text: # Homework Help: Proving something involving real polynomials
  
  1. Mar 21, 2013
  
  ### Whovian
  
  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. Mar 21, 2013
  
  ### 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.
  
  3. Mar 21, 2013
  
  ### Whovian
  
  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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