Proving Equivalence Relations with a Given Condition

AI Thread Summary
The discussion revolves around proving that a defined relation R on the set A = {1,2,3,4,5} X {1,2,3,4,5}, where (x1,y1)R(x2,y2) if x1+y1=x2+y2, is an equivalence relation. Participants clarify the need to demonstrate the reflexive, symmetric, and transitive properties using the condition x1+y1=x2+y2. Examples are provided to illustrate each property, confirming that R satisfies them. The relation R is established as an equivalence relation, distinct from the equality relation on A. The conversation emphasizes the importance of using the defining condition to validate the properties required for equivalence relations.
Dream
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Hi,
Here is my question. I need to prove the following an equivalence relation.
Let A = {1,2,3,4,5} X {1.2,3,4,5} and define a relation R on A by (x1,y1)R{ x2,y2) if x1+y1=x2+y2.

I am bit confused how to use the condition x1+y1=x2+y2 to prove for transitive, symmetric and reflexive properties.
Please help.
Thanks
 
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Can you show that "=" is an equivalence relation on the integers?
 
Thanks for your reply John.
Ya sure. If I have got your question correctly. I have done this.
Let S be a nonempty set of integers and let
equality = be our relation. Then = is an equivalence relation on
S since
(i) a = a for all a S,
(ii) if a = b, then b = a (for all a,b),
(iii) if a = b and a = b then a = c (for all a,b,c ).

I am eager to know whether I am correct on this or not.
 
I think on you probably meant: (iii) If a = b and b = c then a = c. The point is, at any rate, that it's easy to see that x1+y1 and x2+y2 from your original post are both integers.
 
Dream said:
Hi,
Here is my question. I need to prove the following an equivalence relation.
Let A = {1,2,3,4,5} X {1.2,3,4,5} and define a relation R on A by (x1,y1)R{ x2,y2) if x1+y1=x2+y2.

I am bit confused how to use the condition x1+y1=x2+y2 to prove for transitive, symmetric and reflexive properties.
Please help.
Thanks


The relation R that you defined is an equivalence relation on A;
it is not, however, the equality relation on A.

I suggest you state the reflexive, symmetric and transitive properties for binary relations in general.

Then verify that R satisfies all three.
 
Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?
 
Use? Who knows. It's just an example to see if you can prove something is an equivalence relation.

You've not written a proof yet.

So, does (x,y)R(x,y)?
Does (x,y)R(u,v) imply (u,v)R(x,y)?
Does (x,y)R(u,v) and (u,v)R(s,t) imply (x,y)R(s,t)?
 
Dream said:
Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?


Here's some supplementary information to consider.

R is a subset of AxA.

x1+y1 = x1+y2 is the property or predicate (call it P) that defines R, i.e.,
R = {((x1,y1),(x2,y2)) | x1+y1 = x2+y2} (loosely stated).

What use is P? P must be used to establish the three properties in question.
There is no escape here. It must be used.
What I think you'll see is that the properties in question will be inherited
from P itself. Think about '='.

Invariably, for me at least, generating specific examples gives confidence
that something can be established in general.
Here's an example for each property.

Reflexivity:
Ask yourself, is every member of A in relation with itself (all 25 of them)? Of course, yes.
For example, is ((1,2),(1,2)) in R?

Symmetry:
Example: Clearly, ((3,4),(5,2)) is in R. Then we must have that ((5,2),(3,4)) in R and it is.

Transitivity:
Example: Clearly, ((5,1),(4,2)) and ((4,2),(3,3)) are in R. Then we must have ((5,1),(3,3)) in R and it is.
 
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Thanks a lot... :-)
 
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