The Rule That Decides When "Is Related To" Is Trustworthy
Say you build an app that groups people who "know each other". If Aisha knows Ben, does Ben know Aisha? If Ben knows Chen, does Aisha? The answers decide whether your grouping holds together, and each answer has a name: symmetric, transitive, and so on. Those names are the types of relations.
Each type is just a rule about which ordered pairs must be present. Once you can read those rules off a set of pairs, classifying any relation becomes mechanical.
What Is a Relation in Math?
A relation from a set $A$ to a set $B$ is any collection of ordered pairs $(a, b)$ where $a$ comes from $A$ and $b$ comes from $B$. Formally, a relation is a subset of the Cartesian product $A \times B$, the set of every possible pair. When the relation is from a set to itself, it is a subset of $A \times A$.
For example, take the set $A = {1, 2, 3}$. The collection $R = {(1, 2), (2, 3)}$ is a relation on $A$, because each ordered pair uses elements of $A$. A relation is the parent idea behind relations and functions: a function is just a special relation where each input appears in exactly one pair. To work with the sets themselves, sets in math is the companion topic.
What Are the Types of Relations?
Relations are classified by which pairs they are required to contain. Here is the full family at a glance, each stated on a set $A$.
Type | Rule | Example on $A = {1, 2, 3}$ |
|---|---|---|
Empty relation | No element is related to any element; $R = \varnothing$ | $R = {,}$ |
Universal relation | Every element is related to every element; $R = A \times A$ | all $9$ pairs |
Identity relation | Every element is related only to itself | ${(1,1), (2,2), (3,3)}$ |
Inverse relation | Reverse each pair of a relation $R$ | if $R = {(1,2)}$, then $R^{-1} = {(2,1)}$ |
Reflexive | $(a, a) \in R$ for every $a \in A$ | ${(1,1), (2,2), (3,3), (1,2)}$ |
Symmetric | if $(a, b) \in R$ then $(b, a) \in R$ | ${(1,2), (2,1)}$ |
Transitive | if $(a, b) \in R$ and $(b, c) \in R$ then $(a, c) \in R$ | ${(1,2), (2,3), (1,3)}$ |
Antisymmetric | if $(a, b) \in R$ and $(b, a) \in R$ then $a = b$ | ${(1,2), (2,3)}$ |
Equivalence | reflexive and symmetric and transitive together | ${(1,1), (2,2), (3,3)}$ |
Relations can also be sorted by how elements pair up, one-to-one, one-to-many, many-to-one, or many-to-many, which is the lens used most when a relation is a function. The three that matter most for exams, reflexive, symmetric, and transitive, get their own section next.
What Are Reflexive, Symmetric, and Transitive Relations?
These three properties are the building blocks, and every equivalence relation is made from all three at once.
Reflexive. Every element must relate to itself. On $A = {1, 2, 3}$, a relation is reflexive only if it contains $(1,1)$, $(2,2)$, and $(3,3)$. The reflexive property of equality is the everyday case: every number equals itself.
Symmetric. If $a$ relates to $b$, then $b$ must relate back to $a$. So $R = {(1,2), (2,1)}$ is symmetric, while $R = {(1,2)}$ is not, because $(2,1)$ is missing. The symmetric property of equality captures it: if $a = b$ then $b = a$.
Transitive. If $a$ relates to $b$ and $b$ relates to $c$, then $a$ must relate to $c$. So $R = {(1,2), (2,3), (1,3)}$ is transitive, while dropping $(1,3)$ breaks it. Equality is transitive too: if $a = b$ and $b = c$, then $a = c$.
What Is an Equivalence Relation?
An equivalence relation is a relation that is reflexive, symmetric, and transitive all at once. Equality is the model example, since any number equals itself, equality reverses, and it chains through.
The reason equivalence relations matter is that they carve a set into equivalence classes, groups where every member is related to every other member and to nothing outside. "Has the same remainder when divided by $2$" splits the whole numbers into evens and odds. Those two classes never overlap, and together they cover everything, which is exactly what a clean classification needs.
How Is Antisymmetric Different from Symmetric?
This is where students most often mix things up. Symmetric says a two-way link is required: if $(a, b)$ is in $R$, then $(b, a)$ must be too. Antisymmetric says the opposite: the only time both $(a, b)$ and $(b, a)$ may appear is when $a$ and $b$ are the same element.
A relation that is reflexive, symmetric, and transitive is an equivalence relation (it sorts a set into equal groups).
A relation that is reflexive, antisymmetric, and transitive is a partial order (it ranks a set, like $\leq$ or "divides").
The divisibility relation on numbers is antisymmetric: if $a$ divides $b$ and $b$ divides $a$, the two numbers have to be equal. Equality is the one relation that is both symmetric and antisymmetric, because the only pairs it relates in both directions are pairs of a number with itself.
Examples of Types of Relations
The set runs from classifying a small identity relation, through the reflexive-relation trap, up to a full equivalence check and a partial order. Students often call a relation reflexive after seeing a couple of self-loops, without checking that every element of the set has one.
Example 1
Classify $R = {(1,1), (2,2), (3,3)}$ on the set $A = {1, 2, 3}$.
Check each property in turn.
Reflexive: $(1,1)$, $(2,2)$, $(3,3)$ are all present, so yes.
Symmetric: the only pairs are self-loops, and reversing a self-loop gives the same pair, so yes.
Transitive: no two pairs chain into a new requirement, so it holds vacuously.
Final answer: $R$ is reflexive, symmetric, and transitive, so it is an equivalence relation, and it is also the identity relation on $A$.
Example 2
Is $R = {(1,1), (2,2)}$ reflexive on $A = {1, 2, 3}$?
Wrong attempt. A student sees that every pair present is a self-loop, $(1,1)$ and $(2,2)$, and concludes "all self-loops, so reflexive."
Compare against the rule. Reflexive requires $(a, a)$ for every element of $A$, and $A$ contains $3$. The pair $(3,3)$ is nowhere in $R$.
Correct conclusion. A relation is reflexive only when it holds a self-loop for each element of the whole set, not just for the elements that happen to appear.
Final answer: No, $R$ is not reflexive, because $(3,3)$ is missing.
Example 3
Decide whether $R = {(1,2), (2,1), (1,1)}$ on $A = {1, 2}$ is symmetric.
Take each pair and look for its reverse. The reverse of $(1,2)$ is $(2,1)$, which is present. The reverse of $(2,1)$ is $(1,2)$, present. The reverse of $(1,1)$ is itself, present.
Final answer: Yes, $R$ is symmetric, because every pair has its reverse in $R$.
Example 4
Is $R = {(1,2), (2,3)}$ on $A = {1, 2, 3}$ transitive?
The relation has $(1,2)$ and $(2,3)$, which chain from $1$ to $2$ to $3$. Transitivity then demands $(1,3)$.
$$(1,2) \in R \text{ and } (2,3) \in R \Rightarrow (1,3) \text{ required}.$$
But $(1,3)$ is not in $R$.
Final answer: No, $R$ is not transitive; adding $(1,3)$ would make it transitive.
Example 5
Show that "has the same remainder on division by $2$" is an equivalence relation on $A = {1, 2, 3, 4}$.
The relation pairs numbers of equal parity:
$$R = {(1,1), (2,2), (3,3), (4,4), (1,3), (3,1), (2,4), (4,2)}.$$
Reflexive: each of $(1,1), (2,2), (3,3), (4,4)$ is present.
Symmetric: $(1,3)$ pairs with $(3,1)$, and $(2,4)$ pairs with $(4,2)$.
Transitive: $(1,3)$ and $(3,1)$ give $(1,1)$, which is present, and no chain leads outside $R$.
Final answer: $R$ is an equivalence relation, with equivalence classes ${1, 3}$ (odd) and ${2, 4}$ (even).
Example 6
Classify the divisibility relation on $A = {1, 2, 4}$, where $(a, b) \in R$ means $a$ divides $b$.
List the pairs:
$$R = {(1,1), (2,2), (4,4), (1,2), (1,4), (2,4)}.$$
Reflexive: every element divides itself, so all three self-loops appear.
Antisymmetric: no distinct pair appears in both directions, since $a$ dividing $b$ and $b$ dividing $a$ forces $a = b$.
Transitive: $1$ divides $2$ and $2$ divides $4$, and indeed $(1,4)$ is present.
Final answer: $R$ is reflexive, antisymmetric, and transitive, so it is a partial order, not an equivalence relation.
Where Are Types of Relations Used?
"How do you group items that should count as the same?"
Equivalence relations are the mathematics of grouping. Any time you sort things into buckets where everything in a bucket counts as interchangeable, days of the week by their weekday, fractions by their reduced value, records by a matching key, you are using an equivalence relation, and the reflexive, symmetric, transitive rules are what keep the buckets clean.
Where these types earn their place:
Functions. Deciding whether a relation is one-to-one or many-to-one is what separates a genuine function from an arbitrary pairing, the gateway to all of function theory.
Databases and deduplication. "Is the same customer as" needs to be an equivalence relation, or the system merges records it should not.
Order and ranking. Partial orders (reflexive, antisymmetric, transitive) model prerequisites, version histories, and task scheduling, wherever "comes before" matters. The formal theory sits in the study of the equivalence relation and its order-theoretic cousin.
That last idea is the destination worth glimpsing early: the same three properties you test on a six-pair relation are what let computer scientists reason about huge dependency graphs.
Common Mistakes with Types of Relations
Mistake 1: Calling a relation reflexive without checking every element
Where it slips in: Classifying a relation whose set has more elements than the relation happens to mention.
Don't do this: Declare a relation reflexive because the self-loops it does contain look complete.
The correct way: Reflexive means a self-loop $(a, a)$ for every element of the underlying set. The learner who checks only the elements that appear in the pairs misses the ones sitting silently in the set, exactly the gap in Example 2.
Mistake 2: Confusing symmetric with antisymmetric
Where it slips in: A relation that has some two-way pairs and some one-way pairs.
Don't do this: Assume "not symmetric" automatically means "antisymmetric", or that the two are opposites.
The correct way: Symmetric requires the reverse pair every time; antisymmetric forbids a reverse pair unless the elements are equal. A relation can be neither, and equality is the rare one that is both. The memorizer who recites "symmetric means both directions" still has to check whether antisymmetry is being asked instead.
Mistake 3: Assuming a relation is transitive without testing the chains
Where it slips in: A relation with several linked pairs.
Don't do this: Guess transitivity from a couple of pairs that happen to work.
The correct way: For every pair $(a, b)$ and $(b, c)$ in the relation, verify that $(a, c)$ is also present. A single missing shortcut pair breaks transitivity, as ${(1,2), (2,3)}$ shows.
Conclusion
The types of relations are empty, universal, identity, inverse, reflexive, symmetric, transitive, antisymmetric, and equivalence.
A relation is a subset of the Cartesian product, a set of ordered pairs drawn from the underlying sets.
Reflexive, symmetric, and transitive are the three core properties, and together they define an equivalence relation.
Antisymmetric differs from symmetric: with reflexivity and transitivity it makes a partial order instead of an equivalence.
The common slips are skipping elements in the reflexive check, confusing symmetric with antisymmetric, and assuming transitivity without testing the chains.
To take types of relations further with a teacher, explore Bhanzu's algebra tutor, get targeted help with algebra, or join math classes online.
Practice These to Solidify Your Understanding
Classify $R = {(1,1), (2,2), (3,3), (1,2), (2,1)}$ on $A = {1, 2, 3}$ as reflexive, symmetric, transitive, or an equivalence relation.
Is the relation "is less than" ($<$) on ${1, 2, 3}$ reflexive, symmetric, antisymmetric, or transitive?
Write the inverse relation $R^{-1}$ of $R = {(1, 3), (2, 4)}$, and state whether $R$ is symmetric.
If Problem 1 comes out reflexive without you checking $(3,3)$, return to Mistake 1. Want a live Bhanzu trainer to walk through relations and functions with you? Book a free demo class.
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