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Proving something involving real polynomials

  1. Mar 21, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove, for part of a homework problem, that if the ratio of two polynomials ##p## and ##q## with real coefficients is a polynomial, then all of its coefficients are real.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Well, we can first note that for real ##x##, ##p\left(x\right)## and ##q\left(x\right)## are real, and so ##\lim\limits_{a\to x}\left(\dfrac{p\left(a\right)}{q\left(a\right)}\right)## is real (remember that ##\dfrac pq## is a polynomial, and thus ##p## is divisible by ##q##.) I seem to be stuck here proving that any polynomial ##\mathbb{R}\to\mathbb{R}## has real coefficients. Any ideas?

    EDIT: Oh wait. I think maybe induction and differentiation might help?
     
    Last edited: Mar 21, 2013
  2. jcsd
  3. Mar 21, 2013 #2

    Mark44

    Staff: Mentor

    I don't think you need calculus (including limits) here. You have p(x)/q(x) = r(x), where all three functions are polynomials. This implies that p(x) = r(x) * q(x). I haven't taken this any further, but what I would do next is to write the terms of the three polynomials, noting that the sum of the degrees of r and q has to be equal to the degree of p. Then look at how each term of r(x) is computed.
     
  4. Mar 21, 2013 #3
    Hmm. That would be

    $$p_{a+b}\cdot x^{a+b}+p_{a+b-1}\cdot x^{a+b-1}+\ldots+p_0=\left(r_a\cdot x^a+r_{a-1}\cdot x^{a-1}+\ldots+r_0\right)\cdot\left(q_b\cdot x^b+q_{b-1}\cdot x^{b-1}+\ldots+q_0\right)$$

    Examining each term, ##p_0=r_0\cdot q_0##, so ##r_0\in\mathbb{R}## (the projective reals are closed under division.) ##p_1=r_1\cdot q_0+r_0\cdot q_1## (hey, this resembles another approach I took to the problem!) Thus ##r_1\in\mathbb{R}##.

    Ah, I see where this is going. Thanks!

    Though the limits probably still are necessary, otherwise the domain of ##\dfrac pq## is ##\mathbb{R}\setminus\left\{x\ |\ q\left(x\right)=0\right\}##.
     
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