proof of rational root theorem

Let $p(x)\\in\\mathbb{Z}[x]$ . Let $n$ be a positive integer with $\\operatorname{deg}p(x)=n$ . Let $c_{0},\\ldots,c_{n}\\in\\mathbb{Z}$ such that $p(x)=c_{n}x^{n}+c_{n-1}x^{n-1}+\\cdots+c_{1}x+c_{0}$ .

Let $a,b\\in\\mathbb{Z}$ with $\\operatorname{gcd}(a,b)=1$ and $b>0$ such that $\\displaystyle\\frac{a}{b}$ is a root of $p(x)$ . Then

$\\begin{array}[]{ll}0&\\displaystyle=p\\left(\\frac{a}{b}\\right)\\\\\\\\&\\displaystyle=c_{n}\\left(\\frac{a}{b}\\right)^{n}+c_{n-1}\\left(\\frac{a}{b}%\\right)^{n-1}+\\cdots+c_{1}\\cdot\\frac{a}{b}+c_{0}\\\\\\\\&\\displaystyle=c_{n}\\cdot\\frac{a^{n}}{b^{n}}+c_{n-1}\\cdot\\frac{a^{n-1}}{b^{n-1%}}+\\cdots+c_{1}\\cdot\\frac{a}{b}+c_{0}.\\end{array}$

Multiplying through by $b^{n}$ and rearranging yields:

$\\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&%=0\\\\\\\\c_{0}b^{n}&=-c_{n}a^{n}-c_{n-1}a^{n-1}b-\\cdots-c_{1}ab^{n-1}\\\\\\\\c_{0}b^{n}&=a\\left(-c_{n}a^{n-1}-c_{n-1}a^{n-2}b-\\cdots-c_{1}b^{n-1}\\right)\\\\\\end{array}$

Thus, $a|c_{0}b^{n}$ and, by hypothesis, $\\operatorname{gcd}(a,b)=1$ . This implies that $a|c_{0}$ .

Similarly:

$\\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&%=0\\\\\\\\c_{n}a^{n}&=-c_{n-1}a^{n-1}b-\\cdots-c_{1}ab^{n-1}-c_{0}b^{n}\\\\\\\\c_{n}a^{n}&=b\\left(-c_{n-1}a^{n-1}-\\cdots-c_{1}ab^{n-1}-c_{0}b^{n-1}\\right)\\\\\\end{array}$

Therefore, $b|c_{n}a^{n}$ and $b|c_{n}$ .

单词	ProofOfRationalRootTheorem
释义	proof of rational root theorem Let $p(x)\\in\\mathbb{Z}[x]$ . Let $n$ be a positive integer with $\\operatorname{deg}p(x)=n$ . Let $c_{0},\\ldots,c_{n}\\in\\mathbb{Z}$ such that $p(x)=c_{n}x^{n}+c_{n-1}x^{n-1}+\\cdots+c_{1}x+c_{0}$ . Let $a,b\\in\\mathbb{Z}$ with $\\operatorname{gcd}(a,b)=1$ and $b>0$ such that $\\displaystyle\\frac{a}{b}$ is a root of $p(x)$ . Then $\\begin{array}[]{ll}0&\\displaystyle=p\\left(\\frac{a}{b}\\right)\\\\\\\\&\\displaystyle=c_{n}\\left(\\frac{a}{b}\\right)^{n}+c_{n-1}\\left(\\frac{a}{b}%\\right)^{n-1}+\\cdots+c_{1}\\cdot\\frac{a}{b}+c_{0}\\\\\\\\&\\displaystyle=c_{n}\\cdot\\frac{a^{n}}{b^{n}}+c_{n-1}\\cdot\\frac{a^{n-1}}{b^{n-1%}}+\\cdots+c_{1}\\cdot\\frac{a}{b}+c_{0}.\\end{array}$ Multiplying through by $b^{n}$ and rearranging yields: $\\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&%=0\\\\\\\\c_{0}b^{n}&=-c_{n}a^{n}-c_{n-1}a^{n-1}b-\\cdots-c_{1}ab^{n-1}\\\\\\\\c_{0}b^{n}&=a\\left(-c_{n}a^{n-1}-c_{n-1}a^{n-2}b-\\cdots-c_{1}b^{n-1}\\right)\\\\\\end{array}$ Thus, $a\|c_{0}b^{n}$ and, by hypothesis, $\\operatorname{gcd}(a,b)=1$ . This implies that $a\|c_{0}$ . Similarly: $\\begin{array}[]{rl}c_{n}a^{n}+c_{n-1}a^{n-1}b+\\cdots+c_{1}ab^{n-1}+c_{0}b^{n}&%=0\\\\\\\\c_{n}a^{n}&=-c_{n-1}a^{n-1}b-\\cdots-c_{1}ab^{n-1}-c_{0}b^{n}\\\\\\\\c_{n}a^{n}&=b\\left(-c_{n-1}a^{n-1}-\\cdots-c_{1}ab^{n-1}-c_{0}b^{n-1}\\right)\\\\\\end{array}$ Therefore, $b\|c_{n}a^{n}$ and $b\|c_{n}$ .
随便看	BernoullisInequality Bernoulli’s inequality BernsteinInequalities Bernstein inequalities BernsteinPolynomial Bernstein polynomial BerrysParadox Berry’s paradox BertrandsConjecture BertrandsConjectureProofOf BertrandsProblem there are an infinite number of primes ≡1⁢@⁢ERROR(\symoperators)⁢m⁢o⁢d⁢ERROR(\tmspace)+.1667⁢e⁢m⁢ERROR(\tmspace)+.1667⁢e⁢m⁢m there are an infinite number of primes ≡±1⁢@⁢(ERROR(\symoperators)⁢m⁢o⁢d⁢4) ThereAreNoNonsquareDoublyStochasticMatrices there are no non-square doubly stochastic matrices ThereExistAdditiveFunctionsWhichAreNotLinear there exist additive functions which are not linear ThereforeSign therefore sign ThereIsAUniqueReducedFormOfDiscriminant4nOnlyForN12347 there is a unique reduced form of discriminant -4⁢n only for n=1,2,3,4,7 The Rezk completion TheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ the ring of integers of a number field is finitely generated over ℤ TheSetOfAllRealTranscendentalNumbersIsUncountable