Real Numbers and Intervals - Maths: an open, verified university mathematics reference

Why this matters¶

Calculus is about quantities that change without jumps: the position of a car, the temperature of a cup of coffee, the area under a curve. To describe them we need a number system with no gaps, and fractions are not enough.

Draw a square whose sides have length 1. By Pythagoras’ theorem its diagonal $d$ satisfies $d^2 = 1^2 + 1^2 = 2$ . Is $d$ a fraction? We prove below that it is not: no fraction squares to 2. Yet the diagonal certainly has a length, and swinging it down onto the number line marks a definite point on it. The real numbers are the number system in which every point of the line is a number, and so every length is one.

A unit square standing on the number line between 0 and 1, with its diagonal drawn from 0. A dashed arc of radius equal to the diagonal swings it down onto the line, where it lands at a point between 1 and 2. — Figure 1:The diagonal of the unit square has length $\sqrt{2}$ . Swung down onto the number line, it marks a point between 1 and 2 that no fraction reaches.

The second half of this page is a language for sets of real numbers. Calculus keeps asking for which $x$ something holds: $\sqrt{x - 1}$ is defined exactly when $x \ge 1$ , and a function may increase only for $x$ between two values. Intervals and set-builder notation say such things briefly and precisely.

The number systems¶

We use four number systems, each contained in the next:

the natural numbers $\N = \{0, 1, 2, 3, \dots\}$ , for counting. On this site $\N$ contains 0; when we mean the positive integers we write “ $n \ge 1$ ”;
the integers $\Z = \{\dots, -2, -1, 0, 1, 2, \dots\}$ ;
the rational numbers $\Q$ , the fractions $\frac{p}{q}$ with $p$ and $q$ integers and $q \ne 0$ . The same rational number has many such forms: $\frac{1}{2} = \frac{2}{4} = \frac{-3}{-6}$ ;
the real numbers $\R$ , the points of the number line. A real number that is not rational is called irrational.

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

(1)

Every integer $n$ is rational, because $n = \frac{n}{1}$ . The number $\sqrt{2}$ is irrational (Theorem 1 below), and so is $\pi$ , although that proof is harder and we do not need it.

Decimals. Every real number has a decimal expansion, such as $\frac{1}{4} = 0.25$ , $\frac{1}{3} = 0.333\ldots$ or $\sqrt{2} = 1.41421\ldots$ . A bar marks a block of digits that repeats for ever: $0.\overline{3} = 0.333\ldots$ and $0.1\overline{25} = 0.1252525\ldots$ . The decimals tell rational and irrational numbers apart:

The decimal expansion of a rational number $\frac{p}{q}$ terminates or eventually repeats. When we divide $p$ by $q$ by long division, each remainder is one of $0, 1, \dots, q - 1$ , so after at most $q$ steps a remainder comes back, and from then on the digits repeat.
Conversely, a decimal that terminates or eventually repeats is rational: Example 3 shows the method.

So the irrational numbers are exactly the real numbers whose decimals go on for ever without repeating. A decimal approximation such as $\sqrt{2} \approx 1.4142$ is a rational number close to $\sqrt{2}$ , never $\sqrt{2}$ itself.

Order. For any two real numbers $a$ and $b$ , exactly one of $a < b$ , $a = b$ and $a > b$ holds; on the number line, $a < b$ means that $a$ lies to the left of $b$ . We write $a \le b$ for “ $a < b$ or $a = b$ ”.

Every inequality on this site rests on the rules below. Four of them are axioms of the order: like the rules of arithmetic, we take them as defining properties of $\R$ (the rigorous track says more in Remark 5). The others follow from the axioms, and each comes with its reason.

Remark 1 (The order rules)

For all real numbers $a$ , $b$ , $c$ and $d$ :

Trichotomy (axiom): exactly one of $a < b$ , $a = b$ and $a > b$ holds.
Transitivity: if $a < b$ and $b < c$ , then $a < c$ (axiom). The same holds when one of the two inequalities is $\le$ : $a \le b < c$ and $a < b \le c$ each imply $a < c$ . And $a \le b \le c$ implies $a \le c$ . Reason: where a $\le$ is an equality, substitute it; otherwise both inequalities are strict and the axiom applies.
Adding the same number (axiom): if $a < b$ , then $a + c < b + c$ . Also, if $a \le b$ , then $a + c \le b + c$ (reason: for $a = b$ both sides are equal).
Adding two inequalities: if $a < b$ and $c < d$ , then $a + c < b + d$ ; if $a \le b$ and $c \le d$ , then $a + c \le b + d$ ; and if $a \le b$ and $c < d$ , then $a + c < b + d$ . Reason: adding $c$ to $a \le b$ and $b$ to $c < d$ (property 3) gives $a + c \le b + c < b + d$ , and property 2 gives $a + c < b + d$ . The other two cases are the same, with both steps strict or both $\le$ .
Multiplying by a number:
(a) if $a < b$ and $c > 0$ , then $ca < cb$ (axiom);
(b) if $a \le b$ and $c \ge 0$ , then $ca \le cb$ . Even when $a < b$ , the result is then only $\le$ : for $c = 0$ both sides are 0. Reason: for $a < b$ and $c > 0$ this is (a); if $a = b$ or $c = 0$ , both sides are equal;
(c) if $a < b$ and $c < 0$ , then $ca > cb$ : multiplying by a negative number reverses the inequality. Likewise $a \le b$ and $c < 0$ give $ca \ge cb$ . Reason: adding $-c$ to $c < 0$ (property 3) gives $0 < -c$ , so (a) gives $-ca < -cb$ , and adding $ca + cb$ to both sides (property 3) gives $cb < ca$ . For $a = b$ both sides are equal.
Reciprocals of positive numbers: if $a > 0$ , then $\frac{1}{a} > 0$ ; and if $0 < a < b$ , then $\frac{1}{b} < \frac{1}{a}$ . Reason: first, $1 > 0$ , because $1 \ne 0$ , and $1 < 0$ would give $1 \cdot 1 > 1 \cdot 0$ by 5(c), that is $1 > 0$ , against property 1. Next, $\frac{1}{a} \ne 0$ since $a \cdot \frac{1}{a} = 1$ , and $\frac{1}{a} < 0$ would give $a \cdot \frac{1}{a} < a \cdot 0$ by 5(a), that is $1 < 0$ ; so $\frac{1}{a} > 0$ by property 1. Finally, if $0 < a < b$ , then $b > 0$ (property 2), so $ab > 0$ (multiplying $0 < a$ by $b > 0$ , by 5(a)) and $\frac{1}{ab} > 0$ . Multiplying $a < b$ by $\frac{1}{ab} > 0$ (5(a)) gives $\frac{1}{b} < \frac{1}{a}$ .
Negation: “not $a < b$ ” holds exactly when $a \ge b$ , and “not $a \le b$ ” exactly when $a > b$ . Reason: by property 1, $a < b$ fails exactly when $a = b$ or $a > b$ , that is when $a \ge b$ ; and $a \le b$ fails exactly when neither $a < b$ nor $a = b$ holds, that is when $a > b$ .

Sets and intervals¶

A set is a collection of objects, its elements. We write $x \in A$ for “ $x$ is an element of $A$ ” and $x \notin A$ for its negation. A small set can be listed in braces, as in $\{1, 2, 3\}$ . Most sets in calculus are described by a condition instead, in set-builder notation:

\{x \in \R : x > 0\}

(2)

is read “the set of all real numbers $x$ such that $x > 0$ ”. The empty set $\emptyset$ has no elements. For sets $A$ and $B$ :

Notation	Read as	Its elements
$A \subseteq B$	$A$ is a subset of $B$	every element of $A$ is an element of $B$
$A \cap B$	$A$ intersect $B$	the $x$ with $x \in A$ and $x \in B$
$A \cup B$	$A$ union $B$	the $x$ with $x \in A$ or $x \in B$ (or both)
$A \setminus B$	$A$ minus $B$	the $x$ with $x \in A$ and $x \notin B$

The sets of real numbers that calculus uses most are intervals: all the numbers between two endpoints, or all the numbers on one side of a point.

Definition 1 (Interval)

Let $a$ and $b$ be real numbers with $a < b$ . The bounded intervals with endpoints $a$ and $b$ are

\begin{aligned} [a, b] &= \{x \in \R : a \le x \le b\}, & (a, b) &= \{x \in \R : a < x < b\}, \\ [a, b) &= \{x \in \R : a \le x < b\}, & (a, b] &= \{x \in \R : a < x \le b\}. \end{aligned}

(3)

For a real number $c$ , the unbounded intervals are

\begin{aligned} [c, \infty) &= \{x \in \R : x \ge c\}, & (c, \infty) &= \{x \in \R : x > c\}, \\ (-\infty, c] &= \{x \in \R : x \le c\}, & (-\infty, c) &= \{x \in \R : x < c\}, \end{aligned}

(4)

and $(-\infty, \infty) = \R$ . An interval is a set of one of these nine forms. The intervals $(a, b)$ , $(c, \infty)$ , $(-\infty, c)$ and $\R$ are called open, and $[a, b]$ is called a closed interval.

In words. A square bracket means that the endpoint belongs to the interval, a round bracket that it does not. The symbols $\infty$ and $-\infty$ are not real numbers: they only say that the interval has no end on that side, so they always get a round bracket. We ask for $a < b$ so that every interval contains more than one number.

Example. $\{x \in \R : -1 \le x < 3\} = [-1, 3)$ contains ${-1}$ , 0, 2.9 and $\sqrt{2}$ , but not 3. On the number line we draw the included endpoint as a filled dot and the excluded one as a hollow dot (Figure 2).

Figure 2:The interval $[-1, 3)$ (top): the filled dot shows that ${-1}$ belongs to it, the hollow dot that 3 does not. The interval $(2, \infty)$ (bottom) starts just after 2 and has no right end.

Non-example. An interval has no gaps: if two numbers lie in an interval, so does every number between them. The set $\{x \in \R : x \ne 0\}$ is therefore not an interval: it contains ${-1}$ and 1 but not 0, which lies between them. It is the union $(-\infty, 0) \cup (0, \infty)$ of two intervals. The set $\Z$ of integers is not an interval either: it contains 1 and 2 but not 1.5.

Main results¶

We use square roots throughout calculus, and the next pages cite this fact about them.

The proof below uses two facts about whole numbers:

every integer is either even, of the form $2m$ with $m \in \Z$ , or odd, of the form $2m + 1$ with $m \in \Z$ , and not both;
the well-ordering principle, Remark 3.

Proof

We argue by contradiction: we assume that some rational number squares to 2 and deduce something impossible.

Suppose that $r$ is rational and $r^2 = 2$ . Then $r \ne 0$ , and $(-r)^2 = r^2 = 2$ as well, so we may assume that $r > 0$ (otherwise we replace $r$ by $-r$ ). Then $r = \frac{p}{q}$ with positive integers $p$ and $q$ . Among all such ways of writing $r$ , we choose one with the smallest possible denominator $q$ ; there is one, by the well-ordering principle.

From $\frac{p^2}{q^2} = 2$ we get $p^2 = 2q^2$ , so $p^2$ is even.
Then $p$ is even. Otherwise $p$ is odd, $p = 2m + 1$ with $m \in \Z$ , and $p^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1$ is odd, which contradicts step 1.
So $p = 2k$ with $k$ a positive integer. Then $4k^2 = 2q^2$ , that is $q^2 = 2k^2$ , so $q^2$ is even, and the argument of step 2 shows that $q$ is even: $q = 2l$ with $l$ a positive integer.
Now $r = \frac{p}{q} = \frac{2k}{2l} = \frac{k}{l}$ with positive integers $k$ and $l$ , and the denominator $l = \frac{q}{2}$ is smaller than $q$ . This contradicts the choice of $q$ as the smallest possible denominator.

So no rational number squares to 2. Since $\sqrt{2}$ is the positive real number whose square is 2, it is not rational.

Here, as in Remark 2, we have taken from school that $\sqrt{2}$ exists: that there is a positive real number whose square is 2. The rigorous track shows that this is exactly what the completeness of $\R$ guarantees, and what $\Q$ lacks, and proves Remark 2 for every $y \ge 0$ .

Worked examples¶

Example 2 (Find

A \cap B

and

A \cup B

for

A = [-2, 3)

and

B = (1, 5]

)

Goal. Describe the numbers in both intervals, and the numbers in at least one of them.

$x \in A$ means $-2 \le x < 3$ , and $x \in B$ means $1 < x \le 5$ .
Intersection. Both hold when $x$ is larger than ${-2}$ and 1, the larger left endpoint, and smaller than 3 and 5, the smaller right endpoint: $1 < x < 3$ . The endpoint 1 is excluded because $1 \notin B$ , and 3 because $3 \notin A$ .
Union. The intervals overlap (on $(1, 3)$ ), so together they leave no gap: their union runs from the left end of $A$ to the right end of $B$ . Since $-2 \in A$ and $5 \in B$ , both endpoints are included.

\boxed{A \cap B = (1, 3), \qquad A \cup B = [-2, 5]}

(6)

Check. $x = 2$ lies in $A$ and in $B$ , and in $(1, 3)$ . $x = 4$ lies in $B$ only, so it is in the union but not the intersection. $x = 3$ is in $B$ (so in the union) but not in $A$ (so not in the intersection). ✓

Example 3 (Write

0.\overline{36}

as a fraction)

Goal. Find integers $p$ and $q$ with $0.\overline{36} = 0.363636\ldots = \frac{p}{q}$ .

Let $x = 0.363636\ldots$ . The repeating block has two digits, so we multiply by $10^2 = 100$ : $100x = 36.363636\ldots$ .
Subtracting, the repeating tails cancel: $100x - x = 36.3636\ldots - 0.3636\ldots = 36$ , so $99x = 36$ .
Hence $x = \frac{36}{99} = \frac{4}{11}$ , dividing numerator and denominator by 9.

\boxed{0.\overline{36} = \frac{4}{11}}

(7)

Check. Long division of 4 by 11: $40 = 3 \cdot 11 + 7$ , $70 = 6 \cdot 11 + 4$ , and the remainder 4 is back, so $\frac{4}{11} = 0.3636\ldots$ . ✓

Example 4 (The acceptable lengths of a bolt)

Goal. A bolt must be $40\,\text{mm}$ long, with a tolerance of $0.5\,\text{mm}$ : it is accepted if its length differs from $40\,\text{mm}$ by at most $0.5\,\text{mm}$ . Write the set of acceptable lengths as an interval.

Let $L$ be the length in millimetres. “Differs by at most 0.5” means $40 - 0.5 \le L \le 40 + 0.5$ , that is $39.5 \le L \le 40.5$ .
Both endpoints are accepted (“at most”), so both get square brackets.

\boxed{L \in [39.5, 40.5] \text{ (in mm)}}

(8)

Check. A bolt of $40.3\,\text{mm}$ differs from $40\,\text{mm}$ by $0.3\,\text{mm}$ and is accepted; one of $40.6\,\text{mm}$ differs by $0.6\,\text{mm}$ and is rejected, and indeed $40.6 \notin [39.5, 40.5]$ . The interval has length $1\,\text{mm}$ , twice the tolerance. ✓

Common mistakes¶

Rigorous track¶

What makes $\R$ different from $\Q$ ? Both allow the four arithmetic operations, and both are ordered. The difference is that $\R$ has no gaps, and to say this precisely we need upper bounds and suprema.

In words. An upper bound lies on or to the right of every element of $S$ . For example 1, 2 and 100 are upper bounds of $[0, 1)$ , and 0.9 is not, because $0.95 \in [0, 1)$ . Lower bounds and bounded below are defined in the same way with $s \ge u$ , and a set is bounded if it is bounded above and below.

In words. The supremum is the smallest upper bound. It need not belong to $S$ : $\sup\, [0, 1) = 1$ , because 1 is an upper bound and every number $u < 1$ fails to be one (the number $\max\bigl(0, \frac{u + 1}{2}\bigr)$ lies in $[0, 1)$ and is larger than $u$ ). A useful consequence: if $s = \sup S$ and $\eps > 0$ , then $s - \eps$ is smaller than $s$ , so it is not an upper bound, and some element $x \in S$ satisfies $x > s - \eps$ .

Why “non-empty”. Every real number is an upper bound of the empty set, since $\emptyset$ has no element to check. So $\emptyset$ is bounded above, but its upper bounds are all of $\R$ , which has no smallest element: $\emptyset$ has no supremum. A set that is not bounded above, such as $\R$ itself, has no upper bound at all, so no supremum either.

Remark 5 (Why an axiom, and where it comes from)

We do not prove the completeness axiom: we take it, together with the rules of arithmetic and of inequalities, as one of the defining properties of $\R$ . The rational numbers satisfy all the other properties, but not this one: below we find a non-empty set of rationals, bounded above, with no least upper bound among the rationals.

Sketch of where it comes from. The real numbers can be constructed from the rationals. In one construction (Dedekind cuts) a real number is a way of cutting $\Q$ into a lower and an upper part, and the supremum of a non-empty set of cuts that is bounded above is the cut whose lower part is the union of theirs, so completeness becomes a theorem. This sketch leaves out the definitions of the arithmetic operations on cuts and the proofs of their rules; they belong to Real Analysis (ana), which does not exist yet.

Three first consequences of completeness. The first, that $\N$ is not bounded above, is the Archimedean property of $\R$ ; later pages cite it by that name.

Completeness gives

\sqrt{2}

Let $S = \{x \in \R : x \ge 0 \text{ and } x^2 < 2\}$ . It is non-empty ( $1 \in S$ ) and bounded above by 2 (if $x \ge 2$ then $x^2 \ge 4$ ). By Axiom 1 it has a supremum $s$ , and $1 \le s \le 2$ . We show that $s^2 = 2$ by ruling out the other two cases.

If $s^2 < 2$ : let $h = \frac{2 - s^2}{2s + 2}$ . Then $0 < h < 1$ , because $0 < 2 - s^2 < 2 < 2s + 2$ (the first by the case we are in, the others because $s \ge 1$ ). So $h^2 < h$ and $(s + h)^2 = s^2 + 2sh + h^2 < s^2 + h(2s + 2) = 2$ . So $s + h \in S$ , although $s + h > s$ : this contradicts $s$ being an upper bound.
If $s^2 > 2$ : let $h = \frac{s^2 - 2}{2s}$ . Then $h > 0$ , $s - h = \frac{s^2 + 2}{2s} > 0$ and $(s - h)^2 = s^2 - 2sh + h^2 = 2 + h^2 > 2$ . Every $x \in S$ has $x^2 < 2 < (s - h)^2$ , so $x < s - h$ (if $x \ge s - h > 0$ , then multiplying by $x \ge 0$ and by $s - h > 0$ would give $x^2 \ge x(s - h) \ge (s - h)^2$ ). So $s - h$ is an upper bound of $S$ smaller than $s$ : this contradicts condition 2 of Definition 3.

Hence $s^2 = 2$ : the supremum $s$ is the positive square root of 2.

The gap in $\Q$ . The set $S_\Q = \{x \in \Q : x \ge 0 \text{ and } x^2 < 2\}$ is non-empty and bounded above, but no rational number $u$ is its least upper bound among the rationals. Suppose one were. Then $u \ge 1$ (because $1 \in S_\Q$ ), and $u^2 \ne 2$ by Theorem 1. If $u^2 < 2$ , the first case above with $s = u$ gives a rational $h$ and an element $u + h$ of $S_\Q$ larger than $u$ . If $u^2 > 2$ , the second case gives the rational upper bound $u - h = \frac{u^2 + 2}{2u}$ of $S_\Q$ , smaller than $u$ . Both contradict the choice of $u$ . So the argument that produced $\sqrt{2}$ in $\R$ fails in $\Q$ at its very first step: there, the supremum need not exist.

Completeness gives every square root

We prove Remark 2: every $y \ge 0$ has exactly one non-negative square root, and for $s, t \ge 0$ , $s < t \iff s^2 < t^2$ .

Order and uniqueness. If $0 \le s < t$ , then $s^2 < t^2$ : multiplying $s < t$ by $s \ge 0$ gives $s^2 \le st$ , and multiplying it by $t > 0$ gives $st < t^2$ . So two different non-negative numbers have different squares, and $y$ has at most one non-negative square root. In the same way, if $0 \le s \le t$ then $s^2 \le t^2$ . Conversely, let $s, t \ge 0$ with $s^2 < t^2$ . By trichotomy (property 1 of the order rules: exactly one of $s < t$ , $s = t$ and $t < s$ holds), it is enough to rule out the other two. If $s = t$ , then $s^2 = t^2$ ; if $t < s$ , then $t^2 < s^2$ by what we just proved, with $s$ and $t$ swapped. Neither is compatible with $s^2 < t^2$ , so $s < t$ .

Existence (sketch). For $y = 0$ , take $s = 0$ . For $y > 0$ , we repeat the argument for $\sqrt{2}$ with $y$ in place of 2. The set $S_y = \{x \in \R : x \ge 0 \text{ and } x^2 < y\}$ is non-empty ( $0 \in S_y$ ) and bounded above by $1 + y$ (if $x \ge 1 + y$ , then $x \ge 1$ , so $x^2 \ge x > y$ ). By Axiom 1 it has a supremum $s \ge 0$ .

If $s^2 < y$ : let $h$ be the smaller of $\frac{1}{2}$ and $\frac{y - s^2}{2s + 1}$ . Then $0 < h < 1$ , so $h^2 < h$ and $(s + h)^2 < s^2 + h(2s + 1) \le y$ . So $s + h \in S_y$ , although $s + h > s$ .
If $s^2 > y$ : then $s > 0$ (as $s \ge 0$ and $s^2 > 0$ ); let $h = \frac{s^2 - y}{2s}$ . As for $\sqrt{2}$ , $h > 0$ , $s - h = \frac{s^2 + y}{2s} > 0$ and $(s - h)^2 = y + h^2 > y$ . Every $x \in S_y$ has $x^2 < y < (s - h)^2$ , so $x < s - h$ (if $x \ge s - h \ge 0$ , the order step would give $x^2 \ge (s - h)^2$ ). So $s - h$ is an upper bound of $S_y$ smaller than $s$ .

Both cases contradict $s = \sup S_y$ , so $s^2 = y$ .

One more consequence of the Archimedean property, which later pages cite: every real number lies between two consecutive integers.

Remark 7 (The integer part)

For every real number $u$ there is exactly one integer $k$ with $k \le u < k + 1$ . It is called the integer part of $u$ . For example, the integer part of 2.7 is 2, that of 3 is 3, and that of ${-1.5}$ is ${-2}$ (not ${-1}$ , which is greater than ${-1.5}$ ).

Reason. Existence. Let $A$ be the set of integers greater than $u$ . It is non-empty: the Archimedean property gives an integer $n \ge 1$ with $n > u$ . We find its smallest element with the well-ordering principle. The elements of $A$ need not be positive, so we first shift $A$ into the positive integers: the Archimedean property gives an integer $j \ge 1$ with $j > -u$ , and then every $i \in A$ has $i + j > u + j > 0$ (adding $j$ to $i > u$ and $u$ to $j > -u$ , by property 3 of the order rules, then property 2). So the numbers $i + j$ with $i \in A$ form a non-empty set of positive integers, which by the well-ordering principle has a smallest element $m + j$ with $m \in A$ . Then $m$ is the smallest element of $A$ : every $i \in A$ has $i + j \ge m + j$ , so $i \ge m$ (adding $-j$ , property 3). Let $k = m - 1$ . Then $k + 1 = m > u$ . And $k \notin A$ , because $k < m$ and $m$ is the smallest element of $A$ ; so $k$ is not greater than $u$ , that is $k \le u$ by property 7.

Uniqueness. Let $k$ and $l$ be integers with $k \le u < k + 1$ and $l \le u < l + 1$ . Then $k \le u < l + 1$ gives $k < l + 1$ (property 2), and in the same way $l < k + 1$ . Adding $-l$ to the first and $-1 - l$ to the second (property 3) gives $-1 < k - l < 1$ . The only integer strictly between ${-1}$ and 1 is 0, so $k = l$ .

Summary¶

$\N \subset \Z \subset \Q \subset \R$ . The real numbers are the points of the number line; the irrational ones are those whose decimals neither terminate nor repeat.
$\sqrt{2}$ is irrational: no fraction squares to 2 (Theorem 1).
Set-builder notation $\{x \in \R : \text{condition}\}$ describes a set by a condition.
Intervals: a square bracket includes the endpoint, a round one excludes it, and $\pm\infty$ always gets a round bracket:
$[a, b) = \{x \in \R : a \le x < b\}, \qquad (c, \infty) = \{x \in \R : x > c\}.$
(9)
Rigorous track: $\sup S$ is the least upper bound of $S$ , and the completeness axiom says that every non-empty set of reals that is bounded above has one.

Exercises¶

Where this leads¶

These topics build on this one:

Absolute Value and Inequalities
Functions and Their Graphs
Sums and Sigma Notation (coming soon)

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