1. ## The Division Algorithm

OK. I'm clearly in over my head. I bought this intro to number theory, and I figured I'd start working at it now while I'm still taking the lower level undergrad stuff so that when I get to this subject, I've had a really good introduction to it.

Having said that, I want you guys to understand that I can MAYBE do 1 of the problems given in each exercise set after the sections, but I don't want to skip the others, I want to see them worked out - that way I can get a really good grip on what's going on. So, without further adieu, can someone help me with the following?

Let $\displaystyle f(x),g(x)\in{Q}(x)$, so that

$\displaystyle f(x)=a_nx^n+...+a_0,$ $\displaystyle {a}_n\neq0$

$\displaystyle g(x)=b_mx^m+...+b_0,$ $\displaystyle {b}_m\neq0$

where the $\displaystyle a_i$ and $\displaystyle b_i$ are all rational numbers. If $\displaystyle n\geq{m}$, show that the degree of

$\displaystyle h(x)=f(x)-(\frac{a_n}{b_m})x^{n-m}g(x)$ is less than $\displaystyle n$.

I don't even know where to begin. I am sophisticated enough, however, to understand if a proof is valid or not due to my introductory course in logic. Thank you all for looking at this post.

2. I see that

$\displaystyle \displaystyle h(x)=f(x)-\left(\frac{a_n}{b_m}\right)x^{n-m}g(x)$

$\displaystyle \displaystyle h(x)=a_nx^n+...+a_0-\left(\frac{a_nx^n}{b_mx^m}\left(b_mx^m+...+b_0\ri ght)\right)$

$\displaystyle \displaystyle h(x)=a_nx^n+...+a_0-\left(a_nx^n\left(\frac{b_mx^m}{b_mx^m}+...+\frac{ b_0}{b_mx^m}\right)\right)$

$\displaystyle \displaystyle h(x)=a_nx^n+...+a_0-\left(a_nx^n\left(1+...+\frac{b_0}{b_mx^m}\right)\ right)$

$\displaystyle \displaystyle h(x)=a_nx^n+...+a_0-\left(a_nx^n+...+\frac{a_nx^n\times b_0}{b_mx^m}\right)$

$\displaystyle \displaystyle h(x)=a_nx^n+...+a_0-\left(a_nx^n+...+\frac{a_n}{b_m}\times x^{n-m}\times b_0\right)$

Notice the first term of the expression and first term inside the brackets now cancel giving the order of the greatest term being $\displaystyle <n$

3. $\displaystyle \displaystyle \frac{a_n}{b_m}x^{n-m}g(x) = a_nx^n+c_{n-1}x^{n-1} + \cdots$

Thus the leading term from above cancels out the leading term of $\displaystyle \displaystyle f(x)=a_nx^n+...+a_0$.

4. Originally Posted by chiph588@
$\displaystyle \displaystyle \frac{a_n}{b_m}x^{n-m}g(x) = a_nx^n+c_{n-1}x^{n-1} + \cdots$

Thus the leading term from above cancels out the leading term of $\displaystyle \displaystyle f(x)=a_nx^n+...+a_0$.
It's difficult to understand why I didn't see that. I mean, I can mutiply and subtract. I guess it's just that the rest of the problems in this book are so tough that I naturally expected this one to be as well. Thanks.