Reflexive, Symmetric, and Transitive Relations Made Simple

Reflexive, symmetric, and transitive relations are easy to define. The difficulty begins when all three definitions appear in the same question.

You may look at a set of ordered pairs and wonder which pair to check first. You may see a rule such as “a - b is divisible by 4” and know that it feels like an equivalence relation, but still be unsure how to prove it. Transitivity creates the most confusion because it involves three elements and a chain of two related pairs.

There is a much calmer way to handle the topic.

Each property asks a different question:

• Reflexive asks whether every element relates to itself.
• Symmetric asks whether every arrow can be reversed.
• Transitive asks whether every two-step route has its required shortcut.

That difference between proof and disproof is the heart of this chapter. Once it becomes clear, relation questions stop feeling like guesswork.

Begin With the Set, Not the Pairs

Let R be a relation on a set A. This means R is a collection of ordered pairs chosen from A x A.

For example, if:

A = {1, 2, 3}

then:

A x A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}

A relation on A can contain all of these pairs, some of them, or none of them.

This is why you must read the set first. Reflexivity cannot be checked unless you know every element whose self-pair is required. A counterexample is invalid if it uses a number that is not in the given set.

The notation a R b means exactly the same thing as (a,b) in R. Read it as “a is related to b.”

The Four Definitions in One View

Property	Formal test	Picture in an arrow diagram	Fast failure certificate
Reflexive	(a,a) in R for every a in A	Every point has a self-loop	One missing (a,a)
Symmetric	(a,b) in R implies (b,a) in R	Every arrow has a return arrow	(a,b) present, (b,a) absent
Transitive	(a,b) in R and (b,c) in R imply (a,c) in R	Every two-step route has a shortcut	(a,b) and (b,c) present, (a,c) absent
Equivalence	Reflexive, symmetric, and transitive	Separate, internally complete groups	Failure of any one property

Notice the different shapes of the checks:

• Reflexive focuses on one element at a time.
• Symmetric focuses on one ordered pair and its reverse.
• Transitive focuses on two pairs that join through the same middle element.

Keeping these shapes separate prevents most mistakes.

The Proof and Counterexample Rule

The words for every and if matter.

Suppose a relation is claimed to be reflexive. Showing that (1,1) is present proves only that 1 relates to itself. It does not prove that every element does. If the set is {1,2,3}, you must also deal with (2,2) and (3,3).

But to show that the relation is not reflexive, one missing self-pair is enough. If 2 in A and (2,2) not in R, the claim has already failed.

The same logic works for the other properties:

• To prove symmetry, deal with every related pair or prove the reversal using the rule.
• To disprove symmetry, find one related pair whose reverse is absent.
• To prove transitivity, deal with every possible chain or prove the shortcut using the rule.
• To disprove transitivity, find one chain whose shortcut is absent.

Reflexive Relations: The Diagonal Test

A relation R on A is reflexive if:

(a,a) in R for every a in A

If A = {1,2,3,4}, the four compulsory pairs are:

(1,1), (2,2), (3,3), (4,4)

Other pairs do not affect reflexivity. A relation may contain dozens of off-diagonal pairs, but it is not reflexive if even one required self-pair is missing.

Quick finite-set example

Let:

A = {1,2,3}

R = {(1,1), (2,2), (3,3), (1,2)}

All three required diagonal pairs are present. Therefore, R is reflexive.

The extra pair (1,2) neither proves nor destroys reflexivity.

How to disprove reflexivity

Let:

A = {1,2,3}

S = {(1,1), (2,2), (1,3), (3,1)}

Since 3 in A but (3,3) not in S, the relation is not reflexive.

That single sentence is a complete counterexample.

Reflexivity for a rule-defined relation

Suppose R is defined on the integers by:

a R b if a - b is divisible by 5

To test reflexivity, replace b by a:

a - a = 0

Since 0 is divisible by 5, a R a for every integer a. Thus, R is reflexive.

The substitution b = a is the quickest first move in most rule-based questions.

Symmetric Relations: The Reverse-Pair Test

A relation R on A is symmetric if:

(a,b) in R implies (b,a) in R

This condition applies only when (a,b) is already in the relation. It does not say that every possible pair must be present.

Quick finite-set example

Let:

R = {(1,1), (1,2), (2,1), (3,3)}

• (1,2) has the reverse (2,1).
• (2,1) has the reverse (1,2).
• A self-pair is its own reverse.

Therefore, R is symmetric.

How to disprove symmetry

Let:

S = {(1,1), (2,2), (1,2)}

Here (1,2) in S, but (2,1) not in S. Therefore, S is not symmetric.

Do not write only “the reverse pair is missing.” Name both the present pair and the missing reverse.

Symmetry for a rule-defined relation

Return to:

a R b if a - b is divisible by 5

Assume a R b. Then for some integer k:

a - b = 5k

Therefore:

b - a = -5k = 5(-k)

Since -k is also an integer, b - a is divisible by 5. Hence b R a, so the relation is symmetric.

The proof works because reversing a - b changes only its sign, not its divisibility.

Transitive Relations: The Chain-and-Shortcut Test

A relation R on A is transitive if:

(a,b) in R and (b,c) in R imply (a,c) in R

The repeated b is the key. It is the middle element that joins the two pairs.

Think of a journey:

a -> b -> c

If the relation is transitive, the direct shortcut must also exist:

a -> c

Quick failure example

Let:

R = {(1,1), (2,2), (3,3), (1,2), (2,3)}

The pairs (1,2) and (2,3) form the chain:

1 -> 2 -> 3

Transitivity requires (1,3). Since (1,3) not in R, the relation is not transitive.

This is the standard three-part counterexample:

1. State the first pair.
2. State the second pair with the matching middle element.
3. State that the required shortcut pair is absent.

Transitivity for a rule-defined relation

Again let:

a R b if a - b is divisible by 5

Assume a R b and b R c. Then for some integers m and n:

a - b = 5m

b - c = 5n

Add the equations:

a - c = 5m + 5n = 5(m + n)

So a - c is divisible by 5. Hence a R c, and the relation is transitive.

The Most Important Transitivity Trap

Many students believe a relation must contain three linked elements to be transitive. That is not true.

Consider the relation on A = {1,2,3}:

R = {(1,2), (3,1)}

Can the first pair be followed by another pair? It ends at 2, but no pair begins at 2.

Can (3,1) be followed by another pair? Yes. It can be followed by (1,2), producing the chain:

3 -> 1 -> 2

That chain requires (3,2), which is missing. So this particular relation is not transitive.

Now compare it with:

S = {(1,2), (3,2)}

Both arrows end at 2, and no arrow begins at 2. There is no two-step chain at all. Therefore, there is no chain that breaks the condition, and S is transitive.

This may feel strange at first. Transitivity says, “If a two-step chain exists, its shortcut must exist.” When there is no two-step chain, the relation has not violated the rule.

The empty relation

On a nonempty set, the empty relation has no pairs.

• It is not reflexive because the required self-pairs are missing.
• It is symmetric because there is no one-way pair that breaks symmetry.
• It is transitive because there is no two-step chain that breaks transitivity.

This is a useful reminder that missing pairs do not automatically destroy symmetry or transitivity. They destroy those properties only when they complete the correct kind of counterexample.

A Reliable Checklist for a Finite Relation

Suppose a question gives a set and lists the ordered pairs. Use this order.

Step 1: List the compulsory self-pairs

For each a in A, check (a,a).

One missing pair means “not reflexive.” If all are present, the relation is reflexive.

Step 2: Check reverses

For every off-diagonal pair (a,b), look for (b,a).

Self-pairs need no partner because reversing (a,a) gives the same pair.

Step 3: Group chains by their middle element

For each element b:

• find every pair ending at b
• find every pair beginning at b
• combine them to create all chains (a,b), (b,c)
• check whether every resulting (a,c) is present

This middle-element method is safer than staring at the full list and hoping a chain becomes visible.

Step 4: Decide equivalence only at the end

Write three separate conclusions:

Reflexive: yes or no

Symmetric: yes or no

Transitive: yes or no

Only then decide whether the relation is an equivalence relation.

Solved Example 1: A Relation Given as Ordered Pairs

Let:

A = {1,2,3,4}

and

R = {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (1,3), (3,1), (2,3), (3,2)}

Determine whether R is reflexive, symmetric, and transitive.

Reflexive check

All four self-pairs are present:

(1,1), (2,2), (3,3), (4,4)

Therefore, R is reflexive.

Symmetric check

The off-diagonal pairs occur in reverse pairs:

• (1,2) and (2,1)
• (1,3) and (3,1)
• (2,3) and (3,2)

Therefore, R is symmetric.

Transitive check

The elements 1,2,3 are all related to one another, including themselves. Any two-step route among them has its shortcut in the relation.

The element 4 relates only to itself, so chains through 4 also keep producing (4,4), which is present.

Therefore, R is transitive.

Since all three properties hold, R is an equivalence relation.

Its equivalence classes are:

{1,2,3} and {4}

Solved Example 2: One Missing Pair Changes the Answer

Remove (2,3) from the previous relation and call the new relation S.

The relation is still reflexive because all self-pairs remain.

It is not symmetric because (3,2) in S, but (2,3) not in S.

It is also not transitive. For example:

(2,1) in S and (1,3) in S, but (2,3) not in S.

Therefore, S is not an equivalence relation.

This example shows why checking only the diagonal is never enough.

Solved Example 3: Divisibility Creates Equivalence Classes

Define a relation on the integers by:

a R b if 4 divides a - b

We test all three properties.

Reflexive

a - a = 0, and 4 divides 0. So a R a.

Symmetric

If 4 divides a - b, then a - b = 4k for some integer k. Thus b - a = 4(-k), so 4 also divides b - a.

Transitive

If 4 divides a - b and 4 divides b - c, then 4 divides their sum:

(a - b) + (b - c) = a - c

Thus a R c.

The relation is reflexive, symmetric, and transitive. Therefore, it is an equivalence relation.

It separates the integers into four groups according to remainder after division by 4:

• numbers with remainder 0
• numbers with remainder 1
• numbers with remainder 2
• numbers with remainder 3

For instance, 2, 6, 10, and -2 belong to the same class because the difference between any two of them is divisible by 4.

Solved Example 4: Close Does Not Mean Transitive

Define a relation on the real numbers by:

a R b if |a - b| <= 2

Reflexive

|a - a| = 0 <= 2, so the relation is reflexive.

Symmetric

|a - b| = |b - a|, so the relation is symmetric.

Transitive

Take a = 0, b = 2, and c = 4.

|0 - 2| = 2, so 0 R 2.

|2 - 4| = 2, so 2 R 4.

But |0 - 4| = 4, so 0 is not related to 4.

Therefore, the relation is not transitive and hence not an equivalence relation.

The everyday phrase “close to” often creates exactly this pattern. One object can be close to a second, and the second close to a third, while the first and third are too far apart.

Solved Example 5: Perpendicular Lines

On the set of all lines in a plane, define:

l R m if l is perpendicular to m

• It is not reflexive because a line is not perpendicular to itself.
• It is symmetric because if l is perpendicular to m, then m is perpendicular to l.
• It is not transitive because if l is perpendicular to m and m is perpendicular to n, then l and n are parallel, not perpendicular.

Therefore, the relation is symmetric but neither reflexive nor transitive.

Compare this with “is parallel to” on the set of lines. A line is parallel to itself under the relation-based convention, reversing the order changes nothing, and two successive parallel relationships remain parallel. That relation is an equivalence relation.

Solved Example 6: Less Than or Equal To

On the real numbers, define a R b if a <= b.

• Reflexive: a <= a for every real number a.
• Symmetric: false. For example, 2 <= 5, but 5 <= 2 is false.
• Transitive: if a <= b and b <= c, then a <= c.

So <= is reflexive and transitive, but not symmetric. It is not an equivalence relation.

This is also a good reminder not to confuse symmetric with antisymmetric. The relation <= is antisymmetric, but that does not make it symmetric.

Equivalence Relations: Three Locks, One Door

A relation is an equivalence relation only when all three conditions hold.

Check	Question to ask
Reflexive	Does every element belong with itself?
Symmetric	If two elements belong together in one direction, do they belong together in the reverse direction?
Transitive	If the first belongs with the second and the second with the third, must the first belong with the third?

All three are necessary. Two out of three is not enough.

Why is the combination so useful? Because it divides the original set into clean groups called equivalence classes.

Inside one class, every element is related to every other element. Different classes do not overlap. Together, the classes cover the whole set.

You can picture the original set as an archipelago:

• reflexivity places each island within its own group
• symmetry makes every connection mutual
• transitivity completes all indirect connections inside the group
• equivalence classes become separate clusters with no shared island

This is the deeper meaning of an equivalence relation. It is not merely a list of three definitions. It is a consistent rule for deciding when different objects should be treated as belonging to the same category.

How to Find an Equivalence Class

If R is an equivalence relation on A, the equivalence class of a is written [a] and means:

[a] = {x in A : x R a}

Some books write the condition as a R x. For an equivalence relation, symmetry makes the two descriptions agree.

Suppose A = {1,2,3,4,5,6} and a R b means that a and b have the same parity.

Then:

[1] = {1,3,5}

[2] = {2,4,6}

Also:

[3] = [1]

[4] = [2]

Two equivalence classes are either identical or disjoint. They cannot partly overlap.

Arrow Diagrams and Relation Matrices

When the set is small, a picture can make the properties visible.

In an arrow diagram

• Reflexive means every vertex has a loop.
• Symmetric means every arrow between different vertices has an arrow back.
• Transitive means every path of length two has a direct arrow from its start to its end.

An equivalence relation appears as separate complete groups. Within each group, every point connects to every point, including itself.

In a relation matrix

Write the elements of A in the same order along the rows and columns. Put 1 in position (i,j) if (i,j) in R, and 0 otherwise.

• Reflexive means every main diagonal entry is 1.
• Symmetric means the matrix mirrors across its main diagonal.
• Transitive still requires chain checking. A visual mirror or full diagonal alone cannot prove it.

The matrix is excellent for reflexivity and symmetry, but use the middle-element method for a reliable transitivity check.

Common Mistakes and Their Repairs

Mistake 1: Checking only the pairs that look important

Repair: Reflexivity needs every element. Symmetry needs every present pair. Transitivity needs every possible two-step chain.

Mistake 2: Calling a relation symmetric because one reverse pair exists

Repair: Every off-diagonal pair must have its reverse.

Mistake 3: Testing transitivity with pairs that do not join

Pairs (a,b) and (c,d) create a transitivity chain only when b = c.

Mistake 4: Requiring a converse in transitivity

From (a,b) and (b,c), transitivity requires (a,c). It does not require (c,a), (b,a), or (c,b) unless another property demands them.

Mistake 5: Using an element outside the set

If the relation is on {1,2,3}, a counterexample involving 4 proves nothing about the given relation.

Mistake 6: Proving two properties and declaring equivalence

Write all three headings. A relation that is reflexive and symmetric may still fail transitivity.

Mistake 7: Confusing symmetric and antisymmetric

Symmetric says a pair must reverse. Antisymmetric says that if two different directions both occur, the two elements must actually be equal. These ideas are not opposites, and they should not be substituted for one another.

A Clean Answer Format

For a rule-defined relation, this structure works well.

Reflexive

Take any a in A. Show that a R a. Therefore, R is reflexive.

Symmetric

Assume a R b. Use the defining rule to show b R a. Therefore, R is symmetric.

Transitive

Assume a R b and b R c. Combine the two conditions to show a R c. Therefore, R is transitive.

Final conclusion

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

If a property fails, replace the general proof with one valid counterexample and state the failure clearly.

Quick Practice

Try to classify each relation before reading the answers.

Question 1

On A = {1,2}, let:

R = {(1,1), (2,2), (1,2)}

Question 2

On A = {1,2}, let:

R = {(1,2), (2,1)}

Question 3

On the positive integers, define a R b if a divides b.

Question 4

On the integers, define a R b if a^2 = b^2.

Question 5

On A = {1,2,3}, let:

R = {(1,2), (3,2)}

Answers to Quick Practice

1. Reflexive and transitive, but not symmetric. The pair (1,2) has no reverse.
2. Symmetric, but neither reflexive nor transitive. The chain (1,2), (2,1) requires (1,1), which is missing.
3. Reflexive and transitive, but not symmetric. For example, 2 divides 4, but 4 does not divide 2.
4. An equivalence relation. Equality of squares is reflexive and symmetric, and a^2 = b^2 with b^2 = c^2 gives a^2 = c^2.
5. Transitive, but neither reflexive nor symmetric. No pair begins at 2, so there is no two-step chain to violate transitivity.

The Final Counterexample Checklist

Before finishing any question, ask:

For reflexivity

• Did I check every element of the stated set?
• If it fails, did I name an a for which (a,a) is missing?

For symmetry

• Did I reverse every off-diagonal pair?
• If it fails, did I show (a,b) present and (b,a) absent?

For transitivity

• Did I search for pairs whose middle elements match?
• If it fails, did I show (a,b) and (b,c) present but (a,c) absent?
• If no two-step chain exists, did I recognise that the relation is transitive?

For equivalence

• Have I proved all three properties separately?
• If it is an equivalence relation, can I identify the groups it creates?

If you keep one memory picture, use this one:

Loops for reflexive. Reverse arrows for symmetric. Shortcuts for transitive. All three for equivalence.

Once relations feel comfortable, the same habit of reading definitions carefully will also help with one-one, onto, composite, and inverse functions.

Further Reading

• Relations and Functions, Mathematics Part I develops the definitions, ordered-pair examples, divisibility relations, and equivalence classes.
• NCERT Mathematics Laboratory Manual activity on parallel lines turns the three properties into a visual classroom activity.
• MIT OpenCourseWare Basic Analysis connects equivalence relations with equivalence classes and partitions.

Frequently Asked Questions

What is the easiest way to remember reflexive, symmetric, and transitive relations?

Think of loops, reverse arrows, and shortcuts. Reflexive needs a loop at every element. Symmetric needs a reverse for every arrow. Transitive needs a direct shortcut for every two-step route.

What single counterexample disproves reflexivity?

Find an element a in the given set for which (a,a) is not in the relation. One missing required self-pair is enough.

What single counterexample disproves symmetry?

Find (a,b) in the relation while (b,a) is absent. Both facts should be stated in the answer.

What single counterexample disproves transitivity?

Find (a,b) and (b,c) in the relation while (a,c) is absent. The repeated middle element is essential.

Can a relation be transitive if it has very few pairs?

Yes. If its pairs create no two-step chain, there is no chain that violates transitivity. The empty relation is also transitive.

Is every reflexive and symmetric relation an equivalence relation?

No. It must also be transitive. The relation |a - b| <= 2 on the real numbers is reflexive and symmetric but not transitive.

Is every symmetric and transitive relation reflexive?

No. On a nonempty set, the empty relation is symmetric and transitive but not reflexive.

Are self-pairs automatically symmetric?

Yes. Reversing (a,a) gives (a,a) again, so a self-pair supplies its own reverse.

Why is less than or equal to not symmetric?

Because a <= b does not generally imply b <= a. For example, 2 <= 5 is true, but 5 <= 2 is false.

Why does divisibility of a - b often create an equivalence relation?

The difference of an element with itself is zero, reversing the difference only changes its sign, and adding two divisible differences gives another divisible difference. These facts establish reflexivity, symmetry, and transitivity.

What is an equivalence class?

It is the set of all elements related to a chosen element. Equivalence classes form non-overlapping groups that together cover the original set.

What is the safest order for checking a finite relation?

Check diagonal self-pairs first, reverse pairs second, and two-step chains third. Then decide whether all three conditions hold together.