Rational root theorem

Question

In a Complex Analysis course I'm asked to show that the rational root theorem is true, stated as follows.

Show that the following holds true: Let the real rational number $p/q$ (where $p$ and $q$ have no common factor except $\pm 1$ be a solution of $$a_n z^n + a_{n-1}z^{n-1} + \ldots + a_1 z + a_0 = 0, \quad a_j \in \mathbb{Z}.$$ Then $q$ must be a factor of $a_n$ and $p$ must be a factor of $a_0$.

Now, for $n=1$, I can show $p = a_0\frac{-a_1}{q}$ and $q = a_1\frac{-a_0}{p}$.

For $n = 2$, we have $p = \frac{q}{2a_2}(-a_1 \pm \sqrt{a_1^2-4a_2a_0}) =^{?} a_0 (\ldots).$

I suppose we need to somehow use that $a_j \in \mathbb Z$ and $p,q \in \mathbb Z$.

Furthermore, the exercise asks me to show, rather than to prove. But still, can someone point me in the right direction?

@MathematicsStudent1122 I find it quite a non-trivial exercise, in that case. — Mussé Redi, Aug 25 '16 at 19:51
You can find a proof here: http://math.stackexchange.com/questions/658006/why-rational-root-theorem-only-works-with-integers — quid, Aug 25 '16 at 19:56
I'm trying to avoid looking at the proof. First I want to try it myself, can you give a hint? — Mussé Redi, Aug 25 '16 at 19:58
Write the equation as $(a_n z^{n-1} + a_{n-1}z^{n-2} + \ldots + a_1)z = -a_0.$ Plug in $p/q$ and multiply by $q^n$. (This will give you soon one of the two claims the other is similar.) — quid, Aug 25 '16 at 20:02
@quid Was your argument in line with considering the divisors of the LHS and the RHS, or where you using something else? — Mussé Redi, Aug 25 '16 at 21:43
@MusséRedi it's the argument in the answer. You could also have followed the link now. — quid, Aug 25 '16 at 21:46

score 2 · Accepted Answer · answered Aug 25 '16 at 20:42

2

If $z = p/q$ is a rational root $$a_n z^n + a_{n-1}z^{n-1} + \ldots + a_1 z + a_0 = 0$$ reads $$a_n \frac{p^n}{q^n} + \ldots + a_1 \frac{p}{q} + a_0 = 0$$ which multiplied by $q^n$ gives $$a_n p^n + a_{n-1}q p^{n-1} + \ldots + a_1 p q^{n-1} = - a_0 q^n$$

As $p$ divides the lefthand side, p divides the righthand one, and by hypothesis, as $p$ and $q$ cannot have other common factors than $\pm 1$, $p$ divides $a_0$.

But we can also write

$$- a_n p^n = a_{n-1}q p^{n-1} + \ldots + a_1 p q^{n-1} +a_0 q^n$$ so that $q$ dividing the righthand side divides also the lefthand one, and as $p$ and $q$ cannot have other common factors than $\pm 1$, $q$ divides $a_n$.

answered Aug 25 '16 at 20:42

Olórin

12,040

Excellent. The argument of considering $p$ and $q$ as divisors of equations is very neat. – Mussé Redi Aug 25 '16 at 20:45
1

@MusséRedi It is worth emphasizing that this proof depends crucially on FTA = fundamental theorem of arithmetic (existence and uniqueness of prime factorizations), even though that is not explicitly stated. Normally one deduces those divisibilities by using consequences of FTA such as Euclid's Lemma $,(\gcd(a,b)=1,\ a\mid bc,\Rightarrow,a\mid c),$ or closely related results. Without such fundamental results the Rational Root Test need not hold, e.g. it may fail in rings of algebraic integers. – Bill Dubuque Aug 25 '16 at 21:22

Rational root theorem

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