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Integer Implication Root Positive Hypothesis

1	Root
1	Positive
1	Integer
1	Implication
2	Hypothesis

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Proof Of Rational Root Theorem

Let $p(x) \in \mathbb{Z}[x]$ . Let be a positive integer with $\operatorname{deg} p(x)=n$ . Let $c_0, \ldots , c_n \in \mathbb{Z}$ such that $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$ .

Let $a,b \in \mathbb{Z}$ with $\operatorname{gcd}(a,b)=1$ and such that $\displaystyle \frac{a}{b}$ is a root of . Then

$\begin{displaymath}\begin{array}{ll} 0 & \displaystyle =p\left( \frac{a}{b} \rig... ...n-1}}{b^{n-1}} +\cdots +c_1 \cdot \frac{a}{b} +c_0. \end{array}\end{displaymath}$

Multiplying through by and rearranging yields:

$\begin{displaymath}\begin{array}{rl} c_na^n+c_{n-1}a^{n-1}b+\cdots +c_1ab^{n-1} ... ...-1}-c_{n-1}a^{n-2}b -\cdots -c_1b^{n-1} \right) \\ \end{array}\end{displaymath}$

Thus, $a\vert c_0b^n$ and, by hypothesis, $\operatorname{gcd}(a,b)=1$ . This implies that $a\vert c_0$ .

Similarly:

$\begin{displaymath}\begin{array}{rl} c_na^n+c_{n-1}a^{n-1} b+\cdots +c_1ab^{n-1}... ...^{n-1} -\cdots -c_1ab^{n-1} -c_0b^{n-1} \right) \\ \end{array}\end{displaymath}$

Therefore, $b\vert c_na^n$ and $b\vert c_n$ .