Equivalence Classes in PxP for (1,2)

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SUMMARY

The discussion focuses on demonstrating that the relation defined on the set PxP, where (m,n) is equivalent to (p,q) if m*q = p*n, is indeed an equivalence relation. The properties of reflexivity, symmetry, and transitivity are validated through explicit examples. Additionally, three elements in the equivalence class for (1,2) are identified as (3,6), (4,8), and (5,10), confirming the equivalence relation's application.

PREREQUISITES
  • Understanding of equivalence relations in mathematics
  • Familiarity with the notation of ordered pairs
  • Knowledge of reflexive, symmetric, and transitive properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of equivalence relations in more depth
  • Explore examples of equivalence classes in different mathematical contexts
  • Learn about the applications of equivalence relations in set theory
  • Investigate the implications of equivalence relations in modular arithmetic
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Students studying abstract algebra, mathematicians interested in set theory, and educators teaching equivalence relations and their properties.

mamma_mia66
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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) ...
 
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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.
 

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