# Transitive Property

The **transitive property** in its most common form is: when given numbers \(a,\) \(b,\) and \(c,\) \( a = b\) and \( b = c\) being true implies \(a = c.\) Also popular: \( a < b \) and \( b < c \) being true implies that \( a < c .\)

In general, the transitive property indicates that given some comparative relationship \( \preceq ,\) \(a \preceq b\) and \(b \preceq c\) implies that \(a \preceq c.\) The transitive property may seem trivial with equality and inequality, but there are many other relationships that seem transitive but are not. That is, the transitive property is useful to study in order to avoid mistakes in situations where it **doesn't** hold.

There are also formal mathematical structures (like partial ordering) which require transitivity, so transitivity may need to be proven.

## Equality and Inequality

The transitive property of equality is if \( a = b \) and \( b = c ,\) then \( a = c .\)

The transitive property of inequality is if \( a < b \) and \( b < c ,\) then \( a < c .\) Note this also works with the symbols \( > ,\) \( \geq ,\) and \( \leq .\)

## Where Transitivity Fails

Suppose Aki, Brigid, and Caliana are playing tennis. Their Foot Movement, Hand Dexterity, and Stamina are each rated on a scale from 1 to 4. If the rating on two categories of player \(X\) exceed two categories of player \(Y,\) player \(X\) wins the match.

Foot Movement | Hand Dexterity | Stamina | |

Aki | 4 | 2 | 1 |

Brigid | 2 | 1 | 4 |

Caliana | 1 | 3 | 3 |

Notice that Aki beats Brigid (Movement, Dexterity), Brigid beats Caliana (Movement, Stamina), but Caliana beats Aki (Dexterity, Stamina). In symbols, \( A \preceq B ,\) \( B \preceq C ,\) but it is **not** the case that \( A \preceq C ;\) the reverse is true. Therefore this is a non-transitive relationship.

This can happen whenever the "ranking" of things is done in a way that can't be arranged on a single number line. This issue also comes up with voting:

Suppose three people rank the ice cream flavors C, S, and V (chocolate, strawberry, and vanilla) in the following way:

Chet: S > C > V

Omari: C > V > S

Taj: V > S > C

Only one flavor can be bought for a party, so they're going up for a series of runoff votes. That is, two of the flavors will be pitted against each other in a vote, and the winner will face the remaining flavor in a final vote.

For example, if C faces V in a vote, Chet and Omari prefer C to V so that C will win. Then C faces off against S in a vote; since Chet and Taj prefer S to C, S wins the final vote.

If the runoffs can happen in any order, is there a way for chocolate (C) to win the final vote?

## Formal Transitivity

Justifying the transitivity property formally is a matter of showing given a relation \( \preceq ,\) \(a \preceq b\) and \(b \preceq c\) implies that \(a \preceq c.\) This needs to be shown true for **any** choice of \( a,\) \(b,\) and \(c\) in the given set.

Suppose we have a set \( Z = \{1, 2, 3, 4\} \) with a relation defined by the input-output pairs \( \{(1, 3), (1, 2), (3, 2), (2, 4), (4, 1)\} .\) Is this relationship transitive?

While we have a transitivity relationship with \(1 \preceq 3,\) \(3 \preceq 2,\) and \(1 \preceq 2\) all being true, it doesn't hold for every possible trio of \(a,\) \(b,\) and \(c.\) For example, \( 1 \preceq 2 \) and \( 2 \preceq 4 \) but it is

notthe case that \( 1 \preceq 4 .\) Therefore the relation is not transitive.

**Cite as:**Transitive Property.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/transitive-property/