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0.2: Numbers, Sets, and Absolute Values
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0.2: Numbers, Sets, and Absolute Values

Unsurprisingly, numbers are fundamental to everyday life. But how do we manage these numbers, and what types of numbers do mathematicians use? In calculus, it is imperative to be able to describe numbers and construct groups of numbers that satisfy some criteria. We therefore discuss the following topics:

Classifying Numbers

We use numbers to count and measure quantities. An integer is a number with no fractional part: \[\dots \cmaa -3 \cmaa -2 \cmaa -1 \cmaa 0 \cmaa 1 \cmaa 2 \cmaa 3 \cmaa \dots \pd\] A rational number is a fraction of integers, \(m/n,\) where \(m\) and \(n\) are integers such that \(n \ne 0.\) (The name rational comes from ratio.) Examples of rational numbers are \[\tfrac{1}{5} \cmaa 2 = \tfrac{2}{1} \cmaa 1.57 = \tfrac{157}{100} \cmaa 0.333 \ldots = \tfrac{1}{3} \pd\] Thus, all integers are rational numbers, but not all rational numbers are integers. Note that division by \(0\) is undefined, so expressions like \(1/0\) and \(0/0\) are undefined. An irrational number cannot be expressed as a ratio of integers \(m/n\)—for example, \[\sqrt 3 \cmaa \pi \cmaa \sqrt[3]{4} \cmaa \sin 2 \pd\] Decimal expansions of irrational numbers are nonrepeating; the most famous example is pi, \[\pi = 3.14159265 \ldots \cma\] in which the digits have no pattern and do not terminate. By contrast, rational numbers' decimal expansions either terminate or repeat with a pattern. The set of real numbers, denoted \(\RR,\) includes both rational and irrational numbers. In this text, we exclusively work with real numbers.

CLASSIFYING REAL NUMBERS
Real numbers can be classified into three categories.
  1. Integers are numbers that are not fractions \((\dots, -3, -2, -1, 0, 1, 2, 3, \dots).\)
  2. Rational numbers are fractions of integers: \(m/n,\) where \(m\) and \(n\) are integers such that \(n \ne 0.\) Rational numbers include integers.
  3. Irrational numbers cannot be expressed as fractions of integers.
Integers are a subset of rational numbers; rational and irrational numbers together form the real numbers.
EXAMPLE 1
Classify each number as rational or irrational.
  1. \(3 \sqrt{16} - 7\)
  2. \(6.445\)
  3. \(\pi + 3\)
  4. \(\ds \frac{4 \pi}{\pi}\)
  1. Rational. Note that \(\sqrt{16} = 4\) because \(4^2 = 16.\) The expression is therefore equivalent to \[3(4) - 7 = 5 \cma\] which is an integer. Rational numbers include integers.
  2. Rational. The decimal terminates and can therefore be rewritten as \(6445/1000,\) a fraction of integers.
  3. Irrational. It can be proved that the sum of a rational number and an irrational number is irrational.
  4. Rational. The \(\pi\)'s cancel out: \[\frac{4 \cancel{\pi}}{\cancel{\pi}} = 4 \pd\] Because \(4\) is an integer, it is rational.
EXAMPLE 2
Bound the irrational number \(6 + \sqrt{11}\) between two consecutive integers.
Note that \(\sqrt 9\) \(= 3\) and \(\sqrt{16}\) \(= 4.\) So \[ \baat{2} \sqrt{9} &\lt \sqrt{11} &&\lt \sqrt{16} \nl 3 &\lt \sqrt{11} &&\lt 4 \pd \eaat \] Adding \(6\) to every term gives \[\boxed{9 \lt 6 + \sqrt{11} \lt 10}\] These bounds give an idea of how large \(6 + \sqrt{11}\) is since we cannot determine its value exactly.

Sets

A set is a collection of numbers where each number is called an element. An example of a set is \(S = \) \(\{-2, 1, 3, 5, 6\}.\) The notation \(3 \in S\) means \(3\) is an element of \(S.\) Conversely, \(9 \not \in S\) since the number \(9\) is not in \(S.\) We use set-builder notation to produce a set of numbers that satisfy some criterion—for example, \[\{2, 3, 4, 5\} = \{x \mid x \text{ is an integer and } 2 \leq x \leq 5\} \pd\]

The union of two sets \(S\) and \(T\)—that is, \(S \cup T\)—is the set of numbers that are in \(S\) or in \(T\) (or in both \(S\) and \(T\)). Conversely, the intersection of two sets \(S\) and \(T,\) denoted \(S \cap T,\) is the set of numbers that are in both \(S\) and \(T.\) On the other hand, an empty set has no elements and is denoted by the symbol \(\emptyset.\)

EXAMPLE 3
Sets \(A\) and \(B\) are defined by \[ \ba A &= \{-7, -4, 1, 3, 9\} \cma \nl B &= \{-4, -2, 1, 5, 9\} \pd \ea \] Find \(A \cup B\) and \(A \cap B.\)
The union of \(A\) and \(B\) is the set of numbers that are in \(A\) or in \(B\) (or in both): \[A \cup B = \boxed{\{-7, -4, -2, 1, 3, 5, 9\}}\] The intersection of \(A\) and \(B\) is the set of numbers that are in both \(A\) and \(B \col\) \[A \cap B = \boxed{\{-4, 1, 9\}}\]

Intervals An interval is a range of real numbers. The open interval from \(a\) to \(b\) is the set of all real numbers from \(a\) to \(b,\) exclusive: \[(a, b) = \{x \mid a \lt x \lt b \} \pd\] Note the use of parentheses. Conversely, the closed interval from \(a\) to \(b\) is the set of all real numbers from \(a\) to \(b,\) including the endpoints: \[[a, b] = \{x \mid a \leq x \leq b \} \pd\] Note the use of brackets. It's also common to have a half-open interval, in which only one endpoint is included. For example, \([a, b)\) includes \(a\) but excludes \(b,\) and \((a, b]\) includes \(b\) but excludes \(a.\) All interval notations and their corresponding descriptions are given by the following table.

INTERVAL NOTATION
Interval Notation Set-Builder Notation
\((a, b)\) \(\{x \mid a \lt x \lt b\}\)
\([a, b]\) \(\{x \mid a \leq x \leq b\}\)
\([a, b)\) \(\{x \mid a \leq x \lt b\}\)
\((a, b]\) \(\{x \mid a \lt x \leq b\}\)
\((a, \infty)\) \(\{x \mid x \gt a \}\)
\([a, \infty)\) \(\{x \mid x \geq a \}\)
\((-\infty, b)\) \(\{x \mid x \lt b\}\)
\((-\infty, b]\) \(\{x \mid x \leq b\}\)
\((-\infty, \infty)\) \(\{x \mid x \in \RR\}\)

Absolute Values

The absolute value of a number \(a,\) denoted \(\abs a,\) is how far away it is from \(0.\) Thus, all absolute values are nonnegative: \[\abs a \geq 0 \pd\] Simply put, to find the absolute value of a number, remove any negative sign in front of the number. For example, \(\abs{-7} = 7\) and \(\abs 8 = 8.\)

If \(a\) is a negative number, then \(-a\) is a positive number. This fact enables us to rewrite an absolute value expression as a piecewise definition: \begin{equation} \abs a = \bc a \; \; &a \geq 0 \nl -a \; \; &a \lt 0 \pd \ec \label{eq:abs-cases} \end{equation}

EXAMPLE 4
Rewrite the expression \(\abs{5x - 3}\) without absolute values.
Note that \(5x - 3 = 0\) when \(x = 3/5.\) So following \(\eqref{eq:abs-cases},\) we have \[ \ba \abs{5x - 3} &= \bc 5x - 3 \; \; &x \geq \tfrac{3}{5} \nl -(5x - 3) \; \; &x \lt \tfrac{3}{5} \ec \nl &= \boxed{\bc 5x - 3 \; \; &x \geq \tfrac{3}{5} \nl 3 - 5x \; \; &x \lt \tfrac{3}{5} \ec} \ea \]

The following facts about absolute value expressions are useful. You may experiment with numbers to verify that they are true.

EXPRESSIONS WITH ABSOLUTE VALUES
  1. \(\abs x = a\) if and only if \(x = a\) or \(x = -a.\)
  2. \(\abs x \lt a\) if and only if \(-a \lt x \lt a.\)
  3. \(\abs x \leq a\) if and only if \(-a \leq x \leq a.\)
  4. \(\abs x \gt a\) if and only if \(x \gt a\) or \(x \lt -a.\)
  5. \(\abs x \geq a\) if and only if \(x \geq a\) or \(x \leq -a.\)
PROPERTIES OF ABSOLUTE VALUES
Let \(a\) and \(b\) be real numbers and \(n\) be a nonnegative integer.
  1. \(\abs{ab} = \abs a \abs b.\)
  2. \(\ds \abs{\frac{a}{b}} = \frac{\abs a}{\abs b} \cmaa b \ne 0.\)
  3. \(\abs{a^n} = \abs{a}^n.\)

Square Roots of Squares An absolute value expression can be formed by taking the square root of a square; that is, \begin{equation} \sqrt{a^2} = \abs a \pd \label{eq:sqrt-abs} \end{equation} This fact is true because the symbol \(\sqrt{\phantom{x}}\) exclusively means the positive square root and so cannot output a negative value.

EXAMPLE 5
Solve the equation \(x^2 - 81 = 0.\)
Adding \(81\) to both sides gives \[x^2 = 81 \pd\] Taking the square root of both sides and using \(\eqref{eq:sqrt-abs},\) we attain \[ \ba \sqrt{x^2} &= \sqrt{81} \nl \abs x &= 9 \pd \ea \] By (a) of Expressions with Absolute Values, we have \(x = -9\) and \(x = 9.\) We write these solutions in set notation as \[\boxed{\{-9, 9\}}\]
EXAMPLE 6
What values of \(x\) satisfy \(\abs{7x + 5} = 2 \ques\)
By (a) of Expressions with Absolute Values, we establish two cases: \[ \ba 7x + 5 &= 2 \lspace &7x + 5 &= -2 \nl 7x &= -3 \lspace &7x &= -7 \nl x &= -\tfrac{3}{7} \lspace &x &= -1 \pd \ea \] We express both solutions using set notation: \[\boxed{\parbrace{-1, -\tfrac{3}{7}}}\]
EXAMPLE 7
Solve the inequality \(\abs{2x - 3} \gt 1.\)
By (d) of Expressions with Absolute Values, we consider two cases: \[ \ba 2x - 3 &\gt 1 \lspace &2x - 3 &\lt -1 \nl 2x &\gt 4 \lspace &2x &\lt 2 \nl x &\gt 2 \lspace &x &\lt 1 \pd \ea \] Because \(x\) satisfies either inequality, we write our answer in interval notation as \[\boxed{(-\infty, 1) \cup (2, \infty)}\]
EXAMPLE 8
Given \(\abs{x + 1} \lt 1/2,\) bound the value of \(\abs{(3x + 3)^4}.\)
By (c) and (a) of Properties of Absolute Values, we have \[ \ba \abs{(3x + 3)^4} &= \abs{3x + 3}^4 \nl &= \par{3 \abs{x + 1}}^4 \nl &= 3^4 \, \abs{x + 1}^4 \nl &= 81 \, \abs{x + 1}^4 \pd \ea \] Given \(\abs{x + 1} \lt 1/2,\) it follows that \(\abs{x + 1}^4 \lt (1/2)^4\) \(= 1/16.\) We therefore have \[\boxed{\abs{(3x + 3)^4} \lt \tfrac{81}{16}}\]

The final property we discuss is the Triangle Inequality, which enables us to bound the absolute value of a sum. We will use this rule to prove several important laws in calculus.

TRIANGLE INEQUALITY
If \(a\) and \(b\) are real numbers, then the Triangle Inequality gives \begin{equation} \abs{a + b} \leq \abs a + \abs b \pd \label{eq:tri-inequality} \end{equation}

PROOF We begin with the following two inequalities: \[ \baat{2} -\abs a &\leq a &&\leq \abs a \cma \nl - \abs b &\leq b &&\leq \abs b \pd \eaat \] Adding both inequalities gives \[- \par{\abs a + \abs b} \leq a + b \leq \abs a + \abs b \pd\] Using (c) of Expressions with Absolute Values, with \(x\) replaced by \(a + b\) and \(a\) replaced by \(\abs a + \abs b,\) we attain \begin{equation} \abs{a + b} \leq \abs a + \abs b \pd \eqlabel{eq:tri-inequality} \end{equation} \[\qedproof\]

EXAMPLE 9
Suppose that \(\abs{P + 2} \lt 3\) and \(\abs{Q - 8} \lt 5.\) Use the Triangle Inequality to bound \(\abs{P + Q - 6}.\)
Using the Triangle Inequality, as in \(\eqref{eq:tri-inequality},\) with \(a = P + 2\) and \(b = Q - 8\) shows \[ \ba \abs{P + Q - 6} &= \abs{(P + 2) + (Q - 8)} \nl &\leq \abs{P + 2} + \abs{Q - 8} \nl &\lt 3 + 5 = 8 \pd \ea \] Thus, \[\boxed{\abs{P + Q - 6} \lt 8}\]

Classifying Numbers In this text, we work exclusively with real numbers, which can be classified into three categories. Integers are a subset of rational numbers; rational and irrational numbers together form the real numbers.

  1. Integers are numbers that are not fractions \((\dots, -3, -2, -1, 0, 1, 2, 3, \dots).\)
  2. Rational numbers are fractions of integers: \(m/n,\) where \(m\) and \(n\) are integers such that \(n \ne 0.\) Rational numbers include integers.
  3. Irrational numbers cannot be expressed as fractions of integers.

Irrational numbers have decimal expansions that are nonrepeating, whereas rational numbers' decimal expansions either terminate or repeat with a pattern.

Sets A set of numbers is a group of numbers; each number is called an element. The notation \(x \in S\) means \(x\) is an element of the set \(S,\) whereas \(x \not \in S\) means \(x\) is not an element of \(S.\) The union of two sets \(S\) and \(T\) is denoted by \(S \cup T;\) it is the set of numbers that are in \(S\) or in \(T\) (or in both \(S\) and \(T\)). By contrast, the intersection of two sets \(S\) and \(T,\) denoted \(S \cap T,\) is the set of numbers that are in both \(S\) and \(T.\) An empty set has no elements and is denoted by the symbol \(\emptyset.\) Set-builder notation enables us to generate a set of numbers that satisfy some condition. An interval is a range of real numbers between two endpoints. In interval notation, a parenthesis means we exclude the endpoint, whereas a bracket means we include the endpoint. The following table provides all forms of intervals.

Interval Notation Set-Builder Notation
\((a, b)\) \(\{x \mid a \lt x \lt b\}\)
\([a, b]\) \(\{x \mid a \leq x \leq b\}\)
\([a, b)\) \(\{x \mid a \leq x \lt b\}\)
\((a, b]\) \(\{x \mid a \lt x \leq b\}\)
\((a, \infty)\) \(\{x \mid x \gt a \}\)
\([a, \infty)\) \(\{x \mid x \geq a \}\)
\((-\infty, b)\) \(\{x \mid x \lt b\}\)
\((-\infty, b]\) \(\{x \mid x \leq b\}\)
\((-\infty, \infty)\) \(\{x \mid x \in \RR\}\)

Absolute Values The absolute value of a number \(a\) is denoted by \(\abs a;\) it represents how far \(a\) is from \(0.\) Thus, absolute values are nonnegative, so \(\abs a \geq 0.\) We can express an absolute value expression as \begin{equation} \abs a = \bc a \; \; &a \geq 0 \nl -a \; \; &a \lt 0 \pd \ec \eqlabel{eq:abs-cases} \end{equation} Taking the square root of a square returns an absolute value: \begin{equation} \sqrt{a^2} = \abs a \pd \eqlabel{eq:sqrt-abs} \end{equation} The following list provides properties of equations and inequalities with absolute values:

  1. \(\abs x = a\) if and only if \(x = a\) or \(x = -a.\)
  2. \(\abs x \lt a\) if and only if \(-a \lt x \lt a.\)
  3. \(\abs x \leq a\) if and only if \(-a \leq x \leq a.\)
  4. \(\abs x \gt a\) if and only if \(x \gt a\) or \(x \lt -a.\)
  5. \(\abs x \geq a\) if and only if \(x \geq a\) or \(x \leq -a.\)

Let \(a\) and \(b\) be real numbers and \(n\) be a nonnegative integer. Then the following properties are true:

  1. \(\abs{ab} = \abs a \abs b.\)
  2. \(\ds \abs{\frac{a}{b}} = \frac{\abs a}{\abs b} \cmaa b \ne 0.\)
  3. \(\abs{a^n} = \abs{a}^n.\)

The Triangle Inequality enables us to bound the absolute value of a sum: if \(a\) and \(b\) are real numbers, then \begin{equation} \abs{a + b} \leq \abs a + \abs b \pd \eqlabel{eq:tri-inequality} \end{equation}