I'm having a bit of trouble understanding exactly what this question is asking me in my Sets and Proofs homework:
If a polynomial p over $\mathbb{R}$ is an expression of the form $p(x) = a_nx^n + … + a_1x+a_0$, is the relation $p\sim q \text{ if and only if } p(0) = q(0)$ an equivalence relation?
I know for a relation to be an equivalence relation, it must be reflexive symmetric and transitive, but I don't exactly understand how I can test/prove it in this situation.
Thank you!
EDIT: Ok, I think I've figured some of it out, but I'm not sure if my logic is right. So, it's symmetric because $q\sim p$ means $q(0) = p(0)$, and since $p(0) = q(0)$ we know this is true.
And it's reflexive because $p\sim q$ means $p(0) = p(0)$, which is trivially true. What I don't understand is how to prove it is transitive.
Best Answer
The condition $p\sim q$ if and only if $p(0)=q(0)$ implies $p$ and $q$ have the same constant term. As you have already shown reflexivity and symmetry, we have only transitivity to verify. Thus assume $p\sim q$ and $q \sim r$. Then $p(0)=q(0)$ and $q(0)=r(0)$, which implies $p(0)=r(0)$, finishing the result.