These problems concerns polynomials, divisibility, and a similar equivalence rel

Question

These problems 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)=(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]={(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

Reflexive: Let f(x) be a polynomial in R[x], then for the constant polynomial q(x)=1, we have

f(x)=1.f(x)=q(x)f(x)

which shows that f(x)|f(x), hence | is reflexive.

Transitive: Let for the polynomials f(x),g(x) and h(x) in R[x], f(x)|g(x) and g(x)|h(x), then there exist polynomials q(x) and r(x) such that

g(x)=q(x)f(x) and   h(x)=r(x)g(x)

Hence, h(x)=r(x)q(x)f(x)

Hence for the polynomial t(x)=r(x)q(x), we have, h(x)=t(x)f(x), which shows that f(x)|h(x). Hence if f(x)|g(x) and

g(x)|h(x) then f(x)|h(x) implies that the relation | is transitive.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.