The Timpson Lyceum · College Algebra

Lecture 10 (College Algebra)

Overview

This lecture introduces analytic geometry in one dimension — the program, begun by René Descartes, of attaching numbers (coordinates) to geometric objects. Working on a single directed line, we define directed line segments and directed distances, then use coordinates on the number line to compute those distances. From there we prove the basic order properties of inequalities (adding, multiplying, dividing, and transitivity), build up two lemmas about absolute value, and combine them into the one-dimensional triangle inequality. The lecture closes with graphing solution sets on the number line, a short tour of set operations, and several worked inequality problems — including absolute-value inequalities solved via the absolute-value characterization theorem.

Key Concepts

Analytic geometry. The combination of geometry with numbers. In ordinary Euclidean geometry, points such as $A,B,C$ have no coordinates; analytic geometry assigns numbers to points using a coordinate system. The plane version is the Cartesian coordinate plane, named for René Descartes. This lecture restricts everything to one dimension: a single line.

Directed line. A line $\ell$ is directed when one direction is chosen as positive and the other as negative. A segment of a directed line is a directed line segment.

Directed distance. For points $A,B$ on a directed line, $\overline{AB}$ denotes the directed distance from $A$ to $B$ — a signed number (positive if $B$ lies in the positive direction from $A$, negative otherwise). The unsigned length is the absolute value:

• $\lvert AB\rvert = \lvert \overline{AB}\rvert$ (length is always nonnegative).
• $\overline{AB} = -\,\overline{BA}$, equivalently $\overline{AB} + \overline{BA} = 0$.
• Additivity (any three points): for any $A,B,C$ on $\ell$, $\overline{AB} + \overline{BC} = \overline{AC}$, regardless of the order of the points on the line.

Coordinates on the number line. Choose a point as the origin $O$ (coordinate $0$) and a unit length; marking off units gives each point a coordinate. If $P_1$ has coordinate $x_1$ and $P_2$ has coordinate $x_2$, then $\overline{P_1P_2} = x_2 - x_1,$ and the length is $\lvert P_1P_2\rvert = \lvert x_2 - x_1\rvert$. This follows from additivity: $\overline{P_1P_2} = \overline{P_1O} + \overline{OP_2} = -x_1 + x_2$.

Order and inequalities. By definition $b < c \iff c - b > 0$, and $b > c \iff c - b < 0$. The lecture proves the order axioms:

• Add to both sides: $b < c \iff b + h < c + h$.
• Multiply/divide by a positive: if $r > 0$, then $b < c \iff rb < rc$.
• Multiply/divide by a negative flips the sign: if $r < 0$, then $b < c \iff rb > rc$.
• Transitivity: if $b < c$ and $c < d$, then $b < d$.

Absolute value. Defined piecewise: $\lvert x\rvert = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$ Two key results:

• Characterization (lemma): for $d \ge 0$, $\lvert c\rvert \le d \iff -d \le c \le d$. This is the standard tool for removing absolute values from an inequality.
• Bounding lemma: $-\lvert x\rvert \le x \le \lvert x\rvert$ for every $x$.

Triangle inequality (one dimension). For all real $x_1, x_2$, $\lvert x_1 + x_2\rvert \le \lvert x_1\rvert + \lvert x_2\rvert.$ It is named for its Euclidean origin: the distance from $A$ to $C$ never exceeds the distance from $A$ to $B$ plus from $B$ to $C$ — going straight is never longer than detouring.

Graphing a solution set (1-D). The graph of a one-variable numerical statement is the set of points on the number line representing its solution set. Use an open circle for a strict endpoint ($<,>$) and a closed circle for an inclusive endpoint ($\le,\ge$). An and between conditions means intersection of the two solution sets; an or means union.

Set operations. For sets $A,B$:

• Union: $A \cup B = \{x : x \in A \text{ or } x \in B\}$.
• Intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
• Difference: $A \setminus B = \{x : x \in A \text{ and } x \notin B\}$.
• Mutually exclusive (disjoint): $A \cap B = \varnothing$.
• Properties: $A\cup A = A$, $A\cap A = A$, $A\cup\varnothing = A$, $A\cap\varnothing = \varnothing$; $A\cap B \subseteq A$; if $A \subseteq B$ then $A\cup B = B$. Union and intersection are commutative and associative, the distributive laws $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$ and $A\cup(B\cap C) = (A\cup B)\cap(A\cup C)$ hold, and $\subseteq$ is transitive.

Worked Example

Solve an absolute-value inequality. Use the characterization $\lvert u\rvert < d \iff -d < u < d$ (for $d > 0$), then isolate the variable across all three parts.

Solve $\lvert x + 1\rvert < 3$:

Solution set: $\{\,x : -4 < x < 2\,\}$, graphed as open circles at $-4$ and $2$ with the segment between them shaded.

Directed-distance check. With coordinates $a = -5$, $b = 2$, $c = -3$ on a directed line, apply $\overline{P_1P_2} = x_2 - x_1$:

Summary

• Analytic geometry attaches coordinates to geometric points; in one dimension everything happens on a single directed line.
• On a directed line, $\overline{AB}$ is a signed distance: $\overline{AB} = -\overline{BA}$ and $\overline{AB} + \overline{BC} = \overline{AC}$ for any three points.
• Given coordinates, the directed distance is $\overline{P_1P_2} = x_2 - x_1$ and the length is $\lvert x_2 - x_1\rvert$.
• Order rules: you may add the same quantity to both sides and multiply/divide by a positive without changing an inequality; multiplying/dividing by a negative flips it; $<$ is transitive.
• The key absolute-value tool is $\lvert c\rvert \le d \iff -d \le c \le d$ (for $d \ge 0$), which turns an absolute-value inequality into a double inequality you can solve directly.
• Combining the bounding lemma $-\lvert x\rvert \le x \le \lvert x\rvert$ with that tool yields the triangle inequality $\lvert x_1 + x_2\rvert \le \lvert x_1\rvert + \lvert x_2\rvert$.
• Solution sets graph on the number line with open/closed circles; and corresponds to intersection and or to union of solution sets, using the basic set operations $\cup$, $\cap$, and $\setminus$.