Which Relations Satisfy Specific Equivalence Conditions?

Click For Summary

Homework Help Overview

The discussion revolves around identifying relations that satisfy specific equivalence conditions: reflexive, symmetric, and transitive. Participants are exploring examples of relations that meet various combinations of these properties.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of relations, such as reflexivity, symmetry, and transitivity, with attempts to provide examples. Questions arise regarding the distinctions between relations that are only reflexive, only symmetric, or only transitive, as well as combinations of these properties.

Discussion Status

Some participants are offering examples and reasoning about specific relations, while others express confusion about the definitions and differences between the properties. There is an ongoing exploration of various relations and their characteristics, with some guidance being provided through examples.

Contextual Notes

Participants reference a textbook for the course and indicate that the original problem stems from a first course in Abstract Algebra. There is mention of a lack of clarity in the initial explanation of the topic, contributing to the confusion among participants.

vvvidenov
Messages
23
Reaction score
0

Homework Statement


Find relations that satisfying
just Reflexive
just Symmrtic
just Transitive

(R) & (S), but not (T)
(R) & (T), but not (S)
(S) & (T), but not (R)




Homework Equations


S=Z

(a,b) \inR if <=> a>b (T) but, not (S) & (R).

the ex is given in the class, but nothing else was explained. I am confused. Please help if you can.



The Attempt at a Solution



How about some cartisian product.
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
(3,1) (3,2) (3,3)

for (1,1) we have equiv. rel (R, S & T)
 
Physics news on Phys.org
The problem above is first course of Abstarct Algebra. We use textbook (Dan Saracino, second edition).
 
(a,b) is in R iff a>b. Then (a,a) is not in R as a>a is false. (a,b) in R means (b,a) is not in R as if a>b, b>a is false. And if (a,b) and (b,c) are in R, a>b>c implies a>c so (a,c) is in R. Note in this example (1,1) is NOT in R as 1>1 isn't true.

I think a more intuitive notation is aRb to denote 'a is related to b' where R is a relation. Then aRb iff a>b it's clear that if aRb is true, bRa is false since a>b implies b>a is false. Similarly, aRa is false since a>a is false. But aRb, bRc implies a>b, b>c so a>b>c implies a>b and aRc holds. hence > is a relation which is transitive but not symmetric or reflexive.

Try to think up other relations that satisfy, say, aRa but not aRb -> bRa or aRb, bRc -> aRc
 
how about
aRb iff a-b\geq0
not (S), but is (T) and (R)
 
I can't think of example that is only (S) , and only (T). I don't think I understand the differences og being only (R,) only (S), only (T), and when we have (T), but not (R) & (S) and so on.
 
Thank you so much. I found similar questions with answers and I get it now.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 17 ·
Replies
17
Views
11K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K