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Math Help - The Division Algorithm

  1. #1
    No one in Particular VonNemo19's Avatar
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    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 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.
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  2. #2
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    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
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  3. #3
    MHF Contributor chiph588@'s Avatar
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     \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 .
    Last edited by chiph588@; February 3rd 2011 at 03:53 PM.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chiph588@ View Post
     \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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