How do the Properties of Relations work?












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This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.



There are five properties for a relation:



Reflexive - $R rightarrow R$



Symmetrical - $R rightarrow S$ ; $S rightarrow R$



Antisymmetrical - $R rightarrow S$ && ($R rightarrow R$|| $S rightarrow S$)



Asymmetrical -$R rightarrow S$ && !($R rightarrow R$|| $S rightarrow S$)



Transitive - if $R rightarrow S$ && $S rightarrow T$, then $R rightarrow T$



If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.



Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:



1) "Is the same age as" is apparently reflexive, symmetrical and transitive.
2) "Is related to" is also apparently reflexive, symmetrical and transitive.
3) "Is older than" is asymmetric, antisymmetric and transitive.



There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.



I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?










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  • $begingroup$
    @bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
    $endgroup$
    – Charles Stewart
    Jul 27 '10 at 8:24










  • $begingroup$
    @Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
    $endgroup$
    – bryn
    Jul 28 '10 at 1:53










  • $begingroup$
    It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
    $endgroup$
    – timtfj
    Dec 18 '18 at 19:51


















11












$begingroup$


This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.



There are five properties for a relation:



Reflexive - $R rightarrow R$



Symmetrical - $R rightarrow S$ ; $S rightarrow R$



Antisymmetrical - $R rightarrow S$ && ($R rightarrow R$|| $S rightarrow S$)



Asymmetrical -$R rightarrow S$ && !($R rightarrow R$|| $S rightarrow S$)



Transitive - if $R rightarrow S$ && $S rightarrow T$, then $R rightarrow T$



If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.



Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:



1) "Is the same age as" is apparently reflexive, symmetrical and transitive.
2) "Is related to" is also apparently reflexive, symmetrical and transitive.
3) "Is older than" is asymmetric, antisymmetric and transitive.



There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.



I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?










share|cite|improve this question











$endgroup$












  • $begingroup$
    @bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
    $endgroup$
    – Charles Stewart
    Jul 27 '10 at 8:24










  • $begingroup$
    @Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
    $endgroup$
    – bryn
    Jul 28 '10 at 1:53










  • $begingroup$
    It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
    $endgroup$
    – timtfj
    Dec 18 '18 at 19:51
















11












11








11


4



$begingroup$


This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.



There are five properties for a relation:



Reflexive - $R rightarrow R$



Symmetrical - $R rightarrow S$ ; $S rightarrow R$



Antisymmetrical - $R rightarrow S$ && ($R rightarrow R$|| $S rightarrow S$)



Asymmetrical -$R rightarrow S$ && !($R rightarrow R$|| $S rightarrow S$)



Transitive - if $R rightarrow S$ && $S rightarrow T$, then $R rightarrow T$



If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.



Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:



1) "Is the same age as" is apparently reflexive, symmetrical and transitive.
2) "Is related to" is also apparently reflexive, symmetrical and transitive.
3) "Is older than" is asymmetric, antisymmetric and transitive.



There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.



I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?










share|cite|improve this question











$endgroup$




This is simply not clicking for me. I'm currently learning math during the summer vacation and I'm on the chapter for relations and functions.



There are five properties for a relation:



Reflexive - $R rightarrow R$



Symmetrical - $R rightarrow S$ ; $S rightarrow R$



Antisymmetrical - $R rightarrow S$ && ($R rightarrow R$|| $S rightarrow S$)



Asymmetrical -$R rightarrow S$ && !($R rightarrow R$|| $S rightarrow S$)



Transitive - if $R rightarrow S$ && $S rightarrow T$, then $R rightarrow T$



If that's not what you call the properties in English, then please let me know- I have to study it in German, unfortunately, and these are my rough translations.



Continuing on, I just don't know what to do with this information practically. The examples of the book are horrible:



1) "Is the same age as" is apparently reflexive, symmetrical and transitive.
2) "Is related to" is also apparently reflexive, symmetrical and transitive.
3) "Is older than" is asymmetric, antisymmetric and transitive.



There are more useless examples like this. I have no idea how it comes to these conclusions because we're talking about a literal statement. I was hoping perhaps for some real mathematical examples, but the book falls short on those.



I would greatly appreciate it if somebody could explain the above example and perhaps give me a better use for Relations other than... that. Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out?







elementary-set-theory relations






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edited Dec 19 '18 at 0:49









Andrés E. Caicedo

65.8k8160252




65.8k8160252










asked Jul 20 '10 at 19:42









IAEIAE

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512179












  • $begingroup$
    @bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
    $endgroup$
    – Charles Stewart
    Jul 27 '10 at 8:24










  • $begingroup$
    @Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
    $endgroup$
    – bryn
    Jul 28 '10 at 1:53










  • $begingroup$
    It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
    $endgroup$
    – timtfj
    Dec 18 '18 at 19:51




















  • $begingroup$
    @bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
    $endgroup$
    – Charles Stewart
    Jul 27 '10 at 8:24










  • $begingroup$
    @Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
    $endgroup$
    – bryn
    Jul 28 '10 at 1:53










  • $begingroup$
    It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
    $endgroup$
    – timtfj
    Dec 18 '18 at 19:51


















$begingroup$
@bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
$endgroup$
– Charles Stewart
Jul 27 '10 at 8:24




$begingroup$
@bryn: (Deleted old comment) Revised complaint: none of the answers gave an example exposing the difference between symmetry and anti-symmetry.
$endgroup$
– Charles Stewart
Jul 27 '10 at 8:24












$begingroup$
@Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
$endgroup$
– bryn
Jul 28 '10 at 1:53




$begingroup$
@Charles: (Deleted old comment too). To tell you the truth, I felt as though the question didn't make much sense -- it was more like "please explain the entire theory of relations including all relevant distinctions". So I thought I'd at least define the terms properly.
$endgroup$
– bryn
Jul 28 '10 at 1:53












$begingroup$
It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
$endgroup$
– timtfj
Dec 18 '18 at 19:51






$begingroup$
It seems to me that your book has chosen a particularly awkward way to e explain it. $xRy$ as in Bryn's answer—where $R$ stands for whatever relation you're considering—seems easier. Then you replace $R$ by $ge$, $=$, "speaks the same language as" or whatever, and just check which properties $R$ sstisfies.
$endgroup$
– timtfj
Dec 18 '18 at 19:51












7 Answers
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I'd like to change the notation of your definitions, since $R$, $S$ and $T$ would usually be used to stand for the relations themselves (and $x, y$ and $z$ would be more commonly chosen for the objects that might bear the relation to each other).



Reflexive - For all $x: xRx$



Example reflexive relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



Symmetric - For all $x,y$: if $xRy$ then $yRx$



Example symmetric relation: $xRy$ stands for '$x$ and $y$ are $2$ metres apart' (in the set of all people in a particular room)



Antisymmetric - For all $x,y$: if $xRy$ and $yRx$ then $x = y$



Example antisymmetric relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



Asymmetric - For all $x,y$: if $xRy$ then not $yRx$



Example asymmetric relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)



Transitive - For all $x,y,z$: if $xRy$ and $yRz$ then $xRz$



Example transitive relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)






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    Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
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    – timtfj
    Dec 18 '18 at 19:58





















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I think you're thinking about this in the wrong way. Properties don't "work", properties are things that are true for the given relation.



For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.



You could simply say that "∀a : a R a" each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.



Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.






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    I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
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    – timtfj
    Dec 19 '18 at 2:13



















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There are several important kinds of relations, each of which satisfy a different collection of properties:



Equivalence relations:
These are reflexive, symmetric, and transitive. Essentially they're relations that "behave like equality." The most important elementary one is "equivalence modulo m," where say 1 = 6 = 11 modulo 5.



Partial orderings:
These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements). Essentially they're relations that "behave like less than or equal to." An important elementary example is "divides" where we say a|b if the ratio b/a is an integer. Note that 2|6 but 6 does not divide 2. However if 2|4 and 4|8 certainly 2|8.






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    I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
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    – IAE
    Jul 20 '10 at 19:55



















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Asymmetric means simply "not symmetric". So in the binary case, it is NOT the case that if a is related to b, b is related to a.



Antisymmetric means that if a is related to b, and b is related to a, a = b.



To explain your third example:
"is older than" is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. "is older than" is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). "is older than" is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.



So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.






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  • 1




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    Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
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    – Simon Nickerson
    Jul 26 '10 at 4:52






  • 2




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    Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
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    – bryn
    Aug 3 '10 at 3:00



















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It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).



Or is it asking for mathematical relations that have these properties? ">" is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. "greater or equal" is transitive, complete, symmetric, reflexive. This is a weak total order.



"x is a subset of y" is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). "x is a proper subset of y" is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.



Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).



In logic the notion of "x entails y" is transitive, reflexive.



The relation "x has the same integer part as y" over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.






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    The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form.



    There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:




    • Eq = {(x,x) | for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.

    • All = {(x,y) | for all individuals x,y}. The whenever relation, it is also an equivalence relation.

    • Neq = {(x,y) | for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive.

    • Never is the empty set. It is symmetric and transitive, and irreflexive.


    Three basic binary operations on relations, and one unary relation:




    • R∩S = {(x,y) | (x,y) ∈ R and (x,y) ∈ S};

    • R–S = {(x,y) | (x,y) ∈ R and (x,y) ∉ S};

    • R⋅S = {(x,z) | there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition);

    • tr(R) = {(y,x) | (x,y) ∈ R} (transpose).


    Observe that Never = All–All, and Neq = All–Eq.



    We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All.



    Then we can express your five relations:




    1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq;

    2. R is symmetric when R -> tr(R);

    3. R is anti-symmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0.

    4. R is asymmetric when R ∩ tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;

    5. R is transitive when R⋅R -> R.


    Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric. In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.






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    • $begingroup$
      I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
      $endgroup$
      – bryn
      Jul 26 '10 at 8:09










    • $begingroup$
      @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
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      – Charles Stewart
      Jul 28 '10 at 21:14



















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    Other answers (bryn's especially) have done a good job of explaining the different properties. As someone who only recently started studying this stuff (I thought it was high time I learnt some analysis so bought a textbook on it), I'd like to offer some suggestions on how to get comfortable with the properties.



    From what I've seen, introductions to the subject usually want to emphasise straight away that a wide range of things—not just standard mathematical operators—can be relations. So the examples will all be things like "went to the same school as", "has more legs than", "is part of the geographical area of" and so on. This is good, because generality is part of the point. But it's not necessarily the best way to get a quick and firm grasp of the definitions.



    A useful learning principle for anything is: learn one thing at once. How can we do that with reflexive/symmetric/antisymmetric etc. relations?



    I think the answer is: start with what's already familiar to you. Which will probably be operators such as $ge$ , $=$, $subset$ and so on. Use those as your initial examples of relations.



    So my suggestion is:




    • For each property, find a completely familiar operator which has that property: for example, $=$ is reflexive.

    • Write down what the property means for that operator: for example $>$ is asymmetric because $(a>b)$ doesn't allow $(b>a)$.

    • Get a clear picture that way, then move on to the wider examples like "has the same birthday as".


    Once the properties are connected to something that's already totally easy to grasp, it's much easier to understand them and you'll be ready to go on to start using them to define equivalence classes and so on.



    Closing thought



    Some verbally defined relations can need a little extra care in thinking about their properties. For example, if $R$ means "went to the same school as", is it transitive? Does $(aRbRc) Rightarrow (aRc)$? No, because $b$ might have gone to the same primary school as $a$, then the same secondary school as $c$. This is why I think it's easiest to begin with simple operators when trying to understand the meanings of the various properties.






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      7 Answers
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      7 Answers
      7






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      8












      $begingroup$

      I'd like to change the notation of your definitions, since $R$, $S$ and $T$ would usually be used to stand for the relations themselves (and $x, y$ and $z$ would be more commonly chosen for the objects that might bear the relation to each other).



      Reflexive - For all $x: xRx$



      Example reflexive relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Symmetric - For all $x,y$: if $xRy$ then $yRx$



      Example symmetric relation: $xRy$ stands for '$x$ and $y$ are $2$ metres apart' (in the set of all people in a particular room)



      Antisymmetric - For all $x,y$: if $xRy$ and $yRx$ then $x = y$



      Example antisymmetric relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Asymmetric - For all $x,y$: if $xRy$ then not $yRx$



      Example asymmetric relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)



      Transitive - For all $x,y,z$: if $xRy$ and $yRz$ then $xRz$



      Example transitive relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)






      share|cite|improve this answer











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      • $begingroup$
        Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
        $endgroup$
        – timtfj
        Dec 18 '18 at 19:58


















      8












      $begingroup$

      I'd like to change the notation of your definitions, since $R$, $S$ and $T$ would usually be used to stand for the relations themselves (and $x, y$ and $z$ would be more commonly chosen for the objects that might bear the relation to each other).



      Reflexive - For all $x: xRx$



      Example reflexive relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Symmetric - For all $x,y$: if $xRy$ then $yRx$



      Example symmetric relation: $xRy$ stands for '$x$ and $y$ are $2$ metres apart' (in the set of all people in a particular room)



      Antisymmetric - For all $x,y$: if $xRy$ and $yRx$ then $x = y$



      Example antisymmetric relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Asymmetric - For all $x,y$: if $xRy$ then not $yRx$



      Example asymmetric relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)



      Transitive - For all $x,y,z$: if $xRy$ and $yRz$ then $xRz$



      Example transitive relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
        $endgroup$
        – timtfj
        Dec 18 '18 at 19:58
















      8












      8








      8





      $begingroup$

      I'd like to change the notation of your definitions, since $R$, $S$ and $T$ would usually be used to stand for the relations themselves (and $x, y$ and $z$ would be more commonly chosen for the objects that might bear the relation to each other).



      Reflexive - For all $x: xRx$



      Example reflexive relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Symmetric - For all $x,y$: if $xRy$ then $yRx$



      Example symmetric relation: $xRy$ stands for '$x$ and $y$ are $2$ metres apart' (in the set of all people in a particular room)



      Antisymmetric - For all $x,y$: if $xRy$ and $yRx$ then $x = y$



      Example antisymmetric relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Asymmetric - For all $x,y$: if $xRy$ then not $yRx$



      Example asymmetric relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)



      Transitive - For all $x,y,z$: if $xRy$ and $yRz$ then $xRz$



      Example transitive relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)






      share|cite|improve this answer











      $endgroup$



      I'd like to change the notation of your definitions, since $R$, $S$ and $T$ would usually be used to stand for the relations themselves (and $x, y$ and $z$ would be more commonly chosen for the objects that might bear the relation to each other).



      Reflexive - For all $x: xRx$



      Example reflexive relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Symmetric - For all $x,y$: if $xRy$ then $yRx$



      Example symmetric relation: $xRy$ stands for '$x$ and $y$ are $2$ metres apart' (in the set of all people in a particular room)



      Antisymmetric - For all $x,y$: if $xRy$ and $yRx$ then $x = y$



      Example antisymmetric relation: $xRy$ stands for '$x$ is a factor of $y$' (in the set of natural numbers)



      Asymmetric - For all $x,y$: if $xRy$ then not $yRx$



      Example asymmetric relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)



      Transitive - For all $x,y,z$: if $xRy$ and $yRz$ then $xRz$



      Example transitive relation: $xRy$ stands for '$x$ is taller than $y$' (in the set of all people)







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited May 11 '15 at 8:06







      user230715

















      answered Jul 20 '10 at 21:34









      brynbryn

      3,62283130




      3,62283130












      • $begingroup$
        Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
        $endgroup$
        – timtfj
        Dec 18 '18 at 19:58




















      • $begingroup$
        Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
        $endgroup$
        – timtfj
        Dec 18 '18 at 19:58


















      $begingroup$
      Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
      $endgroup$
      – timtfj
      Dec 18 '18 at 19:58






      $begingroup$
      Further example of asymmetry: $x>y$ rules out $y>x$. Further example of antisymmetry: if $xge y$ and $yge x$ then $x=y$.
      $endgroup$
      – timtfj
      Dec 18 '18 at 19:58













      5












      $begingroup$

      I think you're thinking about this in the wrong way. Properties don't "work", properties are things that are true for the given relation.



      For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.



      You could simply say that "∀a : a R a" each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.



      Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
        $endgroup$
        – timtfj
        Dec 19 '18 at 2:13
















      5












      $begingroup$

      I think you're thinking about this in the wrong way. Properties don't "work", properties are things that are true for the given relation.



      For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.



      You could simply say that "∀a : a R a" each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.



      Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
        $endgroup$
        – timtfj
        Dec 19 '18 at 2:13














      5












      5








      5





      $begingroup$

      I think you're thinking about this in the wrong way. Properties don't "work", properties are things that are true for the given relation.



      For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.



      You could simply say that "∀a : a R a" each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.



      Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.






      share|cite|improve this answer











      $endgroup$



      I think you're thinking about this in the wrong way. Properties don't "work", properties are things that are true for the given relation.



      For instance, you say that the a relation has the reflexive property if it satisfies the condition that all elements are related to themselves. Similarly, it has the symmetric property if for all a and b, if a is related to b then b is related to a.



      You could simply say that "∀a : a R a" each time, but because this is something that happens often, this particular property has been given a name, i.e. reflexivity.



      Certain types of specific relations are also given names. For instance, a partial order is reflexive, antisymmetric and transitive, and equivalence relations are reflexive, symmetric and transitive. Again, these are just names that aren't strictly needed, but they make it more convenient talking about these kinds of things.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jul 21 '10 at 16:22

























      answered Jul 21 '10 at 16:09









      Daniel EgebergDaniel Egeberg

      1815




      1815












      • $begingroup$
        I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
        $endgroup$
        – timtfj
        Dec 19 '18 at 2:13


















      • $begingroup$
        I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
        $endgroup$
        – timtfj
        Dec 19 '18 at 2:13
















      $begingroup$
      I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
      $endgroup$
      – timtfj
      Dec 19 '18 at 2:13




      $begingroup$
      I think your point about the names being just convenient labels is really important. First get a grasp of a certain property like $(aRb Rightarrow bRa)$, then add the fact that it's called symmetric. That way, the label is already meaningful when learnt (and it's obvious where symmetric comes from).
      $endgroup$
      – timtfj
      Dec 19 '18 at 2:13











      4












      $begingroup$

      There are several important kinds of relations, each of which satisfy a different collection of properties:



      Equivalence relations:
      These are reflexive, symmetric, and transitive. Essentially they're relations that "behave like equality." The most important elementary one is "equivalence modulo m," where say 1 = 6 = 11 modulo 5.



      Partial orderings:
      These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements). Essentially they're relations that "behave like less than or equal to." An important elementary example is "divides" where we say a|b if the ratio b/a is an integer. Note that 2|6 but 6 does not divide 2. However if 2|4 and 4|8 certainly 2|8.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
        $endgroup$
        – IAE
        Jul 20 '10 at 19:55
















      4












      $begingroup$

      There are several important kinds of relations, each of which satisfy a different collection of properties:



      Equivalence relations:
      These are reflexive, symmetric, and transitive. Essentially they're relations that "behave like equality." The most important elementary one is "equivalence modulo m," where say 1 = 6 = 11 modulo 5.



      Partial orderings:
      These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements). Essentially they're relations that "behave like less than or equal to." An important elementary example is "divides" where we say a|b if the ratio b/a is an integer. Note that 2|6 but 6 does not divide 2. However if 2|4 and 4|8 certainly 2|8.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
        $endgroup$
        – IAE
        Jul 20 '10 at 19:55














      4












      4








      4





      $begingroup$

      There are several important kinds of relations, each of which satisfy a different collection of properties:



      Equivalence relations:
      These are reflexive, symmetric, and transitive. Essentially they're relations that "behave like equality." The most important elementary one is "equivalence modulo m," where say 1 = 6 = 11 modulo 5.



      Partial orderings:
      These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements). Essentially they're relations that "behave like less than or equal to." An important elementary example is "divides" where we say a|b if the ratio b/a is an integer. Note that 2|6 but 6 does not divide 2. However if 2|4 and 4|8 certainly 2|8.






      share|cite|improve this answer









      $endgroup$



      There are several important kinds of relations, each of which satisfy a different collection of properties:



      Equivalence relations:
      These are reflexive, symmetric, and transitive. Essentially they're relations that "behave like equality." The most important elementary one is "equivalence modulo m," where say 1 = 6 = 11 modulo 5.



      Partial orderings:
      These are reflexive, transitive, and (anti-symmetric or maybe asymmetric, I'm having trouble parsing your logical statements). Essentially they're relations that "behave like less than or equal to." An important elementary example is "divides" where we say a|b if the ratio b/a is an integer. Note that 2|6 but 6 does not divide 2. However if 2|4 and 4|8 certainly 2|8.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jul 20 '10 at 19:49









      Noah SnyderNoah Snyder

      7,64232855




      7,64232855












      • $begingroup$
        I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
        $endgroup$
        – IAE
        Jul 20 '10 at 19:55


















      • $begingroup$
        I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
        $endgroup$
        – IAE
        Jul 20 '10 at 19:55
















      $begingroup$
      I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
      $endgroup$
      – IAE
      Jul 20 '10 at 19:55




      $begingroup$
      I'm so sorry about that! I changed the formatting, it's a lot easier to read now! Antisymmetric is treated by my book as, "Points to, but cannot point back. There may be reflexive loops." Asymmetric is the same, except there are no reflexive loops. As you can see, these kinds of explanations are meaningless for me. Yours seems to have some tangent property I can understand! +1
      $endgroup$
      – IAE
      Jul 20 '10 at 19:55











      2












      $begingroup$

      Asymmetric means simply "not symmetric". So in the binary case, it is NOT the case that if a is related to b, b is related to a.



      Antisymmetric means that if a is related to b, and b is related to a, a = b.



      To explain your third example:
      "is older than" is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. "is older than" is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). "is older than" is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.



      So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
        $endgroup$
        – Simon Nickerson
        Jul 26 '10 at 4:52






      • 2




        $begingroup$
        Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
        $endgroup$
        – bryn
        Aug 3 '10 at 3:00
















      2












      $begingroup$

      Asymmetric means simply "not symmetric". So in the binary case, it is NOT the case that if a is related to b, b is related to a.



      Antisymmetric means that if a is related to b, and b is related to a, a = b.



      To explain your third example:
      "is older than" is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. "is older than" is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). "is older than" is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.



      So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
        $endgroup$
        – Simon Nickerson
        Jul 26 '10 at 4:52






      • 2




        $begingroup$
        Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
        $endgroup$
        – bryn
        Aug 3 '10 at 3:00














      2












      2








      2





      $begingroup$

      Asymmetric means simply "not symmetric". So in the binary case, it is NOT the case that if a is related to b, b is related to a.



      Antisymmetric means that if a is related to b, and b is related to a, a = b.



      To explain your third example:
      "is older than" is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. "is older than" is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). "is older than" is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.



      So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.






      share|cite|improve this answer









      $endgroup$



      Asymmetric means simply "not symmetric". So in the binary case, it is NOT the case that if a is related to b, b is related to a.



      Antisymmetric means that if a is related to b, and b is related to a, a = b.



      To explain your third example:
      "is older than" is asymmetric because if Alice is older than Bob, Bob is NOT older than Alice. "is older than" is antisymmetric since if Alice is older than Bob, and Bob is older than Alice, Alice must be Bob because someone must be older (and if this is not the case, Alice simply has two names..). "is older than" is transitive since if Alice is older than Bob, and Bob is older than Charlie, Alice is also older than Charlie.



      So asymmetric and antisymmetric don't cancel out because the first means it's sort of a one-way relation, whereas the second means, loosely, that if it you reverse the operands and both statements are true, the operands must be the same.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jul 20 '10 at 19:55









      Jan GorznyJan Gorzny

      792914




      792914








      • 1




        $begingroup$
        Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
        $endgroup$
        – Simon Nickerson
        Jul 26 '10 at 4:52






      • 2




        $begingroup$
        Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
        $endgroup$
        – bryn
        Aug 3 '10 at 3:00














      • 1




        $begingroup$
        Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
        $endgroup$
        – Simon Nickerson
        Jul 26 '10 at 4:52






      • 2




        $begingroup$
        Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
        $endgroup$
        – bryn
        Aug 3 '10 at 3:00








      1




      1




      $begingroup$
      Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
      $endgroup$
      – Simon Nickerson
      Jul 26 '10 at 4:52




      $begingroup$
      Huh? "Alice must be Bob because someone must be older"?? Two people don't become the same by virtue of having the same age.
      $endgroup$
      – Simon Nickerson
      Jul 26 '10 at 4:52




      2




      2




      $begingroup$
      Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
      $endgroup$
      – bryn
      Aug 3 '10 at 3:00




      $begingroup$
      Asymmetric does not mean "not symmetric" in the context of binary relations. The relation "x is a factor of y" is not symmetric (since 3 is a factor of 6 and not vice versa). But it is not asymmetric either by any standard definition of asymmetry.
      $endgroup$
      – bryn
      Aug 3 '10 at 3:00











      2












      $begingroup$

      It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).



      Or is it asking for mathematical relations that have these properties? ">" is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. "greater or equal" is transitive, complete, symmetric, reflexive. This is a weak total order.



      "x is a subset of y" is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). "x is a proper subset of y" is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.



      Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).



      In logic the notion of "x entails y" is transitive, reflexive.



      The relation "x has the same integer part as y" over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).



        Or is it asking for mathematical relations that have these properties? ">" is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. "greater or equal" is transitive, complete, symmetric, reflexive. This is a weak total order.



        "x is a subset of y" is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). "x is a proper subset of y" is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.



        Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).



        In logic the notion of "x entails y" is transitive, reflexive.



        The relation "x has the same integer part as y" over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).



          Or is it asking for mathematical relations that have these properties? ">" is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. "greater or equal" is transitive, complete, symmetric, reflexive. This is a weak total order.



          "x is a subset of y" is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). "x is a proper subset of y" is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.



          Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).



          In logic the notion of "x entails y" is transitive, reflexive.



          The relation "x has the same integer part as y" over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.






          share|cite|improve this answer









          $endgroup$



          It's not clear what the question is asking. Is it asking for logical relations among the properties? e.g. a relation that is transitive and symmetric MUST be reflexive (unless the relation holds between no two objects).



          Or is it asking for mathematical relations that have these properties? ">" is transitive, asymmetric and total or complete (for all distinct x,y either xRy or yRx). It is a strict total order. "greater or equal" is transitive, complete, symmetric, reflexive. This is a weak total order.



          "x is a subset of y" is reflexive and antisymmetric but not necessarily transitive, symmetric (although it can be either of those things if you restrict yourself to the right sort of sets). "x is a proper subset of y" is asymmetric and irreflexive (for all x, it is not the case that xRx), and can be transitive if you restrict yourself to the right sort of sets.



          Note that in both these cases, one can define one of each pair of relations in terms of the other (plus a notion of identity or non-identity).



          In logic the notion of "x entails y" is transitive, reflexive.



          The relation "x has the same integer part as y" over the real numbers is transitive, reflexive, symmetric. This is called an equivalence relation.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 21 '10 at 15:41









          SeamusSeamus

          1,93321934




          1,93321934























              1












              $begingroup$

              The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form.



              There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:




              • Eq = {(x,x) | for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.

              • All = {(x,y) | for all individuals x,y}. The whenever relation, it is also an equivalence relation.

              • Neq = {(x,y) | for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive.

              • Never is the empty set. It is symmetric and transitive, and irreflexive.


              Three basic binary operations on relations, and one unary relation:




              • R∩S = {(x,y) | (x,y) ∈ R and (x,y) ∈ S};

              • R–S = {(x,y) | (x,y) ∈ R and (x,y) ∉ S};

              • R⋅S = {(x,z) | there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition);

              • tr(R) = {(y,x) | (x,y) ∈ R} (transpose).


              Observe that Never = All–All, and Neq = All–Eq.



              We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All.



              Then we can express your five relations:




              1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq;

              2. R is symmetric when R -> tr(R);

              3. R is anti-symmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0.

              4. R is asymmetric when R ∩ tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;

              5. R is transitive when R⋅R -> R.


              Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric. In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
                $endgroup$
                – bryn
                Jul 26 '10 at 8:09










              • $begingroup$
                @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
                $endgroup$
                – Charles Stewart
                Jul 28 '10 at 21:14
















              1












              $begingroup$

              The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form.



              There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:




              • Eq = {(x,x) | for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.

              • All = {(x,y) | for all individuals x,y}. The whenever relation, it is also an equivalence relation.

              • Neq = {(x,y) | for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive.

              • Never is the empty set. It is symmetric and transitive, and irreflexive.


              Three basic binary operations on relations, and one unary relation:




              • R∩S = {(x,y) | (x,y) ∈ R and (x,y) ∈ S};

              • R–S = {(x,y) | (x,y) ∈ R and (x,y) ∉ S};

              • R⋅S = {(x,z) | there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition);

              • tr(R) = {(y,x) | (x,y) ∈ R} (transpose).


              Observe that Never = All–All, and Neq = All–Eq.



              We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All.



              Then we can express your five relations:




              1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq;

              2. R is symmetric when R -> tr(R);

              3. R is anti-symmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0.

              4. R is asymmetric when R ∩ tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;

              5. R is transitive when R⋅R -> R.


              Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric. In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
                $endgroup$
                – bryn
                Jul 26 '10 at 8:09










              • $begingroup$
                @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
                $endgroup$
                – Charles Stewart
                Jul 28 '10 at 21:14














              1












              1








              1





              $begingroup$

              The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form.



              There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:




              • Eq = {(x,x) | for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.

              • All = {(x,y) | for all individuals x,y}. The whenever relation, it is also an equivalence relation.

              • Neq = {(x,y) | for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive.

              • Never is the empty set. It is symmetric and transitive, and irreflexive.


              Three basic binary operations on relations, and one unary relation:




              • R∩S = {(x,y) | (x,y) ∈ R and (x,y) ∈ S};

              • R–S = {(x,y) | (x,y) ∈ R and (x,y) ∉ S};

              • R⋅S = {(x,z) | there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition);

              • tr(R) = {(y,x) | (x,y) ∈ R} (transpose).


              Observe that Never = All–All, and Neq = All–Eq.



              We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All.



              Then we can express your five relations:




              1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq;

              2. R is symmetric when R -> tr(R);

              3. R is anti-symmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0.

              4. R is asymmetric when R ∩ tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;

              5. R is transitive when R⋅R -> R.


              Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric. In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.






              share|cite|improve this answer











              $endgroup$



              The calculus of relations is an algebra of operations over sets of pairs of individuals, where for any relation R, we can express the relation in the usual infix manner: x R y iff (x,y) ∈ R. This allows all of the properties of relations to be expressed in equational form.



              There are four fundamental relations we want to define, each of which make non-useless examples of three of your five properties of relations, plus one other useful property, irreflexivity:




              • Eq = {(x,x) | for each individual x}. This is the diagonal or equality relation that holds only between something and itself. It is reflexive, symmetric and transitive, which is to say it is an equivalence relation.

              • All = {(x,y) | for all individuals x,y}. The whenever relation, it is also an equivalence relation.

              • Neq = {(x,y) | for all individuals x,y for which x ≠ y}: the inequality relation that holds between anything different. It is symmetric and irreflexive, but not transitive.

              • Never is the empty set. It is symmetric and transitive, and irreflexive.


              Three basic binary operations on relations, and one unary relation:




              • R∩S = {(x,y) | (x,y) ∈ R and (x,y) ∈ S};

              • R–S = {(x,y) | (x,y) ∈ R and (x,y) ∉ S};

              • R⋅S = {(x,z) | there is some y such that (x,y) ∈ R and (y,z) ∈ S} (composition);

              • tr(R) = {(y,x) | (x,y) ∈ R} (transpose).


              Observe that Never = All–All, and Neq = All–Eq.



              We say R -> S when S holds whenever R holds. This is the same as saying either that R is a subset of S, or that R∩S=R, or that R-S=Never. So Never -> Eq, and Eq -> All.



              Then we can express your five relations:




              1. R is reflexive when Eq -> R; dually we have an additional property of a relation, where R is anti-reflexive when R -> Neq;

              2. R is symmetric when R -> tr(R);

              3. R is anti-symmetric when R ∩ tr(R) -> Eq, or equivalently when R ∩ tr(R) ∩ Neq = 0.

              4. R is asymmetric when R ∩ tr(R)=Never: a relation is asymmetric when it is anti-symmetric and anti-reflexive;

              5. R is transitive when R⋅R -> R.


              Also, how can a relation be a- and antisymmetrical at the same time? Don't they cancel each other out? — Look at Eq: it is both symmetric and anti-symmetric, although it is not asymmetric. In fact, all asymmetric relations are anti-symmetric, but not vice versa: the difference is that asymmetric and anti-symmetric differ in what they assert about the diagonal: anti-symmetric doesn't care about what pairs there might be along the diagonal, whilst asymmetric insists that there are no pairs along the diagonal, which is irreflexivity.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jul 28 '10 at 21:12

























              answered Jul 26 '10 at 1:05









              Charles StewartCharles Stewart

              3,3892329




              3,3892329












              • $begingroup$
                I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
                $endgroup$
                – bryn
                Jul 26 '10 at 8:09










              • $begingroup$
                @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
                $endgroup$
                – Charles Stewart
                Jul 28 '10 at 21:14


















              • $begingroup$
                I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
                $endgroup$
                – bryn
                Jul 26 '10 at 8:09










              • $begingroup$
                @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
                $endgroup$
                – Charles Stewart
                Jul 28 '10 at 21:14
















              $begingroup$
              I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
              $endgroup$
              – bryn
              Jul 26 '10 at 8:09




              $begingroup$
              I think you have mixed up the definitions of asymmetric and antisymmetric in this -- could you edit your post to fix this? (or indicate a reference which uses the words in the way you have).
              $endgroup$
              – bryn
              Jul 26 '10 at 8:09












              $begingroup$
              @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
              $endgroup$
              – Charles Stewart
              Jul 28 '10 at 21:14




              $begingroup$
              @bryn: fixed, at last. I find the regular terminology counter-intuitive – anti-symmetric ought to be more completely incompatible with symmetric than asymmetric, but the terminology is completely standard so far as I know.
              $endgroup$
              – Charles Stewart
              Jul 28 '10 at 21:14











              0












              $begingroup$

              Other answers (bryn's especially) have done a good job of explaining the different properties. As someone who only recently started studying this stuff (I thought it was high time I learnt some analysis so bought a textbook on it), I'd like to offer some suggestions on how to get comfortable with the properties.



              From what I've seen, introductions to the subject usually want to emphasise straight away that a wide range of things—not just standard mathematical operators—can be relations. So the examples will all be things like "went to the same school as", "has more legs than", "is part of the geographical area of" and so on. This is good, because generality is part of the point. But it's not necessarily the best way to get a quick and firm grasp of the definitions.



              A useful learning principle for anything is: learn one thing at once. How can we do that with reflexive/symmetric/antisymmetric etc. relations?



              I think the answer is: start with what's already familiar to you. Which will probably be operators such as $ge$ , $=$, $subset$ and so on. Use those as your initial examples of relations.



              So my suggestion is:




              • For each property, find a completely familiar operator which has that property: for example, $=$ is reflexive.

              • Write down what the property means for that operator: for example $>$ is asymmetric because $(a>b)$ doesn't allow $(b>a)$.

              • Get a clear picture that way, then move on to the wider examples like "has the same birthday as".


              Once the properties are connected to something that's already totally easy to grasp, it's much easier to understand them and you'll be ready to go on to start using them to define equivalence classes and so on.



              Closing thought



              Some verbally defined relations can need a little extra care in thinking about their properties. For example, if $R$ means "went to the same school as", is it transitive? Does $(aRbRc) Rightarrow (aRc)$? No, because $b$ might have gone to the same primary school as $a$, then the same secondary school as $c$. This is why I think it's easiest to begin with simple operators when trying to understand the meanings of the various properties.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Other answers (bryn's especially) have done a good job of explaining the different properties. As someone who only recently started studying this stuff (I thought it was high time I learnt some analysis so bought a textbook on it), I'd like to offer some suggestions on how to get comfortable with the properties.



                From what I've seen, introductions to the subject usually want to emphasise straight away that a wide range of things—not just standard mathematical operators—can be relations. So the examples will all be things like "went to the same school as", "has more legs than", "is part of the geographical area of" and so on. This is good, because generality is part of the point. But it's not necessarily the best way to get a quick and firm grasp of the definitions.



                A useful learning principle for anything is: learn one thing at once. How can we do that with reflexive/symmetric/antisymmetric etc. relations?



                I think the answer is: start with what's already familiar to you. Which will probably be operators such as $ge$ , $=$, $subset$ and so on. Use those as your initial examples of relations.



                So my suggestion is:




                • For each property, find a completely familiar operator which has that property: for example, $=$ is reflexive.

                • Write down what the property means for that operator: for example $>$ is asymmetric because $(a>b)$ doesn't allow $(b>a)$.

                • Get a clear picture that way, then move on to the wider examples like "has the same birthday as".


                Once the properties are connected to something that's already totally easy to grasp, it's much easier to understand them and you'll be ready to go on to start using them to define equivalence classes and so on.



                Closing thought



                Some verbally defined relations can need a little extra care in thinking about their properties. For example, if $R$ means "went to the same school as", is it transitive? Does $(aRbRc) Rightarrow (aRc)$? No, because $b$ might have gone to the same primary school as $a$, then the same secondary school as $c$. This is why I think it's easiest to begin with simple operators when trying to understand the meanings of the various properties.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Other answers (bryn's especially) have done a good job of explaining the different properties. As someone who only recently started studying this stuff (I thought it was high time I learnt some analysis so bought a textbook on it), I'd like to offer some suggestions on how to get comfortable with the properties.



                  From what I've seen, introductions to the subject usually want to emphasise straight away that a wide range of things—not just standard mathematical operators—can be relations. So the examples will all be things like "went to the same school as", "has more legs than", "is part of the geographical area of" and so on. This is good, because generality is part of the point. But it's not necessarily the best way to get a quick and firm grasp of the definitions.



                  A useful learning principle for anything is: learn one thing at once. How can we do that with reflexive/symmetric/antisymmetric etc. relations?



                  I think the answer is: start with what's already familiar to you. Which will probably be operators such as $ge$ , $=$, $subset$ and so on. Use those as your initial examples of relations.



                  So my suggestion is:




                  • For each property, find a completely familiar operator which has that property: for example, $=$ is reflexive.

                  • Write down what the property means for that operator: for example $>$ is asymmetric because $(a>b)$ doesn't allow $(b>a)$.

                  • Get a clear picture that way, then move on to the wider examples like "has the same birthday as".


                  Once the properties are connected to something that's already totally easy to grasp, it's much easier to understand them and you'll be ready to go on to start using them to define equivalence classes and so on.



                  Closing thought



                  Some verbally defined relations can need a little extra care in thinking about their properties. For example, if $R$ means "went to the same school as", is it transitive? Does $(aRbRc) Rightarrow (aRc)$? No, because $b$ might have gone to the same primary school as $a$, then the same secondary school as $c$. This is why I think it's easiest to begin with simple operators when trying to understand the meanings of the various properties.






                  share|cite|improve this answer











                  $endgroup$



                  Other answers (bryn's especially) have done a good job of explaining the different properties. As someone who only recently started studying this stuff (I thought it was high time I learnt some analysis so bought a textbook on it), I'd like to offer some suggestions on how to get comfortable with the properties.



                  From what I've seen, introductions to the subject usually want to emphasise straight away that a wide range of things—not just standard mathematical operators—can be relations. So the examples will all be things like "went to the same school as", "has more legs than", "is part of the geographical area of" and so on. This is good, because generality is part of the point. But it's not necessarily the best way to get a quick and firm grasp of the definitions.



                  A useful learning principle for anything is: learn one thing at once. How can we do that with reflexive/symmetric/antisymmetric etc. relations?



                  I think the answer is: start with what's already familiar to you. Which will probably be operators such as $ge$ , $=$, $subset$ and so on. Use those as your initial examples of relations.



                  So my suggestion is:




                  • For each property, find a completely familiar operator which has that property: for example, $=$ is reflexive.

                  • Write down what the property means for that operator: for example $>$ is asymmetric because $(a>b)$ doesn't allow $(b>a)$.

                  • Get a clear picture that way, then move on to the wider examples like "has the same birthday as".


                  Once the properties are connected to something that's already totally easy to grasp, it's much easier to understand them and you'll be ready to go on to start using them to define equivalence classes and so on.



                  Closing thought



                  Some verbally defined relations can need a little extra care in thinking about their properties. For example, if $R$ means "went to the same school as", is it transitive? Does $(aRbRc) Rightarrow (aRc)$? No, because $b$ might have gone to the same primary school as $a$, then the same secondary school as $c$. This is why I think it's easiest to begin with simple operators when trying to understand the meanings of the various properties.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 19 '18 at 1:53

























                  answered Dec 19 '18 at 0:25









                  timtfjtimtfj

                  2,503420




                  2,503420






























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