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Determine whether the relation is reflexive, symmetric, antisymmetric, transitive, and/or a partial order.

$ (x,y) \in R $ if $ x \ge y $ when defined on the set of positive integers.

I'm not sure how to even start this problem.

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  • $\begingroup$ Do you know the definitions of the properties in question? Because if you do, checking them for this particular relation is very, very easy. $\endgroup$ – Brian M. Scott Mar 27 '13 at 1:32
  • $\begingroup$ I do know the definitions but I am having a hard time applying them to the problem. My professor didn't give us much practice with this and my notes are unclear. $\endgroup$ – ctzdev Mar 27 '13 at 1:34
  • $\begingroup$ without specifying what the set is where $x,y$ come from, and without specifying what it means that $x\ge y$ it is impossible to answer the question. A relation is always defined on a set. Specifying that set is part of specifying what the relation is, and can affect properties the relation may or may not have. $\endgroup$ – Ittay Weiss Mar 27 '13 at 1:37
  • $\begingroup$ Sorry, I forgot to include that part. It's fixed now. $\endgroup$ – ctzdev Mar 27 '13 at 1:39
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I’ll check two of the properties to give you the idea. I’m assuming that $R$ is a relation on the set of real numbers.

  • Reflexivity: $R$ is reflexive if $\langle x,x\rangle\in R$ for every real number $x$. By the definition of $R$, $\langle x,x\rangle\in R$ if and only if $x\ge x$; is this true for every real number $x$? Definitely, so $R$ is reflexive.

  • Symmetry: $R$ is symmetric if it has the following property: for any real numbers $x$ and $y$, if $\langle x,y\rangle\in R$, then $\langle y,x\rangle\in R$. For this specific relation that property says: for any real numbers $x$ and $y$, if $x\ge y$, then $y\ge x$. Is that true? Of course not: take $x=2$ and $y=1$, and we certainly have $x\ge y$, since $2\ge 1$, but it’s clearly not true that $y\ge x$, because $1\not\ge 2$. Thus, $R$ is not symmetric.

I’ll leave transitivity to you, just reminding you of the definitions.

  • Antisymmetry: For any real numbers $x$ and $y$, if $\langle x,y\rangle\in R$ and $\langle y,x\rangle\in R$, then $x=y$. Is this true for this relation? Just translate $\langle x,y\rangle\in R$ and $\langle y,x\rangle\in R$ into more familiar terms, and it should be very clear.

  • Transitivity: For any real numbers $x,y$, and $z$, if $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, then $\langle x,z\rangle\in R$. Again, if you translate the hypothesis into more familiar terms, you should have no trouble deciding whether the statement is true of this relation or not.

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  • $\begingroup$ Thanks. I'm starting to understand. But, since the relation is (x,y), how come we dont check y for reflexivity? Why only x if the relation is x and y; (x,y)? $\endgroup$ – ctzdev Mar 27 '13 at 1:56
  • $\begingroup$ @NotoriousArab: $x$ and $y$ are just names, place-holding symbols; they have no actual identities. We could just as well describe $R$ by saying that $\langle u,v\rangle\in R$ if and only if $u\ge v$. Or we could say that $\langle\alpha,\beta\rangle\in R$ if and only if $\alpha\ge\beta$. Or that $\langle\beta,\alpha\rangle\in R$ if and only if $\beta\ge\alpha$. The statement that $R$ is reflexive is just the statement that every real number is related to itself, that $\langle\text{thing},\text{thing}\rangle\in R$ no matter what real number $\text{thing}$ might be. $\endgroup$ – Brian M. Scott Mar 27 '13 at 2:06
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    $\begingroup$ @NotoriousArab: For all $x$ and for all $y$, i.e., for all possible pairs of $x$ and $y$ (including those in which $x=y$). $\endgroup$ – Brian M. Scott Mar 27 '13 at 2:13
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    $\begingroup$ @NotoriousArab: No, there’s no there exists in there anywhere. $R$ is transitive if it has the following property: if $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, where $x,y$, and $z$ are real numbers, then $\langle x,z\rangle\in R$. It doesn’t say that there are any real numbers such that $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, though for this relation there are; it just says that if three real numbers have that property, then $\langle x,z\rangle\in R$ as well. $\endgroup$ – Brian M. Scott Mar 27 '13 at 2:23
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    $\begingroup$ @NotoriousArab: No. Antisymmetry says that if $\langle x,y\rangle\in R$ and $\langle y,x\rangle\in R$, then $x=y$. Now $\langle 2,1\rangle\in R$, but $\langle 1,2\rangle\notin R$, so the choice $x=2,y=1$ can’t tell you anything about antisymmetry of $R$ one way or the other. Antisymmetry doesn’t say anything about two numbers $x$ and $y$ unless they are related in both directions, $\langle x,y\rangle\in R$ and $\langle y,x\rangle\in R$, in which case it says that they have to be equal. $\endgroup$ – Brian M. Scott Mar 27 '13 at 2:37

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