# I need a relation which is not reflexive, not symmetric, and not transitive

I need an example of a relation which is simultaneously not reflexive, not symmetric, and not transitive. Any accessible examples? Thanks in advance.

• $a,b\in \mathbb{Z}$, $a\sim b$ iff $b=a+1$. – Dario Jul 1 '14 at 11:59
• For any people $x, y$, try "$x$ loves $y$" :) – Shaun Jul 1 '14 at 12:22
• Not all people are narcissists, @anorton $\ddot\smile$ – Shaun Jul 1 '14 at 15:37
• @Shaun whoops. I was getting "not reflexive" confused with "irreflexive" – apnorton Jul 1 '14 at 15:38
• A fun follow-on to your question is: what is the size of the smallest set on which you can define such a relation? – Eric Lippert Jul 1 '14 at 19:56

Here's a non-mathematical one: "is the father of".
You are not your own father. You are not your father's father. Your father's father is not your father.

• It's actually stronger than "not reflexive, symmetric or transitive", it's irreflexive ($\forall x, x \not\sim x$), asymmetric ($x \sim y \Rightarrow y \not\sim x$), and antitransitive ($(x \sim y \wedge y \sim z) \Rightarrow x \not\sim z$). – M. Vinay Jul 1 '14 at 12:36
• – Emily Jul 1 '14 at 15:57
• “Luke, I am your ….” :) – Scott Jul 1 '14 at 17:33
• I'm not complaining, just observing: this is in some sense the same answer as Dario's comment to the question. – Steve Jessop Jul 1 '14 at 18:11
• @SteveJessop Ah, I see "$a$ is the predecessor of $b$". Good observation. – M. Vinay Jul 2 '14 at 2:37

Think of three points $u, v, w$ with relation $R = \{(u, v), (v, w) \}$. So $u$ is related to $v$ and $v$ is related to $w$. This is not reflexive since $(u,u) \notin R$, not symmetric because $(v, u) \notin R$ and not transitive because $(u, w) \notin R$.

What beats what in Roshambo or "Rock, Paper, Scissors" is such a relation.

1. not reflexive: rock does not beat rock.

2. not symmetric: rock beats scissors, but scissors does not beat rock.

3. not transitive: rock beats scissors and scissors beats paper, but rock does not beat paper.

The same is true of "Rock, Paper, Scissors, Lizard, Spock".

A simple one is : Define $R$ on $\mathbb{Z}$ by $(x,y)\in R$ if and only if $x-y=10$.

On $\mathbb{N}$, consider $$a \sim b \iff a +2b = 5,$$ then

• $1\nsim 1$,

• $3\sim 1$ but $1 \nsim 3$,

• $3\sim 1, 1 \sim 2$ but $3 \nsim 2$.

• Please don't use a symbol that's a variant of "$=$" for a relation that has none of the properties of equality! – David Richerby Jul 1 '14 at 15:33
• @DavidRicherby It has been edited. – Surb Jul 1 '14 at 15:55

How about: "is the square of", defined on the set of positive integers? In other words, $$a \sim b \iff a=b^2$$

This relation is not reflexive (most numbers are not their own square), not symmetric (if $a$ is the square of $b$ then in most cases $b$ is not the square of $a$) and not transitive (if $a$ is the square of $b$ and $b$ is the square of $c$ then in general $a$ will not be the square of $c$).

Minimal example:

$X = \{0,1,2\}$

$R = \{(0,1),(1,2)\}$

Less than, but close: $$a \sim b \iff a < b \ \ \text{ but } \ \ a > (b-1)$$

So

• $1\nsim 1$,

• $0.5\sim 1$ but $1 \nsim 0.5$,

• $0\sim 0.5$, $0.5 \sim 1$, but $0 \nsim 1$.

Why not $a\sim b$ if and only if either $a=3$ and $b=4$ or $a=4$ and $b=5$?

• Whoops, right you are! I’ll add another point. – Lubin Jul 2 '14 at 12:37

Take any directed acyclic graph amd the arcs form an irreflexive, asymmetric antitransitive relation of its nodes. Then add some loops (not to all nodes), back-arcs (not to all of them) and some skip-forward arcs (not to all directed paths) and you have a more general relation with your restrictions.

Ex: 1) Strong version: a->b, b->c, c->d, a->e 2) Then add: a->a, b->a,a->d

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