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.

  • 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

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 $p$.

  • $\begingroup$ 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

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