Exercise 1.1 · Q14: Parallel lines and the class of $y=2x+4$

Basics you need

Solution

1. Fix the meaning of "parallel", including the convention

   Convention. In this question, as is standard when parallelism is treated as a relation, a line is regarded as parallel to itself (coincident lines count as parallel). So "$L_1$ is parallel to $L_2$" means: $L_1$ and $L_2$ are the same line, or they never meet.

   Now express this through slopes. Give each line $M$ a direction $d(M)$⁠: its slope $m$ if $M$ is $y = mx + c$⁠, or the word "vertical" if $M$ is $x = k$⁠. Then

   $$(L_1, L_2)\in R \iff d(L_1) = d(L_2).$$

   To see why, check each case.

   • Both non-vertical, same slope: $y = mx + c_1$ and $y = mx + c_2$⁠. A common point would need $mx + c_1 = mx + c_2$⁠, that is $c_1 = c_2$⁠. So either $c_1 = c_2$ and they are the same line, or $c_1 \ne c_2$ and they never meet. Parallel in both cases.
   • Both non-vertical, different slopes $m_1 \ne m_2$⁠: solving $m_1x + c_1 = m_2x + c_2$ gives $x = \frac{c_2 - c_1}{m_1 - m_2}$⁠, so they meet in exactly one point. They are different lines that meet: not parallel.
   • Both vertical: $x = k_1$ and $x = k_2$ are the same line if $k_1 = k_2$ and never meet otherwise. Parallel.
   • One vertical, one not: $x = k$ meets $y = mx + c$ at the single point $(k, mk + c)$⁠. Not parallel.
   "Parallel" is a geometric word, and the textbook does not spell out what happens when the two lines are the same. That case decides reflexivity, so it must be settled first. The answer key confirms the convention: it lists all lines $y = 2x + c$⁠, $c\in\R$⁠, and $c = 4$ is the line $y = 2x + 4$ itself. Turning "parallel" into "same direction" then makes the three properties properties of equality.
   Each line has exactly one direction: a vertical line has no slope, and a non-vertical line has exactly one slope. So $d(M)$ is well defined for every line $M$⁠.

2. Reflexive: is every line parallel to itself?

   Take any line $L_1$⁠. Then $d(L_1) = d(L_1)$⁠, so by Step 1, $(L_1,L_1)\in R$⁠. (Directly: $L_1$ is the same line as itself, which the convention counts as parallel.) Since $L_1$ was arbitrary, $R$ is reflexive.

   Reflexivity is about every line, so we take an arbitrary one. This is the one step that depends on the convention: under the stricter reading "parallel means distinct lines that never meet", no line would be parallel to itself.
   Equality is reflexive: every slope (or "vertical") equals itself.

3. Symmetric: if $L_1$ is parallel to $L_2$⁠, is $L_2$ parallel to $L_1$⁠?

   Suppose $(L_1,L_2)\in R$⁠. Then $d(L_1) = d(L_2)$⁠, hence $d(L_2) = d(L_1)$⁠, so $(L_2,L_1)\in R$⁠. Hence $R$ is symmetric.

   We start from the "if" part and use it. Geometrically, "the two lines never meet, or are the same line" does not depend on which line is named first.
   Equality is symmetric: $a = b \Rightarrow b = a$⁠.

4. Transitive: does $L_1 \parallel L_2$ and $L_2 \parallel L_3$ give $L_1 \parallel L_3$⁠?

   Suppose $(L_1,L_2)\in R$ and $(L_2,L_3)\in R$⁠. Then

   $$\begin{gathered} d(L_1) = d(L_2) \qquad\text{and}\qquad d(L_2) = d(L_3),\\ \text{so}\qquad d(L_1) = d(L_3). \end{gathered}$$

   By Step 1, $(L_1,L_3)\in R$⁠. Hence $R$ is transitive.

   Spelled out in the two possible cases: if $L_2$ is vertical, then $L_1$ and $L_3$ are both vertical, so they are parallel. If $L_2$ has slope $m$⁠, then $L_1$ and $L_3$ both have slope $m$⁠, so they are parallel.

   The conclusion does not mention $L_2$⁠, so we remove it: $d(L_1)$ and $d(L_3)$ are both equal to $d(L_2)$⁠. Notice that $L_1$ and $L_3$ may turn out to be the same line (for example $L_3 = L_1$⁠); the convention makes that case harmless.
   Equality is transitive: $a = b$ and $b = c$ give $a = c$⁠.

5. Conclude, and see why the convention is needed

   $R$ is reflexive, symmetric and transitive, so $R$ is an equivalence relation in $L$.

   Without the convention the result would be false. Suppose "parallel" meant "distinct and never meeting". Take $L_1 : y = 2x + 4$ and $L_2 : y = 2x$⁠. They are distinct and never meet, so $(L_1,L_2)$ and $(L_2,L_1)$ would both be in $R$⁠. Transitivity (with $L_3 = L_1$⁠) would then demand $(L_1,L_1)\in R$⁠, but a line is not distinct from itself. So under that reading $R$ is neither reflexive nor transitive (it is still symmetric).

   This explicit counterexample shows the convention is not a technicality: it is exactly what makes the question's claim true. Compare Example 3, where the textbook shows perpendicularity is symmetric but neither reflexive nor transitive.
   One explicit chain where transitivity fails is enough to show a relation is not transitive (Basic 4). Basics: proving and disproving

6. The set of all lines related to $y = 2x + 4$

   Call the line $\ell : y = 2x + 4$⁠. It is non-vertical with slope $d(\ell) = 2$⁠. Let

   $$[\ell] = \{M \in L : (M,\ell)\in R\} = \{M \in L : d(M) = 2\}.$$

   We show $[\ell] = \{\text{lines } y = 2x + c : c\in\R\}$⁠, proving both inclusions.

   • Every line related to $\ell$ has the form $y = 2x + c$⁠. If $(M,\ell)\in R$⁠, then $d(M) = 2$⁠. In particular $M$ is not vertical (a vertical line has no slope), so $M$ has an equation $y = mx + c$⁠, and its slope is $m = 2$⁠. So $M$ is $y = 2x + c$⁠, where $c$ is its $y$-intercept.
   • Every line $y = 2x + c$ is related to $\ell$⁠. For any real $c$⁠, the line $y = 2x + c$ has slope 2, which equals $d(\ell)$⁠. So it is parallel to $\ell$⁠.

   Hence the set of all lines related to $y = 2x + 4$ is $\{y = 2x + c : c \in \R\}$. Different values of $c$ give different lines (they cross the $y$-axis at different points $(0,c)$⁠), and $c = 4$ gives $\ell$ itself.

   The class must contain all lines parallel to $\ell$ and nothing else, so both inclusions are needed. The first inclusion relies on vertical lines being excluded: a vertical line has no slope, so it is not parallel to $\ell$⁠. The second uses only that the slope of $y = 2x + c$ can be read off as the coefficient of $x$⁠.
   For a line written as $y = mx + c$⁠, the coefficient of $x$ is the slope and the constant is the $y$-intercept. Basics: equivalence classes

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