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Math Help - The Division Algorithm 2

  1. #1
    No one in Particular VonNemo19's Avatar
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    The Division Algorithm 2

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

    f(x)=g(x)q(x)+r(x)

    and either

    r(x)\equiv0, or

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

    Subcase (1) \text{degree }g(x)>\text{degree }f(x)
    Prove directly
    Subcase (2) \text{degree }f(x)\geq\text{degree }g(x)
    Form polynomial 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 h(x) was discussed in the previous problem here ---> http://www.mathhelpforum.com/math-he...hm-170136.html
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    The cases a) doesn't really have much to it.

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

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

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

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

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

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

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

    But this doesn't matter right?
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