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

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    $\begingroup$ $a,b\in \mathbb{Z}$, $a\sim b$ iff $b=a+1$. $\endgroup$
    – Dario
    Commented Jul 1, 2014 at 11:59
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    $\begingroup$ For any people $x, y$, try "$x$ loves $y$" :) $\endgroup$
    – Shaun
    Commented Jul 1, 2014 at 12:22
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    $\begingroup$ Not all people are narcissists, @anorton $\ddot\smile$ $\endgroup$
    – Shaun
    Commented Jul 1, 2014 at 15:37
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    $\begingroup$ @Shaun whoops. I was getting "not reflexive" confused with "irreflexive" $\endgroup$
    – apnorton
    Commented Jul 1, 2014 at 15:38
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    $\begingroup$ A fun follow-on to your question is: what is the size of the smallest set on which you can define such a relation? $\endgroup$ Commented Jul 1, 2014 at 19:56

10 Answers 10

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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.

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    $\begingroup$ 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 ($(x \sim y \wedge y \sim z) \Rightarrow x \not\sim z$). $\endgroup$
    – M. Vinay
    Commented Jul 1, 2014 at 12:36
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    $\begingroup$ Note: does not work for grandfathers. $\endgroup$
    – Emily
    Commented Jul 1, 2014 at 15:57
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    $\begingroup$ “Luke, I am your ….” :) $\endgroup$ Commented Jul 1, 2014 at 17:33
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    $\begingroup$ I'm not complaining, just observing: this is in some sense the same answer as Dario's comment to the question. $\endgroup$ Commented Jul 1, 2014 at 18:11
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    $\begingroup$ @SteveJessop Ah, I see "$a$ is the predecessor of $b$". Good observation. $\endgroup$
    – M. Vinay
    Commented Jul 2, 2014 at 2:37
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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 $(u, w) \notin R$.

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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".

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

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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 $3 \nsim 2$.

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  • $\begingroup$ Please don't use a symbol that's a variant of "$=$" for a relation that has none of the properties of equality! $\endgroup$ Commented Jul 1, 2014 at 15:33
  • $\begingroup$ @DavidRicherby It has been edited. $\endgroup$
    – Surb
    Commented Jul 1, 2014 at 15:55
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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 $c$).

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Minimal example:

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

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

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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 $0 \nsim 1$.

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

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  • $\begingroup$ Whoops, right you are! I’ll add another point. $\endgroup$
    – Lubin
    Commented Jul 2, 2014 at 12:37
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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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  • $\begingroup$ Velcome to our site! $\endgroup$ Commented Jul 3, 2014 at 7:50

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