Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

The first part of my assignment was creating 9 relations based on the set M := {A,B,C} itself with ascending cardinality

I ended up having this, I was assured that this is correct:

R1= {(A,A)}

R2= {(A,A),(A,B)}

R3= {(A,A),(A,B),(A,C)}

R4= {(A,A),(A,B),(A,C),(B,A)}

R5= {(A,A),(A,B),(A,C),(B,A),(B,B)}

R6= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C)}

R7= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A)}

R8= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B)}

R9= {(A,A),(A,B),(A,C),(B,A),(B,B),(B,C),(C,A),(C,B),( C,C)}

Ok now I need to check R1, R2, R4 for properties.

R1= {(A,A)}

R1 is not reflexive, because it does not contain (B,B) and (C,C)

R1 not sure about transitivity. Can I use (A,A) for the formal definiton aRb ^ bRc => aRc? I dont know how far I can reach out with the definition.

Do I need 3 different pairs or is one pair also enough to prove it? Like: I have (A,A) and (A,A). Therefore I also have (A,A)?

R1 is symmetrical (A,A) is the same as (A,A)

R1 is also anti symmetrical (even though I struggle with the definition, but I know that in this case it is because A and A are identical.)

R2= {(A,A),(A,B)}

R2 is not reflexive, because it does not contain (B,B) and (C,C)

R2 is transitive? Almost same case like above. Can I prove like that? (A,A) and (A,B), therefore (A,B)

R2 is not symmetrical?

R2 is anti symmetrical because (A,A) is also a part of (A,B)

R4= {(A,A),(A,B),(A,C),(B,A)}

R4 is not reflexive, because it does not contain (B,B) and (C,C)

R4 totally no clue at all if it is transitive.

R4 is not symmetrical ?

R4 is anti symmetrical because (A,A) is also in (A,B),(A,C),(B,A)?

This took some time to wrote down, I hope most of it is correct.

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

You are mostly correct.

Any one element relation is transitive by default. So $\displaystyle R_1$ is.

$\displaystyle R_4$ is **not** anti symmetrical, $\displaystyle (A,B) \wedge (B,A) \not{\Rightarrow~} A = B$

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Thank you very much!

Could you please be so kind and explain anti-symmetry for a dumbo like me? Is this arguement for example "(A,A) is part of (B,A)" correct and a solid proof?

And why is R2/R4 not symmetrical? I know it makes sense that it is not, but how would I argue?

Is R4 transitive or not?

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Quote:

Originally Posted by

**Cyganek** Could you please be so kind and explain anti-symmetry for a dumbo like me? Is this arguement for example "(A,A) is part of (B,A)" correct and a solid proof?

And why is R2/R4 not symmetrical? Is R4 transitive or not?

There is no point in re-inventing the wheel.

Have a look at this page and at this one.

BTW $\displaystyle R_4$ is transitive.

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

transitive means, as you say "if aRb and bRc then aRc". Here, where you have "ARB" symbolized by (A, B) that is "if (X, Y) and (Y, Z) then (X, Z)".

For 1) {(A, A)} that is only "if (A, A) and (A, A) then (A, A)"

Of course "if (X, X) and (X, X) then (X, X)" is always true as long as (X, X) is in the relation so we don't often worry about that and can require that "X" is not the same as "Y" and that "Y" is not the same as "Z". But in a case where there are **no** such pairs, we can use the basic logical fact that "if X is false then the statement 'if X then Y' is TRUE for all Y.

Either way we get that the relation is transitive.

I hate to contradict Plato but R4= {(A,A),(A,B),(A,C),(B,A)} **is not** transitive because

We have (B, A) and (A, C) but do NOT have (B, C).

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Quote:

Originally Posted by

**Plato** BTW $\displaystyle R_4$ is transitive.

I disagree. $\displaystyle (B,A),(A,B) \in R_4$ but $\displaystyle (B,B) \notin R_4$

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Thank you both!

One (maybe stupid question) about the last thing you wrote:

In R2 I said (A,A) ^ (A,B) => (A,B)

Cant I say the same here?

(A,A) ^ (A,B) => (A,B)

and

(A,A) ^ (A,C) => (A,C)

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

To show transitivity, you must show all pairs that satisfy the pattern yield the correct outcome. So, it is not a question of "can you say the same". You must say them to prove it is transitive.

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Quote:

Originally Posted by **HallsofIvy;804737 I hate to contradict Plato but R4= {(A,A),(A,B),(A,C),(B,A)} [b**

is not[/b] transitive because

We have (B, A) and (A, C) but do NOT have (B, C).

Quote:

Originally Posted by

**SlipEternal** I disagree. $\displaystyle (B,A),(A,B) \in R_4$ but $\displaystyle (B,B) \notin R_4$

You are both correct. Actually I don't know which one I looked at. That what I get not replying with quote.

Re: Checking relations for reflexivity, transitivity, symmetry and anti symmetry.

Okay thank you very much guys! To make it clear and sum it up:

R1= {(A,A)}

R1 is not reflexive

R1 is transitive

R1 is symmetrical

R1 is anti symmetrical

R2= {(A,A),(A,B)}

R2 is not reflexive

R2 is transitive

R2 is not symmetrical

R2 is anti symmetrical

R4= {(A,A),(A,B),(A,C),(B,A)}

R4 is not reflexive

R4 is not transitive

R4 is not symmetrical

R4 is not anti symmetrical

Is this correct?