Let R[x] denote the set of polynomials in x with real coefficients. Fix p(x) ? R

Question

Let R[x] denote the set of polynomials in x with real coefficients. Fix p(x) ? R[x]. Define a relation ~ on R[x] by f ~ g if f(x) - g(x) is a multiple of p(x).

Prove it is a equivalence relation.

Explanation / Answer

Reflexive: ~ is clearly reflexive, as f(x)-f(x)=0=0*p(x). Symmetric: Let f(x)~g(x). Then f(x)-g(x)=p(x)*r(x), where r(x) is an element of R[x] ==> -1[f(x)-g(x)]=-p(x)*r(x) ==> g(x)-f(x) = p(x)*s(x); where s(x)=-r(x). Then g(x)~f(x), and ~ is Symmetric. Transitive. Let, f(x), g(x) and h(x) be elements of R[x] such that f(x)~g(x) and g(x)~h(x). Then f(x)-g(x)=p(x)*r(x), and g(x)-h(x)=p(x)*s(x), for some r(x), s(x) in R[x]. Then g(x)=p(x)*s(x)+h(x). Thus f(x)-g(x)=p(x)*r(x) ==> f(x)-[p(x)*s(x)+h(x)]=p(x)*r(x) ==> f(x)-h(x)-p(x)*s(x)=p(x)*r(x) ==> f(x)-h(x)=p(x)[r(x)+s(x)]. Therefore f(x)~h(x). Therefre ~ is an equivalence relation.

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