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Given a set {0,1,2}, is the relation below a equivalence relationship?

{(0, 0),(1, 1),(2, 2)}

It is self evident that this relation is reflexive, as it conforms this rule: ∀x ∈ A, x R x.

Looking at the symmetric rule : R is symmetric if, and only if, ∀x, y ∈ A, if x R y then y R x and the transitive rule R is transitive if, and only if, ∀x, y, z ∈ A, if x R y and y R z then x R z.

I am still confuse on how those 2 rules suit the relation above, please help?

Asaf Karagila
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sctts-lol
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  • Suppose $xRy$ and you happen to know, say, that $x=2$. Then what do you know about $y$? Is it necessarily the case that $yRx$, given that $x=2$ and whatever you've been able to figure out about $y$? Does this help you see what's going on? Using this example, can you generalize to arbitrary $x \in { 0, 1, 2 }$? In other words, don't overthink this. Just walking through the definitions gets you where you need to be. Use concrete examples to help develop your intuition. – Robert Shore Apr 30 '21 at 03:44
  • @RobertShore From what I see its just 3 reflexive relations in the set, I still find a hard time finding the symmetric and transitive relation, could you give more hint please ? – sctts-lol Apr 30 '21 at 03:52
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    @sctts-lol: The whole set is one relation; it just happens to contain three ordered pairs. It is the whole relation that is reflexive, transitive, and symmetric, not the individual ordered pairs. It is reflexive, for instance, because it contains the pair $\langle x,x\rangle$ for every $x$ in the set ${0,1,2}$. It is symmetric because if $x$ and $y$ are in ${0,1,2}$, and $x\mathrel{R}y$, then $x=y$ (because $R$ contains no ordered pair $\langle x,y\rangle$ with $x\ne y$), and therefore $y\mathrel{R}x$ as well. – Brian M. Scott Apr 30 '21 at 04:22
  • Remember that: $x\mathrel R y$ means $\langle x,y\rangle\in R$. – Graham Kemp Apr 30 '21 at 05:04
  • @sctts-lol Try answering my question: If you happen to know that $x=2$ and $xRy$, then what do you know about $y$? – Robert Shore Apr 30 '21 at 09:15

1 Answers1

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Symmetric:
$R\subset A\times A$ is symmetric if $\forall x,y\in A, (x,y)\in R\implies (y,x)\in R $ $\tag 1$
Its negation will be: $R$ is not symmetric if $\exists x,y\in A$ such that $(x,y)\in R$ and $(y,x)\notin R $ $\tag 2$

Note that $(2)$ is not true for your $R$ and hence its negation must be true that is $R$ is symmetric.

Transitivity:
$R$ is transitive if $\forall x,y,z \in A, xRy \land yRz\implies xRz$ $\tag 3$
Its negation will be: $\exists x,y,z\in A, (xRy\land yRz)$ and $(x\not R z)$ $\tag 4$

Note that $(4)$ is not true for your $R$ and hence its negation must be true that is $R$ is transitive.

Koro
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