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My understanding of relations is fair enough for now, however I am unsure how to determine the answer and provide a brief explanation.

I can't say I am a fan (yet) of the notation used either.

Here is an example taken from the homework:

$$xRy \Leftrightarrow y|x \mbox{ is an integer}$$

Any hints, tips, and pointers into the right direction would be greatly appreciated! =) Also, if it is not too much to ask, may you go easy on using formal notation to explain? I am not the most confident reader of it, yet.

user134768
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    Do try something! For instance, is $xRx$ for any $x\in$... by the way, you didn't write which set the relation is defined on. – alex Mar 11 '14 at 21:03
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    Are you sure that's how the example was written? Usually the definition of $|$ is $a|b\iff a/b$ is an integer. $y|x$ is not a number, it's true or false, so $y|x$ is never an integer :P

    http://mathworld.wolfram.com/Divides.html

     – JustAskin Mar 11 '14 at 21:13
  • @alex there was no set provided – user134768 Mar 11 '14 at 21:21

2 Answers2

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Equivalence relations and partial orders each have three defining properties:

(1) They are both transitive. That is, for a relation R, $aRb\land bRc\implies aRc$ (if aRb and bRc, aRc).

(2) They are both reflexive. That is, $aRa$ is true.

(3) Equivalence relations are symmetric; i.e. $aRb\iff bRa$. In contrast, partial orders are ANTIsymmetric, i.e. $(aRb\land bRa)\iff a=b$ (aRb and bRa can only be BOTH true if a=b, otherwise either aRb or bRa (or neither) but not both)

So, in order to tell whether a relation is a partial order or an equivalence relation, you just need to check if it's symmetric or antisymmetric. In your case, take 2 and 4. $2R4$, does $4R2$ ?

(I have intentionally used some symbols and explained them so you can hopefully get comfortable with them)

JustAskin
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  • So, with your example of 2 and 4...since 2 is in 4, but 4 is not in 2, it can not be symmetric? I know there is a chance it would be neither symmetric or anti-symmetric... and hopefully my view on anti-symmetric is right (still 'iffy' with that one) but in this case it would not be anti-symmetric either? So it would neither partial or an equivalence relation. (Edit: Also, thank you for your answer) – user134768 Mar 11 '14 at 21:28
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    @user134768 You are correct: Because $2R4$ and $\neg(4R2)$ (4 does not R 2), it is not symmetric.

    Now think, are there any numbers such that $aRb\land bRa$ but $b\neq a$?

    The answer, for this example, is "It depends on the set." If it's just on positive integers, then no, there are no numbers $a$ and $b$ such that $aRb\land bRa\land b\neq a$. But if it's the integers, take for example $a=4,b=-4$. $aRb$ is true (it's -1, which is an integer), $bRa$ is true (-1, integer), and $a\neq b$. Thus it is NOT antisymmetric on the integers. This is troubling because they didn't specify a set :/

     – JustAskin Mar 11 '14 at 21:41
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By definition given relation $R$ is partial order if and only if it satisfies certain conditions. So what you want to do is to go down the list of the conditions and check if they are all satisfied for $R$.

borg
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