{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW3B - q n p n a n-1 p n-1 q a 1 pq n-1 a q n = 0 Rewrite...

This preview shows page 1. Sign up to view the full content.

Math 103B Homework Solutions HW3 Jacek Nowacki April 30, 2007 Chapter 17 Problem 4. Suppose that f ( x ) = x n + a n - 1 x n - 1 + . . . + a 0 Z [ x ]. If r is a rational number and x - r divides f ( x ), show that r is an integer. Proof. x - r divides f ( x ) implies that f ( x ) = ( x - r ) g ( x ) for some g ( x ) Q [ x ]. It is easy to see that f ( r ) = 0. Since r Q , we can write r = p q for p and q relatively prime. This way we get f ( p q ) = ( p q ) n + a n - 1 ( p q ) n - 1 + . . . + a 1 ( p q ) + a 0 = 0 Multiply everything in the above equation by
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: q n p n + a n-1 p n-1 q + . . . + a 1 pq n-1 + a q n = 0 Rewrite the equation q ± a n-1 p n-1 + . . . + a 1 pq n-2 + a q n-1 ² =-p n Hence, q = 1 as otherwise any prime dividing q would also divide p (Euclid’s Theorem). This shows that r must be an integer. 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online