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Chapter 1 · Relations and Functions

Example 3: Perpendicular lines: symmetric only

ExamplesTypes of relations · Section 1.2
The question

Let L be the set of all lines in a plane and RR be the relation in L defined as R={(L1,L2):L1 is perpendicular to L2}R = \{(L_1, L_2) : L_1 \text{ is perpendicular to } L_2\}⁠. Show that RR is symmetric but neither reflexive nor transitive.

Exam answer

Not reflexive: A line is not perpendicular to itself, so (L1,L1)∉R(L_1, L_1) \notin R for every line L1L_1⁠. Thus RR is not reflexive.

Symmetric: Let (L1,L2)∈R(L_1, L_2) \in R⁠. Then L1⊥L2⇒L2⊥L1L_1 \perp L_2 \Rightarrow L_2 \perp L_1⁠, so (L2,L1)∈R(L_2, L_1) \in R⁠. Thus RR is symmetric.

Not transitive: Take L1:y=0L_1 : y = 0⁠, L2:x=0L_2 : x = 0 and L3:y=1L_3 : y = 1⁠. Then L1⊥L2L_1 \perp L_2 and L2⊥L3L_2 \perp L_3⁠, so (L1,L2),(L2,L3)∈R(L_1, L_2), (L_2, L_3) \in R⁠. But L1∥L3L_1 \parallel L_3⁠, so (L1,L3)∉R(L_1, L_3) \notin R⁠. Thus RR is not transitive.

Hence RR is symmetric but neither reflexive nor transitive.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Perpendicular" means the two lines meet at a right angle. A right angle between two lines does not care which line is named first, so the relation is symmetric. But a line makes an angle of 0°, not 90°, with itself, so it is not reflexive. And turning through a right angle twice brings you back to the original direction: two lines perpendicular to the same line are parallel to each other, not perpendicular, so it is not transitive. Proving the "yes" needs a general argument; each "no" needs only one explicit counterexample.

Given, to show, and the plan
Given
L = the set of all lines in a plane, and the rule (L1,L2)∈R  ⟺  L1⊥L2(L_1,L_2)\in R \iff L_1 \perp L_2⁠.
To show
(a) RR is not reflexive; (b) RR is symmetric; (c) RR is not transitive.
Plan
Recall what each property demands and what it takes to prove or disprove it. Then: (a) exhibit a line not perpendicular to itself; (b) argue for arbitrary lines; (c) exhibit three specific lines that break the transitive rule, and explain why such a triple always appears.

Basics you need

1Perpendicular and parallel lines

L1⊥L2L_1 \perp L_2 means the lines meet and the angle between them is 90∘90^\circ⁠. In coordinates, two non-vertical lines with slopes m1,m2m_1, m_2 are perpendicular exactly when m1m2=−1m_1 m_2 = -1⁠; a horizontal and a vertical line are also perpendicular. Lines are parallel when they have the same direction (equal slopes).

2The three properties

Reflexive: (a,a)∈R(a,a)\in R for every aa⁠. Symmetric: (a1,a2)∈R⇒(a2,a1)∈R(a_1,a_2)\in R \Rightarrow (a_2,a_1)\in R⁠. Transitive: (a1,a2)∈R(a_1,a_2)\in R and (a2,a3)∈R⇒(a1,a3)∈R(a_2,a_3)\in R \Rightarrow (a_1,a_3)\in R⁠. Basics: reflexive, symmetric, transitive

3Proving versus disproving

Each property is a "for all" statement. To prove one, argue for arbitrary elements. To disprove one, a single counterexample is enough: one line for reflexive, one triple of lines for transitive. Basics: how to prove and disprove

4Two lines perpendicular to the same line are parallel

If L1⊥L2L_1 \perp L_2 and L3⊥L2L_3 \perp L_2⁠, then L1L_1 and L3L_3 make equal corresponding angles (both 90∘90^\circ⁠) with the transversal L2L_2⁠, so L1∥L3L_1 \parallel L_3⁠. This is a standard fact of plane geometry and is the heart of part (c).

Part (a): not reflexive

  1. What reflexive would require

    For RR to be reflexive we would need (L1,L1)∈R(L_1, L_1)\in R⁠, that is L1⊥L1L_1 \perp L_1⁠, for every line L1L_1 in the plane. To show it fails, one line that is not perpendicular to itself is enough.

    Reflexivity is a "for every" claim, so it is destroyed by a single exception. In fact here it fails for every line, which is even stronger than we need.
    Negation of "for every aa⁠, (a,a)∈R(a,a)\in R" is "there is some aa with (a,a)∉R(a,a)\notin R". Basics: how to prove and disprove

  2. A line is never perpendicular to itself

    Take any line L1L_1⁠, for example the xx-axis y=0y = 0⁠. The angle between L1L_1 and itself is 0∘0^\circ⁠: the two "copies" point in exactly the same direction. Since 0∘≠90∘0^\circ \ne 90^\circ⁠, L1L_1 is not perpendicular to L1L_1⁠. In slope language, a line of slope mm would need m⋅m=m2=−1m\cdot m = m^2 = -1⁠, which no real number satisfies (and a vertical line is not perpendicular to a vertical line). Hence

    (L1,L1)∉R,(L_1, L_1) \notin R,

    so RR is not reflexive.

    This is the textbook's reason: "a line L1L_1 can not be perpendicular to itself". The slope check makes it airtight: m2≥0m^2 \ge 0 for every real mm⁠, so m2=−1m^2 = -1 is impossible.
    The square of a real number is never negative. Perpendicular slopes multiply to −1-1 (Basic 1).

Part (b): symmetric

  1. If L1⊥L2L_1 \perp L_2⁠, then L2⊥L1L_2 \perp L_1

    Let L1,L2L_1, L_2 be any lines with (L1,L2)∈R(L_1, L_2)\in R⁠. Then L1L_1 and L2L_2 meet at a right angle. The angle at their crossing point is the same angle whichever line we name first, so L2L_2 and L1L_1 also meet at a right angle:

    (L1,L2)∈R  ⇒  L1⊥L2  ⇒  L2⊥L1  ⇒  (L2,L1)∈R.(L_1,L_2)\in R \;\Rightarrow\; L_1 \perp L_2 \;\Rightarrow\; L_2 \perp L_1 \;\Rightarrow\; (L_2,L_1)\in R.

    In slope form: m1m2=−1m_1 m_2 = -1 gives m2m1=−1m_2 m_1 = -1⁠, because multiplication does not depend on order. Hence RR is symmetric.

    Symmetry is an "if … then" claim, so we start from (L1,L2)∈R(L_1,L_2)\in R and derive (L2,L1)∈R(L_2,L_1)\in R⁠. It holds because the defining condition (a right angle at the crossing, or m1m2=−1m_1m_2=-1⁠) treats the two lines in exactly the same way. Swapping their names cannot change it.
    Multiplication is commutative: m1m2=m2m1m_1m_2 = m_2m_1⁠. Symmetric: (a1,a2)∈R⇒(a2,a1)∈R(a_1,a_2)\in R \Rightarrow (a_2,a_1)\in R (Definition 3(ii)).

Part (c): not transitive

  1. What transitive would require

    Transitivity would say: whenever L1⊥L2L_1 \perp L_2 and L2⊥L3L_2 \perp L_3⁠, also L1⊥L3L_1 \perp L_3⁠. To disprove it, we need just three lines with (L1,L2)∈R(L_1,L_2)\in R⁠, (L2,L3)∈R(L_2,L_3)\in R but (L1,L3)∉R(L_1,L_3)\notin R⁠.

    Again a single counterexample kills a "for all" statement. Choosing simple lines (horizontal and vertical) makes every claim about them easy to check.

  2. An explicit counterexample

    Take

    L1: y=0,L2: x=0,L3: y=1.L_1:\ y = 0, \qquad L_2:\ x = 0, \qquad L_3:\ y = 1.
    • L1L_1 is horizontal and L2L_2 is vertical, so L1⊥L2L_1 \perp L_2⁠: (L1,L2)∈R(L_1,L_2)\in R⁠.
    • L2L_2 is vertical and L3L_3 is horizontal, so L2⊥L3L_2 \perp L_3⁠: (L2,L3)∈R(L_2,L_3)\in R⁠.
    • L1L_1 and L3L_3 are both horizontal, so they are parallel (they never meet). Parallel lines are not perpendicular: (L1,L3)∉R(L_1,L_3)\notin R⁠.

    So the transitive rule fails for this triple, and RR is not transitive.

    L1: y = 0L3: y = 1L2: x = 0 90°90° L1 and L3 are parallel:they never meet, soL1 is not perpendicular to L3 (L1, L2) ∈ R and (L2, L3) ∈ R, but (L1, L3) ∉ R
    This is the textbook's Fig 1.1 with coordinates attached. Each blue line meets the gold line at a right angle, yet the two blue lines are parallel.
    This concrete triple is a complete, rigorous disproof. Coordinates remove any doubt: "horizontal" and "vertical" are exact descriptions, and two distinct horizontal lines never cross.
    A horizontal line and a vertical line are perpendicular. Two different horizontal lines are parallel. A pair of parallel lines does not meet, so it has no angle of 90∘90^\circ (Basic 1).

  3. Why it always fails: L1L_1 is in fact parallel to L3L_3

    The textbook says more: if L1⊥L2L_1 \perp L_2 and L2⊥L3L_2 \perp L_3⁠, then L1L_1 can never be perpendicular to L3L_3⁠; in fact L1∥L3L_1 \parallel L_3⁠. Here is why. Both L1L_1 and L3L_3 make a 90∘90^\circ angle with the same line L2L_2⁠. Equal corresponding angles with a transversal mean the lines are parallel (Basic 4), so L1L_1 and L3L_3 have the same direction. Two lines with the same direction meet at an angle of 0∘0^\circ (or not at all), never at 90∘90^\circ⁠.

    With slopes, using a tilted example: let L2:y=−12xL_2: y = -\tfrac12 x⁠. Lines perpendicular to it have slope mm with m⋅(−12)=−1m\cdot(-\tfrac12) = -1⁠, so m=2m = 2⁠. Thus L1:y=2xL_1: y = 2x and L3:y=2x+3L_3: y = 2x + 3 are both perpendicular to L2L_2⁠, and both have slope 2, so they are parallel. For them to be perpendicular we would need 2⋅2=−12 \cdot 2 = -1⁠, which is false.

    The single counterexample in Step 2 of this part already finishes the proof. This step explains the pattern behind it, so you see the failure is not an accident of the lines we chose: every chain L1⊥L2⊥L3L_1 \perp L_2 \perp L_3 ends with L1L_1 parallel to L3L_3⁠. (It can even happen that L3=L1L_3 = L_1⁠, and a line is not perpendicular to itself, by Part (a).)
    Two lines perpendicular to the same line are parallel (Basic 4). Perpendicular slopes: m1m2=−1m_1 m_2 = -1⁠. Basics: reflexive, symmetric, transitive
Result

Not reflexive: no line is perpendicular to itself, so (L1,L1)∉R(L_1,L_1)\notin R⁠. Symmetric: L1⊥L2⇒L2⊥L1L_1 \perp L_2 \Rightarrow L_2 \perp L_1⁠. Not transitive: y=0⊥x=0y=0 \perp x=0 and x=0⊥y=1x=0 \perp y=1⁠, but y=0y=0 and y=1y=1 are parallel, not perpendicular. So RR is symmetric but neither reflexive nor transitive.

Check it with numbers

Slopes: L1:y=2xL_1: y = 2x (slope 2), L2:y=−12xL_2: y = -\tfrac12x (slope −12-\tfrac12⁠), L3:y=2x+3L_3: y = 2x+3 (slope 2). Then 2⋅(−12)=−12\cdot(-\tfrac12) = -1 ✓ (perpendicular), (−12)⋅2=−1(-\tfrac12)\cdot 2 = -1 ✓ (perpendicular, and the reverse order gives the same, as symmetry says), but 2⋅2=4≠−12\cdot 2 = 4 \ne -1⁠, so L1L_1 is not perpendicular to L3L_3⁠. The site's verification script also tests all 72 equations ax+by=cax + by = c (they describe 32 different lines) with a,b∈{−2,…,2}a, b \in\{-2,\dots,2\}⁠, c∈{−1,0,1}c\in\{-1,0,1\}⁠: none is perpendicular to itself, symmetry never fails, and every chain L1⊥L2⊥L3L_1\perp L_2\perp L_3 ends with L1∥L3L_1 \parallel L_3⁠.

Common mistakes
  1. Calling RR "not transitive" without a counterexample. Saying "it doesn't feel transitive" is not a proof. Name three lines and check each pair.
  2. Thinking parallel lines are perpendicular "at infinity". Parallel lines never meet, so there is no right angle between them.
  3. Proving symmetry with one example. Showing that the axes are perpendicular both ways checks one pair only. The argument must work for arbitrary L1,L2L_1, L_2⁠.
  4. Mixing up "not reflexive" and "irreflexive". Not reflexive needs just one line with (L,L)∉R(L,L)\notin R⁠. Here every line fails, which is more than required.
What to take away
  1. A property can hold while the others fail: perpendicularity is the standard example of "symmetric only".
  2. Prove a property with arbitrary elements; disprove it with one explicit, fully checked counterexample.
  3. Two lines perpendicular to the same line are parallel. That one geometric fact is why "perpendicular to" is not transitive, and why "parallel to" (Exercise 1.1 Q14) behaves differently.

Every fact used, and where it comes from

FactUsed inSource
Reflexive, symmetric, transitiveAll partsDefinition 3, Section 1.2 (Lesson: types of relations)
One counterexample disproves a "for all" statementPart (a) Step 1; Part (c) Steps 1, 2Logic of quantifiers (Basics: how to prove and disprove)
Perpendicular slopes: m1m2=−1m_1 m_2 = -1⁠; horizontal ⊥ verticalPart (a) Step 2; Part (b) Step 1; Part (c) Step 3Coordinate geometry of straight lines (Class XI)
m2≥0m^2 \ge 0 for real mmPart (a) Step 2Squares of real numbers are non-negative
Two lines perpendicular to the same line are parallelPart (c) Step 3Plane geometry: corresponding angles
The picture of L1,L2,L3L_1, L_2, L_3Part (c) Step 2Fig 1.1, Example 3
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