Results 1 to 3 of 3

Thread: The Division Algorithm 2

  1. #1
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,849

    The Division Algorithm 2

    The division algorithm has an analoguen in $\displaystyle Q(x)$ that asserts: given two polynomials $\displaystyle f(x),g(x),g(x)$ not identically 0, there exists $\displaystyle q(x),r(x)\in{Q}(x)$ such that

    $\displaystyle f(x)=g(x)q(x)+r(x)$

    and either

    $\displaystyle r(x)\equiv0$, or

    $\displaystyle 0\leq\text{degree of }r(x)<\text{degree of }g(x)$.

    Prove this. One route is to proceed by the following steps:

    a) If $\displaystyle f(x)\equiv0$ or the degree of $\displaystyle f(x)$ is 0, prove the assertion.
    b) Proceed by induction on the degree of $\displaystyle f(x)$ and assume results for all cases where $\displaystyle f(x)$ is of degree less than $\displaystyle n$. Let $\displaystyle f(x)$ be of degree $\displaystyle n$.

    Subcase (1) $\displaystyle \text{degree }g(x)>\text{degree }f(x)$
    Prove directly
    Subcase (2) $\displaystyle \text{degree }f(x)\geq\text{degree }g(x)$
    Form polynomial $\displaystyle h(x)=f(x)-(\frac{a_n}{b_n})x^{n-m}g(x)$ and use inductive hypothesis.


    OK. If someone could just do the first part of this, a), then I think that MAYBE I can do the rest. Please don't just go and post an entire solution. I really want to try and do some of this. I just can't get started.

    Thanks to anyone willing to walk me through this.

    Note: The polynomial $\displaystyle h(x)$ was discussed in the previous problem here ---> http://www.mathhelpforum.com/math-he...hm-170136.html
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    The cases a) doesn't really have much to it.

    If $\displaystyle f(x) \in \mathbb{Q}[x]$ and has degree 0 this means that

    $\displaystyle f(x)=m$ for some $\displaystyle m \in \mathbb{Q}$

    If the degree of $\displaystyle g(x)$ is larger than 0

    Then $\displaystyle f(x)=0\cdot g(x)+m$

    if the degree of $\displaystyle g(x)$ is 0 then $\displaystyle g(x)=c,\quad c \in \mathhbb{Q}$

    Then $\displaystyle \displaystyl f(x)=\left( \frac{m}{c}\right)c+0=g(x)q(x)+r(x)$

    All of this "division" is taking place in the field or rational numbers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,849
    So, since we are just establishing existence, it doesn't matter that when $\displaystyle g(x)$ has degree 0, and $\displaystyle f(x)=\frac{m}{c}c+0$, that this is not unique. Because it isn't necessary that $\displaystyle r(x)=0$ because we could just as easily have

    $\displaystyle f(x)=mc+r$, $\displaystyle r\neq0$.

    But this doesn't matter right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Division Algorithm.....
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Jun 26th 2010, 06:53 AM
  2. Division algorithm/mod
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Feb 13th 2010, 12:51 PM
  3. Division Algorithm
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jan 18th 2009, 03:11 PM
  4. Division Algorithm...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Jul 8th 2008, 08:33 PM
  5. division algorithm
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Sep 7th 2007, 01:39 PM

Search Tags


/mathhelpforum @mathhelpforum