Discrete Math: Proving p(x)|(p1(x)-p2(x)) is Equiv. Rel.

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SUMMARY

The discussion centers on proving that the relation defined by \( p(x) | (p_1(x) - p_2(x)) \) is an equivalence relation for polynomials \( p_1(x) \) and \( p_2(x) \) in the polynomial ring \( F[x] \). The key properties of an equivalence relation—reflexivity, symmetry, and transitivity—are essential for this proof. The user expresses confusion about applying these properties, particularly in demonstrating the equivalence between polynomials. The solution involves showing that if \( p_1 \sim p_2 \), then \( p | (p_1 - p_2) \) holds true, confirming the equivalence relation.

PREREQUISITES
  • Understanding of polynomial functions in the context of \( F[x] \)
  • Knowledge of equivalence relations and their properties: reflexivity, symmetry, and transitivity
  • Familiarity with divisibility in polynomials
  • Basic proof techniques in discrete mathematics
NEXT STEPS
  • Study the properties of equivalence relations in detail
  • Learn about polynomial divisibility and its implications in \( F[x] \)
  • Explore examples of equivalence relations in algebraic structures
  • Practice proving equivalence relations with various polynomial examples
USEFUL FOR

Students of discrete mathematics, particularly those struggling with polynomial equivalence relations, as well as educators seeking to clarify these concepts for learners.

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Homework Statement


Let p(x) be a polynomial in F[x].

Show that p1(x)≈p2(x) if and only if p(x)|(p1(x)-p2(x)) is an equivalence relation

The Attempt at a Solution


To be completely honest, I have no idea where to begin. This class has been a nightmare and this has been, by far, the worst professor I have ever had. No one in the class has any idea what is going on. I don't even really understand and to show the the second part of this is an equivalence relation.

Thanks in advance for the help. I won't be offended if you speak to me like I'm a small child as I am so lost in this class.
 
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Well what are the three properties of an equivalence relation?
 
tt2348 said:
Well what are the three properties of an equivalence relation?

In order for an equivalence relation to exist it must be symmetric, transitive and reflexive, but I don't know how to apply those.
 
Start out with showing p1~p1... That is, p|(p1-p1)... p1~p2 => p|(p1-p2) => p|-(p2-p1) (assuming p=/=-1) => p|(p2-p1)
Also p1~p2 and p2~p3.. What would p1-p2+(p2-p3) look like?
 

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