• Let S be a set and ~ a relation between any two objects in the set:
• Reflexivity For every a in S, a ~ a
• Symmetry For any a, b in S, if a ~ b -> b ~ a
• Transitivity For any a, b, c in S, if a ~ b and b ~ c –> a ~ c

Transitivity: links chain together

The core idea, in plain words

A relation $\sim$ on a set $S$ is transitive when you can always take a shortcut:

If you can get from $a$ to $b$, and from $b$ to $c$, then $a$ is already directly linked to $c$. Chains of any length collapse to one step (apply the rule repeatedly).

An everyday analogy: ancestors

If Ana is an ancestor of Ben and Ben is an ancestor of Cara, Ana is an ancestor of Cara. Transitive. But “is a parent of” is not: Ana is Cara’s grandparent, not her parent. Adding “transitively” to “parent of” (following chains) is exactly how you build “ancestor of”. Mathematicians call that the transitive closure.

A small worked example

On $S = \{1, 2, 3\}$:

• $R_1 = \{(1,2), (2,3), (1,3)\}$: transitive ✔. The only chain is $1 \to 2 \to 3$, and the shortcut $(1,3)$ is present.
• $R_2 = \{(1,2), (2,3)\}$: not transitive ✘, since $(1,3)$ is missing.
• $R_3 = \{(1,2), (2,1)\}$: not transitive ✘! The chain $1 \to 2 \to 1$ demands $(1,1)$, and it’s missing. A sneaky one: people forget chains can loop back.

On the integers: $<$ and $\le$ ✔, $a \mid b$ ✔ (if $b = ka$ and $c = mb$ then $c = (mk)a$), $a \equiv b \pmod n$ ✔, “$a \ne b$” ✘ ($1 \ne 2$, $2 \ne 1$, but $1 = 1$).

Counter-examples worth remembering

• Transitive but not symmetric: $<$. $1 < 2$ but not $2 < 1$.
• Transitive but not reflexive: $<$ again; nothing is less than itself.
• Symmetric but not transitive: “within 1 of”: $|1-2| \le 1$, $|2-3| \le 1$, but $|1-3| = 2$. See Symmetry.
• Real-world non-transitivity: Rock beats scissors, scissors beats paper, paper beats rock. “Beats” is not transitive, which is what makes the game work.

A picture in words: every two-step path has a shortcut

In the arrow picture, transitive means whenever you see a path of two arrows $a \to b \to c$, the direct arrow $a \to c$ is already drawn.

graph LR
  1((1)) --> 2((2))
  2 --> 3((3))
  1 --> 3

Here the shortcut $1 \to 3$ is present, so the relation is transitive. Delete it and transitivity breaks.

Common misconceptions

• “Transitivity needs three different elements.” No: $a, b, c$ may coincide. That’s why $\{(1,2),(2,1)\}$ fails: $a = c = 1$.
• “If a relation has no chains, it isn’t transitive.” The opposite: with no two-step chains, the rule is never tested, so it holds vacuously. $\{(1,2)\}$ is transitive.
• “Preferences are always transitive.” Economists assume they are; real people (and majority votes, see Condorcet’s paradox) can be cyclic.

How it connects

• Transitivity is the one property shared by both equivalence relations and partial orders. See Set Relations, Reflexivity, Symmetry.
• It is the hidden engine of logical chains: “if $p \to q$ and $q \to r$ then $p \to r$” is transitivity of implication, a cousin of Modus Ponens.

The truth revealed: transitivity is what lets small steps add up, and spotting where it fails (“close to”, “beats”, “friend of”) is half of clear thinking.