The real line: order, intervals and absolute value

Summary

This lesson studies the real numbers as points of a line. It deduces from three axioms the rules for working with inequalities, proves with the Archimedean property that the rational numbers are dense in the line, proves with completeness that every nonnegative real number has an $n$ th root, defines intervals and characterizes them as the sets without gaps, and introduces the absolute value as the measure of distance, with its fundamental lemma on inequalities and the triangle inequality. The problems apply these tools to bounding products, solving equations and inequalities with absolute values, controlling the error of an approximation and minimizing a sum of distances.

Learning objectives

Deduce from the order axioms the rules for adding, multiplying and inverting inequalities, and apply them without multiplying by numbers of unknown sign.
Apply the Archimedean property and prove with it that between two distinct real numbers there is always a rational number.
Describe sets of real numbers by means of intervals, operate with them and recognize when a set is an interval.
Interpret the absolute value as a distance, translate conditions with absolute values into intervals and apply the triangle inequality.

The real line

Two points are chosen on a line: one represents $0$ and another, located to its right, represents $1$ . The length of the line segment joining them serves as the unit, and to each real number $x$ there corresponds a point: if $x$ is positive, the one located to the right of the origin at distance $x$ ; if it is negative, the one located to the left at distance $-x$ . The rational numbers occupy points that can be constructed with straightedge and compass, and their decimal expansion, studied in the lesson on the decimal expansion of rational numbers, indicates in which decimal subdivision of the unit each of them falls. Now, in the lesson on number systems it was seen that the rational numbers leave gaps, such as the one occupied by $\sqrt{2}$ , and it was admitted that the real numbers form a complete ordered field, whose completeness axiom fills those gaps. That is why it is admitted that the correspondence between real numbers and points is bijective: to each real number there corresponds a point and to each point, a real number. This identification is not proved, because the line of geometry is not defined here independently of the numbers; it is the geometric reading of completeness, and it is called the real line.

The order axioms

The relation “being to the left of” is translated into the order relation between real numbers, which is defined from the set of positive numbers. Among the axioms that the lesson on number systems admits for the real numbers, together with the field axioms and the completeness axiom, the order axioms refer to that set, and it is worth recalling them:

Trichotomy: for each real number $x$ exactly one of the three statements $x > 0$ , $x = 0$ or $-x > 0$ holds.
The sum of two positive numbers is positive.
The product of two positive numbers is positive.

A real number $x$ is negative if $-x$ is positive. We write $x < y$ , or $y > x$ , when the difference $y - x$ is positive, and $x \leq y$ when $x < y$ or $x = y$ . Applied to $y - x$ , trichotomy says that for any two real numbers exactly one of the statements $x < y$ , $x = y$ or $y < x$ holds: two distinct points of the line always lie one to the left of the other. Moreover, in solved problem 5 of the same lesson it was proved, using only these axioms and the rule of signs, that the square of every nonzero real number is positive; in particular, $1 = 1^2 > 0$ . The rules for working with inequalities are consequences of these axioms and of the field axioms; the lesson on number systems deduced them in its lemma on the rules of order, and here they are gathered together and interpreted on the line.

TheoremThe properties of order

For all real numbers $x$ , $y$ and $z$ : if $x < y$ and $y < z$ , then $x < z$ ; if $x < y$ , then $x + z < y + z$ ; if $x < y$ and $z > 0$ , then $xz < yz$ ; if $x < y$ and $z < 0$ , then $xz > yz$ ; if $x > 0$ , its inverse is positive; and if $0 < x < y$ , then $0 < \frac{1}{y} < \frac{1}{x}$ . Each statement reduces to deciding the sign of a difference, as in the lesson on the integers, where these rules were proved for the integers; the proof, which repeats the one in the lesson on number systems, follows from the axioms alone, and the rules therefore hold for all real numbers. The figure represents each property on the line: translating preserves the order of two points, so does stretching from the origin, and reflecting or inverting reverses it.

Proof

$z - x = \resaltar{(z - y) + (y - x)} > 0$
Transitivity: if $x < y$ and $y < z$ , the differences $y - x$ and $z - y$ are positive, and by the second axiom so is their sum, which is $z - x$ .
$(y + z) - (x + z) = \resaltar{y - x} > 0$
Compatibility with addition: adding the same number $z$ to both sides does not change the difference.
$\dato{pos}{yz - xz = \resaltar{(y - x) z} > 0}$
Product by a positive number: if $z > 0$ , the difference $yz - xz$ is the product of two positive numbers, and it is positive by the third axiom.
$\dato{neg}{xz - yz = (y - x)\resaltar{(-z)} > 0}$
Product by a negative number: if $z < 0$ , its opposite $-z$ is positive, and the previous step, applied to $-z$ , gives a positive difference in the reverse order.
$\dato{inv}{x > 0 \rightarrow \frac{1}{x} > 0}$
Sign of the inverse: let $x > 0$ . If its inverse were negative, multiplying $0 < x$ by it would reverse the inequality, and we would obtain $0 > 1$ , which is false; and the inverse is not zero, because its product by $x$ equals $1$ .
$\frac{1}{x} - \frac{1}{y} = \resaltar{\frac{y - x}{xy}} > 0$
Inverses: if $0 < x < y$ , the difference of the inverses is the product of $y - x$ by the inverse of $xy$ , both positive. Taking inverses of two positive numbers reverses their order.

From these rules follow others that are used constantly. Two inequalities in the same direction can be added: if $a < b$ and $c < d$ , then $a + c < b + c < b + d$ . Two inequalities between nonnegative numbers can be multiplied: if $0 \leq a \leq b$ and $0 \leq c \leq d$ , then $ac \leq bc \leq bd$ . On the other hand, they cannot be subtracted or divided side by side, and they cannot be multiplied if some side is negative, as solved problem 1 shows. Finally, between nonnegative numbers squaring preserves and reflects order, as was proved in the lesson on number systems for powers:

0 \leq p \wedge 0 \leq q \rightarrow \left(p < q \leftrightarrow p^2 < q^2\right)

The density of the rational numbers in the line

The completeness axiom, stated in the lesson on number systems, asserts that every nonempty set of real numbers that is bounded above has a supremum, that is, a least upper bound. In the lesson on the decimal expansion of rational numbers the Archimedean property was deduced from it: for every real number there exists a natural number greater than it, so that the points $1, 2, 3, \ldots$ advance along the line without any bound. From this follows a form that is often used: if $h > 0$ and $x$ is any real number, there exists a natural number $n$ greater than $\frac{x}{h}$ , and multiplying by $h$ , which is positive, gives $nh > x$ . Geometrically, a line segment, however small, repeated enough times, exceeds any other; this is how Archimedes used it, and it takes its name from him.

h > 0 \rightarrow (\forall x \in \mathbb{R})(\exists n \in \mathbb{N})(nh > x)

In the lesson on the rational numbers it was proved that between two distinct rational numbers there are infinitely many rational numbers. With the Archimedean property something stronger can be proved, the fact that the lesson on number systems left admitted: the rational numbers are spread over the whole real line, with no interval, however short, that does not contain one.

TheoremDensity of the rational numbers in the real line

Between two distinct real numbers there is always a rational number: if $x < y$ , there exists a rational number $q$ such that $x < q < y$ . The idea is to enlarge the interval between $x$ and $y$ until its length exceeds $1$ , so that it contains an integer, and then to shrink it back. The figure shows the original interval above and the enlarged one below.

Proof

$\dato{n}{ny > nx + 1}$
By the Archimedean property, applied to $h = y - x > 0$ , there exists a natural number $n$ such that $n(y - x) > 1$ : enlarged $n$ times, the interval measures more than the unit.
$\dato{m}{m - 1 \leq nx < m}$
Let $m$ be the least integer greater than $nx$ . It exists: by the Archimedean property there is a natural number $k$ with $k > -nx$ , so that $nx + k > 0$ ; the set of natural numbers greater than $nx + k$ is not empty, again by the Archimedean property, and by the well-ordering principle it has a least element $j$ . The integer $m = j - k$ is greater than $nx$ , and $m - 1$ is not, because $j - 1$ is either zero or a natural number that does not exceed $nx + k$ .
$nx < m \leq \resaltar{nx + 1} < ny$
Adding $1$ to the first inequality and combining it with the one from the first step, the integer $m$ lies strictly between $nx$ and $ny$ .
$x < \resaltar{\frac{m}{n}} < y$
Dividing by $n$ , which is positive, the inequalities are preserved: the quotient of $m$ by $n$ , which is rational, lies between $x$ and $y$ .

Applied again to the interval between $x$ and the rational number obtained, the theorem gives another rational number, and so on: between two distinct real numbers there are infinitely many rational numbers. The same happens with the irrational numbers, as proposed problem 4 asks you to prove. Rational and irrational numbers are therefore intermingled along the whole line, even though their decimal expansions are of such different nature.

The existence of roots

In the lesson on number systems the principal $n$ th root of a nonnegative real number $a$ was defined as the unique nonnegative real number whose $n$ th power is $a$ ; there uniqueness was proved and existence was left admitted, because its proof requires an argument with suprema. With completeness and the rules of order that argument can now be given. We use a bound for the difference of two powers: if $0 \leq c \leq b$ , when the following product is expanded the intermediate terms cancel, as in the sum of a geometric progression, and each of the $n$ terms of the second factor is at most $b^{n - 1}$ .

b^n - c^n = (b - c)\left(b^{n - 1} + b^{n - 2} c + \cdots + c^{n - 1}\right) \leq (b - c) \cdot n b^{n - 1}

TheoremExistence of the

n

th root

Let $n \geq 2$ be a natural number. For every real number $a \geq 0$ there exists a real number $s \geq 0$ such that $s^n = a$ . If $a = 0$ , it suffices to take $s = 0$ ; in what follows, $a > 0$ . The desired root is obtained as the supremum of the nonnegative numbers whose $n$ th power does not exceed $a$ , and the two strict inequalities are ruled out separately. The figure draws the curve $y = x^n$ and the line at height $a$ : the set $S$ is the interval of the axis under the part of the curve that does not rise above the line, and each false assumption about $s$ is refuted by a point that falls on the wrong side.

Proof

$x > 1 + a \rightarrow x^n \geq x > a$
The set $S$ of nonnegative real numbers whose $n$ th power does not exceed $a$ contains $0$ . It is bounded by $1 + a$ : if $x > 1 + a$ , then $x > 1$ , and its power is at least $x$ , which already exceeds $a$ .
$\dato{s}{x \in S \rightarrow x \leq s}$
By completeness, $S$ has a supremum $s$ , which is nonnegative because $0$ belongs to $S$ .
$\dato{h}{h = \frac{a - s^n}{a + n(s + 1)^{n - 1}}}$
Suppose that $s^n < a$ . We take a number $h$ that is positive and less than $1$ , because its numerator is less than its denominator.
$(s + h)^n - s^n \leq h \cdot n(s + 1)^{n - 1} \leq \resaltar{a - s^n}$
By the bound on the difference of powers, with $b = s + h \leq s + 1$ and $c = s$ , the power of $s + h$ does not exceed $a$ ; so $s + h$ belongs to $S$ and is greater than its supremum, which is impossible.
$\dato{k}{h = \frac{s^n - a}{n s^{n - 1}}}$
Suppose that $s^n > a$ . Then $s > 0$ , and we take another number $h$ , positive and less than $s$ , because $h \cdot n s^{n - 1} = s^n - a < s^n \leq n s^n$ .
$s^n - (s - h)^n \leq h \cdot n s^{n - 1} = \resaltar{s^n - a}$
With $b = s$ and $c = s - h$ , the bound shows that the power of $s - h$ is at least $a$ .
$x \in S \rightarrow x \leq s - h < s$
If some $x$ in $S$ were greater than $s - h$ , the preservation of order by powers would give $x^n > (s - h)^n \geq a$ , contrary to the definition of $S$ . Thus $s - h$ would be an upper bound of $S$ less than $s$ , which is impossible.
$s^n = a$
By trichotomy, only equality remains: $s$ is the principal $n$ th root of $a$ , unique by what was proved in the lesson on number systems.

With this theorem and the density of the rational numbers, the identification that the lesson on number systems left admitted is also completed: the supremum of the set $A$ of rational numbers whose square is less than $2$ is $\sqrt{2}$ . Indeed, $\sqrt{2}$ is an upper bound of $A$ , because if a rational number $x$ were greater than $\sqrt{2}$ , it would be positive and its square would exceed $2$ . And no smaller bound works: if $u < \sqrt{2}$ , density, applied to the interval that goes from the larger of the numbers $u$ and $0$ up to $\sqrt{2}$ , gives a rational number $q$ with $0 < q < \sqrt{2}$ and $u < q$ ; its square is less than $2$ , because between nonnegative numbers powers preserve order, so $q$ belongs to $A$ and exceeds $u$ .