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Precalculus JumpStart

Section 2.7 Absolute Value

The absolute value \(|x|\) of a real number \(x\) provides its magnitude, i.e. how large it is without regard to being positive or negative. Geometrically, it represents the distance the number is from zero along the real number line.
Image of the real number line and the absolute value of a number depicting distance from zero.
We can express this precisely with the conditional expression
\begin{equation} |x| = \begin{cases} x, \quad &\text{if $x \geq 0$}\\ -x, \quad & \text{if $x \lt 0$} \end{cases}\tag{2.7.1} \end{equation}
For example,
\begin{equation*} |4|=4, \quad \text{because $4 \geq 0$}, \end{equation*}
while the absolute value of \(-4\) is the opposite of \(-4\text{:}\)
\begin{equation*} |-4| = -(-4) = 4, \quad \text{because $4 \lt 0$.} \end{equation*}
Note that \(|0| = 0\text{.}\) Formulas like Equation (2.7.1) which involving multiple cases are common in engineering. Take some time to carefully understand the notation used above.

Distance between numbers.

We can measure the distance between two real numbers \(a\) and \(b\) on the real line by subtracting them and then computing the absolute value to obtain a distance. This ensures that we do not need to worry about the order the subtraction was performed.
\begin{equation*} \left(\text{distance between $a$ and $b$}\right) = \left| a-b\right| = \left|b-a\right|. \end{equation*}

Example 2.7.1.

If we divide the portion of the real line between \(-6\) and \(8\) into ten equal width parts, what is the width of each part?
Solution.
The distance between \(-6\) and \(8\) is
\begin{equation*} \left|-6-8\right| = \left|-14\right| = 14. \end{equation*}
We then divide this by \(10\) to obtain a width of \(14/10 =1.4\) for each part.

Example 2.7.2. Order of Operations.

Order of operations requires us to evaluate expressions within the absolute value completely before combining with other operations. In other words, treat absolute values like you would parentheses. For example,
\begin{align*} -5\left|8-10 \right| + 3 \amp=\quad -5 \left|-2\right| + 3 \\ \amp=\quad -5 \left(-(-2)\right) + 3 \\ \amp=\quad -5 \cdot 2 + 3 \\ \amp=\quad -10 + 3 \\ \amp=\quad -7 \end{align*}

Example 2.7.3.

Evaluate the expression
\begin{equation*} \left|\left|2-5\right|-4\right|^3 \end{equation*}
Solution.
\begin{align*} \left|\left|2-5\right|-4\right|^3 \amp=\left|\left|-3\right|-4\right|^3\\ \amp=\left|3-4\right|^3\\ \amp=\left|-1\right|^3\\ \amp=1^3\\ \amp=1 \end{align*}

Checkpoint 2.7.5.

Evaluate
\begin{equation*} \left|\left|-2\right|-\left|3\right|\right| \end{equation*}
Answer.
\(\left|\left|-2\right|-\left|3\right|\right| = 1\)