Given the Set S ={1,2,3,4}. Define a relation on S that

[knowLittle]

Homework Statement

a.) Is symmetric and transitive, but not reflective:

b.) consists of exactly 8 ordered pairs and is symmetric and transitive:

The Attempt at a Solution

If the question asks me to define some relation, do I need to define some math property like power of some number or something like that or can I just state some set R that satisfies what they require?
R = { (1,2), ( 1,3 ), (1,4 ) , ( 2,3) , (2,1) , (2,4) , ( 3,1) , (3,2) , (3,4) , (4,1), (4,2), (4,3) }
Is this correct?

About part
b.) Any idea

 

[PeroK]

For b) R cannot be reflexive, but you could still have aRa for some a, just not all.

Does that help?

 

[pasmith]

Homework Statement

a.) Is symmetric and transitive, but not reflective:

b.) consists of exactly 8 ordered pairs and is symmetric and transitive:

The Attempt at a Solution

If the question asks me to define some relation, do I need to define some math property like power of some number or something like that or can I just state some set R that satisfies what they require?
R = { (1,2), ( 1,3 ), (1,4 ) , ( 2,3) , (2,1) , (2,4) , ( 3,1) , (3,2) , (3,4) , (4,1), (4,2), (4,3) }
Is this correct?

You have (1,2) and (2,1) in that set, so if the relation is transitive then (1,1) should be in there, but it isn't. And in fact transitivity requires that all ordered pairs be in that set, so the relation is reflexive.

 

[knowLittle]

You have (1,2) and (2,1) in that set, so if the relation is transitive then (1,1) should be in there, but it isn't. And in fact transitivity requires that all ordered pairs be in that set, so the relation is reflexive.

But, the problem states that we don't want transitivity. Should the solution be that there does not exist such relation, then?

 

[pasmith]

For b) R cannot be reflexive, but you could still have aRa for some a, just not all.

Actually a symmetric and transitive relation on $\{1,2,3,4\}$ which consists of exactly eight ordered pairs must be reflexive.

There are only four diagonal pairs, so you must have aRb for some distinct a and b.

Symmetry requires that if aRb for distinct a and b then bRa, and hence transitivity requires aRa and bRb. That's four ordered pairs.

If you add aRc for $c \notin \{a,b\}$ then you must also add cRa, cRc, bRc and cRb, which is another five.

 

[knowLittle]

So, you are saying that there exists a relation of 9 ordered pairs, but not 8?

 

[pasmith]

But, the problem states that we don't want transitivity. Should the solution be that there does not exist such relation, then?

Part (a) states that you don't want reflexivity.

There are many relations on {1,2,3,4} which are symmetric and transitive but not reflexive.

 

[pasmith]

So, you are saying that there exists a relation of 9 ordered pairs, but not 8?

That's not what I'm saying.

Also, part (b) does not require that your relation not be reflexive.

 

[knowLittle]

For part a.)
Symmetry does not imply reflexivity?
R= { (1,2) , (2,1 ) , (1,1), ...}

 

[pasmith]

For part a.)
Symmetry does not imply reflexivity?

R= { (1,2) , (2,1 ) , (1,1), ...}

What's the "..."?

 

[knowLittle]

What's the "..."?

I was just trying to show the simplest case, when if symmetry and transitivity is required, then reflexivity is implied. So, this is my question. Does symmetry and transitivity imply reflexivity?
The dots meant some other pairs I didn't want to think about.

 

[knowLittle]

Nevermind, the properties are defined only when every $x \in S$ is defined in R.

 

[knowLittle]

Sorry, but my concepts were off.

The solution to both could be this:

$R = \{ (1,2), (1,1) , (1,3) , (2,1), (2,3) , (2,2) , (3,1), (3,2) \}$

Is this correct?

 

[pasmith]

I was just trying to show the simplest case, when if symmetry and transitivity is required, then reflexivity is implied. So, this is my question. Does symmetry and transitivity imply reflexivity?

No. {(1,1)} is a relation on S which is symmetric and transitive but is not reflexive.

Sorry, but my concepts were off.

The solution to both could be this:

$R = \{ (1,2), (1,1) , (1,3) , (2,1), (2,3) , (2,2) , (3,1), (3,2) \}$

Is this correct?

No, because $(3,3) \notin R$ so R is not transitive: 3R1 and 1R3 require 3R3.

Start with R = {(1,1), (1,2), (2,1), (2,2)}. How can you add four more pairs while keeping R symmetric and transitive? You can't add (1,3) or (2,3) or (3,1) or (3,2), because adding anyone of those to R requires adding the other three and (3,3) to R. That's an extra five pairs.

 

[PeroK]

No. {(1,1)} is a relation on S which is symmetric and transitive but is not reflexive.

An interesting option would be the relation represented by the empty set! That would be symmetric and transitive but not reflexive.