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The Division Algorithm 2
The division algorithm has an analoguen in )
that asserts: given two polynomials ,g(x),g(x))
not identically 0, there exists ,r(x)\in{Q}(x))
such that
=g(x)q(x)+r(x))
and either
\equiv0)
, or
<\text{degree of }g(x))
.
Prove this. One route is to proceed by the following steps:
a) If \equiv0)
or the degree of )
is 0, prove the assertion.
b) Proceed by induction on the degree of and assume results for all cases where is of degree less than 
. Let be of degree .
Subcase (1) >\text{degree }f(x))
Prove directly
Subcase (2) \geq\text{degree }g(x))
Form polynomial =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 )
was discussed in the previous problem here ---> http://www.mathhelpforum.com/math-he...hm-170136.html
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The cases a) doesn't really have much to it.
If ![f(x) \in \mathbb{Q}[x]](http://latex.codecogs.com/png.latex?f(x) \in \mathbb{Q}[x])
and has degree 0 this means that
=m)
for some 
If the degree of )
is larger than 0
Then =0\cdot g(x)+m)
if the degree of is 0 then =c,\quad c \in \mathhbb{Q})
Then =\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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So, since we are just establishing existence, it doesn't matter that when has degree 0, and =\frac{m}{c}c+0)
, that this is not unique. Because it isn't necessary that =0)
because we could just as easily have
=mc+r)
, 
.
But this doesn't matter right?
