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Questions about proof of the division article.

  1. Dec 21, 2009 #1
    (Thread should be named: Question about proof of the division algorithm, sorry about that)

    I am reading this proof of the division article:
    http://xmlearning.maths.ed.ac.uk/lecture_notes/polynomials/division_algorithm/division_algorithm.php" [Broken]

    I will write some of it here in case you don't want to click on any links.
    Theorem: Let [tex]f,g \in R[X] [/tex] be polynomials with [tex]g \neq [/tex].
    Then there exist unique polynomials [tex]q,r \in R[X] [/tex] such that [tex] f=gq+r [/tex] and [tex]deg(r)<deg(g) [/tex].

    Proof: We first prove existence of [tex]q,r[/tex].
    If [tex]deg(g)>deg(f)[/tex] then we set [tex]q=0[/tex] and [tex]r=f[/tex].
    Otherwise, [tex]deg(f) \geq deg(g)[/tex]

    [tex] f=a_0+...+a_mX^m[/tex] where [tex]a_m \neq 0[/tex].
    [tex] g=b_0+...+b_nX^n[/tex] where [tex]b_n \neq 0[/tex].

    Define the integer [tex]d=m-n \geq 0[/tex]. We will use induction in [tex]d[/tex].

    Base step: Let [tex]d=0[/tex], then [tex]m=n[/tex]. We set [tex] q=\frac{a_m}{b_m}[/tex].

    I will stop there.
    How come they set [tex] q=\frac{a_m}{b_m}[/tex]?
    Since [tex]f(x)=q(x)g(x)+r(x)[/tex] isn't then
    [tex]q(x)=\frac{f(x)}{g(x)} - \frac{r(x)}{g(x)}[/tex] ?

    Thank you for your time.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 21, 2009 #2


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    In this "base case", d= 0, the two polynomials have the same degree and so their quotient is a number, not a function of x. For example, if [itex]f(x)= ax^2+ bx+ c[/itex] and [itex]g(x)= ex^2+ fx+ g[/itex] then
    [tex]\frac{ax^2+ bx+ c}{ex^2+ fx+ g}[/tex]
    is just the number a/e with a first degree remainder.
    Last edited by a moderator: May 4, 2017
  4. Dec 21, 2009 #3
    Thank you :blushing:

    One more question (kind of the same):
    The article goes on with the induction step:

    Now we assume that this is true whenever [tex]d<k[/tex] and let [tex]d=k[/tex], so that
    [tex]m=n+k[/tex]. Let [tex]f_1=f-(\frac{a_m}{b_n}x^{m-n}g)[/tex].

    I do not understand this last step.
    Since [tex]\frac{a_m}{b_n}x^{m-n}[/tex] would be the first step of long division, does [tex]f_1[/tex] somehow refer to this?

    Feeling quite stupid here. Good thing it's Christmas soon, will drown my sorrows in food and drink.

    Thanks again!
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