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Thread: 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 $\displaystyle \frac{p}{q}$, it evaluates to 0. You have to prove that q divides the coefficient of $\displaystyle x^n$ and p divides the constant.


    Replace x by $\displaystyle \frac{p}{q}$. 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

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

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle a_np^n.$

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

    You should also show that $\displaystyle p$ divides $\displaystyle a_0$. 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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