4
$\begingroup$

This is an exercise from Spivak's "Calculus" 4th edition.

18.a. Prove that if $x$ satisfies $$x^n + a_{n-1}x^{n-1}+ \dots + a_0 = 0$$ for some integers $a_{n-1}, \dots, a_0$, then $x$ is irrational unless $x$ is an integer.

The solution is given as

Suppose $x = p/q$ where $p$ and $q$ are natural numbers with no common factor. Then $$\frac{p^n}{q^n} + a_{n-1}\frac{p^{n-1}}{q^{n-1}}+ \dots + a_0 = 0$$, so $$\tag{*} p^n + a_{n-1}p^{n-1}q + \dots + a_0 q^n = 0$$ Now if $q \neq \pm 1$, then $q$ has some prime number as a factor. This prime factor divides every term of (*) other than $p^n$, so it must divide $p^n$ also. Therefore it divides $p$, a contradiction. So $q = \pm 1$, which means that $x$ is an integer.

I am confused about the conclusions in bold above. How does every other term of (*) being divisible by the prime factor affect $p^n$? Also, I've been struggling to prove that if $k$ is prime and $k|p^n$, then $k|p$, which seems to be the lemma used above.

$\endgroup$
6
  • 2
    $\begingroup$ $p^n =- a_{n-1}p^{n-1}q - \dots - a_0 q^n$ and that factor of $q$ divides every term on the right-hand side. – See also “rational root theorem”. $\endgroup$ – Martin R Jun 13 '20 at 18:39
  • $\begingroup$ @MartinR Thanks! That makes sense now. $\endgroup$ – Iyeeke Jun 13 '20 at 18:43
  • $\begingroup$ Great, elegant solution! $\endgroup$ – UmbQbify Jun 13 '20 at 19:02
  • $\begingroup$ Sorry but I don't understand the bar notation, to my understanding it means, if k divides pⁿ then k also divides p? $\endgroup$ – UmbQbify Jun 13 '20 at 19:04
  • $\begingroup$ @AshWhole Yes, that is correct. $\endgroup$ – Iyeeke Jun 13 '20 at 19:34
2
$\begingroup$

Every other terms has a $q$ factor explicitly besides $p^n$.

Since $$p^n = -q(a_{n-1}p^{n-1} + \ldots + a_0q^{n-1})$$

Hence $q$ must divide $p^n$.

As for your second quesiton, let a prime factor of $q$ be $q_0$. Consider the prime factorization of $p=\prod_{i=1}^m q_i^{r_i} $, then $p^n=\prod_{i=1}^{m} q_i^{nr_i} $, if $q_0$ divide $p^n$ is a prime, then $q_0$ must be one of the $q_i$ with positive $r_i$ power. Hence $q_0$ divide $p$.

$\endgroup$
3
  • $\begingroup$ Thanks for your response! Is there a more formal argument for the second question? I don't think that $p$ has to be prime in this case. It was given as a natural number. $\endgroup$ – Iyeeke Jun 13 '20 at 18:48
  • $\begingroup$ Edited the answer. $\endgroup$ – Siong Thye Goh Jun 14 '20 at 2:46
  • $\begingroup$ Thank you! I understand the second part now. $\endgroup$ – Iyeeke Jun 14 '20 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.