \infty, - 3) \cup (1, \infty)$ .
Sanity check at three points: $x = 0$ gives $∣ - 3 ∣ = 3 < 0$ , false ✓ (and $0$ is excluded); $x = 2$ gives $1 < 4$ , true ✓; $x = - 4$ gives $7 < 8$ , true ✓.
Note the squaring step was legitimate only because both sides were known non-negative. Squaring an inequality with a possibly negative side reverses truth values and is the standard error here.

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Remark (What an ordered field cannot give you). Everything so far holds in $Q$ verbatim. So no amount of algebra and order will produce $2$ , prove that a bounded increasing sequence converges, or guarantee that a continuous function changing sign has a root. Each of those is an existence claim, and existence is what the first two groups of axioms never assert.

This is the precise sense in which analysis is not algebra: the interesting theorems are about existence, and they all come from the third axiom.

Exercises for §1.2 →

1.3Completeness, and why the rationals fail it

Definition 1.7 (Upper bound, supremum, infimum). Let $S \subseteq R$ be non-empty. A number $M$ is an upper bound for $S$ if $s \leq M$ for every $s \in S$ . The supremum $sup S$ is the least upper bound: an upper bound $M$ such that no $M^{'} < M$ is an upper bound. Infimum is defined dually as the greatest lower bound.

Proposition 1.8 (The supremum is unique, and characterised by an epsilon condition). A set has at most one supremum. Moreover $M = sup S$ if and only if $M$ is an upper bound and, for every $ε > 0$ , there exists $s \in S$ with $s > M - ε$ .

Proof. Uniqueness. If $M$ and $M^{'}$ are both least upper bounds, then $M$ is an upper bound and $M^{'}$ is least, so $M^{'} \leq M$ ; symmetrically $M \leq M^{'}$ . Hence $M = M^{'}$ .

Characterisation. Suppose $M = sup S$ and let $ε > 0$ . Then $M - ε < M$ , so $M - ε$ is not an upper bound, meaning some $s \in S$ exceeds it. Conversely, suppose $M$ is an upper bound with the stated property and let $M^{'} < M$ . Put $ε = M - M^{'} > 0$ ; the property supplies $s > M - ε = M^{'}$ , so $M^{'}$ is not an upper bound. Hence $M$ is least.∎

The epsilon characterisation is the form used in practice: it says you can always find an element of $S$ as close to the supremum as you like, which is what makes suprema usable in limit arguments.

Example 1.9 (Proving a supremum from the definition). Let $S = {2 - 3/ n : n \in N}$ . Prove $sup S = 2$ directly from the definition.

Solution.

Upper bound. For every $n$ , $3/ n > 0$ , so $2 - 3/ n < 2$ . Hence $2$ is an upper bound.
Least. Let $ε > 0$ . By the proposition The supremum is unique, and characterised by an epsilon condition it suffices to produce an element exceeding $2 - ε$ . Solve $2 - 3/ n > 2 - ε$ , which rearranges to $3/ n < ε$ , that is $n > 3/ ε$ . Such an $n$ exists by the Archimedean property proved later in this chapter.
So for every $ε > 0$ there is an element of $S$ above $2 - ε$ , and $2$ is the least upper bound.
The supremum is not attained: $2 - 3/ n = 2$ would force $3/ n = 0$ , impossible.
Sanity check on the smallest elements: $n = 1$ gives $- 1$ , $n = 2$ gives $0.5$ , $n = 3$ gives $1$ — increasing toward $2$ and never reaching it ✓. Note $in f S = - 1$ , which is attained.

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Theorem 1.10 (The rationals are not complete). The set $S = {q \in Q : q > 0, q^{2} < 2}$ is non-empty and bounded above in $Q$ , but has no least upper bound in $Q$ .

Proof. $S$ is non-empty ( $1 \in S$ ) and bounded above by $2$ , since $q > 2$ would give $q^{2} > 4$ .

Suppose $M \in Q$ were a least upper bound. By the irrationality of $2$ — proved below — $M^{2} \neq = 2$ , so either $M^{2} < 2$ or $M^{2} > 2$ .

If $M^{2} < 2$ , we exhibit a larger element of $S$ , contradicting that $M$ is an upper bound. Take $h = \frac{2 - M ^{2}}{2 ( M + 1 )}$ , which is rational and satisfies $0 < h < 1$ . Then

(M + h)^{2} = M^{2} + 2 M h + h^{2} < M^{2} + 2 M h + h = M^{2} + h (2 M + 1) < M^{2} + 2 h (M + 1) = M^{2} + (2 - M^{2}) = 2,

using $h^{2} < h$ and $2 M + 1 < 2 M + 2$ . So $M + h \in S$ and $M + h > M$ .

If $M^{2} > 2$ , we exhibit a smaller upper bound, contradicting leastness. Take $h = \frac{M ^{2} - 2}{2 M} > 0$ . Then

(M - h)^{2} = M^{2} - 2 M h + h^{2} > M^{2} - 2 M h = M^{2} - (M^{2} - 2) = 2,

so $M - h$ is still an upper bound for $S$ (any $q$ exceeding it would have $q^{2} > 2$ ), and $M - h < M$ .

Both cases are impossible, so no rational least upper bound exists.∎

Proposition 1.11 (Irrationality of the square root of two). There is no rational number whose square is $2$ .

Proof. Suppose $p / q = 2$ with $p, q$ integers, $q \neq = 0$ , and the fraction in lowest terms. Then $p^{2} = 2 q^{2}$ , so $p^{2}$ is even, so $p$ is even (an odd number squared is odd). Write $p = 2 k$ ; then $4 k^{2} = 2 q^{2}$ , so $q^{2} = 2 k^{2}$ and $q$ is even by the same argument. But then $p$ and $q$ share the factor $2$ , contradicting lowest terms.∎

The two results together are the whole motivation for the chapter. The rationals have a hole exactly where $2$ should be, and the hole is invisible to the field and order axioms — you cannot detect it by arithmetic, only by noticing that a bounded set fails to have a least upper bound.

Exercises for §1.3 →

1.4Constructing the reals by Cauchy sequences

The axioms describe the reals. They do not produce them, and a description satisfied by nothing is worth little. Two constructions build a complete ordered field out of $Q$ alone.

Definition 1.12 (Cauchy sequence of rationals). A sequence $(q_{n})$ of rationals is Cauchy if for every rational $ε > 0$ there is $N$ with $∣ q_{n} - q_{m} ∣ < ε$ for all $n, m \geq N$ : the terms eventually cluster together, without reference to any limit.

That last clause is the point. "Converges" requires a limit to converge to, which may not exist in $Q$ ; "Cauchy" is a condition on the sequence alone. The construction exploits the gap between the two.

Definition 1.13 (The Cauchy construction). Call two Cauchy sequences $(q_{n})$ and $(r_{n})$ equivalent if $∣ q_{n} - r_{n} ∣ \to 0$ . Define $R$ to be the set of equivalence classes, with

[(q_{n})] + [(r_{n})] = [(q_{n} + r_{n})], [(q_{n})] \cdot [(r_{n})] = [(q_{n} r_{n})],

and $[(q_{n})] < [(r_{n})]$ if some rational $ε > 0$ and index $N$ have $r_{n} - q_{n} > ε$ for all $n \geq N$ .

Proposition 1.14 (The relation is an equivalence relation). Equivalence of Cauchy sequences is reflexive, symmetric and transitive.

Proof. Reflexivity and symmetry are immediate from $∣ q_{n} - q_{n} ∣ = 0$ and $∣ q_{n} - r_{n} ∣ = ∣ r_{n} - q_{n} ∣$ .

For transitivity, suppose $∣ q_{n} - r_{n} ∣ \to 0$ and $∣ r_{n} - s_{n} ∣ \to 0$ . Given $ε > 0$ , choose $N_{1}$ beyond which the first is under $ε /2$ and $N_{2}$ beyond which the second is. For $n \geq max (N_{1}, N_{2})$ , the triangle inequality gives

∣ q_{n} - s_{n} ∣ \leq ∣ q_{n} - r_{n} ∣ + ∣ r_{n} - s_{n} ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Remark (Why equivalence classes rather than sequences). Two different sequences can describe the same number: $1, 1.4, 1.41, 1.414, \dots$ and $2, 1.5, 1.42, 1.415, \dots$ both close in on $2$ , and neither has a better claim. Taking equivalence classes says exactly that the approximation scheme is irrelevant and only the target matters.

$Q$ embeds by sending $q$ to the class of the constant sequence, and one checks this embedding preserves arithmetic and order. The new elements are the classes of Cauchy sequences that fail to converge in $Q$ — one for each hole.

Example 1.15 (A rational Cauchy sequence with no rational limit). Define $q_{1} = 1$ and $q_{n + 1} = \frac{q _{n}}{2} + \frac{1}{q _{n}}$ . Show the terms are rational, compute four of them, and identify the obstruction to convergence in $Q$ .

Solution.

Rationality is immediate by induction: $q_{1} = 1$ is rational, and the recurrence uses only $+$ , $\div$ and $\times$ on rationals with $q_{n} \neq = 0$ .
Computing: $q_{2} = \frac{1}{2} + 1 = \frac{3}{2} = 1.5$ ; $q_{3} = \frac{3}{4} + \frac{2}{3} = \frac{17}{12} = 1.41 \overline{6}$ ; $q_{4} = \frac{17}{24} + \frac{12}{17} = \frac{577}{408} = 1.4142156 \dots$
The terms are closing in on $1.4142135 \dots$ , and the fixed point of the recurrence satisfies $x = x /2 + 1/ x$ , that is $x^{2} = 2$ .
So the sequence is Cauchy — successive terms agree to rapidly increasing precision — but its only possible limit is $2$ , which by the proposition Irrationality of the square root of two is not rational.
Sanity check on the convergence rate: $q_{4}$ already matches $2$ to six decimal places, and the error squares at each step. This is Newton's method applied to $x^{2} - 2$ , which is why it is so fast.
This single sequence is the entire motivation for the construction: it is manifestly a legitimate object built from rationals alone, it manifestly should converge, and in $Q$ it does not. The construction declares such a sequence to be a number.

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Exercises for §1.4 →

1.5Constructing the reals by Dedekind cuts

The second construction takes the opposite view. Rather than adding limits, it fills gaps in the order directly, by identifying a real number with the set of rationals below it.

Definition 1.16 (The Dedekind construction). A cut is a set $A ⊊ Q$ , non-empty, with no greatest element, and downward closed: if $a \in A$ and $q < a$ then $q \in A$ . Define $R$ to be the set of cuts, ordered by inclusion: $A < B$ means $A ⊊ B$ .

Example 1.17 (The cut that is not a rational). Exhibit the cut corresponding to $2$ and verify it is a cut.

Solution.

Take $A = {q \in Q : q < 0} \cup {q \in Q : q \geq 0, q^{2} < 2}$ .
It is non-empty ( $- 1 \in A$ ) and not all of $Q$ ( $2 \in / A$ , since $4 > 2$ ).
It is downward closed: if $a \in A$ and $q < a$ , then either $q < 0$ , putting $q$ in the first piece, or $0 \leq q < a$ with $a^{2} < 2$ , giving $q^{2} \leq a^{2} < 2$ .
It has no greatest element — this is precisely the computation in the proof of the theorem The rationals are not complete, where any $M \in A$ was shown to admit $M + h \in A$ with $h > 0$ .
Sanity check on what makes this work: the cut is a perfectly good set of rationals, so it exists without assuming $2$ does. The construction manufactures the missing number out of the rationals below it, which is available all along.

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Intuition. A Dedekind cut is the shadow a number casts on the rational line. Every real number $x$ determines the set of rationals below it, and conversely that set determines $x$ — so one may as well define the number to be the set.

The move is characteristic of foundational mathematics: rather than asking what $2$ is, identify it with something built from what is already available that behaves the same way. The rationals below $2$ form a perfectly concrete object; declaring that object to be $2$ costs nothing and supplies everything.

The condition "no greatest element" is a tie-breaking convention. Without it, both ${q < 1}$ and ${q \leq 1}$ would be cuts at the same place, and the rational $1$ would have two representatives.

Exercises for §1.5 →

1.6The two constructions agree

Two constructions, built on entirely different ideas, and a natural worry: do they give the same thing? The answer is that they must, and the reason is the axioms rather than any correspondence between sequences and cuts.

Theorem 1.18 (Uniqueness of the complete ordered field). Any two complete ordered fields are isomorphic: there is a bijection between them preserving $+$ , $\cdot$ and $<$ , and it is unique.

Proof. We sketch the argument, whose details are routine but long. Let $F$ and $F^{'}$ be complete ordered fields. Any field homomorphism must send $0 \mapsto 0^{'}$ and $1 \mapsto 1^{'}$ ; by additivity this forces $n \mapsto n^{'}$ for every positive integer, then for every integer, then $p / q \mapsto p^{'} / q^{'}$ for every rational. So the map is already determined on the copy of $Q$ inside each field, and no choice has been made.

Extend to all of $F$ by completeness: for $x \in F$ set

φ (x) = sup^{'} {q^{'} : q \in Q, q < x},

the supremum taken in $F^{'}$ , which exists because the set is non-empty and bounded above — both facts requiring the Archimedean property, itself a consequence of completeness as shown in the next section. Checking that $φ$ preserves the operations and order, and is a bijection, is a sequence of applications of the density of $Q$ .

Uniqueness follows because $φ$ was forced at every step: determined on $Q$ by the field axioms, and on the rest by density and order-preservation.∎

So "the real numbers" is legitimate. The Cauchy construction and the cut construction both produce complete ordered fields, and the theorem says those are isomorphic, so nothing depends on which was used. This is why no analysis textbook, having built the reals one way, ever mentions the construction again.

Remark (Two constructions, two generalisations). The constructions are not redundant, because they generalise in different directions.

The Cauchy method needs only a notion of distance, so it completes any metric space — a construction used constantly, and the one behind the $p$ -adic numbers, which complete $Q$ with respect to a different distance and produce something utterly unlike $R$ .

The Dedekind method needs only an order, so it completes any ordered set, with no arithmetic required. Which construction to reach for depends on which structure the situation supplies.

Exercises for §1.6 →

1.7The least upper bound property

Theorem 1.19 (Least upper bound property). Every non-empty subset of $R$ that is bounded above has a least upper bound in $R$ . Dually, every non-empty subset bounded below has a greatest lower bound.

Proof. The first statement is the completeness axiom itself. The dual follows by reflection: given $S$ non-empty and bounded below, put $- S = {- s : s \in S}$ . A lower bound $m$ for $S$ gives an upper bound $- m$ for $- S$ , so $- S$ is bounded above and $M = sup (- S)$ exists. Then $- M$ is a lower bound for $S$ , and it is the greatest: any larger lower bound $m^{'} > - M$ would make $- m^{'} < M$ an upper bound for $- S$ , contradicting leastness.∎

Pitfall. The supremum need not belong to the set. $sup (0, 1) = 1$ and $1 \in / (0, 1)$ ; $sup [0, 1] = 1$ and $1 \in [0, 1]$ . When the supremum is attained it is the maximum, and the distinction matters — the Extreme Value Theorem is precisely the assertion that in certain circumstances the supremum is attained, which would be vacuous if suprema always were.

Note also that the axiom requires the set to be non-empty. The empty set is bounded above by every real number, so it has no least upper bound; excluding it is not fussiness.

Exercises for §1.7 →

1.8Suprema in use, and the existence of roots

Example 1.20 (Computing suprema and infima). Find $sup$ and $in f$ of each set, and say whether each is attained: (a) ${1 - 1/ n : n \in N}$ ; (b) ${(- 1)^{n} (1 + 1/ n) : n \in N}$ ; (c) ${q \in Q : q^{2} < 2}$ .

Solution.

(a) The terms increase toward $1$ . Every term is below $1$ , so $1$ is an upper bound; and for any $ε > 0$ the term $1 - 1/ n$ exceeds $1 - ε$ once $n > 1/ ε$ , so by the proposition The supremum is unique, and characterised by an epsilon condition, $sup = 1$ — not attained. The infimum is $0$ , attained at $n = 1$ .
(b) Even $n$ gives $1 + 1/ n$ , decreasing to $1$ from above, with largest value $3/2$ at $n = 2$ . Odd $n$ gives $- (1 + 1/ n)$ , largest in magnitude at $n = 1$ giving $- 2$ . So $sup = 3/2$ attained, $in f = - 2$ attained.
(c) $sup = 2$ , not attained, and not even a member of $Q$ — this is the theorem The rationals are not complete restated. The infimum is $- 2$ .
Sanity check on (a): a set can have a supremum it never reaches, and the epsilon characterisation is what detects it — there are elements arbitrarily close to $1$ , and none equal to it.

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Theorem 1.21 (Existence of square roots). For every real $a > 0$ there is a unique real $x > 0$ with $x^{2} = a$ .

Proof. Let $S = {t > 0 : t^{2} < a}$ . It is non-empty — if $a \geq 1$ then $1/2 \in S$ ... more carefully, $min (a, 1) /2$ has square at most $min (a, 1)^{2} /4 < a$ . It is bounded above by $a + 1$ , since $t > a + 1 > 1$ gives $t^{2} > t > a$ . By the theorem Least upper bound property, $x = sup S$ exists, and $x > 0$ .

Now rule out both inequalities, by exactly the computation used for $Q$ — but this time the conclusion is not a contradiction with rationality, it is a contradiction with $x$ being a supremum.

If $x^{2} < a$ , put $h = min (1, \frac{a - x ^{2}}{2 ( x + 1 )}) > 0$ . Then $h \leq 1$ gives $h^{2} \leq h$ and

(x + h)^{2} = x^{2} + 2 x h + h^{2} \leq x^{2} + h (2 x + 1) < x^{2} + 2 h (x + 1) \leq x^{2} + (a - x^{2}) = a,

so $x + h \in S$ , contradicting that $x$ is an upper bound.

If $x^{2} > a$ , put $h = \frac{x ^{2} - a}{2 x} > 0$ , which satisfies $h < x$ . Then

(x - h)^{2} = x^{2} - 2 x h + h^{2} > x^{2} - 2 x h = a,

so every $t > x - h$ has $t^{2} > a$ and lies outside $S$ ; hence $x - h$ is an upper bound smaller than $x$ , contradicting leastness.

Therefore $x^{2} = a$ . For uniqueness, if $0 < x < y$ then $x^{2} < x y < y^{2}$ , so the squaring map is injective on positives.∎

The comparison with the earlier proof is the whole point of the chapter. The same algebra that showed $Q$ has no least upper bound for that set shows that in $R$ the least upper bound must square to exactly $2$ . The algebra was never the issue; the availability of the supremum was.

Exercises for §1.8 →

1.9Completeness has several equivalent faces

The least upper bound property is one of several conditions that each express "no gaps", and in an Archimedean ordered field they are equivalent. Knowing the list matters because different arguments want different forms.

Theorem 1.22 (Monotone convergence). Every increasing sequence of reals that is bounded above converges, and its limit is the supremum of its terms.

Proof. Let $(a_{n})$ be increasing and bounded above, and let $L = sup {a_{n}}$ , which exists by the theorem Least upper bound property. Given $ε > 0$ , the epsilon characterisation supplies $N$ with $a_{N} > L - ε$ . For $n \geq N$ , monotonicity gives $a_{n} \geq a_{N} > L - ε$ , and $a_{n} \leq L$ since $L$ is an upper bound. Hence $∣ a_{n} - L ∣ < ε$ for all $n \geq N$ , which is convergence to $L$ .∎

Theorem 1.23 (Nested interval property). Let $I_{n} = [a_{n}, b_{n}]$ be closed bounded intervals with $I_{1} \supseteq I_{2} \supseteq \dots$ . Then $⋂_{n} I_{n} \neq = \emptyset$ . If in addition $b_{n} - a_{n} \to 0$ , the intersection is a single point.

Proof. Nesting means $(a_{n})$ is increasing and $(b_{n})$ decreasing, with $a_{n} \leq b_{m}$ for all $n, m$ : if $n \leq m$ then $a_{n} \leq a_{m} \leq b_{m}$ , and if $n > m$ then $a_{n} \leq b_{n} \leq b_{m}$ .

So every $b_{m}$ is an upper bound for ${a_{n}}$ , and $a = sup {a_{n}}$ exists with $a \leq b_{m}$ for every $m$ . Since also $a \geq a_{n}$ for every $n$ , the point $a$ lies in every $I_{n}$ , so the intersection is non-empty.

If $b_{n} - a_{n} \to 0$ and $x < y$ both lay in the intersection, then $b_{n} - a_{n} \geq y - x > 0$ for every $n$ , contradicting convergence to $0$ .∎

Figure 1.24. Nested closed intervals whose lengths shrink to zero. Each contains the next, and the left endpoints increase while the right endpoints decrease; completeness guarantees the supremum of the left endpoints lies in every interval, so the intersection is non-empty. The same picture over

Q

can have empty intersection — intervals closing in on

2

trap nothing at all — which is the property's entire content.

Remark (The equivalent forms, and one trap). In an ordered field the following are equivalent: the least upper bound property; monotone convergence; the nested interval property together with the Archimedean property; Bolzano-Weierstrass; and Cauchy completeness together with the Archimedean property.

The parenthetical conditions are not decoration. There exist non-Archimedean ordered fields in which Cauchy sequences all converge and yet the least upper bound property fails — so "every Cauchy sequence converges" is strictly weaker than completeness in the sense of this chapter. That is why the next section proves the Archimedean property from the least upper bound property rather than assuming it: over $R$ it comes free, and it is doing real work.

Exercises for §1.9 →

1.10The Archimedean property

Theorem 1.25 (Archimedean property). The following equivalent statements hold in $R$ :

$N$ is not bounded above.
For every real $x$ there is $n \in N$ with $n > x$ .
For every $ε > 0$ there is $n \in N$ with $1/ n < ε$ .
For all reals $a > 0$ and $b$ , there is $n \in N$ with $na > b$ .

Proof. Statements 1 and 2 are restatements of each other. For 2 $\Rightarrow$ 3, given $ε > 0$ choose $n > 1/ ε$ ; then $1/ n < ε$ by the proposition Basic order facts. For 2 $\Rightarrow$ 4, choose $n > b / a$ and multiply by $a > 0$ . Each of 3 and 4 gives back 2 by specialising.∎

The name is historical and the content is not obvious, despite the statements looking like truisms. "No real number exceeds every integer" is exactly the assertion that there are no infinite elements, and there are perfectly consistent ordered fields containing some.

Exercises for §1.10 →

1.11Proving it from completeness

Theorem 1.26 (Completeness implies the Archimedean property). In a complete ordered field, $N$ is not bounded above.

Proof. Suppose it were. Then $N$ is non-empty and bounded above, so by the theorem Least upper bound property it has a least upper bound $M = sup N$ .

Since $M - 1 < M$ and $M$ is the least upper bound, $M - 1$ is not an upper bound. So some $n \in N$ satisfies $n > M - 1$ , whence

n + 1 > M .

But $n + 1 \in N$ , so $M$ fails to be an upper bound for $N$ — a contradiction.∎

That proof is three lines and it is the hinge of the whole subject. It converts a statement about arbitrary bounded sets into the concrete tool — "choose $n$ with $1/ n < ε$ " — used in virtually every epsilon argument that follows, in this chapter and in every later one.

Remark (Why this cannot be proved from the field and order axioms alone). Consider the field of rational functions $R (t)$ , ordered by declaring $f > 0$ when the leading coefficient of $f$ (written in lowest terms, comparing behaviour as $t \to \infty$ ) is positive. This is a genuine ordered field, and in it the element $t$ exceeds every constant — so $N$ is bounded above, by $t$ .

The Archimedean property therefore genuinely requires completeness, and the existence of this example is what makes the three-line proof above worth writing down rather than waving at.

Exercises for §1.11 →

1.12Consequences of the Archimedean property

Corollary 1.27 (The floor function exists). For every real $x$ there is a unique integer $n$ with $n \leq x < n + 1$ .

Proof. By the theorem Archimedean property the set ${k \in Z : k \leq x}$ is non-empty (some integer is below $x$ , since $- x$ is exceeded by some natural number) and bounded above by $x$ . A non-empty set of integers bounded above has a greatest element, $n$ . Then $n \leq x$ , and $n + 1 > x$ since otherwise $n + 1$ would belong to the set and exceed its maximum. Uniqueness holds because two such integers would differ by less than $1$ .∎

Example 1.28 (The Archimedean property doing ordinary work). Show that $in f {1/ n : n \in N} = 0$ , and that $lim_{n \to \infty} 1/ n = 0$ .

Solution.

Every $1/ n > 0$ , so $0$ is a lower bound.
Suppose $m > 0$ were a greater lower bound. By part 3 of the theorem Archimedean property there is $n$ with $1/ n < m$ , contradicting that $m$ bounds the set below. So $0$ is the greatest lower bound.
For the limit: given $ε > 0$ , the same statement supplies $N$ with $1/ N < ε$ , and then $1/ n \leq 1/ N < ε$ for all $n \geq N$ . That is exactly convergence to $0$ .
Sanity check on what was used: nothing but the Archimedean property, which in turn used only completeness. In a non-Archimedean ordered field the sequence $1/ n$ does not tend to zero — it stays above every infinitesimal — so this familiar fact is not a triviality about fractions.

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Exercises for §1.12 →

1.13Density of the rationals

Theorem 1.29 (Density of the rationals). Between any two distinct reals there is a rational: if $x < y$ then there exists $q \in Q$ with $x < q < y$ .

The statement is worth reading carefully. The rationals are countable and the reals are not, as the last section shows — so $Q$ is a vanishingly small part of $R$ , and yet it is spread so finely that no interval, however short, avoids it. Smallness of cardinality and density in the order are unrelated properties.

Exercises for §1.13 →

1.14Proving density from the Archimedean property

Proof. Assume $x < y$ , so $y - x > 0$ . By part 3 of the theorem Archimedean property choose $n \in N$ with

\frac{1}{n} < y - x, equivalently n y - n x > 1.

So the interval $(n x, n y)$ has length greater than $1$ , and an interval of length exceeding $1$ must contain an integer. Precisely: let $m = ⌊ n x ⌋ + 1$ , which exists by the corollary The floor function exists. Then

m > n x and m \leq n x + 1 < n y,

the second inequality using $n x + 1 < n y$ from the choice of $n$ . Dividing the chain $n x < m < n y$ by $n > 0$ gives

x < \frac{m}{n} < y,

and $m / n$ is rational.∎

Example 1.30 (Finding the rational explicitly). Find a rational strictly between $2 = 1.41421 \dots$ and $2 + 0.001$ .

Solution.

Here $y - x = 0.001$ , so the proof asks for $n$ with $1/ n < 0.001$ , that is $n \geq 1001$ . Take $n = 1001$ .
$n x = 10012 = 1415.6278 \dots$ , so $m = ⌊ n x ⌋ + 1 = 1416$ .
The rational is $m / n = 1416/1001 = 1.414585 \dots$ , and indeed $1.414214 < 1.414585 < 1.415214$ ✓.
Sanity check on the mechanism: making $n$ large enough forces the grid of fractions ${m / n}$ to have spacing finer than the gap, so the grid cannot step over the interval. That is all the proof does, and it is why the Archimedean property is exactly what is needed — it is the guarantee that the spacing can be made arbitrarily fine.

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Exercises for §1.14 →

1.15Density of the irrationals

Corollary 1.31 (The irrationals are dense too). Between any two distinct reals there is an irrational number.

Proof. Let $x < y$ . Apply the theorem Density of the rationals to the pair $x - 2 < y - 2$ , obtaining a rational $q$ with

x - 2 < q < y - 2 .

Adding $2$ throughout gives $x < q + 2 < y$ . And $q + 2$ is irrational: if it were rational, then $2 = (q + 2) - q$ would be a difference of rationals and hence rational, contradicting the proposition Irrationality of the square root of two.∎

Corollary 1.32 (Every real is a limit of rationals). For every real $x$ there is a sequence of rationals converging to $x$ .

Proof. For each $n$ , apply density to the interval $(x - 1/ n, x)$ to obtain a rational $q_{n}$ with $x - 1/ n < q_{n} < x$ . Then $∣ q_{n} - x ∣ < 1/ n \to 0$ by the example above, so $q_{n} \to x$ .∎

The last corollary is what makes the Cauchy construction inevitable in hindsight: every real is approximable by rationals, so the rationals already contain enough information to specify every real — the only thing missing was a way to name the target, which is what an equivalence class of Cauchy sequences supplies.

Exercises for §1.15 →

1.16Dedekind cuts in detail

Definition 1.33 (Dedekind cut). A Dedekind cut is a set $A \subseteq Q$ satisfying

$A \neq = \emptyset$ and $A \neq = Q$ ;
$A$ is downward closed: $a \in A$ and $q < a$ imply $q \in A$ ;
$A$ has no greatest element.

Proposition 1.34 (Cuts are totally ordered by inclusion). For any two cuts $A, B$ , exactly one of $A ⊊ B$ , $A = B$ , $B ⊊ A$ holds.

Proof. Suppose $A \neq \subseteq B$ ; we show $B ⊊ A$ . Pick $a \in A ∖ B$ . For any $b \in B$ we cannot have $b \geq a$ , since downward closure of $B$ would then put $a \in B$ . So $b < a$ for every $b \in B$ , and downward closure of $A$ puts every such $b$ in $A$ . Hence $B \subseteq A$ , and the inclusion is strict because $a \in A ∖ B$ .

Thus either $A \subseteq B$ or $B ⊊ A$ , and in the first case either $A = B$ or $A ⊊ B$ . The three cases are mutually exclusive by definition.∎

Figure 1.35. The cut at

2

. Every rational is on one side or the other, and the set

A

of those below has no largest member while its complement has no smallest — that is the gap in

Q

. A cut at a rational point, shown at

1

, looks the same except that the complement does have a least element. Defining a real number to *be* the left-hand set turns the gap into an object.

Exercises for §1.16 →

1.17Arithmetic on cuts

Definition 1.36 (Addition and order for cuts). For cuts $A, B$ define

A + B = {a + b : a \in A, b \in B}, A < B ⟺ A ⊊ B .

The zero element is $0^{*} = {q \in Q : q < 0}$ , and the negative of $A$ is

- A = {q \in Q : - q is an upper bound for A that A does not reach},

more precisely ${q : \exists r > 0 with - q - r \in / A}$ .

Proposition 1.37 (The sum of two cuts is a cut). If $A$ and $B$ are cuts then so is $A + B$ .

Proof. Non-empty and proper. $A + B$ is non-empty since both are. For properness, pick $p \in / A$ and $q \in / B$ ; then $p + q \in / A + B$ , because any $a + b$ with $a \in A$ , $b \in B$ has $a < p$ and $b < q$ by downward closure, so $a + b < p + q$ .

Downward closed. Let $a + b \in A + B$ and $r < a + b$ . Then $r - b < a$ , so $r - b \in A$ by downward closure of $A$ , and $r = (r - b) + b \in A + B$ .

No greatest element. Given $a + b \in A + B$ , use that $A$ has no greatest element to find $a^{'} \in A$ with $a^{'} > a$ ; then $a^{'} + b > a + b$ lies in $A + B$ .∎

Remark (Where multiplication becomes awkward). Addition is painless because it respects the order uniformly. Multiplication is not: the naive definition ${ab}$ fails, because if $A$ and $B$ contain arbitrarily negative rationals then their products are arbitrarily large positive, and the set is all of $Q$ .

The standard repair defines multiplication first for cuts greater than $0^{*}$ , by

A \cdot B = {q : q < ab for some a \in A, b \in B with a, b > 0},

and then extends to the remaining sign combinations by cases using $(- A) B = - (A B)$ . It works, and the case analysis is tedious — which is a fair summary of why the Cauchy construction is often preferred despite being less geometrically transparent.

Exercises for §1.17 →

1.18Why cuts form a complete ordered field

The payoff of the construction is that completeness becomes almost trivial, because the supremum of a set of cuts is simply their union.

Theorem 1.38 (The cuts are complete). Let $S$ be a non-empty collection of cuts that is bounded above by some cut $B$ . Then

C = A \in S ⋃ A

is a cut, and $C = sup S$ .

Proof. $C$ is a cut. It is non-empty since $S$ is and its members are. It is proper because every $A \in S$ satisfies $A \subseteq B$ , so $C \subseteq B \neq = Q$ . It is downward closed: if $q \in C$ then $q \in A$ for some $A \in S$ , and any $r < q$ lies in that same $A$ , hence in $C$ . It has no greatest element: given $q \in C$ , pick $A ∋ q$ ; since $A$ has no greatest element there is $q^{'} \in A$ with $q^{'} > q$ , and $q^{'} \in C$ .

$C$ is an upper bound. Every $A \in S$ satisfies $A \subseteq C$ by construction, which is exactly $A \leq C$ in the inclusion order.

$C$ is least. Let $D$ be any upper bound, so $A \subseteq D$ for every $A \in S$ . Then the union of those sets is contained in $D$ , that is $C \subseteq D$ , so $C \leq D$ .∎

Intuition. That proof is three short paragraphs, and compared with everything else in the chapter it is effortless. The reason is worth noticing: the cut construction was designed so that completeness would be free, by identifying a real number with a downward-closed set and the order with inclusion. Suprema of sets of sets are unions, and unions always exist.

The cost was paid elsewhere — in the case analysis multiplication requires. Every construction of the reals pays somewhere, because the object being built is genuinely harder than the rationals; the constructions differ in where they put the difficulty, not in whether there is any.

Exercises for §1.18 →

1.19Countable and uncountable

The reals are complete. They are also, in a way completeness does not by itself reveal, enormously larger than the rationals.

Definition 1.39 (Countable and uncountable). A set is countable if it is finite or can be put in bijection with $N$ — equivalently, if its elements can be listed as $x_{1}, x_{2}, x_{3}, \dots$ with every element appearing. A set that is not countable is uncountable.

Proposition 1.40 (The rationals are countable). $Q$ is countable.

Proof. Every positive rational is $p / q$ in lowest terms with $p, q \in N$ . Assign to it the height $p + q$ . For each height $h \geq 2$ there are only finitely many pairs $(p, q)$ with $p + q = h$ , hence finitely many rationals of that height.

List the positive rationals by increasing height, ordering the finitely many within each height by numerator. Every positive rational appears exactly once, and the list is indexed by $N$ . Interleave with the negatives and $0$ to list all of $Q$ .∎

Example 1.41 (Listing the rationals by height). List the first eight positive rationals in the height ordering of the proof above, and say which height contains $5/7$ .

Solution.

Height $2$ : only $1/1$ . Height $3$ : $1/2$ , $2/1$ . Height $4$ : $1/3$ , $3/1$ — note $2/2$ is excluded because it is not in lowest terms.
Height $5$ : $1/4$ , $2/3$ , $3/2$ , $4/1$ .
So the first eight are $1, \frac{1}{2}, 2, \frac{1}{3}, 3, \frac{1}{4}, \frac{2}{3}, \frac{3}{2}$ .
For $5/7$ : the fraction is already in lowest terms since $g cd (5, 7) = 1$ , so its height is $5 + 7 = 12$ .
Sanity check that the listing is exhaustive: any positive rational in lowest terms has some finite height $p + q$ , and it appears in that finite block. Nothing can escape the list by being "too far along", because every height is reached after finitely many steps.
Note what this does not do: it gives no order-preserving list. The rationals cannot be listed in increasing order, since between any two there is another — density and countability coexist precisely because the listing ignores the order.

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A set being countable is a strong statement — it says the set is no larger than $N$ despite $Q$ being dense and $N$ being sparse. Density and cardinality are independent, and the next result shows the reals differ from the rationals in the second respect, not the first.

Exercises for §1.19 →

1.20Cantor's diagonal argument

Theorem 1.42 (The reals are uncountable). The interval $[0, 1]$ cannot be put in bijection with $N$ . Hence neither can $R$ .

Proof. Suppose some list $x_{1}, x_{2}, x_{3}, \dots$ contained every element of $[0, 1]$ . Write each in decimal,

x_{n} = 0. d_{n 1} d_{n 2} d_{n 3} \dots,

choosing the non-terminating representation whenever a number has two (so $0.5$ is written $0.4999 \dots$ ), which makes the expansion unique.

Define a new number $y = 0. e_{1} e_{2} e_{3} \dots$ by

e_{k} = {5, 6, d_{kk} \neq = 5, d_{kk} = 5.

Then $y \in [0, 1]$ . But $y \neq = x_{k}$ for every $k$ , because $y$ and $x_{k}$ differ in the $k$ -th decimal place by construction. The digits $5$ and $6$ were chosen to avoid $0$ and $9$ , so $y$ has no alternative representation that could secretly coincide with some $x_{k}$ .

So $y$ is an element of $[0, 1]$ missing from the list, contradicting the assumption that the list was exhaustive. No such list exists.∎

Pitfall. The argument does not say "the list was badly chosen, pick a better one". It takes an arbitrary list and manufactures a missing element, so it defeats every list at once. Responding with "just add $y$ to the list" misses the structure of the argument: the new list is again a list, and the diagonal construction applied to it produces a new missing number.

The restriction to non-terminating decimals is not pedantry either. Without it, $0.4999 \dots$ and $0.5000 \dots$ are the same real with two expansions, and a diagonal construction could produce a $y$ that differs from some $x_{k}$ digit-wise while equalling it numerically. Choosing digits $5$ and $6$ sidesteps the issue entirely.

Figure 1.43. The diagonal construction. The listed numbers run down the page and their decimal expansions across it; the highlighted diagonal digits

d_{11}, d_{22}, d_{33}, \dots

are the ones the new number is built to disagree with. Whatever list is offered, changing every diagonal digit produces a number differing from the

k

-th entry in the

k

-th place, hence from every entry.

Exercises for §1.20 →

1.21The cardinality of the continuum

Theorem 1.44 (The reals have the cardinality of the power set of the naturals).

∣ R ∣ = ∣ P (N) ∣ = 2^{ℵ_{0}}

Proof. We exhibit injections both ways and invoke the Schroder-Bernstein theorem, which states that injections $X \to Y$ and $Y \to X$ produce a bijection.

$P (N) \to R$ . Send a subset $S \subseteq N$ to the real number whose binary expansion has a $1$ in position $n$ exactly when $n \in S$ — or, to avoid the dual-representation problem, use base $3$ with digits $0$ and $1$ only. Distinct subsets give distinct digit strings and hence distinct reals, since no number has two base- $3$ expansions using only the digits $0$ and $1$ .

$R \to P (Q)$ . Send $x$ to the set ${q \in Q : q < x}$ — its Dedekind cut. Distinct reals give distinct cuts by the theorem Density of the rationals: if $x < y$ there is a rational strictly between, belonging to one cut and not the other. Since $Q$ is countable, $P (Q)$ has the same cardinality as $P (N)$ .

Schroder-Bernstein converts the two injections into a bijection.∎

Remark (The continuum hypothesis). Cantor's theorem gives $∣ N ∣ < ∣ P (N) ∣ = ∣ R ∣$ , so there are strictly more reals than naturals. The natural follow-up — is there a set strictly between? — is the continuum hypothesis.

It is neither provable nor refutable from the standard axioms of set theory: Gödel showed in 1940 that it cannot be disproved, and Cohen showed in 1963 that it cannot be proved. So the question of how many real numbers there are, asked precisely, has no answer the usual axioms determine. That is dealt with properly in Set Theory.

A more immediately unsettling consequence of this section: the algebraic numbers — all roots of polynomials with integer coefficients — form a countable set, so almost every real number is transcendental. Yet exhibiting even one transcendental number took until 1844, and proving $π$ transcendental took until 1882. Almost all real numbers are objects no formula will ever name.

Example 1.45 (Counting the algebraic numbers). Show that the set of algebraic numbers — roots of non-zero polynomials with integer coefficients — is countable, and deduce that transcendental numbers exist.

Solution.

For each pair $(n, H)$ of positive integers, consider polynomials of degree at most $n$ whose coefficients are integers of absolute value at most $H$ . There are at most $(2 H + 1)^{n + 1}$ such polynomials — a finite number.
Each non-zero polynomial of degree $d$ has at most $d$ roots, so the set of roots arising from that finite collection is finite.
Every algebraic number appears in one of these finite collections, and the collections are indexed by pairs $(n, H)$ — a countable index set. A countable union of finite sets is countable.
Since $R$ is uncountable by the theorem The reals are uncountable, and the algebraic numbers are countable, the complement is non-empty. So transcendental numbers exist.
Sanity check on the strength of the conclusion: this proves existence without exhibiting a single example. Cantor's argument is famously non-constructive here, which is why Liouville's explicit transcendental of 1844 was a separate achievement.
In fact the argument proves much more: the transcendentals are uncountable, so they vastly outnumber the algebraic numbers. Every number anyone writes down by a formula involving roots is in the countable, hence vanishingly small, part.

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Exercises for §1.21 →

Common Mistakes

Treating completeness as obvious. It is the one axiom $Q$ fails, and every existence theorem in analysis depends on it.
Assuming the supremum belongs to the set. $sup (0, 1) = 1 \in / (0, 1)$ . A supremum that is attained is a maximum; the distinction is the content of the Extreme Value Theorem.
Forgetting the non-emptiness hypothesis. Every real number bounds $\emptyset$ above, so it has no least upper bound.
Taking the Archimedean property for granted. It fails in perfectly consistent ordered fields; over $R$ it is a theorem, proved from completeness in three lines.
Confusing density with size. $Q$ is dense in $R$ and countable; the irrationals are dense and uncountable. Neither property implies anything about the other.
Thinking the diagonal argument can be patched by adding the missing number. It defeats every list, including the repaired one.
Ignoring dual decimal representations. $0.4999 \dots = 0.5$ , which is why the diagonal construction must avoid the digits $0$ and $9$ .
Believing the constructions matter afterwards. They establish existence; every later argument uses the axioms alone, which is what the uniqueness theorem licenses.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter