Solve Equivalence Relation Homework: Prove ~ is Equivalent

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Homework Help Overview

The problem involves proving that a defined relation ~ on the set of ordered pairs of natural numbers, N^2, is an equivalence relation. The relation is defined such that (n, m) ~ (k, l) if and only if n + l = m + k. The task requires demonstrating that this relation is reflexive, symmetric, and transitive.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to prove the reflexive and transitive properties of the relation. There is confusion about the notation ~ and its dual role as both the name of the relation and as part of the relation itself. Some participants suggest checking specific cases to clarify the definitions.

Discussion Status

Some guidance has been provided regarding the reflexivity and transitivity proofs, with participants encouraging the original poster to apply the definition of the relation directly. There is an ongoing exploration of how to articulate the equivalence classes formed by the relation.

Contextual Notes

Participants express confusion regarding the notation and its implications, indicating a need for further clarification on the concept of equivalence relations and their properties.

thenoob
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Homework Statement


We define a relation ~ for N^2 by:

(n, m) ~(k, l) <=> n + l = m + k

Show that ~ is a equivalence relation


Homework Equations



A relation R on a set A is equivalent if R is:
reflexive if x R x for all x that is an element of A
symmetric if x R y implies y R x, for all x,y that is an element of A
transitive if x R y and y R z imply x R z, for all x,y,z that is an element of A

The Attempt at a Solution


I have to prove this by showing that the relation is reflexive, symmetric and transitive. The symmetric part i understand, the fact that we can turn the relation around and it will be the same: n + l = m + k vs m + k = n + l.

But how do I prove the refexive and transitive part? I am really confused by the ~ and do not quite understand why it appears both as the name of the relation and as an element of the relation itself.
 
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To prove reflexivity, you must check whether (x,y) ~ (x,y), i.e. whether any element a of N^2 satisfies a ~ a . By using the definition of ~ that you gave, can you check that (x,y) ~ (x,y)?

For transitivity, you must check whether (a,b) ~ (c,d) and (e,f) ~ (g,h) implies (a,b) ~ (g,h). Again, its just a matter of using the definition of ~.
 
Last edited:
dx said:
To prove reflexivity, you must check whether (x,y) ~ (x,y), i.e. whether any element a of N^2 satisfies a ~ a . By using the definition of ~ that you gave, can you check that (x,y) ~ (x,y)?

You mean by writing (x, y) ~ (x, y) <=> x + y = x + y ?

dx said:
For transitivity, you must check whether (a,b) ~ (c,d) and (e,f) ~ (g,h) implies (a,b) ~ (g,h). Again, its just a matter of using the definition of ~.

I think my problem is that I get really confused by the ~ and what it means.
 
~ is a relation, just like < or >. Just like you can check whether a < b, you can check whether (a,b) ~ (c,d) using the definition of ~ that you gave.
 
Welcome to PF!

Hi thenoob! Welcome to PF! :smile:

Maybe this will help … :smile: (and maybe it won't! :redface:)

An equivalence relation separates N^2 into mutually exclusive sets.

Anything in one set ~ anything else in that set, but ~ nothing in any other set.

Everything is in exactly one set.

So how would you describe the set (the equivalence class) that (1,0) is in (if necessary, pick a few examples, and find a pattern)? :smile:
 

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