This problem concerns polynomials, divisibility, and a similar equivalence relat

Question

This problem concerns polynomials, divisibility, and a similar equivalence relation to that above. A polynomial of degree n in a variable x with real coefficients is the set of expressions of the form
f(x)=S(j=0)^n ajx^j =a0+a1x+…+anx^n
Where ne{0,1,2,…}=N, ajeR for je{0,1,2,…,n} and an is not equal to 0}. The set of all polynomials is the set R[x]={S(j=0)^n ajx^j =a0+a1x+…+anx^n :neN, ajeR for je{0,1,2,…,n} and an is not equal to 0}

For polynomials f(x), g(x) R[x], say that f(x)|g(x) (that is f(x) divides g(x)) if there is a polynomials q(x) such that g(x)=f(x)q(x). Thus the remainder of g(x) upon division by f(x) is 0. Show that | is a reflexive, transitive relation on R[x].

Explanation / Answer

A given binary relation ~ on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. Equivalently, for all a, b and c in A: a ~ a. (Reflexivity) if a ~ b then b ~ a. (Symmetry) if a ~ b and b ~ c then a ~ c. (Transitivity) For polynomials f(x), g(x) eR[x], f(x)~f(x) as x^2+1 divides 0 (Reflexivity) f(x)~g(x) then x^2+1 divides f(x)-g(x) and hence it also divides g(x)-f(x)Hence g(x)~f(x) (Symmetry) f(x)~g(x),g(x)~p(x) then as x^2+1 divides f(x)-g(x),g(x)-p(x) it also divides their sum which is f(x)-p(x) .Hence f(x)~p(x) (Transitivity)

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