Difference between revisions of "Chapter 17 Question 4" - Rhea

Latest revision as of 16:28, 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)

Nick- Thanks so much for posting! This problem makes sense now!