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Trying to understand antisymmetry

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2

$begingroup$

I'm trying to understand antisymmetry in relations and I'm really confused.

I know that the defintion of antisymmetry is as follows: if $xRy$ and $yRx$ then $ x = y$.

I'm aware that the contrapositive exists: $xneq y Rightarrow left( x,yright) notin R$ or $(y,x) notin R $

Now let's take an example: the relation $R = (a , a), (b , c), (c , b), (d , d) $ on $X = a, b, c, d $ is not antisymmetric because both $(b,c)$ and $(c,b)$ are in $R$.

I am not sure to understand the justification for it being not antisymmetric. If we take the first definition of antisymmetry we see that we have $xRy$ and $yRx$ therefore we should have $x = y$. However that's not the case because if we set $b = c$ we wouldn't need to have two elements. Therefore it's not antisymmetric. Is my understanding correct?

It doesn't seem that I can use the contrapositive here.

discrete-mathematics relations

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edited 8 hours ago

J.G.

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2

$begingroup$

I'm trying to understand antisymmetry in relations and I'm really confused.

I know that the defintion of antisymmetry is as follows: if $xRy$ and $yRx$ then $ x = y$.

I'm aware that the contrapositive exists: $xneq y Rightarrow left( x,yright) notin R$ or $(y,x) notin R $

Now let's take an example: the relation $R = (a , a), (b , c), (c , b), (d , d) $ on $X = a, b, c, d $ is not antisymmetric because both $(b,c)$ and $(c,b)$ are in $R$.

I am not sure to understand the justification for it being not antisymmetric. If we take the first definition of antisymmetry we see that we have $xRy$ and $yRx$ therefore we should have $x = y$. However that's not the case because if we set $b = c$ we wouldn't need to have two elements. Therefore it's not antisymmetric. Is my understanding correct?

It doesn't seem that I can use the contrapositive here.

discrete-mathematics relations

share|cite|improve this question

edited 8 hours ago

J.G.

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asked 8 hours ago

WindBreezeWindBreeze

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$endgroup$

add a comment
 | 

$begingroup$

I'm trying to understand antisymmetry in relations and I'm really confused.

I know that the defintion of antisymmetry is as follows: if $xRy$ and $yRx$ then $ x = y$.

I'm aware that the contrapositive exists: $xneq y Rightarrow left( x,yright) notin R$ or $(y,x) notin R $

Now let's take an example: the relation $R = (a , a), (b , c), (c , b), (d , d) $ on $X = a, b, c, d $ is not antisymmetric because both $(b,c)$ and $(c,b)$ are in $R$.

I am not sure to understand the justification for it being not antisymmetric. If we take the first definition of antisymmetry we see that we have $xRy$ and $yRx$ therefore we should have $x = y$. However that's not the case because if we set $b = c$ we wouldn't need to have two elements. Therefore it's not antisymmetric. Is my understanding correct?

It doesn't seem that I can use the contrapositive here.

discrete-mathematics relations

share|cite|improve this question

edited 8 hours ago

J.G.

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asked 8 hours ago

WindBreezeWindBreeze

18988 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

J.G.

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asked 8 hours ago

WindBreezeWindBreeze

18988 bronze badges

share|cite|improve this question

edited 8 hours ago

J.G.

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asked 8 hours ago

WindBreezeWindBreeze

18988 bronze badges

edited 8 hours ago

J.G.

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edited 8 hours ago

J.G.

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asked 8 hours ago

WindBreezeWindBreeze

18988 bronze badges

asked 8 hours ago

WindBreezeWindBreeze

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2 Answers
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Whoever wrote that example probably expected you to assume the explicit elements $a,,b,,c,,d$ of $X$ are pairwise distinct (so we can talk about relations on a general size-$4$ set). If they're not, the situation is very different as you've noted. But if they are, $R$ as defined above is an antisymmetric relation on such an $X$.

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answered 8 hours ago

J.G.J.G.

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The difference between not symmetric and antisymmetric is that for not symmetric you only need one pair such that $aRb$ but not $bRa$ but you may have other pairs for which the relations go both ways.

For antisymmetric you are not allowed to have any pair which goes both sides unless the pair is $(a,a)$

For example $$R= (1,2), (2,1), (3,4)$$ is not symmetric but it is not antisymmetric while $$S=(1,1),(2,2),(3,3)$$ is antisymmetric and symmetric.

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answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

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3

$begingroup$

Whoever wrote that example probably expected you to assume the explicit elements $a,,b,,c,,d$ of $X$ are pairwise distinct (so we can talk about relations on a general size-$4$ set). If they're not, the situation is very different as you've noted. But if they are, $R$ as defined above is an antisymmetric relation on such an $X$.

share|cite|improve this answer

answered 8 hours ago

J.G.J.G.

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$endgroup$

add a comment
 | 

3

$begingroup$

Whoever wrote that example probably expected you to assume the explicit elements $a,,b,,c,,d$ of $X$ are pairwise distinct (so we can talk about relations on a general size-$4$ set). If they're not, the situation is very different as you've noted. But if they are, $R$ as defined above is an antisymmetric relation on such an $X$.

share|cite|improve this answer

answered 8 hours ago

J.G.J.G.

49.5k33 gold badges4343 silver badges6666 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Whoever wrote that example probably expected you to assume the explicit elements $a,,b,,c,,d$ of $X$ are pairwise distinct (so we can talk about relations on a general size-$4$ set). If they're not, the situation is very different as you've noted. But if they are, $R$ as defined above is an antisymmetric relation on such an $X$.

share|cite|improve this answer

answered 8 hours ago

J.G.J.G.

49.5k33 gold badges4343 silver badges6666 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

J.G.J.G.

49.5k33 gold badges4343 silver badges6666 bronze badges

answered 8 hours ago

J.G.J.G.

49.5k33 gold badges4343 silver badges6666 bronze badges

answered 8 hours ago

J.G.J.G.

49.5k33 gold badges4343 silver badges6666 bronze badges

3

$begingroup$

The difference between not symmetric and antisymmetric is that for not symmetric you only need one pair such that $aRb$ but not $bRa$ but you may have other pairs for which the relations go both ways.

For antisymmetric you are not allowed to have any pair which goes both sides unless the pair is $(a,a)$

For example $$R= (1,2), (2,1), (3,4)$$ is not symmetric but it is not antisymmetric while $$S=(1,1),(2,2),(3,3)$$ is antisymmetric and symmetric.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

55.3k44 gold badges2727 silver badges7676 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

The difference between not symmetric and antisymmetric is that for not symmetric you only need one pair such that $aRb$ but not $bRa$ but you may have other pairs for which the relations go both ways.

For antisymmetric you are not allowed to have any pair which goes both sides unless the pair is $(a,a)$

For example $$R= (1,2), (2,1), (3,4)$$ is not symmetric but it is not antisymmetric while $$S=(1,1),(2,2),(3,3)$$ is antisymmetric and symmetric.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

55.3k44 gold badges2727 silver badges7676 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The difference between not symmetric and antisymmetric is that for not symmetric you only need one pair such that $aRb$ but not $bRa$ but you may have other pairs for which the relations go both ways.

For antisymmetric you are not allowed to have any pair which goes both sides unless the pair is $(a,a)$

For example $$R= (1,2), (2,1), (3,4)$$ is not symmetric but it is not antisymmetric while $$S=(1,1),(2,2),(3,3)$$ is antisymmetric and symmetric.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

55.3k44 gold badges2727 silver badges7676 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

55.3k44 gold badges2727 silver badges7676 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

55.3k44 gold badges2727 silver badges7676 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

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