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So, I have a set $A=\{1,2,3\}$, and I have to do a bunch of different relations for it, but I can't seem to be able to grasp as to what exactly I'm supposed to do here. One of the asked questions is like so:

Form a relation to set $A$ so that the relation is reflexive, but is not symmetrical nor transitive.

The logic I can understand well enough, and I realize that I have to form a relation that fulfills

$xRx$, for $x\in A$

$xRy\not =yRx$ for $x, y \in A$

$xRy \land yRz \not => xRz$ for $x, y, z \in A$

but I can't figure out how I can actually form the relation, apart from writing out the rules of the relation.

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    $\begingroup$ It's a bid misleading to call $A$ group, which has a separate meaning. It's better just to call $A = \{1, 2, 3\}$ a /set/. $\endgroup$ – Travis Willse Mar 20 '17 at 15:18
  • $\begingroup$ Changed to more appropriate form $\endgroup$ – Grak Mar 20 '17 at 15:25
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The second logical condition you've written down is not quite correct---in fact (since $A$ is nonempty), the special case $y = x$ contradicts the first contradiction. Symmetry means that $x R y$ iff $y R x$ for all $x, y \in A$, and the negation of this condition is that there is some choice of $x, y \in A$ such that $x R y$ but not $y R x$.

The third condition is also not quite correctly written, for basically the same reason that the second is not.

As for finding a relation, one can, as you say, simply construct one from scratch, and there are many choices. I recommend doing this once if you haven't already---the process may be illuminating, even if the result is now.

Here's a more intuitive source for constructing a relation.

Hint Consider a relation that defines the rules of the familiar game rock-paper-scissors.

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  • $\begingroup$ Stupid question, but... What exactly can I imagine "$R$" being? I can't wrap my head around how I should handle $R$ in my logical thinking here. $\endgroup$ – Grak Mar 20 '17 at 18:40
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    $\begingroup$ Let's suggestively relabel our set elements $\{r, p, s\}$. Then we need to decide whether each of the nine possible pairs $(x, y)$, $x, y \in A$, are in the relation (regarded as a subset of $A \times A$), that is whether $xRy$ is true or false for each pair $(x, y)$. Reflexivity means that $xRx$ is true for all $x \in A$, so $rRr, pRp, sRs$ are all true. Now, we have to define whether the other six statements, $rRp, rRs, pRr, \ldots$ are true. $\endgroup$ – Travis Willse Mar 21 '17 at 16:54
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    $\begingroup$ The rules for R.P.S. amount to saying which object wins in any particular pairing, so we can define the truth value of these remaining statements by declaring $xRy$ is true iff $x$ defeats $y$ in rock-paper-scissors. For example, rock defeats paper, so $rRp$ is true but $pRr$ is false. This already shows that the relation $R$ we've defined is not symmetric. Can you now show that $R$ is not transitive either? $\endgroup$ – Travis Willse Mar 21 '17 at 16:57
  • $\begingroup$ I suppose transitivity could be proven false by pointing out that while rRp is true and pRs is true, rRs is not true. But anyway, thanks for the great examples, really helps understanding the logic at grass root level $\endgroup$ – Grak Mar 22 '17 at 19:21
  • $\begingroup$ Yes, and you only need one counterexample to show that transitivity doesn't hold. $\endgroup$ – Travis Willse Mar 23 '17 at 0:33

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