Mathematical-Reasoning · Unit 28 · Video 4 · Interactive Practice

Equal or Disjoint: Why Classes Never Partly Overlap

IKey Formulas

Statement	Used for
$z \in [x] \iff z \sim x$	the membership test, used in both directions
(1) $x \in [x]$	reflexivity alone
(2) $x \sim y \iff [x] = [y]$	symmetry and transitivity, plus part (1) for the converse
(3) $[x] = [y]$ or $[x] \cap [y] = \emptyset$	one shared element $z$ gives $[x] = [z] = [y]$ by part (2)
$\mathbb{Z} = [0] \cup [1] \cup \cdots \cup [n-1]$ , $\ [j] \cap [k] = \emptyset$ for $j \ne k$	the corollary for congruence modulo $n$

Key Insight: If two equivalence classes share even one element, they are the same class.

IIThe Proof, One Move at a Time

Each line of the proof uses the membership test or exactly one property of $\sim$ .

Part Step 1

IIIClasses of $x \sim y \iff x^2 - 4 = y^2 - 4$

Each class is the set of inputs where one horizontal line meets the parabola $f(x) = x^2 - 4$ .

x 2.5 y 1

IVQuiz Questions

Problem 1 · Part (1)

Which property of the equivalence relation $\sim$ proves part (1), $x \in [x]$ for every $x \in A$ ?

✅ Correct!

x \sim x

by reflexivity, and the membership test turns that into

x \in [x]

❌ Not quite. Symmetry needs a statement

x \sim y

to start from; part (1) has none.

❌ Not quite. Transitivity chains two given statements; part (1) needs only

x \sim x

❌ Not quite. Part (1) is a one-line proof that uses a single property.

Show solution

Let $x \in A$ . Since $\sim$ is reflexive, $x \sim x$ . By the definition $[x] = \{ z \in A \mid z \sim x \}$ , this says $x \in [x]$ . Only reflexivity was used.

Problem 2 · Two Statements Ending at $y$

Let $\sim$ be an equivalence relation on $A$ and suppose $w \sim y$ and $x \sim y$ . What can be concluded?

✅ Correct! Symmetry turns

x \sim y

into

y \sim x

; then

w \sim y

and

y \sim x

chain to

w \sim x

❌ Not quite. They do not chain as written, but symmetry flips one of them so that transitivity applies.

❌ Not quite. Equivalent elements need not be equal:

1 \equiv 4 \pmod 3

but

1 \ne 4

❌ Not quite. Use symmetry first.

Show solution

Transitivity needs a statement ending at $y$ followed by one starting at $y$ . By symmetry, $x \sim y$ gives $y \sim x$ . Now $w \sim y$ and $y \sim x$ , so transitivity gives $w \sim x$ . In class language: $w$ and $x$ both lie in $[y]$ , and part (2) of the theorem shows $[w] = [y] = [x]$ .

Problem 3 · A Class of Real Numbers

On $\mathbb{R}$ define $x \sim y$ when $x^2 - 4 = y^2 - 4$ . Find $[-3]$ .

✅ Correct!

z^2 - 4 = 5

exactly when

z^2 = 9

, so

z = \pm 3

❌ Not quite.

3^2 - 4 = 5 = (-3)^2 - 4

, so

3 \sim -3

as well.

❌ Not quite.

5 = f(-3)

is the height of the line, an output; the class holds the inputs

z

with

f(z) = 5

❌ Not quite. Solve

z^2 - 4 = (-3)^2 - 4

Show solution

$z \in [-3]$ exactly when $z^2 - 4 = (-3)^2 - 4 = 5$ , that is, $z^2 = 9$ . The solutions are $z = 3$ and $z = -3$ , so

[-3] = \{-3, 3\} = [3].

Problem 4 · Equal or Disjoint on $\mathbb{R}$

On $\mathbb{R}$ define $x \sim y$ when $x - y \in \mathbb{Z}$ . This is an equivalence relation.

How are $[\tfrac{1}{3}]$ and $[\tfrac{7}{3}]$ related?

How are $[\tfrac{1}{3}]$ and $[\tfrac{1}{2}]$ related?

✅ Correct!

\tfrac{7}{3} - \tfrac{1}{3} = 2 \in \mathbb{Z}

, while

\tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} \notin \mathbb{Z}

; part (3) leaves no third option.

❌ Check the difference.

\tfrac{7}{3} - \tfrac{1}{3} = 2

, an integer, so part (2) applies.

❌ Check the difference.

\tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}

; and "share some but not equal" is exactly what part (3) rules out.

Show solution

Since $\tfrac{7}{3} - \tfrac{1}{3} = 2 \in \mathbb{Z}$ , we have $\tfrac{7}{3} \sim \tfrac{1}{3}$ , so $[\tfrac{1}{3}] = [\tfrac{7}{3}]$ by part (2).

Since $\tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6} \notin \mathbb{Z}$ , $\tfrac{1}{2} \not\sim \tfrac{1}{3}$ , so $[\tfrac{1}{2}] \ne [\tfrac{1}{3}]$ by part (2), and part (3) forces $[\tfrac{1}{3}] \cap [\tfrac{1}{2}] = \emptyset$ . Concretely, $[\tfrac{1}{3}] = \{\tfrac{1}{3} + k \mid k \in \mathbb{Z}\}$ and $[\tfrac{1}{2}] = \{\tfrac{1}{2} + k \mid k \in \mathbb{Z}\}$ share nothing.

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Equal or Disjoint: Why Classes Never Partly Overlap

IKey Formulas

IIThe Proof, One Move at a Time

IIIClasses of x∼y ⟺ x2−4=y2−4x \sim y \iff x^2 - 4 = y^2 - 4x∼y⟺x2−4=y2−4

IVQuiz Questions

IIIClasses of $x \sim y \iff x^2 - 4 = y^2 - 4$