
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 $\displaystyle h(x)=f(x)(\frac{a_n}{b_n})x^{nm}g(x)$ 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/mathhe...hm170136.html

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.

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?