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Math Help - If r is a rational solution r = p/q and p and q are coprime, show that q|an and p|a0.

  1. #1
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    If r is a rational solution r = p/q and p and q are coprime, show that q|an and p|a0.

    Suppose that r is a solution of the equation:

    anxn + a(n−1)x(n−1) + . . . + a1x + a0 = 0

    where the coefficients ak belongs to Z for k = 0, 1, . . . n, and n is greater or equal to 1. If r is a rational solution r = p/q, where p, q belong to Z and p and q are
    coprime, show that q|an and p|a0.

    Im not even sure where to begin, im so confused, what am i trying to prove? and how do i prove it.
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  2. #2
    Member Traveller's Avatar
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    You have been given a polynomial in which if you substitute x by \frac{p}{q}, it evaluates to 0. You have to prove that q divides the coefficient of x^n and p divides the constant.


    Replace x by \frac{p}{q}. Take LCM etc. , multiply both sides by whatever required to get an expression only in terms of integers. Can you see something after you do all this ?
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  3. #3
    Member HappyJoe's Avatar
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    Your current situation is that you have a polynomial

    a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,

    where all coefficients are integers, a_n\neq 0, and n\geq 1, and you know that the rational number r=\frac{p}{q} is a root of this polynomial, where p and q are coprime.

    Your problem in this setup is to prove that the numerator p of r divides a_0, and that the denominator q of r divides a_n, the coefficient of x^n.

    Here's how one could go about doing it. Since r is a root of the polynomial, we have:

    a_nr^n+a_{n-1}r^{n-1}+\cdots+a_1r+a_0=0.

    We know that r=\frac{p}{q}, so replace r by \frac{p}{q} in the expression to get:

    a_n(\frac{p}{q})^n+a_{n-1}(\frac{p}{q})^{n-1}+\cdots+a_1\frac{p}{q}+a_0=0.

    Multiply each term of this equation by q^n (check this carefully) to get:

    a_np^n+a_{n-1}qp^{n-1}+\cdots+a_1q^{n-1}p+q^na_0=0.

    I will refer to this equation as (*). In (*), take all but the first term and move them to the right hand side:

    a_np^n=-a_{n-1}qp^{n-1}-\cdots-a_1q^{n-1}p-q^na_0.

    Notice how q is a factor of each of the terms on the right hand side, which is why q divides the entire right hand side. But the right hand side is equal to the left hand side, and so q also divides the expression

    a_np^n.

    However, the integers p and q were chosen to be coprime, which means that they have no common factor. So for q to divide a_np^n, it must divide a_n. This is part of what you were supposed to show.

    You should also show that p divides a_0. To do this, start out with (*) again, and make some slight changes to what just happened. This will lead you to what you want.
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  4. #4
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    Wow, thanks so much, i understand it now, and was able to find the solution for p divides ao. Thanks alot!
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