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Equivalence Relation Problem

by Dream
Tags: equivalence, relation
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Dream
#1
Jun4-07, 01:51 AM
P: 4
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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JohnDuck
#2
Jun4-07, 02:36 AM
P: 74
Can you show that "=" is an equivalence relation on the integers?
Dream
#3
Jun4-07, 02:48 AM
P: 4
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.

JohnDuck
#4
Jun4-07, 03:04 AM
P: 74
Equivalence Relation Problem

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.
fopc
#5
Jun4-07, 03:15 AM
P: 90
Quote Quote by Dream View Post
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.
Dream
#6
Jun4-07, 04:00 AM
P: 4
Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?
matt grime
#7
Jun4-07, 05:34 AM
Sci Advisor
HW Helper
P: 9,396
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)?
fopc
#8
Jun4-07, 06:55 AM
P: 90
Quote Quote by Dream View Post
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.
Dream
#9
Jun5-07, 02:10 AM
P: 4
Thanks a lot... :-)


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