The real numbers

The rationals and the reals obey the same rules of arithmetic and the same rules for inequalities, yet the rationals can be listed and the reals cannot, so something separates them that no rule of algebra mentions.

The previous lesson found that split by comparing infinite sets through bijections, and left open what the reals have that the rationals lack. This lesson names it, the completeness axiom, and uses it to pay debts that earlier lessons deferred to "a later lesson": that $ℕ$ has no upper bound in $ℝ$ , used without proof in the lessons on sets and on order; that the rationals with square less than $2$ have no least upper bound among the rationals, stated in the lesson on order; and that some real number has square $2$ , claimed in the lesson on quantifiers. It assumes upper bounds and the supremum: $u = \sup A$ when $u$ is an upper bound of $A$ lying below every other upper bound. As throughout the course, $ℕ = {1, 2, 3, \dots}$ .

The gap where $\sqrt{2}$ should be

Work inside $ℚ$ and let $T = {q \in ℚ : q^{2} < 2}$ . It is nonempty, since $1 \in T$ , and any positive $u$ with $u^{2} > 2$ is an upper bound, because an element $q > u$ would have $q^{2} > u^{2} > 2$ . The claim deferred by the lesson on order is that no rational is the least upper bound. The proof rests on two pieces of algebra that push a candidate up or down.

The first says a number whose square is below $2$ is never the top of $T$ . Let $u \geq 1$ with $u^{2} < 2$ , and set $h = \frac{2 - u^{2}}{2 u + 1}$ . Then $h > 0$ , and $h < 1$ because $2 - u^{2} \leq 1 < 3 \leq 2 u + 1$ . Since $0 < h < 1$ gives $h^{2} < h$ ,

(u + h)^{2} = u^{2} + 2 u h + h^{2} < u^{2} + h (2 u + 1) = 2

So $u + h$ is larger than $u$ with square still below $2$ . From $u = 1.4$ it gives $h = \frac{1}{95}$ , and $(1.4 + \frac{1}{95})^{2} \approx 1.9896$ .

The second says a number whose square is above $2$ is never the least upper bound. Let $u > 0$ with $u^{2} > 2$ , and set $k = \frac{u^{2} - 2}{2 u} > 0$ . Then $u - k = \frac{u^{2} + 2}{2 u}$ is positive, and

(u - k)^{2} = u^{2} - 2 u k + k^{2} > u^{2} - 2 u k = 2

So $u - k$ is a smaller upper bound. From $u = \frac{3}{2}$ it gives $\frac{17}{12}$ , with square $\frac{289}{144}$ . Both $h$ and $k$ are built from $u$ by arithmetic, so if $u$ is rational, so are they.

Example. Prove that $T = {q \in ℚ : q^{2} < 2}$ has no least upper bound in $ℚ$ .

Suppose, for contradiction, that $u \in ℚ$ is the least upper bound of $T$ . Since $1 \in T$ , $u \geq 1$ . Since $\sqrt{2}$ is irrational, as an earlier lesson proved, $u^{2} \neq 2$ , so either $u^{2} < 2$ or $u^{2} > 2$ . If $u^{2} < 2$ , the rational $u + h$ above lies in $T$ and exceeds $u$ , contradicting that $u$ is an upper bound. If $u^{2} > 2$ , the rational $u - k$ is an upper bound of $T$ smaller than $u$ , contradicting that $u$ is the least. Both cases are impossible, so $T$ has no least upper bound in $ℚ$ .

Now you. Prove that ${q \in ℚ : q^{2} < 3}$ has no least upper bound in $ℚ$ , using the irrationality of $\sqrt{3}$ .

Answer

Suppose $u \in ℚ$ is the least upper bound. Since $1$ is in the set, $u \geq 1$ , and since $\sqrt{3}$ is irrational, $u^{2} \neq 3$ . If $u^{2} < 3$ , let $h = \frac{3 - u^{2}}{2 u + 1}$ , rational, positive, and less than $1$ because $3 - u^{2} \leq 2 < 3 \leq 2 u + 1$ . Then $(u + h)^{2} < u^{2} + h (2 u + 1) = 3$ , so $u + h$ is in the set and exceeds $u$ , a contradiction. If $u^{2} > 3$ , let $k = \frac{u^{2} - 3}{2 u}$ . Then $u - k = \frac{u^{2} + 3}{2 u} > 0$ and $(u - k)^{2} > u^{2} - 2 u k = 3$ , so $u - k$ is a smaller upper bound, a contradiction.

The ordered field axioms

That proof used only arithmetic, order and the irrationality of $\sqrt{2}$ . To say what the reals have in addition, their familiar properties must be written down as a list, so that it is clear which ones the rationals share.

Addition and multiplication are commutative and associative, multiplication distributes over addition, there are elements $0 \neq 1$ with $x + 0 = x$ and $x \cdot 1 = x$ , every $x$ has a negative $- x$ , and every $x \neq 0$ has a reciprocal $\frac{1}{x}$ : these make $ℝ$ a field. The order $\leq$ is total and respects the arithmetic: $x \leq y$ implies $x + z \leq y + z$ , and $0 \leq x$ with $0 \leq y$ implies $0 \leq x y$ . A field with such an order is an ordered field. Every rule of school algebra, from $0 \cdot x = 0$ to "multiplying by a negative number reverses an inequality", follows from these axioms, and here they are taken as given.

The trouble is that $ℚ$ satisfies every one of them. So no argument from these axioms alone can produce a number whose square is $2$ , since it would work equally well in $ℚ$ , where there is none. Something else must be assumed.

The completeness axiom

It concerns least upper bounds, exactly where the rationals fail.

Completeness axiom. Every nonempty set of real numbers that is bounded above has a least upper bound in $ℝ$ .

The reals are an ordered field satisfying this axiom, a complete ordered field, and that is all this course assumes about them. The rationals are an ordered field that does not: $T$ is nonempty and bounded above, and has no rational least upper bound. Both conditions in the axiom are needed. The empty set has every real as an upper bound and so no least one, and $ℕ$ , as a later section proves, has no upper bound at all.

The idea took its modern form in 1872, in two independent constructions. Richard Dedekind, in Braunschweig, published Stetigkeit und irrationale Zahlen ("Continuity and irrational numbers"), defining a real number as a cut of the rationals into a lower and an upper class; he dated the idea to November 1858, when, teaching calculus in Zurich, he could find no rigorous proof that a bounded increasing quantity approaches a limit. The same year Georg Cantor, in Halle, built the reals from sequences of rationals whose terms eventually cluster together. In both constructions completeness is a theorem, and any two complete ordered fields turn out to be the same up to relabelling, so taking the axiom as a starting point describes $ℝ$ exactly.

Working with the supremum

The definition of $\sup A$ quantifies over every upper bound. Proofs usually want a form that tests only numbers just below $s$ .

Example. Prove that $s = \sup A$ if and only if $s$ is an upper bound of $A$ and, for every $ε > 0$ , some $a \in A$ satisfies $a > s - ε$ .

Suppose first that $s = \sup A$ . Then $s$ is an upper bound by definition. Let $ε > 0$ . Since $s - ε < s$ and $s$ lies below every upper bound, $s - ε$ is not an upper bound, so some $a \in A$ has $a > s - ε$ .

Conversely, suppose $s$ is an upper bound with the $ε$ property, and let $v$ be any upper bound of $A$ . Suppose, for contradiction, that $v < s$ , and take $ε = s - v > 0$ . Some $a \in A$ has $a > s - ε = v$ , contradicting that $v$ is an upper bound. So $s \leq v$ for every upper bound $v$ , and $s = \sup A$ .

The second half chose $ε$ to fit the contradiction. Picking $ε$ after the thing to be defeated is known is the move the final lesson makes over and over.

Lower bounds need no second axiom. If $A$ is nonempty and bounded below by $m$ , then $- A = {- a : a \in A}$ is nonempty and bounded above by $- m$ , so it has a supremum, and reversing every inequality shows that $- \sup (- A)$ is the greatest lower bound of $A$ . So every nonempty set bounded below has an infimum, $\inf A = - \sup (- A)$ .

Now you. Prove that $t = \inf A$ if and only if $t$ is a lower bound of $A$ and, for every $ε > 0$ , some $a \in A$ satisfies $a < t + ε$ .

Answer

Suppose $t = \inf A$ . It is a lower bound by definition. Let $ε > 0$ . Since $t + ε > t$ and $t$ lies above every lower bound, $t + ε$ is not a lower bound, so some $a \in A$ has $a < t + ε$ . Conversely, suppose $t$ is a lower bound with this property, and let $m$ be any lower bound. If $m > t$ , take $ε = m - t$ : some $a \in A$ has $a < t + ε = m$ , contradicting that $m$ is a lower bound. So $m \leq t$ for every lower bound $m$ , and $t = \inf A$ .

The Archimedean property

The lessons on sets and on order promised a proof that $ℕ$ has no upper bound in $ℝ$ . It does not follow from the ordered field axioms: there are ordered fields containing elements larger than every natural number. In $ℝ$ it follows from completeness.

Example. Prove the Archimedean property: for every real $x$ there is a natural number $n$ with $n > x$ .

Suppose, for contradiction, that some real $x$ has $n \leq x$ for every $n \in ℕ$ . Then $ℕ$ is nonempty and bounded above, so by completeness it has a supremum $s$ . By the characterisation with $ε = 1$ , some $n \in ℕ$ has $n > s - 1$ . Then $n + 1 > s$ , and $n + 1 \in ℕ$ , contradicting that $s$ is an upper bound of $ℕ$ .

The form the earlier lessons used follows at once: for every $ε > 0$ some $n$ has $\frac{1}{n} < ε$ . Take $n > \frac{1}{ε}$ and multiply by the positive number $\frac{ε}{n}$ . For $ε = 0.03$ the condition is $n > 33.3$ , so $n = 34$ is the first that works, with $\frac{1}{34} \approx 0.0294$ .

Now you. Prove that $\inf {\frac{1}{n} : n \in ℕ} = 0$ .

Answer

Every $\frac{1}{n}$ is positive, so $0$ is a lower bound. Let $ε > 0$ . By the Archimedean property some $n \in ℕ$ has $n > \frac{1}{ε}$ , so $\frac{1}{n} < ε = 0 + ε$ . By the characterisation of the infimum, the infimum is $0$ , which is not itself in the set.

The square root of two exists

The lesson on quantifiers said that "some $x$ has $x^{2} = 2$ " is false over the rationals and true over the reals. The real half needs completeness, and the proof is the rational argument run again.

Let $S = {x \in ℝ : x^{2} < 2}$ . It is nonempty, since $1 \in S$ , and bounded above by $2$ , since $x \geq 2$ gives $x^{2} \geq 4$ . By completeness it has a supremum $s$ , and $s \geq 1$ . The two pieces of algebra from the first section never used rationality, so they apply to $s$ . If $s^{2} < 2$ , then $s + h$ lies in $S$ and exceeds $s$ , impossible for an upper bound. If $s^{2} > 2$ , then $s - k$ is a smaller upper bound, impossible for the least one. So $s^{2} = 2$ . No other positive number qualifies, since $t > 0$ and $t^{2} = 2$ give $(s - t) (s + t) = 0$ with $s + t > 0$ . This $s$ is $\sqrt{2}$ .

Over $ℚ$ , the algebra ruled out both inequalities, irrationality ruled out equality, and no supremum was left. Over $ℝ$ , completeness supplies the supremum, the same algebra rules out both inequalities, and equality is what remains. The gap in the rationals is exactly where the reals have a number.

The rationals are dense

For all their gaps, the rationals are everywhere: between any two reals $a < b$ lies a rational. By the Archimedean property choose $n$ with $\frac{1}{n} < b - a$ , so $n b - n a > 1$ . The integers greater than $n a$ form a nonempty set bounded below (the Archimedean property, applied to $n a$ and to $- n a$ , gives an integer on each side), so by well-ordering it has a least element $m$ . Then $m - 1 \leq n a < m$ , so $n a < m \leq n a + 1 < n b$ , and dividing by $n$ gives $a < \frac{m}{n} < b$ . Between $0.3$ and $0.31$ , the proof takes $n = 101$ and $m = 31$ , and $\frac{31}{101} \approx 0.3069$ .

Density settles the last claim of the lesson on order, that $T$ has supremum $\sqrt{2}$ in $ℝ$ . Every $q \in T$ has $q < \sqrt{2}$ , and for any $b < \sqrt{2}$ there is a rational $q$ strictly between $\max {b, 0}$ and $\sqrt{2}$ , which lies in $T$ and exceeds $b$ . The irrationals are dense too: a rational $q$ between $a - \sqrt{2}$ and $b - \sqrt{2}$ gives the irrational $q + \sqrt{2}$ between $a$ and $b$ .

Where completeness is needed next

The Archimedean property, the density of the rationals and the existence of $\sqrt{2}$ are facts calculus uses without comment, and each came from one axiom. The first place calculus needs completeness itself is its foundation, the limit of a sequence. The decimal truncations $1$ , $1.4$ , $1.41$ , $1.414$ , and so on, of $\sqrt{2}$ form an increasing bounded sequence of rationals with nothing in $ℚ$ to converge to. The final lesson writes convergence with $ε$ and $N$ , proves limits with the methods of the whole course, and shows that in $ℝ$ every bounded increasing sequence converges, the theorem Dedekind found no proof of in 1858.