Testing the Properties of Relations

Hello! Last lesson, you represented a relation as ordered pairs or as arrows in a directed graph. We will now ask what patterns those pairs follow. You will test four properties—reflexivity, symmetry, antisymmetry, and transitivity—and, just as importantly, learn what evidence is enough to establish or refute each one.

Start with the set, then ask four questions

A relation $R$ on a set $S$ selects pairs from $S\times S$. The set matters: to decide whether $R$ is reflexive, for example, you must know every element that needs a self-pair.

Here are the four tests. Read “$aRb$” as “$(a,b)$ belongs to $R$.”

The first three tests are illustrated in Neso Academy’s Types of Relations (Part 1). Watch the indicated segments for how the definitions are checked against actual pairs; we will examine transitivity separately.

A useful habit is to separate two kinds of reasoning. To show that a property fails, find one precise counterexample. To show that it holds, account for every case—or give an argument that works for arbitrary elements.

Reversing a pair is not the same as forbidding its reverse

Recall the factor relation from last lesson, on $S=\{1,2,3\}$:

It is reflexive: the three required self-pairs are present. It is not symmetric: $(1,2)\in R$, but $(2,1)\notin R$.

What about antisymmetry? The one-way connection from $1$ to $2$ is perfectly acceptable. So are all three loops. To break antisymmetry, we would need $(a,b)$ and $(b,a)$ for some $a\ne b$. No such distinct pair occurs in $R$, so $R$ is antisymmetric.

This is the distinction most worth holding onto:

• Symmetry requires a reverse whenever a pair is present.
• Antisymmetry forbids having both directions between distinct elements.

They are not opposites. A relation containing only self-pairs is both symmetric and antisymmetric: reversing a self-pair changes nothing, and there are no two-way connections between distinct elements.

The image also shows why “not reflexive” does not mean “no loops.” The left-hand graph has some loops, just not one at every vertex.

For a closer look at the symmetry–antisymmetry distinction and the graph tests, use the following targeted parts of 7.2: Properties of Relations by Mathematics LibreTexts.

Transitivity checks connected pairs

Reflexivity concerns loops; symmetry and antisymmetry compare directions. Transitivity asks a different question: whenever $(a,b)$ and $(b,c)$ are present, is $(a,c)$ also present?

The middle entry must match. The pairs $(1,2)$ and $(2,3)$ create a transitivity requirement for $(1,3)$. The pairs $(1,2)$ and $(3,2)$ do not create one together, because the first pair ends at $2$ while the second begins at $3$.

Here is a relation on $S=\{1,2,3\}$ that makes several tests visible at once:

It is reflexive because all three self-pairs are present. It is symmetric because each pair between distinct elements has its reverse. It is not antisymmetric, witnessed by $(1,2)$ and $(2,1)$, since $1\ne 2$.

It is also not transitive: $(1,2)$ and $(2,3)$ belong to $Q$, but $(1,3)$ does not. That single connected pair of arrows is enough to settle the question; you need not check every other route once you have found a failure.

To establish transitivity rather than refute it, the burden is broader. Return to the factor relation $R$. You could inspect every connected pair in its five-pair roster. But its defining rule gives a shorter argument: if $a$ is a factor of $b$, and $b$ is a factor of $c$, then $a$ is a factor of $c$. This works for every $a,b,c\in S$, so $R$ is transitive. Thus the factor relation has three of our four properties: reflexive, antisymmetric, and transitive, but not symmetric.

On a very small set, systematic inspection is often quickest. For a relation specified by a rule, such as “is a factor of,” try reasoning directly from the rule. In either method, remember that $a$, $b$, and $c$ need not be distinct: a route can involve a loop or return to its starting point.

A dependable testing routine

When you are given a new relation, work in this order:

1. Identify the set and the selected pairs. For a finite rule-defined relation, listing its pairs may prevent missed cases.
2. Check self-pairs for reflexivity.
3. Check reverses for symmetry, then look specifically for a two-way connection between distinct elements for antisymmetry.
4. Check connected pairs for transitivity. Either find a missing required pair or justify why every required pair is present.

Neso Academy’s Types of Relations (Solved Problem) applies all four questions to the rule $x+y=0$. This is a useful contrast with checking a finite roster: the relation is given by a condition, so counterexamples and general reasoning do the work.

The central habit is to state your evidence, not just your verdict. “Not symmetric” becomes a clear argument when you name a pair that is present and its reverse that is absent. “Not transitive” becomes clear when you name both connected pairs and the missing required pair.

You can now test a relation using its roster, rule, or directed graph: look for all loops, reversed pairs, forbidden two-way connections between distinct elements, and completed two-step routes. A single counterexample defeats a property; establishing one calls for a complete check or a general argument.

Next, you will use some of these tests to identify equivalence relations, then determine the groups of elements—equivalence classes—that they form.