This deck defines relations as subsets of a product and covers the reflexive, symmetric, antisymmetric, and transitive properties, then equivalence relations and their classes, and the fundamental theorem that equivalence relations on a set correspond exactly to partitions of it. It targets the classic misconceptions: that symmetry and transitivity together imply reflexivity, that a partition is the same thing as an arbitrary cover, and that classes of related points are distinct.
Subject: Foundations of Higher Mathematics · 107 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. State precisely what a relation is and read off its domain and range.
2. Test a relation for the reflexive, symmetric, antisymmetric, and transitive properties, with genuine non-examples.
3. Prove a given relation is (or is not) an equivalence relation, and describe its equivalence classes.
4. State and prove both directions of the correspondence between equivalence relations and partitions, and form the quotient set.
5. Recognize the kernel of a function and congruence mod n as the archetypal equivalence relations.
Warm-up
Discussion prompt
Before we open Relations, Equivalence Relations & Partitions: without looking back, what was the main idea of Sets, Operations & the Boolean Algebra of Sets, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck treats sets as the ambient language of mathematics. It separates membership from the subset relation and covers set-builder notation, the empty set, power sets, the Boolean-algebra laws for union, intersection, and complement - which are identical to those of propositional logic - and Cartesian products, then shows why naive comprehension collapses into Russell's paradox. It targets the classic confusion between element-of and subset-of, the belief that a universal set exists, and the slip between the empty set and the set containing the empty set.
Concept
A relation from a set A to a set B records which elements of A stand in some connection to which elements of B. Nothing more.
We make that precise by identifying the relation with the set of pairs that are connected.
\[ R \subseteq A \times B \]
relation from A to B — Any subset of the Cartesian product of A and B. We write a R b, or a is related to b, to mean the pair (a, b) belongs to R.
\[ a \mathrel{R} b \;\iff\; (a,b) \in R \]
Counterexample
Discussion prompt
A relation from a set A to a set B records which elements of A stand in some connection to which elements of B. Nothing more.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
We make that precise by identifying the relation with the set of pairs that are connected.
Intuition
Draw the elements of A on the left and B on the right. Draw an arrow from a to b exactly when they are related. The relation is the collection of arrows.
There is no rule, no formula, no meaning required. Any set of arrows at all is a legal relation, including the empty set of arrows and the set of all possible arrows.
This is the CS view: a relation is a lookup table of which pairs return true. A predicate on two arguments.
Analogy
Discussion prompt
Explain A relation is just a set of arrows by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Draw the elements of A on the left and B on the right. Draw an arrow from a to b exactly when they are related. The relation is the collection of arrows.
Concept
Two derived sets pick out the elements that actually participate on each side.
domain — The set of first coordinates that appear in R: every a that is related to at least one b.
\[ \operatorname{dom} R = \{\, a \in A : \exists b \in B,\; (a,b) \in R \,\} \]
range — The set of second coordinates that appear in R: every b that is related to from at least one a.
\[ \operatorname{ran} R = \{\, b \in B : \exists a \in A,\; (a,b) \in R \,\} \]
The domain can be a proper subset of A, and the range a proper subset of B. A relation is not required to touch every element.
Definition probe
Sort into buckets
Every line below is part of the definition of relation from A to B or of domain — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Reading off a concrete relation into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Test each of the 9 candidate pairs.
Worked example
Let A be the set below and B the set below, with R the divides relation from A to B.
\[ A = \{2,3,4\}, \quad B = \{4,6,9\}, \quad a \mathrel{R} b \iff a \mid b \]
List the pairs where a divides b
Why: Test each of the 9 candidate pairs. 2 divides 4 and 6; 3 divides 6 and 9; 4 divides 4.
\[ R = \{(2,4),(2,6),(3,6),(3,9),(4,4)\} \]
Read off the domain
Why: Which first coordinates appear? 2, 3, and 4 all appear, so the domain is all of A.
\[ \operatorname{dom} R = \{2,3,4\} \]
Read off the range
Why: Which second coordinates appear? 4, 6, and 9 all appear, so the range is all of B.
\[ \operatorname{ran} R = \{4,6,9\} \]
Verify the pair count against the product
Why: The product has 9 pairs; we selected exactly the 5 that satisfy the divides test, and each listed pair genuinely divides. The answer is the 5-element set above.
\[ |A \times B| = 9, \quad |R| = 5 \]
Picture it
Animation
Shows: Each line of the worked example "Reading off a concrete relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The product has 9 pairs; we selected exactly the 5 that satisfy the divides test, and each listed pair genuinely divides. The answer is the 5-element set above.
Concept
When the two sides are the same set, we speak of a relation on that set. This is the case that matters for the rest of the deck.
\[ R \subseteq A \times A \]
Now a and b are drawn from the same pool, so it becomes meaningful to ask whether every element relates to itself, whether relatedness is two-way, and whether it chains. Those are the four structural properties we study next.
Explain it
Discussion prompt
Explain A relation on a set to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
When the two sides are the same set, we speak of a relation on that set. This is the case that matters for the rest of the deck.
Concept
The first property asks whether every element is related to itself.
reflexive — A relation on A in which every element is related to itself, with no exceptions across the whole set A.
\[ \forall a \in A,\; (a,a) \in R \]
Example: equality is reflexive. Non-example: the strictly-less-than relation is not reflexive, because no number is strictly less than itself.
\[ a = a \;\text{holds}, \qquad a < a \;\text{fails} \]
Concept
The second property asks whether relatedness is always mutual.
symmetric — A relation in which whenever a is related to b, b is also related to a. The arrows always come in matched pairs.
\[ \forall a,b \in A,\; (a,b) \in R \implies (b,a) \in R \]
Example: has the same birthday as is symmetric. Non-example: less-than-or-equal is not symmetric.
\[ 2 \le 3 \;\text{but}\; 3 \not\le 2 \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of relation from A to B, domain, range, reflexive, symmetric as Relations, Equivalence Relations & Partitions uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
Antisymmetry is the near-opposite of symmetry, and the two are easy to confuse. It says the only way both directions can hold is if the two elements are equal.
antisymmetric — A relation in which a related to b and b related to a can only happen when a and b are the same element.
\[ \forall a,b \in A,\; \big((a,b) \in R \wedge (b,a) \in R\big) \implies a = b \]
Example: less-than-or-equal is antisymmetric. This is the property behind partial orders, and it is not the logical negation of symmetric.
\[ a \le b \wedge b \le a \implies a = b \]
Concept
The fourth property asks whether the relation chains: if you can step from a to b and from b to c, can you step directly from a to c?
transitive — A relation in which any two-step chain can be shortcut to a single step: a related to b and b related to c forces a related to c.
\[ \forall a,b,c \in A,\; \big((a,b) \in R \wedge (b,c) \in R\big) \implies (a,c) \in R \]
Example: less-than is transitive. Non-example: is a parent of is not transitive, since a grandparent is not a parent.
\[ a < b \wedge b < c \implies a < c \]
Intuition
Each property is a shape you can spot in the arrow diagram.
Reflexive: every node has a self-loop. Symmetric: every arrow has a twin pointing back. Transitive: every two-hop path already has its shortcut arrow drawn in.
Antisymmetric: the only two-way pairs allowed are self-loops. These are checkable, mechanical conditions, not vague vibes. To disprove one, you only need a single offending witness.
Step zero
Discussion prompt
Testing all four on the divides relation — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check reflexive
Answer:
Worked example
Take the divides relation on the set below.
\[ A = \{1,2,3,6\}, \qquad a \mathrel{R} b \iff a \mid b \]
Check reflexive
Why: Every integer divides itself, since a equals 1 times a. So every self-pair is present.
\[ \forall a \in A,\; a \mid a \;\checkmark \]
Check symmetric
Why: Look for a two-way failure. 2 divides 6 but 6 does not divide 2. One witness is enough to kill symmetry.
\[ 2 \mid 6 \;\text{but}\; 6 \nmid 2 \implies \text{not symmetric} \]
Check antisymmetric
Why: If a divides b and b divides a with positive integers, then a and b have the same size, so they are equal. No distinct two-way pair exists.
\[ a \mid b \wedge b \mid a \implies a = b \;\checkmark \]
Check transitive
Why: If a divides b and b divides c then b equals a times k and c equals b times m, so c equals a times km. Divisibility chains.
\[ a \mid b \wedge b \mid c \implies a \mid c \;\checkmark \]
Verify the verdict against a concrete witness
Why: Divides on this set is reflexive, antisymmetric, transitive, and NOT symmetric. Sanity check antisymmetry: the pair (2,6) is present but (6,2) is absent, so no distinct back-and-forth pair violates it. This is exactly the profile of a partial order.
\[ \text{reflexive},\ \text{antisymmetric},\ \text{transitive},\ \lnot\,\text{symmetric} \]
Picture it
Animation
Shows: Each line of the worked example "Testing all four on the divides relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every integer divides itself, since a equals 1 times a. So every self-pair is present.
Estimation
Predict first
Consider being within 1 of, on the integers: a and b are related when they differ by at most 1.
Commit before you compute: what does A relation that is symmetric but not transitive come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the explicit witness triple
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The triple (1,2,3) satisfies both hypotheses of transitivity yet violates the conclusion, so R is reflexive and symmetric but NOT transitive.
Worked example
Consider being within 1 of, on the integers: a and b are related when they differ by at most 1.
\[ a \mathrel{R} b \iff |a-b| \le 1 \]
Check reflexive
Why: The distance from a to itself is 0, which is at most 1.
\[ |a-a| = 0 \le 1 \;\checkmark \]
Check symmetric
Why: Absolute difference is unchanged when you swap the arguments.
\[ |a-b| = |b-a| \;\checkmark \]
Check transitive
Why: Chain 1 to 2 to 3: each consecutive pair differs by 1, but the endpoints differ by 2. The shortcut fails.
\[ 1 \mathrel{R} 2 \wedge 2 \mathrel{R} 3 \;\text{but}\; |1-3| = 2 > 1 \]
Verify with the explicit witness triple
Why: The triple (1,2,3) satisfies both hypotheses of transitivity yet violates the conclusion, so R is reflexive and symmetric but NOT transitive. This is precisely why nearness is not an equivalence relation, a fact we will lean on shortly.
\[ (1,2),(2,3) \in R,\quad (1,3) \notin R \]
Picture it
Animation
Shows: Each line of the worked example "A relation that is symmetric but not transitive", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The distance from a to itself is 0, which is at most 1.
Trap
A tempting fake proof: suppose a is related to b. By symmetry b is related to a. By transitivity, from a related to b and b related to a we get a related to a. Therefore reflexive.
\[ a \mathrel{R} b \xRightarrow{\text{sym}} b \mathrel{R} a \xRightarrow{\text{trans}} a \mathrel{R} a \;? \]
The hidden assumption: that a is related to some b in the first place. If a is related to nothing, the argument never starts, and reflexivity at a can simply fail.
\[ R = \varnothing \text{ on } A = \{1\}: \text{ symmetric } \checkmark,\ \text{transitive } \checkmark,\ \text{reflexive } \times \]
The empty relation on a nonempty set is vacuously symmetric and vacuously transitive, but it is not reflexive: element 1 is not related to itself.
\[ R = \varnothing,\quad A = \{1\},\quad (1,1) \notin R \]
Reflexivity is an independent axiom precisely because it is the one property that fails when a point is isolated. That is why an equivalence relation must demand all three separately.
\[ \text{Sym} \wedge \text{Trans} \;\not\Rightarrow\; \text{Refl} \]
Concept
An equivalence relation packages the three good properties that make a relation behave like a generalized notion of sameness.
equivalence relation — A relation on a set that is reflexive, symmetric, and transitive, all three at once. Often written with a tilde.
\[ \text{reflexive} \;\wedge\; \text{symmetric} \;\wedge\; \text{transitive} \]
Notice antisymmetry is deliberately absent. Antisymmetry pushes toward orderings; symmetry pushes toward sameness. A relation that is both symmetric and antisymmetric can only relate equal elements.
\[ a \sim b \;\text{read as}\; a \text{ is equivalent to } b \]
Intuition
Every equivalence relation is secretly the phrase agrees with, after you forget some detail. Same last digit. Same remainder. Same color. Same connected component.
Reflexive: a thing agrees with itself. Symmetric: agreement is mutual. Transitive: agreement chains. These are exactly the laws equality obeys, minus the demand that agreeing things be literally identical.
So an equivalence relation is equality after deliberately blurring your vision. That blurring is the single most important construction in this course: it is how quotients are born.
Missing information
Discussion prompt
On the integers, relate two numbers when 5 divides their difference, equivalently when they leave the same remainder on division by 5.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The difference of a with itself is 0, and 5 divides 0.
Worked example
On the integers, relate two numbers when 5 divides their difference, equivalently when they leave the same remainder on division by 5.
\[ a \sim b \iff 5 \mid (a - b) \]
Reflexive
Why: The difference of a with itself is 0, and 5 divides 0.
\[ 5 \mid (a - a) = 0 \;\checkmark \]
Symmetric
Why: If 5 divides a minus b, it divides the negative b minus a as well.
\[ 5 \mid (a-b) \implies 5 \mid -(a-b) = (b-a) \;\checkmark \]
Transitive
Why: If 5 divides a minus b and b minus c, it divides their sum, which telescopes to a minus c.
\[ 5 \mid (a-b) \wedge 5 \mid (b-c) \implies 5 \mid (a-c) \;\checkmark \]
Verify all three hold, so it is an equivalence relation
Why: Reflexive, symmetric, and transitive are each confirmed by the divisibility algebra above. Concretely 7 and 12 are related since their difference 5 is divisible by 5, and indeed both leave remainder 2. The relation is an equivalence relation.
\[ 7 \sim 12 \text{ since } 5 \mid 5,\quad 7 = 5\cdot 1 + 2,\ 12 = 5\cdot 2 + 2 \]
Picture it
Animation
Shows: Each line of the worked example "Same remainder mod 5 is an equivalence relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Reflexive, symmetric, and transitive are each confirmed by the divisibility algebra above. Concretely 7 and 12 are related since their difference 5 is divisible by 5, and indeed both leave remainder 2. The relation is an equivalence relation.
Concept
The same construction works for any fixed positive modulus n, not just 5. This relation is the backbone of number theory and of modular arithmetic in code.
congruent modulo n — Two integers are congruent mod n when n divides their difference. This is an equivalence relation on the integers for every fixed n at least 1.
\[ a \equiv b \pmod{n} \iff n \mid (a-b) \]
The proof is identical to the mod 5 case with 5 replaced by n. What makes it special, beyond being an equivalence relation, is that it also respects addition and multiplication, which is why arithmetic survives the quotient. We return to that at the end.
\[ a \equiv b,\ c \equiv d \pmod n \implies a+c \equiv b+d \pmod n \]
Concept
Once you have an equivalence relation, each element drags along everything equivalent to it. That bundle is the central object.
equivalence class of a — The set of all elements equivalent to a, written with square brackets. It is a subset of the underlying set, gathering everything that agrees with a.
\[ [a] = \{\, x \in X : x \sim a \,\} \]
The element a is called a representative of its class. By reflexivity a always lies in its own class, so no class is ever empty.
\[ a \in [a] \quad\text{since}\quad a \sim a \]
Intuition
Picture sorting every element into buckets so that two things land in the same bucket exactly when they are equivalent. Each bucket is an equivalence class.
Any element in a bucket can serve as the name on the bucket. The class does not care which name you pick; 7 and 12 name the very same mod-5 bucket. This is why well-definedness will matter so much later.
The punchline we prove next: these buckets never partly overlap. Two buckets are either identical or completely disjoint. There is no in-between.
Fill the middle
Fill in the blanks
From The classes of same remainder mod 5 — finish the line. Write what belongs on the right of the equals sign before you look.
[0] = \{\dots,-10,-5,0,5,10,\dots\}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. These are the integers with remainder 0, the multiples of 5.
Worked example
Continue with congruence mod 5 on the integers. Group the integers by remainder.
\[ a \sim b \iff 5 \mid (a-b) \]
Collect everything equivalent to 0
Why: These are the integers with remainder 0, the multiples of 5.
\[ [0] = \{\dots,-10,-5,0,5,10,\dots\} \]
Collect the classes of 1, 2, 3, 4
Why: Each remainder gives one class; shifting a class by 1 slides to the next remainder.
\[ [1],\,[2],\,[3],\,[4] \text{ are the remainder-1,2,3,4 integers} \]
Note that 5 different starting points would repeat
Why: The class of 5 equals the class of 0, since 5 and 0 differ by a multiple of 5. There are exactly 5 distinct classes.
\[ [5] = [0],\quad [6] = [1],\ \dots \]
Verify the classes cover the integers without overlap
Why: Every integer has exactly one remainder among 0,1,2,3,4, so it lands in exactly one class, and the five classes together exhaust the integers. This set of 5 classes is the quotient, written as the integers mod 5.
\[ \mathbb{Z} = [0] \sqcup [1] \sqcup [2] \sqcup [3] \sqcup [4] \]
Picture it
Animation
Shows: Each line of the worked example "The classes of same remainder mod 5", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every integer has exactly one remainder among 0,1,2,3,4, so it lands in exactly one class, and the five classes together exhaust the integers. This set of 5 classes is the quotient, written as the integers mod 5.
Concept
Everything about the fundamental theorem rests on one clean equivalence: being related is the same as having the same class.
\[ a \sim b \;\iff\; [a] = [b] \]
This is the formal version of the buckets-do-not-partly-overlap slogan. Its contrapositive is equally useful: if a and b are not related, their classes are disjoint sets.
\[ a \not\sim b \;\iff\; [a] \cap [b] = \varnothing \]
Step zero
Discussion prompt
Proving a related to b iff their classes are equal — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Forward, assume a related to b, show the classes are equal
Answer:
Worked example
Assume a tilde is an equivalence relation on X. Prove the biconditional both ways.
\[ a \sim b \iff [a] = [b] \]
Forward, assume a related to b, show the classes are equal
Why: Take any x in the class of a, so x is related to a. Since a is related to b, transitivity gives x related to b, hence x lies in the class of b. This shows one inclusion.
\[ x \sim a \wedge a \sim b \implies x \sim b \]
Get the reverse inclusion for free by symmetry
Why: From a related to b, symmetry gives b related to a, so the identical argument with roles swapped shows the class of b sits inside the class of a. Two inclusions give equality.
\[ [a] \subseteq [b] \ \text{and}\ [b] \subseteq [a] \implies [a] = [b] \]
Backward, assume the classes are equal, show a related to b
Why: By reflexivity a lies in the class of a. If that class equals the class of b, then a lies in the class of b, which means exactly that a is related to b.
\[ a \in [a] = [b] \implies a \sim b \]
Verify with the mod 5 instance
Why: Both directions are established, so the biconditional holds. Check it: 7 and 12 are related mod 5, and indeed the class of 7 and the class of 12 are the same remainder-2 bucket. The lemma checks out on a concrete case.
\[ 7 \sim 12 \iff [7] = [12] = [2] \]
Picture it
Animation
Shows: Each line of the worked example "Proving a related to b iff their classes are equal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both directions are established, so the biconditional holds. Check it: 7 and 12 are related mod 5, and indeed the class of 7 and the class of 12 are the same remainder-2 bucket. The lemma checks out on a concrete case.
Trap
A common slip when counting classes: writing the class of 2 and the class of 7 as two separate mod-5 classes because 2 and 7 are different numbers.
\[ [2] \ne [7] \;?\quad \text{because } 2 \ne 7 \]
This double-counts. The number of classes gets inflated, and any function defined on classes risks being declared ill-defined when it is actually fine.
\[ \text{miscount: } 5 \text{ classes reported as more} \]
Distinct representatives can name the same class. Since 2 and 7 differ by 5, they are congruent, so by the key lemma their classes are literally the same set.
\[ 2 \sim 7 \pmod 5 \implies [2] = [7] \]
Always test equality of classes by the relation, never by the labels. A class has many names and only one identity.
\[ [a] = [b] \iff a \sim b, \ \text{not} \ a = b \]
Concept
We now define the geometric object that classes secretly are: a way of slicing a set into disjoint, non-empty, exhaustive pieces.
partition — A collection of non-empty subsets of X, called blocks, that are pairwise disjoint and whose union is all of X. Every element belongs to exactly one block.
\[ \bigcup_{P \in \mathcal{P}} P = X, \qquad P \ne Q \implies P \cap Q = \varnothing, \qquad P \ne \varnothing \]
The three clauses in order: cover, disjoint, non-empty. Drop any one and the structure breaks. Together they force each element into precisely one block.
Picture it
Figure (svg): A rectangle representing a set of nine dots divided into three non-overlapping boxes: a left box with four dots, a middle box with three dots, and a right box with two dots, illustrating a partition into three blocks.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Here a 9-element set is carved into three blocks. Every dot sits in exactly one box, the boxes never overlap, and no box is empty.
Concept
Here a 9-element set is carved into three blocks. Every dot sits in exactly one box, the boxes never overlap, and no box is empty.
Figure (svg): A rectangle representing a set of nine dots divided into three non-overlapping boxes: a left box with four dots, a middle box with three dots, and a right box with two dots, illustrating a partition into three blocks.
This picture is the whole content of the fundamental theorem. An equivalence relation draws exactly these boxes, and any such boxing defines an equivalence relation. The two ideas are the same idea seen from two sides.
Trap
Claiming the two sets below partition the numbers 1 through 6, because together they contain everything.
\[ \{1,2,3,4\} \ \text{and}\ \{3,4,5,6\} \]
They do cover the set, but 3 and 4 sit in both blocks. This is a cover, not a partition, and it does not come from any equivalence relation.
\[ \{1,2,3,4\} \cap \{3,4,5,6\} = \{3,4\} \ne \varnothing \]
A partition demands pairwise-disjoint blocks. Repair the overlap by giving each element a single home.
\[ \{1,2\},\ \{3,4\},\ \{5,6\} \]
Now every element is in exactly one block. Overlapping blocks would force an element to be equivalent to two mutually non-equivalent things, breaking transitivity. Disjointness is not decoration; it is what makes the correspondence work.
\[ \text{cover} + \text{disjoint} + \text{non-empty} = \text{partition} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
A partition demands pairwise-disjoint blocks. Repair the overlap by giving each element a single home.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
Here is the central result the whole deck has been building toward.
Theorem. For a fixed set X, equivalence relations on X correspond bijectively to partitions of X. Given an equivalence relation, its classes form a partition; given a partition, related-means-same-block is an equivalence relation; and these two constructions are mutually inverse.
\[ \{\text{equivalence relations on } X\} \;\longleftrightarrow\; \{\text{partitions of } X\} \]
We prove each direction as a worked example, then confirm they invert each other. This is the same structural pattern you will meet again as the First Isomorphism Theorem: a structure-preserving map plus a quotient.
Estimation
Predict first
Let a tilde be an equivalence relation on X. Show the set of equivalence classes is a partition: non-empty, covering, and pairwise disjoint or equal.
Commit before you compute: what does Direction 1: equivalence relation gives a partition come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the three partition clauses are met
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Non-empty, covering, and pairwise-disjoint-or-equal are all established, which is exactly the definition of a partition.
Worked example
Let a tilde be an equivalence relation on X. Show the set of equivalence classes is a partition: non-empty, covering, and pairwise disjoint or equal.
\[ \mathcal{P} = \{\, [a] : a \in X \,\} \]
Each class is non-empty
Why: By reflexivity a is related to itself, so a lies in the class of a. No class is empty.
\[ a \in [a] \implies [a] \ne \varnothing \]
The classes cover X
Why: Every element a belongs to its own class, and that class is one of the blocks, so the union of all classes is all of X.
\[ \forall a \in X,\; a \in [a] \implies \bigcup_{a} [a] = X \]
Two classes are either identical or disjoint
Why: Suppose the classes of a and b share an element x. Then x is related to a and to b, so by symmetry and transitivity a is related to b, and the key lemma forces the classes to be equal. So no partial overlap is possible.
\[ [a] \cap [b] \ne \varnothing \implies a \sim b \implies [a] = [b] \]
Verify the three partition clauses are met
Why: Non-empty, covering, and pairwise-disjoint-or-equal are all established, which is exactly the definition of a partition. Sanity check on mod 5: the five classes are non-empty, union to the integers, and no two of them share an element. Direction 1 holds.
\[ \mathcal{P} \text{ is a partition of } X \]
Picture it
Animation
Shows: Each line of the worked example "Direction 1: equivalence relation gives a partition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Non-empty, covering, and pairwise-disjoint-or-equal are all established, which is exactly the definition of a partition. Sanity check on mod 5: the five classes are non-empty, union to the integers, and no two of them share an element. Direction 1 holds.
Missing information
Discussion prompt
Let the collection of blocks below be a partition of X. Define a relation by same block and prove it is an equivalence relation.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Because the blocks cover X, every a lies in some block P, and then a shares block P with itself.
Worked example
Let the collection of blocks below be a partition of X. Define a relation by same block and prove it is an equivalence relation.
\[ a \sim b \iff \exists P \in \mathcal{P},\; a \in P \wedge b \in P \]
Reflexive
Why: Because the blocks cover X, every a lies in some block P, and then a shares block P with itself.
\[ a \in P \implies a \sim a \]
Symmetric
Why: Sharing a block is stated symmetrically in a and b, so swapping them changes nothing.
\[ a,b \in P \implies b,a \in P \]
Transitive
Why: If a and b share block P, and b and c share block Q, then b lies in both P and Q. Disjointness forces P equal to Q, so a and c share that one block.
\[ b \in P \cap Q \implies P = Q \implies a,c \in P \]
Verify all three hold and identify the classes
Why: Reflexive, symmetric, and transitive all follow from cover plus disjointness, so same block is an equivalence relation. Moreover its classes are exactly the original blocks. Transitivity is precisely where disjointness of the partition was needed, which is why an overlapping cover fails. Direction 2 holds.
\[ [a] = P \text{ where } a \in P \]
Picture it
Animation
Shows: Each line of the worked example "Direction 2: a partition gives an equivalence relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Reflexive, symmetric, and transitive all follow from cover plus disjointness, so same block is an equivalence relation. Moreover its classes are exactly the original blocks. Transitivity is precisely where disjointness of the partition was needed, which is why an overlapping cover fails. Direction 2 holds.
Concept
The final piece: the two directions undo each other, so the correspondence is a genuine bijection, not just a pair of maps.
Start with an equivalence relation, take its classes, then relate points that share a class: you recover the original relation, by the key lemma. Start with a partition, form same-block, then take its classes: you recover the original blocks, as shown in Direction 2.
\[ \sim \;\longmapsto\; \mathcal{P}_\sim \;\longmapsto\; \sim_{\mathcal{P}_\sim} \;=\; \sim \]
So counting equivalence relations on a finite set is the same as counting partitions of it: the Bell numbers. On a 3-element set there are exactly 5.
\[ |\{\text{equiv. rel. on a 3-set}\}| = B_3 = 5 \]
Concept
Once the classes form a partition, we can treat each class as a single new point. The set of these points is the quotient.
quotient set — The set whose elements are the equivalence classes of X under a tilde. It is X with equivalent points glued together, written X mod tilde.
\[ X/{\sim} \;=\; \{\, [x] : x \in X \,\} \]
canonical projection — The map sending each element to its own equivalence class. It is always surjective, and it forgets exactly the information the relation blurred.
\[ \pi : X \to X/{\sim}, \qquad \pi(x) = [x] \]
This is the universal move of the whole course. Number systems, modular arithmetic, quotient groups and rings, and function spaces up to almost-everywhere are all built by choosing an equivalence relation and passing to the quotient.
Concept
Equivalence relations are not exotic. Every function generates one automatically, by relating inputs that share an output.
kernel of f — For a function f from X to Y, the relation on X that relates two inputs exactly when f sends them to the same output.
\[ a \sim_f b \iff f(a) = f(b) \]
This is reflexive, symmetric, and transitive because equality of outputs is. Its classes are the fibers of f, the preimages of single points, and the quotient is a perfect copy of the image of f. Every equivalence relation arises this way from its own projection map.
\[ [a]_{\sim_f} = f^{-1}(\{f(a)\}), \qquad X/{\sim_f} \;\cong\; \operatorname{im} f \]
Ranking
Put in order
Put the moves of The kernel of the remainder map recovers mod n into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Two integers are kernel-related exactly when they have the same remainder, which is the same-remainder-mod-5 relation from earlier.
Worked example
Let f send each integer to its remainder on division by 5, a function into the five possible remainders.
\[ f : \mathbb{Z} \to \{0,1,2,3,4\}, \qquad f(a) = a \bmod 5 \]
Write down the kernel relation
Why: Two integers are kernel-related exactly when they have the same remainder, which is the same-remainder-mod-5 relation from earlier.
\[ a \sim_f b \iff (a \bmod 5) = (b \bmod 5) \]
Identify the fibers
Why: The preimage of remainder r is exactly the class of numbers leaving remainder r, one of the five mod-5 classes.
\[ f^{-1}(\{2\}) = [2] = \{\dots,-3,2,7,12,\dots\} \]
Match the quotient to the image
Why: There are 5 fibers and 5 remainders, and the projection followed by the induced map is a bijection between them.
\[ \mathbb{Z}/{\sim_f} \;\cong\; \{0,1,2,3,4\} \]
Verify against the direct definition of mod 5
Why: Same remainder mod 5 and n divides the difference define the identical relation, since two numbers share a remainder exactly when their difference is a multiple of 5. The kernel construction reproduces congruence mod 5 exactly, confirming every equivalence relation is a kernel.
\[ (a \bmod 5) = (b \bmod 5) \iff 5 \mid (a-b) \]
Picture it
Animation
Shows: Each line of the worked example "The kernel of the remainder map recovers mod n", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Same remainder mod 5 and n divides the difference define the identical relation, since two numbers share a remainder exactly when their difference is a multiple of 5. The kernel construction reproduces congruence mod 5 exactly, confirming every equivalence relation is a kernel.
Step zero
Discussion prompt
Differ by an integer on the reals gives the circle — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Confirm it is an equivalence relation
Answer:
Worked example
On the real line, relate two reals when their difference is a whole number.
\[ x \sim y \iff x - y \in \mathbb{Z} \]
Confirm it is an equivalence relation
Why: The difference of x with itself is 0, an integer; if x minus y is an integer so is its negative; and a sum of integers is an integer. Reflexive, symmetric, transitive.
\[ 0 \in \mathbb{Z},\ \ -(x-y)\in\mathbb{Z},\ \ (x-y)+(y-z)\in\mathbb{Z} \]
Describe a single class
Why: The class of x is all reals reachable by adding integers to x, a copy of the integers shifted by x.
\[ [x] = \{\, x + n : n \in \mathbb{Z} \,\} \]
Pick a canonical representative in the unit interval
Why: Every class contains exactly one real in the half-open unit interval, namely the fractional part. So classes correspond to points of that interval with the ends identified.
\[ x \sim \{x\} \in [0,1), \qquad 0 \sim 1 \]
Verify the quotient is a circle
Why: Gluing 0 to 1 in the unit interval bends it into a loop, so the quotient is the circle. Check a class: the reals 0.3, 1.3, and negative 0.7 all differ by integers and share the representative 0.3, one point of the circle. The reals mod the integers is the circle.
\[ \mathbb{R}/\mathbb{Z} \;\cong\; S^1 \]
Picture it
Animation
Shows: Each line of the worked example "Differ by an integer on the reals gives the circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Gluing 0 to 1 in the unit interval bends it into a loop, so the quotient is the circle. Check a class: the reals 0.3, 1.3, and negative 0.7 all differ by integers and share the representative 0.3, one point of the circle. The reals mod the integers is the circle.
Trap
Define a relation on people by shares at least one hobby with. It feels like a sameness relation, so it feels like an equivalence relation.
\[ a \sim b \iff \text{hobbies}(a) \cap \text{hobbies}(b) \ne \varnothing \]
But sharing is not transitive. Ana shares chess with Bo, Bo shares tennis with Cy, yet Ana and Cy may share nothing. The two-step chain has no shortcut.
\[ a \sim b \wedge b \sim c \;\not\Rightarrow\; a \sim c \]
Reflexive and symmetric are not enough; you must actually verify transitivity with a would-be counterexample in mind.
\[ \{\text{chess}\},\ \{\text{chess},\text{tennis}\},\ \{\text{tennis}\} \]
The middle person bridges two others who need not be related, so the relation only connects, it does not classify. Compare same favorite hobby, a kernel relation, which is genuinely transitive because equality is.
\[ \text{overlap: not transitive} \quad\text{vs}\quad \text{equality of value: transitive} \]
Intuition
When A equals B, the arrow picture is literally a directed graph: the elements are vertices and each related pair is a directed edge.
In this language the three equivalence properties become graph slogans. Reflexive: every vertex has a self-loop. Symmetric: every edge is bidirectional. Transitive: every directed path of length two already has its shortcut edge.
So an equivalence relation is a graph that splits into disjoint fully-connected clumps, each clump with all self-loops and all internal edges present. Those clumps are exactly the equivalence classes, and the graph view makes the partition visible at a glance.
Concept
Every relation can be reversed by flipping every arrow. This gives a compact way to restate symmetry.
inverse relation — The relation obtained by swapping the two coordinates of every pair in R. It reverses the direction of every arrow.
\[ R^{-1} = \{\, (b,a) : (a,b) \in R \,\} \]
A relation is symmetric precisely when it equals its own inverse. So symmetry is the statement that reversing the arrows changes nothing.
\[ R \text{ symmetric} \iff R = R^{-1} \]
Concept
Relations chain the same way functions do: follow one arrow, then another. The composite records where two-step journeys can end up.
composition — For relations R then S, the composite relates a to c whenever some intermediate b has a related to b by R and b related to c by S.
\[ S \circ R = \{\, (a,c) : \exists b,\; (a,b)\in R \wedge (b,c)\in S \,\} \]
Transitivity is now a single containment: a relation is transitive exactly when composing it with itself produces nothing new. This is the clean, structural reading of the transitive axiom.
\[ R \text{ transitive} \iff R \circ R \subseteq R \]
Estimation
Predict first
Let R on the three-element set below be given by this exact list of pairs. Decide whether it is an equivalence relation and find the classes.
Commit before you compute: what does Checking an explicitly listed relation come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify and read off the classes
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. All three properties hold, so R is an equivalence relation.
Worked example
Let R on the three-element set below be given by this exact list of pairs. Decide whether it is an equivalence relation and find the classes.
\[ A=\{1,2,3\},\ \ R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\} \]
Check reflexive
Why: All three self-pairs (1,1), (2,2), and (3,3) are in the list, so every element is related to itself.
\[ (1,1),(2,2),(3,3) \in R \;\checkmark \]
Check symmetric
Why: The only non-loop pairs are (1,2) and (2,1), and each has its reverse present, so no one-way arrow exists.
\[ (1,2)\in R \wedge (2,1)\in R \;\checkmark \]
Check transitive
Why: The only composable two-step chains use the pair 1 to 2 and 2 to 1, giving 1 to 1 and 2 to 2, both already present. No chain forces a missing pair.
\[ (1,2),(2,1)\Rightarrow(1,1)\in R;\ \ (2,1),(1,2)\Rightarrow(2,2)\in R \;\checkmark \]
Verify and read off the classes
Why: All three properties hold, so R is an equivalence relation. Since 1 and 2 are related but neither is related to 3, the classes are the pair one-two and the singleton three. Cross-check: two blocks, non-empty, disjoint, covering all of A.
\[ [1]=[2]=\{1,2\}, \quad [3]=\{3\} \]
Picture it
Animation
Shows: Each line of the worked example "Checking an explicitly listed relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: All three self-pairs (1,1), (2,2), and (3,3) are in the list, so every element is related to itself.
Concept
If a relation lacks a property, we can add exactly the missing pairs to force it, and no more. The result is called a closure.
The reflexive closure adds all self-loops; the symmetric closure adds every reversed arrow; the transitive closure adds every shortcut, repeatedly, until chains have nowhere new to reach.
\[ R \cup \Delta, \qquad R \cup R^{-1}, \qquad R \cup R^2 \cup R^3 \cup \cdots \]
Doing all three and iterating yields the smallest equivalence relation containing R, the equivalence closure. In graph terms it is reachability: two vertices end up related exactly when a path connects them. This is Warshall's algorithm in the CS world.
Pattern
1. To prove an equivalence relation, check exactly three things
Why: Reflexive: every a related to itself. Symmetric: a related to b forces b related to a. Transitive: a to b and b to c forces a to c. Check all three; none implies another.
2. To disprove one property, exhibit a single explicit witness
Why: One self-pair that is missing kills reflexive; one one-way pair kills symmetric; one bad triple kills transitive. A universal claim dies to one counterexample.
3. To find the classes, group by the relation and pick representatives
Why: Collect everything related to a fixed a; that set is the class of a. Equal classes may wear different labels, so test with the relation, not the names.
4. Cross-check with the partition view
Why: The classes must be non-empty, cover the set, and be pairwise disjoint. If they overlap without being equal, you made an error, because equivalence classes never partially overlap.
5. To build a quotient, project each point to its class
Why: The quotient set X mod tilde has the classes as its points, and the canonical projection sends x to its class. Any equivalence relation is the kernel of that projection.
Real world
Discussion prompt
Outside this lesson: where does Relations, Equivalence Relations & Partitions actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of The recipe: verify, classify, quotient is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck defines relations as subsets of a product and covers the reflexive, symmetric, antisymmetric, and transitive properties, then equivalence relations and their classes, and the fundamental theorem that equivalence relations on a set correspond exactly to partitions of it. It targets the classic misconceptions: that symmetry and transitivity together imply reflexivity, that a partition is the same thing as an arbitrary cover, and that classes of related points are distinct.
Concept
Among all equivalence relations on a set, two sit at the boundaries and are worth knowing by name.
The finest relates nothing but equal elements; every class is a singleton and the quotient is a faithful copy of X.
\[ \Delta = \{\,(x,x) : x \in X\,\}, \qquad [x] = \{x\} \]
The coarsest relates everything to everything; there is a single class and the quotient is one point.
\[ X \times X, \qquad [x] = X \ \text{for all } x \]
Every other equivalence relation lands strictly between these two. More related pairs means fewer, bigger classes: a coarser partition.
Intuition
Choosing an equivalence relation is choosing what to ignore. Relate integers by their last digit and you have decided to forget everything but the last digit.
The quotient is the world that remains after that deliberate forgetting. The finest relation forgets nothing; the coarsest forgets everything. Interesting mathematics lives in the choices between.
This is why quotients are the engine of abstraction: modular arithmetic forgets multiples of n, projective geometry forgets scale, and homology forgets boundaries. Same move every time.
Explain it
Discussion prompt
Explain A quotient is controlled forgetting to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Choosing an equivalence relation is choosing what to ignore. Relate integers by their last digit and you have decided to forget everything but the last digit.
Concept
Reflexive and transitive are shared by two great families of relations. The fork is the third axiom.
Add symmetry and you get an equivalence relation, a notion of sameness whose job is to classify. Add antisymmetry instead and you get a partial order, a notion of ranking whose job is to compare.
\[ \text{R} + \text{S} + \text{T} = \text{equivalence}, \qquad \text{R} + \text{antisym} + \text{T} = \text{partial order} \]
A relation that is both symmetric and antisymmetric can only relate an element to itself, which is exactly the finest equivalence relation, the identity. Sameness and ranking meet only at equality.
Analogy
Discussion prompt
Explain Equivalence versus partial order by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Reflexive and transitive are shared by two great families of relations. The fork is the third axiom.
Hypothesis
Predict first
Same parity is an equivalence relation is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Recognize it as a kernel relation
Why: Same parity means the parity function assigns them equal values, so this is the kernel of the mod 2 map and is automatically an equivalence relation.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
On the integers, relate two numbers when they have the same parity, both even or both odd.
\[ a \sim b \iff 2 \mid (a - b) \]
Recognize it as a kernel relation
Why: Same parity means the parity function assigns them equal values, so this is the kernel of the mod 2 map and is automatically an equivalence relation.
\[ a \sim b \iff (a \bmod 2) = (b \bmod 2) \]
List the classes
Why: There are exactly two remainders mod 2, so exactly two classes: the evens and the odds.
\[ [0] = 2\mathbb{Z}, \qquad [1] = 2\mathbb{Z} + 1 \]
Verify the classes partition the integers
Why: Every integer is even or odd but not both, so the two classes are non-empty, disjoint, and cover the integers. Check: 4 and negative 10 are both even, hence related, and both sit in the class of 0. Same parity is an equivalence relation with two classes.
\[ \mathbb{Z} = [0] \sqcup [1] \]
Picture it
Animation
Shows: Each line of the worked example "Same parity is an equivalence relation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every integer is even or odd but not both, so the two classes are non-empty, disjoint, and cover the integers. Check: 4 and negative 10 are both even, hence related, and both sit in the class of 0. Same parity is an equivalence relation with two classes.
Step zero
Discussion prompt
Enumerating all equivalence relations on a three-element set — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Count the all-separate and all-together partitions
Answer:
Worked example
By the fundamental theorem, listing equivalence relations on the set below is the same as listing its partitions.
\[ X = \{1,2,3\} \]
Count the all-separate and all-together partitions
Why: Three singletons give the finest relation; one block of all three gives the coarsest. That is 2 so far.
\[ \{\{1\},\{2\},\{3\}\}, \qquad \{\{1,2,3\}\} \]
Count the one-pair-together partitions
Why: Choose which single pair shares a block, leaving the third alone. There are three such choices.
\[ \{\{1,2\},\{3\}\},\ \{\{1,3\},\{2\}\},\ \{\{2,3\},\{1\}\} \]
Verify the total is the Bell number
Why: Two extremes plus three middle partitions gives five, matching the third Bell number. There is no partition with a block of size 2 and another of size 2, since only three elements are available, so the list is complete.
\[ 2 + 3 = 5 = B_3 \]
Picture it
Animation
Shows: Each line of the worked example "Enumerating all equivalence relations on a three-element set", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two extremes plus three middle partitions gives five, matching the third Bell number. There is no partition with a block of size 2 and another of size 2, since only three elements are available, so the list is complete.
Concept
To define a function on classes by naming a representative, you must check the answer does not depend on which representative you picked. This is the well-definedness obligation.
\[ \bar{f}([a]) = g(a) \ \text{is well defined} \iff \big(a \sim b \implies g(a) = g(b)\big) \]
In kernel language: a rule on classes is well defined exactly when it is constant on each class. Skip this check and you may write down a formula that secretly gives two different outputs for the same class.
Counterexample
Discussion prompt
To define a function on classes by naming a representative, you must check the answer does not depend on which representative you picked. This is the well-definedness obligation.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Ranking
Put in order
Put the moves of Addition is well-defined on the integers mod n into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Suppose we picked different representatives of the same two classes.
Worked example
Define a sum on mod-n classes using representatives, then prove the choice of representative does not matter.
\[ [a] + [b] := [a + b] \]
Set up the well-definedness test
Why: Suppose we picked different representatives of the same two classes. We must show the resulting sum-class is unchanged.
\[ a \equiv a', \quad b \equiv b' \pmod n \;\Rightarrow\; a + b \equiv a' + b'? \]
Use the divisibility hypotheses
Why: By assumption n divides a minus a prime and b minus b prime; adding these, n divides the combined difference.
\[ n \mid (a - a') \ \wedge\ n \mid (b - b') \implies n \mid (a+b) - (a'+b') \]
Conclude the sum-classes agree
Why: Since n divides the difference of the two sums, the sums are congruent, so they name the same class. The operation ignores representative choice.
\[ [a+b] = [a'+b'] \]
Verify with a concrete representative swap
Why: Mod 5, use 2 plus 3 to get class 0, then swap to representatives 7 and 8: their sum 15 is again in class 0. The answer survived the swap, confirming addition descends to the quotient. This compatibility is what makes congruence a congruence, not merely an equivalence.
\[ [2]+[3]=[5]=[0], \quad [7]+[8]=[15]=[0] \]
Picture it
Animation
Shows: Each line of the worked example "Addition is well-defined on the integers mod n", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Mod 5, use 2 plus 3 to get class 0, then swap to representatives 7 and 8: their sum 15 is again in class 0. The answer survived the swap, confirming addition descends to the quotient. This compatibility is what makes congruence a congruence, not merely an equivalence.
Trap
On the integers mod 5, try to define a map by sending a class to the class of its representative squared, but pick the rule using the actual representative rather than the class.
\[ h([a]) := a \ \ (\text{the plain integer, not its class}) \]
This is not a function on classes at all: 2 and 7 name the same class, yet the rule returns 2 for one representative and 7 for the other.
\[ [2] = [7] \ \text{but}\ h([2]) = 2 \ne 7 = h([7]) \]
A legal map out of a quotient must give equal outputs for equal classes. Squaring into a class does respect the relation and is well defined.
\[ s([a]) := [a^2], \qquad a \equiv b \implies a^2 \equiv b^2 \pmod 5 \]
Check: the class of 2 squared is the class of 4, and the class of 7 squared is the class of 49, which is also the class of 4. Equal classes, equal output. Always verify the rule is constant on each class before calling it a function.
\[ [2^2] = [4], \quad [7^2] = [49] = [4] \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Elimination
Eliminate the wrong options
Which list correctly states which of reflexive, symmetric, transitive the empty relation satisfies?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Symmetry and transitivity are vacuously true because their hypotheses about existing pairs are never triggered. Reflexivity fails because some element is not related to itself. This is the exact counterexample to symmetric-plus-transitive implying reflexive.
Check
Consider the empty relation on a non-empty set: no element is related to anything, including itself.
\[ R = \varnothing \ \text{on}\ A \ne \varnothing \]
Check your understanding
Which list correctly states which of reflexive, symmetric, transitive the empty relation satisfies?
Answer: A
Why: Symmetry and transitivity are vacuously true because their hypotheses about existing pairs are never triggered. Reflexivity fails because some element is not related to itself. This is the exact counterexample to symmetric-plus-transitive implying reflexive.
Prediction
Predict first
Which pair of integers lies in the SAME equivalence class?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 23 and 2
Why: Two integers share a class exactly when 7 divides their difference. Here 23 minus 2 is 21, which is 7 times 3, so 23 and 2 are congruent mod 7 and share the class of 2.
Check
Work in the integers under congruence mod 7.
\[ a \sim b \iff 7 \mid (a-b) \]
Check your understanding
Which pair of integers lies in the SAME equivalence class?
Answer: A
Why: Two integers share a class exactly when 7 divides their difference. Here 23 minus 2 is 21, which is 7 times 3, so 23 and 2 are congruent mod 7 and share the class of 2.
Check
Which of the following is a genuine partition of the set below?
\[ X = \{1,2,3,4,5,6\} \]
Check your understanding
Which collection of subsets is a partition of X?
Answer: A
Why: A partition needs non-empty blocks that are pairwise disjoint and together cover X. Option A has three disjoint non-empty blocks whose union is all six elements, so it satisfies every clause.
Check
Using the correspondence with partitions, count the equivalence relations on a two-element set.
\[ X = \{a, b\} \]
Check your understanding
How many distinct equivalence relations are there on a two-element set?
Answer: A
Why: Equivalence relations correspond to partitions. A two-element set has exactly two partitions: two singletons, or one block containing both. So there are two equivalence relations, matching the second Bell number.
Commit first
Predict first
Is this rule a well-defined function on the classes mod 6, and why?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Yes: if a and a' are congruent mod 6 then 3a and 3a' are congruent mod 6, so the output class is independent of the representative.
Why: If 6 divides a minus a prime, then 6 divides 3 times that difference, which is 3a minus 3 a prime. So 3a and 3 a prime are congruent mod 6 and name the same class; the rule respects the relation and is well-defined.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
On the integers mod 6, consider defining a map on classes by a rule stated with a representative.
\[ f([a]) := [3a] \quad \text{on} \ \mathbb{Z}/6\mathbb{Z} \]
Check your understanding
Is this rule a well-defined function on the classes mod 6, and why?
Answer: A
Why: If 6 divides a minus a prime, then 6 divides 3 times that difference, which is 3a minus 3 a prime. So 3a and 3 a prime are congruent mod 6 and name the same class; the rule respects the relation and is well-defined.
Check
Each option is a relation on the real numbers. Which one is transitive?
Check your understanding
Which of these relations on the reals is transitive?
Answer: A
Why: Strict less-than is transitive: if a is less than b and b is less than c then a is less than c. The others each admit a two-step chain whose endpoints break the relation, so they fail transitivity.
Check
A relation from A to B is given explicitly below.
\[ R = \{(1,5),(1,6),(3,5)\} \ \text{from}\ A=\{1,2,3\} \ \text{to}\ B=\{5,6,7\} \]
Check your understanding
What are the domain and range of R?
Answer: A
Why: The domain is the set of first coordinates that actually appear, namely 1 and 3, and the range is the set of second coordinates that appear, namely 5 and 6. Element 2 and value 7 never occur in any pair.
Prediction
Predict first
A relation R is symmetric if and only if which condition holds?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: R equals its inverse R inverse
Why: Symmetry says every pair (a,b) in R has its reverse (b,a) also in R, which is exactly the statement that R and its inverse are the same set of pairs.
Check
Recall the inverse relation flips every pair.
\[ R^{-1} = \{(b,a) : (a,b) \in R\} \]
Check your understanding
A relation R is symmetric if and only if which condition holds?
Answer: A
Why: Symmetry says every pair (a,b) in R has its reverse (b,a) also in R, which is exactly the statement that R and its inverse are the same set of pairs.
Elimination
Eliminate the wrong options
According to the fundamental theorem, the equivalence classes of an equivalence relation on X always form what?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The classes are non-empty by reflexivity, they cover X because each element lies in its own class, and any two classes are equal or disjoint. That is exactly the definition of a partition.
Check
State the content of the fundamental theorem precisely.
Check your understanding
According to the fundamental theorem, the equivalence classes of an equivalence relation on X always form what?
Answer: A
Why: The classes are non-empty by reflexivity, they cover X because each element lies in its own class, and any two classes are equal or disjoint. That is exactly the definition of a partition.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The recipe: verify, classify, quotient · A relation is a set of ordered pairs · A relation is just a set of arrows · Domain and range of a relation · A relation on a set. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A relation is just a set of ordered pairs. On a single set, the properties reflexive, symmetric, antisymmetric, and transitive are each a mechanical check that dies to one counterexample.
An equivalence relation is reflexive, symmetric, and transitive together, none implied by the others. Its equivalence classes never partially overlap, so two classes are equal or disjoint.
The fundamental theorem: equivalence relations on a set correspond exactly to partitions of it, and the two constructions invert each other. Passing to the quotient is controlled forgetting, and every equivalence relation is the kernel of its own projection.
| Idea | What it says |
|---|---|
| equivalence relation | reflexive, symmetric, transitive |
| equivalence class | everything related to a chosen point |
| partition | non-empty, disjoint, covering blocks |
| fundamental theorem | equiv. relations = partitions |
| quotient X mod tilde | classes as new points, glue equivalents |
Next stops that reuse this exact machinery: modular arithmetic and the ring of integers mod n, quotient groups and the First Isomorphism Theorem, and quotient constructions throughout algebra and topology.
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