The Division Algorithm

**VonNemo19** · February 3rd 2011, 01:15 PM

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

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

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

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

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

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.

**pickslides** · February 3rd 2011, 01:39 PM

I see that

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

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

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

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

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

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

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

**chiph588@** · February 3rd 2011, 01:42 PM

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

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

**VonNemo19** · February 3rd 2011, 04:52 PM

Originally Posted by chiph588@

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

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

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.

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