Mathematical-Reasoning · Unit 27 · Video 6 · Interactive Practice

Equivalence Relations: Angles, Lines and Set Sizes

IKey Formulas

Relation	Definition	Related objects share
Identity $I_A$ on $A$	$I_A = \{ (x, x) \mid x \in A \}$	the element itself
Angles on $\mathbb{R}$	$x \sim y \iff x - y = 2k\pi$ for some $k \in \mathbb{Z}$	the point on the unit circle, so $\sin$ and $\cos$
Lines in the plane	$\ell_1 \mathrel{P} \ell_2 \iff \ell_1 \parallel \ell_2$ or $\ell_1 = \ell_2$	the direction
Subsets of a finite $U$	$A \approx B \iff \operatorname{card}(A) = \operatorname{card}(B)$	the number of elements

Key Insight: Whenever two objects are related exactly when some quantity takes the same value on both, reflexive, symmetric and transitive are inherited from equality.

IIAngles a Whole Number of Turns Apart

Two angles are related exactly when they end at the same point of the unit circle.

x 7π/3 y π/3

IIISorting Subsets by Size

Under $A \approx B \iff \operatorname{card}(A) = \operatorname{card}(B)$ , a subset is related to exactly the subsets in its own column.

IVQuiz Questions

Problem 1 · Related to π/3

Given: on $\mathbb{R}$ , $x \sim y$ if and only if $x - y = 2k\pi$ for some $k \in \mathbb{Z}$ . Which angle is related to $\tfrac{\pi}{3}$ ?

✅ Correct!

\tfrac{\pi}{3} - \left(-\tfrac{5\pi}{3}\right) = 2\pi = 2 \cdot 1 \cdot \pi

, so

k = 1

❌ Not quite.

\tfrac{4\pi}{3} - \tfrac{\pi}{3} = \pi

, half a turn: the opposite point of the circle.

❌ Not quite.

\tfrac{\pi}{3} - \left(-\tfrac{\pi}{3}\right) = \tfrac{2\pi}{3}

, not a multiple of

2\pi

. The two angles share a cosine but not a sine.

❌ Not quite.

\sin \tfrac{2\pi}{3} = \sin \tfrac{\pi}{3}

, but

\tfrac{2\pi}{3} - \tfrac{\pi}{3} = \tfrac{\pi}{3}

is not

2k\pi

; the points on the circle differ.

❌ Not quite. Subtract and divide by

2\pi

Show solution

Compute $\dfrac{\pi/3 - y}{2\pi}$ for each option and ask whether it is an integer:

$y = \tfrac{4\pi}{3}$ : $\dfrac{-\pi}{2\pi} = -\tfrac{1}{2}$ .
$y = -\tfrac{5\pi}{3}$ : $\dfrac{2\pi}{2\pi} = 1$ , an integer.
$y = -\tfrac{\pi}{3}$ : $\dfrac{2\pi/3}{2\pi} = \tfrac{1}{3}$ .
$y = \tfrac{2\pi}{3}$ : $\dfrac{-\pi/3}{2\pi} = -\tfrac{1}{6}$ .

So $\tfrac{\pi}{3} \sim -\tfrac{5\pi}{3}$ : one full turn apart, the same point of the unit circle.

Problem 2 · Parallel Without the Equality Clause

Given: on the set of lines in the plane, $\ell_1 \mathrel{Q} \ell_2$ if and only if $\ell_1 \parallel \ell_2$ , where parallel lines are distinct lines that never meet. Which properties does $Q$ fail?

✅ Correct! No line is parallel to itself, and

\ell_1 \parallel \ell_2

\ell_2 \parallel \ell_1

would then force

\ell_1 \parallel \ell_1

: a there-and-back chain with no loop.

❌ Close, but… Take

z = x

in the definition of transitive:

\ell_1 \mathrel{Q} \ell_2

and

\ell_2 \mathrel{Q} \ell_1

would require

\ell_1 \mathrel{Q} \ell_1

, which is false.

❌ Not quite. A line is not parallel to itself under this definition, so

Q

is not reflexive.

❌ Not quite.

Q

also fails reflexivity: no line is related to itself.

❌ Not quite. Check each property, including chains that return to their start.

Show solution

Reflexive: fails, since parallel lines are distinct, so $\ell \not\parallel \ell$ for every line $\ell$ .

Symmetric: holds: if $\ell_1 \parallel \ell_2$ , then $\ell_2 \parallel \ell_1$ .

Transitive: fails. Let $\ell_1 \parallel \ell_2$ be two parallel lines. Then $\ell_1 \mathrel{Q} \ell_2$ and $\ell_2 \mathrel{Q} \ell_1$ , but $\ell_1 \mathrel{Q} \ell_1$ is false.

Adding "or $\ell_1 = \ell_2$ " repairs both: it supplies every loop, and with it the relation $P$ is an equivalence relation.

Problem 3 · Counting Related Subsets

Given: $U = \{1, 2, 3, 4\}$ and $A \approx B \iff \operatorname{card}(A) = \operatorname{card}(B)$ on $\mathcal{P}(U)$ . How many subsets $B$ of $U$ satisfy $\{1, 2\} \approx B$ ?

✅ Correct! Every

2

-element subset of

U

qualifies,

\{1, 2\}

itself included: there are

6

❌ Close, but… The relation is reflexive, so

\{1, 2\} \approx \{1, 2\}

counts too.

❌ Not quite.

4

is the number of subsets of

\{1, 2\}

. Here

B

ranges over all subsets of

U

with

2

elements.

❌ Not quite.

16

is the size of

\mathcal{P}(U)

; only subsets with exactly

2

elements are related to

\{1, 2\}

❌ Not quite. List the

2

-element subsets of

U

Show solution

$\{1, 2\} \approx B$ exactly when $\operatorname{card}(B) = 2$ . The $2$ -element subsets of $\{1, 2, 3, 4\}$ are

\{1, 2\},\ \{1, 3\},\ \{1, 4\},\ \{2, 3\},\ \{2, 4\},\ \{3, 4\},

six in all. By size, the $16$ subsets of $U$ split into groups of $1, 4, 6, 4, 1$ .

Problem 4 · Equal Sines

Given: on $\mathbb{R}$ , $x \mathrel{S} y$ if and only if $\sin x = \sin y$ . Which statement is true?

✅ Correct! Equal values of

\sin

inherit all three properties from equality, and

\sin 0 = \sin \pi = 0

❌ Not quite. Being different from the

2k\pi

relation does not stop

S

from being an equivalence relation; check the three properties for

S

itself.

❌ Not quite.

0 \mathrel{S} \pi

, but

\pi - 0 = \pi

is not

2k\pi

. Related angles have equal sines, but equal sines do not make angles related.

❌ Not quite. If

\sin x = \sin y

, then

\sin y = \sin x

❌ Not quite.

S

is defined by equality of one quantity.

Show solution

Reflexive: $\sin x = \sin x$ . Symmetric: $\sin x = \sin y$ gives $\sin y = \sin x$ . Transitive: $\sin x = \sin y$ and $\sin y = \sin z$ give $\sin x = \sin z$ . So $S$ is an equivalence relation.

It is a different relation from $x - y = 2k\pi$ : $\sin 0 = \sin \pi = 0$ , so $0 \mathrel{S} \pi$ , while $\pi - 0 = \pi$ is not of the form $2k\pi$ . Every pair related by the $2k\pi$ relation is related by $S$ , but not conversely.

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