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Transitive Relations: Special case

Very Simple RelationsTransitive Relations on a setWhy is $(1,2),(3,4)$ a transitive relation?How can you elegantly show that the relation is transitive?Partial Order relation conditions (transitive)Are mathematical relations intrinsically transitive?How to find the greatest and minimal element in the set of all transitive relations over $1,2,3,4$?Definition of a transitive relation.

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

5

$begingroup$

Suppose I have a set $A = 1,2,3,4$ and I define a relation $R$ on $A$ as $R = (1,2),(3,4)$. Is this relation transitive or not?

My book says a relation is transitive if $(x,y) in R$ and $(y,z) in R$ implies $(x,z) in R$ for all $x,y,z in R$. I cannot conclude what that means for the example above.

So is a relation called transitive or not when there are no ordered pairs like $(x,y),(y,z)$ that belong to $R$?

discrete-mathematics elementary-set-theory relations

share|cite|improve this question

edited 8 hours ago

Taroccoesbrocco

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asked 9 hours ago

Ram KeswaniRam Keswani

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is transitive.
  $endgroup$
  – Berci
  9 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

Suppose I have a set $A = 1,2,3,4$ and I define a relation $R$ on $A$ as $R = (1,2),(3,4)$. Is this relation transitive or not?

My book says a relation is transitive if $(x,y) in R$ and $(y,z) in R$ implies $(x,z) in R$ for all $x,y,z in R$. I cannot conclude what that means for the example above.

So is a relation called transitive or not when there are no ordered pairs like $(x,y),(y,z)$ that belong to $R$?

discrete-mathematics elementary-set-theory relations

share|cite|improve this question

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is transitive.
  $endgroup$
  – Berci
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Suppose I have a set $A = 1,2,3,4$ and I define a relation $R$ on $A$ as $R = (1,2),(3,4)$. Is this relation transitive or not?

My book says a relation is transitive if $(x,y) in R$ and $(y,z) in R$ implies $(x,z) in R$ for all $x,y,z in R$. I cannot conclude what that means for the example above.

So is a relation called transitive or not when there are no ordered pairs like $(x,y),(y,z)$ that belong to $R$?

discrete-mathematics elementary-set-theory relations

share|cite|improve this question

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

share|cite|improve this question

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

edited 8 hours ago

Taroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

asked 9 hours ago

Ram KeswaniRam Keswani

51855 silver badges1616 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

Sure, the binary relation $R = (1,2), (3,4)$ on the set $A = 1,2,3,4$ is transitive.

Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) in R$ and $(y,z) in R$, there is nothing to check.

share|cite|improve this answer

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Yes, your relation is indeed transitive. The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter. Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation.

Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place) you can take from one element to another you can also get there by using a single step instead. Since all of your directed paths are of length $1$ anyways, there is nothing more to check.

share|cite|improve this answer

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Or one can determine the truth value of the proposition : $forall x,y,z in X : ( x rmR y land yrmR z )$ $implies$ $xrmRz $.

Let P= $( x rmR y land yrmR z )$ , Q= $xrmRz$

Suppose we have $ xrmRy$, so its value is true, the truth value of $ yrmRz$ is always false. So the truth value of $P$ is false, based on the truth value table of $land$.
On the other hand, the truth value of $Q$ is false, so the truth value of $Pimplies Q$ will be true, based on the truth value of $ implies$.It means the definition holds, so the transitive property of the relation is satisfied.

share|cite|improve this answer

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

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$endgroup$

add a comment | 

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4

$begingroup$

Sure, the binary relation $R = (1,2), (3,4)$ on the set $A = 1,2,3,4$ is transitive.

Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) in R$ and $(y,z) in R$, there is nothing to check.

share|cite|improve this answer

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Sure, the binary relation $R = (1,2), (3,4)$ on the set $A = 1,2,3,4$ is transitive.

Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) in R$ and $(y,z) in R$, there is nothing to check.

share|cite|improve this answer

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

$endgroup$

add a comment | 

$begingroup$

Sure, the binary relation $R = (1,2), (3,4)$ on the set $A = 1,2,3,4$ is transitive.

Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) in R$ and $(y,z) in R$, there is nothing to check.

share|cite|improve this answer

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

answered 9 hours ago

TaroccoesbroccoTaroccoesbrocco

7,21988 gold badges1818 silver badges4343 bronze badges

2

$begingroup$

Yes, your relation is indeed transitive. The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter. Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation.

Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place) you can take from one element to another you can also get there by using a single step instead. Since all of your directed paths are of length $1$ anyways, there is nothing more to check.

share|cite|improve this answer

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Yes, your relation is indeed transitive. The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter. Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation.

Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place) you can take from one element to another you can also get there by using a single step instead. Since all of your directed paths are of length $1$ anyways, there is nothing more to check.

share|cite|improve this answer

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

add a comment | 

$begingroup$

Yes, your relation is indeed transitive. The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter. Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation.

Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place) you can take from one element to another you can also get there by using a single step instead. Since all of your directed paths are of length $1$ anyways, there is nothing more to check.

share|cite|improve this answer

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

answered 9 hours ago

JMoravitzJMoravitz

53.7k44 gold badges4343 silver badges9393 bronze badges

answered 9 hours ago

JMoravitzJMoravitz

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1

$begingroup$

Or one can determine the truth value of the proposition : $forall x,y,z in X : ( x rmR y land yrmR z )$ $implies$ $xrmRz $.

Let P= $( x rmR y land yrmR z )$ , Q= $xrmRz$

Suppose we have $ xrmRy$, so its value is true, the truth value of $ yrmRz$ is always false. So the truth value of $P$ is false, based on the truth value table of $land$.
On the other hand, the truth value of $Q$ is false, so the truth value of $Pimplies Q$ will be true, based on the truth value of $ implies$.It means the definition holds, so the transitive property of the relation is satisfied.

share|cite|improve this answer

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Or one can determine the truth value of the proposition : $forall x,y,z in X : ( x rmR y land yrmR z )$ $implies$ $xrmRz $.

Let P= $( x rmR y land yrmR z )$ , Q= $xrmRz$

Suppose we have $ xrmRy$, so its value is true, the truth value of $ yrmRz$ is always false. So the truth value of $P$ is false, based on the truth value table of $land$.
On the other hand, the truth value of $Q$ is false, so the truth value of $Pimplies Q$ will be true, based on the truth value of $ implies$.It means the definition holds, so the transitive property of the relation is satisfied.

share|cite|improve this answer

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges

$endgroup$

add a comment | 

$begingroup$

Or one can determine the truth value of the proposition : $forall x,y,z in X : ( x rmR y land yrmR z )$ $implies$ $xrmRz $.

Let P= $( x rmR y land yrmR z )$ , Q= $xrmRz$

Suppose we have $ xrmRy$, so its value is true, the truth value of $ yrmRz$ is always false. So the truth value of $P$ is false, based on the truth value table of $land$.
On the other hand, the truth value of $Q$ is false, so the truth value of $Pimplies Q$ will be true, based on the truth value of $ implies$.It means the definition holds, so the transitive property of the relation is satisfied.

share|cite|improve this answer

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges

answered 8 hours ago

Maryam AjorlouMaryam Ajorlou

1655 bronze badges