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 wit?h 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}
The notion of degree allows a division algorithm to be defined on R[x]. Namely, given polynomials f(x), and g(x), with deg(f(x))<deg(g(x)), there are unique polynomials q(x) and r(x) such that g(x)=q(x)f(x)+r(x) where deg(r(x))=deg(f(x)), and deg(q(x))*deg(f(x))=deg(g(x)). The polynomial f(x) is called the divisor, g(x) is called the dividend, q(x) is called the quotient, and r(x) is called the remainder.

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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