(New page: <math>\scriptstyle f(x)\ =\ x^n+a_{n-1}x^{n-1}+\ldots+a_0\ =\ \in\ Z[x],\ (x-r)\mid f(x),\ r=\frac{p}{q},\ p,q\in\mathbb{Z},\ q\neq0</math> <math>\scriptstyle\Rightarrow\ f(x)\ =\ (x-\fra...) |
|||
| Line 1: | Line 1: | ||
| − | <math>\scriptstyle f(x)\ =\ x^n+a_{n-1}x^{n-1}+\ldots+a_0\ =\ \in\ Z[x],\ (x-r)\mid f(x),\ r=\frac{p}{q},\ p,q\in\mathbb{Z},\ q\neq0</math> | + | <math>\scriptstyle f(x)\ =\ x^n+a_{n-1}x^{n-1}+\ldots+a_0\ =\ \in\ Z[x],\ (x-r)\mid f(x),\ r=\frac{p}{q},\ p,q\in\mathbb{Z},\ gcd(p,q)=1,\ q\neq0</math> |
| + | |||
| + | Note that we can assume that <math>\scriptstyle gcd(p,q)=1</math> because in any representation <math>\scriptstyle r\ =\ \frac{s}{t}\ \mid s,t\in\mathbb{Z},\ t\neq0</math> where <math>\scriptstyle s</math> and <math>\scriptstyle t</math> are not coprime we can find a coprime <math>\scriptstyle p</math> and <math>\scriptstyle q</math> such that <math>\scriptstyle r\ =\ \frac{p\cdot gcd(s,t)}{q\cdot gcd(s,t)}\ =\ \frac{p}{q}</math>. | ||
<math>\scriptstyle\Rightarrow\ f(x)\ =\ (x-\frac{p}{q})(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0),\ \ b_{n-2},\ldots,b_0\in\mathbb{Q}</math> | <math>\scriptstyle\Rightarrow\ f(x)\ =\ (x-\frac{p}{q})(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0),\ \ b_{n-2},\ldots,b_0\in\mathbb{Q}</math> | ||
| Line 15: | Line 17: | ||
<math>\scriptstyle\Rightarrow\ p^n\ =\ q\cdot(-a_{n-1}p^{n-1}-a_{n-2}p^{n-2}q-\ldots-a_1pq^{n-2}-a_0q^{n-1})</math> | <math>\scriptstyle\Rightarrow\ p^n\ =\ q\cdot(-a_{n-1}p^{n-1}-a_{n-2}p^{n-2}q-\ldots-a_1pq^{n-2}-a_0q^{n-1})</math> | ||
| − | <math>\scriptstyle\Rightarrow\ q\mid p^n\ \ | + | <math>\scriptstyle\Rightarrow\ q\mid p^n</math> |
| + | |||
| + | Since <math>\scriptstyle gcd(p,q)=1</math>, this implies that <math>\scriptstyle q\ =\ 1</math>. | ||
| − | <math>\scriptstyle | + | <math>\scriptstyle\Rightarrow\ r\ =\ \frac{p}{1}\ =\ p</math> |
<math>\scriptstyle\Rightarrow\ r\ \in\ \mathbb{Z}</math>. <math>\scriptstyle\Box</math> | <math>\scriptstyle\Rightarrow\ r\ \in\ \mathbb{Z}</math>. <math>\scriptstyle\Box</math> | ||
:--[[User:Narupley|Nick Rupley]] 03:37, 8 April 2009 (UTC) | :--[[User:Narupley|Nick Rupley]] 03:37, 8 April 2009 (UTC) | ||
Revision as of 15:12, 8 April 2009
$ \scriptstyle f(x)\ =\ x^n+a_{n-1}x^{n-1}+\ldots+a_0\ =\ \in\ Z[x],\ (x-r)\mid f(x),\ r=\frac{p}{q},\ p,q\in\mathbb{Z},\ gcd(p,q)=1,\ q\neq0 $
Note that we can assume that $ \scriptstyle gcd(p,q)=1 $ because in any representation $ \scriptstyle r\ =\ \frac{s}{t}\ \mid s,t\in\mathbb{Z},\ t\neq0 $ where $ \scriptstyle s $ and $ \scriptstyle t $ are not coprime we can find a coprime $ \scriptstyle p $ and $ \scriptstyle q $ such that $ \scriptstyle r\ =\ \frac{p\cdot gcd(s,t)}{q\cdot gcd(s,t)}\ =\ \frac{p}{q} $.
$ \scriptstyle\Rightarrow\ f(x)\ =\ (x-\frac{p}{q})(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0),\ \ b_{n-2},\ldots,b_0\in\mathbb{Q} $
$ \scriptstyle f(r)\ =\ f(\frac{p}{q})\ =\ (\frac{p}{q}-\frac{p}{q})(\textstyle\cdots\cdots\scriptstyle)\ =\ 0 $.
$ \scriptstyle\Rightarrow\ (\frac{p}{q})^n+a_{n-1}(\frac{p}{q})^{n-1}+\ldots+a_1(\frac{p}{q})+a_0\ =\ 0 $.
Now, multiply both sides by $ \scriptstyle q^n $:
$ \scriptstyle p^n+a_{n-1}p^{n-1}q+\ldots+a_1pq^{n-1}+a_0q^n\ =\ 0 $
$ \scriptstyle\Rightarrow\ p^n\ +\ q\cdot(a_{n-1}p^{n-1}+a_{n-2}p^{n-2}q+\ldots+a_1pq^{n-2}+a_0q^{n-1})\ =\ 0 $
$ \scriptstyle\Rightarrow\ p^n\ =\ q\cdot(-a_{n-1}p^{n-1}-a_{n-2}p^{n-2}q-\ldots-a_1pq^{n-2}-a_0q^{n-1}) $
$ \scriptstyle\Rightarrow\ q\mid p^n $
Since $ \scriptstyle gcd(p,q)=1 $, this implies that $ \scriptstyle q\ =\ 1 $.
$ \scriptstyle\Rightarrow\ r\ =\ \frac{p}{1}\ =\ p $
$ \scriptstyle\Rightarrow\ r\ \in\ \mathbb{Z} $. $ \scriptstyle\Box $
- --Nick Rupley 03:37, 8 April 2009 (UTC)
