# The Division Algorithm 2

• Feb 3rd 2011, 05:16 PM
VonNemo19
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
• Feb 4th 2011, 04:12 PM
TheEmptySet
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.
• Feb 5th 2011, 12:18 PM
VonNemo19
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?