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Proof of Rational Root Theorem

  1. Nov 14, 2005 #1
    I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:

    Say we have a polynomial:

    [tex] F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0} [/tex]

    where all of the coefficients are integers and [tex] a_{0} [/tex] does not equal zero.

    I want to prove that there is a factor of [tex]a_{n}[/tex], call it q, and a factor of [tex]a_{0}[/tex], call it p, so that:

    [tex] F(\frac{p}{q}) = 0 [/tex]

    p and q are prime numbers. So I want to prove that p is a factor of [tex]a_{0}[/tex] and q is a factor of [tex]a_{n}[/tex]. Next, I insert p/q into F(x):

    [tex] F(\frac{p}{q}) = a_{n}\left(\frac{p}{q}\right)^{n} + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + ... + a_{1}\left(\frac{p}{q}\right) + a_{0} = 0 [/tex]

    multiplying by [tex] q^{n} [/tex] we get:

    [tex] a_{n}p^{n} + a_{n-1}p^{n-1}q + ... + a_{1}p q^{n-1} + a_{0}q^{n} = 0 [/tex]

    and factorizing out p:

    [tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

    As p, q are prime numbers, the Greatest Common Devisor of p and q is 1, i.e. GCD(p, q) = 1. And hence, p devides [tex]a_{0}[/tex] and is then a factor of [tex]a_{0}[/tex].

    As the same kind of argument applies to q being a factor of [tex]a_{n}[/tex], I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in:

    [tex] p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n} [/tex]

    how do I prove that: [tex] a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1} [/tex] is an integer?
     
    Last edited: Nov 14, 2005
  2. jcsd
  3. Nov 14, 2005 #2
    You won't be able to prove that. Exercise: find a polynomial with integer coefficients which doesn't have any rational roots.

    The "rational root" theorem that /I'm/ familiar with states that if p/q is a rational root of F and gcd(p, q) = 1, then q is a factor of a_n and p is a factor of a_0. (Note that this theorem doesn't guarantee that F actually has any rational roots).

    A finite sum of integers is clearly an integer itself.
     
  4. Nov 14, 2005 #3

    matt grime

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    Firstly, as stated your trying to prove something that is false. There is no reason to suppose that an integer polynomial has *any* rational roots.


    Then there is the issue of what a rational root might look like even if it had one: consider 2x-1, its rational root is 1/2, and not of the form p/q where p and q are primes,

    However, I don't see that issue you have is at all an issue: all of the a_r are integers, p and q are integers, and multiplying and adding together integers yields integers.
     
  5. Nov 14, 2005 #4
    Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n.

    The statement: "A finite sum of integers is clearly an integer itself." hit the nail =)

    Thanks guys, this helped a lot!
     
  6. Nov 14, 2005 #5

    matt grime

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    You asserted they were primes. They are at best coprime, or prime to each other.
     
  7. Sep 29, 2008 #6

    ila

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    My maths isn't that strong. Im having trouble with this proof as well. Basically its the logic of saying that p divides a0 and hence is a factor. Why couldn't q be factored out of the equation (instead of p) and then divide the a0*q^n on the RHS by q and just say that q is a factor of a0 (in addition to a factor of q^n).

    Lastly, I don't understand why just because p divides [tex]a_{0}[/tex] means that it is a factor of a0. The logic seems to be that just because q^n can't be factored by p therefor it implies that a0 can be factored by p.
     
  8. Sep 30, 2008 #7

    HallsofIvy

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    I would approach it from the other end. IF polynomial p(x) has root m/n, then (x- m/n) is a factor: p(x)= (x- m/n)q(x). So np(x)= (nx- m)q(x)= (nx-m)(an-1xn-1+ ....+ a0). It should be obvious from that, that the constant term has m as a factor. It is a little harder, because of that "n" in "np(x)", to see that the leading coefficient has m as a factor.
     
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