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Generalize Theorem 1.3.9 by proving that every rational solution of a polyno- mial equation x" + an-1a"-1 + + a1x + = 0, | with integer coefficients ak, is an integer solution.

Generalize Theorem 1.3.9 by proving that every rational solution of a polyno- mial equation x" + an-1a"-1 + + a1x + = 0, | with integer coefficients ak, is an integer solution.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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12. Generalize Theorem 1.3.9 by proving that every rational solution of a polyno- mial equation -1x"-1 + «1x + a• = 0, +... with integer coefficients a k, is an integer solution.
Transcribed Image Text:12. Generalize Theorem 1.3.9 by proving that every rational solution of a polyno- mial equation -1x"-1 + «1x + a• = 0, +... with integer coefficients a k, is an integer solution.
Theorem 1.3.9. If k is an integer and the equation x = k has a rational solution, then that solution is actually an integer. Proof. Suppose r is a ratiomal number such that r2 = k. Let r = as a fraction in which n and m have no common factors. Then, er expressed 2 n k and so n? = m²k. m This equationm implies that m divides n2. However, if m 7 1, then m can be expressed as a product of primes, and each of these primes must also divide n2. However, if a prime number divides n2, it must also divide n (Exercise 1.3.14). Thus, each prime factor of m divides n. Since n and m have no common factors, this is impossible. We conclude that m = 1 and, hence, that r = n is an integer.
Transcribed Image Text:Theorem 1.3.9. If k is an integer and the equation x = k has a rational solution, then that solution is actually an integer. Proof. Suppose r is a ratiomal number such that r2 = k. Let r = as a fraction in which n and m have no common factors. Then, er expressed 2 n k and so n? = m²k. m This equationm implies that m divides n2. However, if m 7 1, then m can be expressed as a product of primes, and each of these primes must also divide n2. However, if a prime number divides n2, it must also divide n (Exercise 1.3.14). Thus, each prime factor of m divides n. Since n and m have no common factors, this is impossible. We conclude that m = 1 and, hence, that r = n is an integer.
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