Equivalence Classes in PxP for (1,2)

Click For Summary

Homework Help Overview

The problem involves defining an equivalence relation on the set PxP using the notation (m,n)≈(p,q) if m*q=p*n. The task is to demonstrate that this relation satisfies the properties of reflexivity, symmetry, and transitivity, and to identify three elements in the equivalence class for (1,2).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to begin the problem, with one suggesting to list pairs to understand the relation better. Another participant outlines the properties of reflexivity, symmetry, and transitivity, providing explicit checks for these properties.

Discussion Status

The discussion includes attempts to verify the properties of the equivalence relation, with one participant providing a detailed explanation of the reflexive, symmetric, and transitive properties. There is also a mention of specific elements in the equivalence class for (1,2), indicating some progress in the exploration of the problem.

Contextual Notes

Participants express confusion regarding the notation used in the equivalence relation and the initial steps needed to approach the problem. There is an emphasis on explicitly checking the properties of the relation.

mamma_mia66
Messages
51
Reaction score
0

Homework Statement



On set PxP, define (m,n)\approx(p,q) if m*q=p*n
Show that \approx is an equivalence relation on PxP and list three elements in equivalence class for (1,2)



Homework Equations







The Attempt at a Solution


I will appreciate any help how to start this problem. I now I have to show R, S and T properties, but I am confused from the notation above m*q=p*n

do I have to start with listing some pairs like
(0,0) (0,1) (1,0) (0,2) (1,1) (2,0) (0,3) (1,2) (2,1) (3,0) ...
 
Physics news on Phys.org
For reflexive, show that (m, n) ~ (m, n) is true. For symmetric, show that if (m, n) ~ (p, q) then (p, q) ~ (m, n). Just check these explicitly to see if they work out.
 
How is that?

Reflexive: b/c m*n=m*n then (m,n)\approx(m,n)

Symmetric: if (m,n)\approx(p,q)

then m*q=p*n and p*n=m*q

=> (p,q)\approx(m,n)

Transitive: if (m,n) \approx(p,q) and (p,q)\approx(r,s)

then m*q=p*n and p*s=r*q

m/n=p/q and p/q=r/s

m/n=r/s

m*s=r*n

(m,n)\approx(r,s)

the three elements:
(1,2)= (3,6) (1,2)= (4,8) (1,2)= (5,10)
 
Looks fine to me.
 

Similar threads

Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K