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?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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