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

hammonjj
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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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