Mathematical-Reasoning · Unit 27 · Video 4 · Interactive Practice

Equivalence Relations: All Three Properties

IKey Formulas

Notion	Statement	How a proof starts
Equivalence relation $\sim$ on $A$	reflexive on $A$ , symmetric and transitive	three separate proofs
Reflexive	$(\forall x \in A)\,(x \sim x)$	take an arbitrary $x \in A$ and show $x \sim x$
Symmetric, transitive	$x \sim y \Rightarrow y \sim x$ ; $\ (x \sim y \wedge y \sim z) \Rightarrow x \sim z$	assume the hypothesis, derive the conclusion
On $\mathbb{Q}$	$x \sim y \iff x - y \in \mathbb{Z}$	an equivalence relation

Key Insight: Write the relation as a statement about integers, use a fact about integers, and translate back: $x - x = 0$ , $\ y - x = -(x - y)$ , $\ x - z = (x - y) + (y - z)$ .

IIRationals That Differ by a Whole Number

Two rationals are related when steps of length one lead from one to the other.

IIIWhich Relations Pass All Three?

On $\{1, \dots, 8\}$ , an equivalence relation fills the diagonal, mirrors across it, and closes every chain.

Relation

IVQuiz Questions

Problem 1 · Less Than or Equal

Given: the relation $\le$ on $\mathbb{Z}$ . Which statement explains why it is not an equivalence relation?

✅ Correct!

\le

is reflexive and transitive, and one pair whose reverse fails is enough to rule out symmetry.

❌ Not quite.

3 \le 3

is true, since

\le

allows equality;

\le

is reflexive.

❌ Not quite.

1 \le 3

is true. In fact

x \le y

and

y \le z

always give

x \le z

❌ Not quite. Reflexivity and transitivity both hold for

\le

; only symmetry fails.

❌ Not quite. Check each property separately.

Show solution

Reflexive: $x \le x$ for every integer $x$ .

Symmetric: fails, with witness $1 \le 2$ and $2 \not\le 1$ .

Transitive: if $x \le y$ and $y \le z$ , then $x \le z$ .

Like "is a multiple of", the relation $\le$ is reflexive and transitive but not symmetric, so it is not an equivalence relation.

Problem 2 · Related to Three Quarters

Given: on $\mathbb{Q}$ , $x \sim y$ if and only if $x - y \in \mathbb{Z}$ . Which number is related to $\tfrac{3}{4}$ ?

✅ Correct!

\tfrac{3}{4} - \left(-\tfrac{5}{4}\right) = \tfrac{8}{4} = 2 \in \mathbb{Z}

❌ Not quite.

\tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}

, not an integer.

❌ Not quite.

\tfrac{3}{2} - \tfrac{3}{4} = \tfrac{3}{4}

, not an integer. Related numbers differ by a whole number, not by a factor.

❌ Not quite.

\tfrac{3}{4} - \left(-\tfrac{3}{4}\right) = \tfrac{3}{2}

, not an integer.

❌ Not quite. Subtract and test for an integer.

Show solution

$\tfrac{3}{4} - \left(-\tfrac{5}{4}\right) = 2 \in \mathbb{Z}$ , so $\tfrac{3}{4} \sim -\tfrac{5}{4}$ .
$\tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}$ ; $\ \tfrac{3}{4} - \tfrac{3}{2} = -\tfrac{3}{4}$ ; $\ \tfrac{3}{4} - \left(-\tfrac{3}{4}\right) = \tfrac{3}{2}$ : none is an integer.

The numbers related to $\tfrac{3}{4}$ are exactly $\tfrac{3}{4} + k$ with $k \in \mathbb{Z}$ : $\dots, -\tfrac{5}{4}, -\tfrac{1}{4}, \tfrac{3}{4}, \tfrac{7}{4}, \tfrac{11}{4}, \dots$

Problem 3 · The Transitive Step

Given: in the proof that $\sim$ on $\mathbb{Q}$ is transitive, $x - y = a$ and $y - z = b$ with $a, b \in \mathbb{Z}$ . Which equation finishes the proof?

✅ Correct!

(x - y) + (y - z) = x - z

, so

x - z = a + b

, and a sum of integers is an integer.

❌ Not quite. Subtracting gives

(x - y) - (y - z) = x - 2y + z

; the

y

terms cancel only when you add.

❌ Not quite. Products appeared in the "multiple of" proof, where

x = yp

and

y = zq

. Here the relation is about differences, so add.

❌ Not quite. Nothing says the two differences are equal; that is why they get different names

a

and

b

❌ Not quite. Combine the two equations so that

y

cancels.

Show solution

Assume $x \sim y$ and $y \sim z$ , so $x - y = a$ and $y - z = b$ for some integers $a$ and $b$ . Adding the equations,

x - z = (x - y) + (y - z) = a + b.

Since the integers are closed under addition, $a + b \in \mathbb{Z}$ , so $x \sim z$ .

Problem 4 · Within Distance One

Given: the relation $R$ on $\mathbb{R}$ with $x \mathrel{R} y$ if and only if $|x - y| \le 1$ . Which statement is true?

✅ Correct!

R

is reflexive and symmetric, but distances of at most

1

can add up to more than

1

❌ Not quite. Try a chain:

0 \mathrel{R} 1

and

1 \mathrel{R} 2

, but

0 \not\mathrel{R} 2

❌ Not quite. The relation uses the absolute value, and

|y - x| = |x - y|

, so

R

is symmetric.

❌ Not quite.

0 \le 1

, so

x \mathrel{R} x

for every

x

R

is reflexive.

❌ Not quite. Test each property on small numbers.

Show solution

Reflexive: $|x - x| = 0 \le 1$ .

Symmetric: $|y - x| = |-(x - y)| = |x - y|$ , so $x \mathrel{R} y$ gives $y \mathrel{R} x$ .

Transitive: fails. $|0 - 1| = 1 \le 1$ and $|1 - 2| = 1 \le 1$ , but $|0 - 2| = 2 \gt 1$ .

So $R$ is not an equivalence relation.

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