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Why does the rational root theorem only work when the polynomial has integer coefficients? Specifically, why would it not apply if the coefficients were rational?

Recall the rational root theorem (or rational root test, rational zero theorem, rational zero test or $p/q$ theorem) states a constraint on rational solutions of a polynomial equation

$$ a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0$$ with integer coefficients ${\displaystyle a_{i}\in \mathbb {Z}}$.

The theorem states that each rational solution $\,x = p⁄q\,$ written in lowest terms (so $p$ and $q$ are relatively prime), satisfies:

  • $p$ is an integer factor of the constant term $a_0$, and
  • $q$ is an integer factor of the leading coefficient $a_n$.
Martin Sleziak
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Supernova
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  • what would you prefer? – Will Jagy Jan 31 '14 at 03:12
  • The theorem refers to the numerator and denominator of a possible rational root, saying these divide the constant term and leading term. If you allow noninteger coefficients, at least the constant term and lead term would have to be integers, or it wouldn't make sense to look for numerator and denominator being divisors of them. Also maybe one could cook one up with a rational root, where it violates the root theorem, if one is allowed to have intermediate noninteger coefficients. – coffeemath Jan 31 '14 at 03:13
  • Like $p(x) = x^3 - (1 + 2 \sqrt 2) x^2 + (2 + 2 \sqrt 2) x - 2 = (x - \sqrt 2)^2(x - 1)$ is not a polynomial with integer coefficients but has rational roots as described by the rational root theorem. – Neil W Jan 31 '14 at 03:33
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    Cause divisibility only makes sense in integers – kingW3 Mar 02 '15 at 16:09
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    @kingW3 Not true - divisibility can be relative to any subring - not only $\Bbb Z,,$ e.g. see here and here. – Bill Dubuque Feb 28 '21 at 14:04

2 Answers2

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The Rational Root Test (RRT) proof depends on the polynomial coefs being integers. Let's recall it.

Theorem $ $ If $\,f(x) = f_n x^n + \cdots + f_0\,$ is a polynomial with $\,\color{#c00}{{\rm integer\ coefs}\ \,f_i\in \Bbb Z}\,$ and $\,f(x)\,$ has a rational root $\,x = a/b,\ \color{#0a0}{\gcd(a,b)=1},\,$ then $\,a\mid f_0\,$ and $\,b\mid f_n.\,$ Further if $f$ is primitive (gcd of all coefs $= 1)$ and squarefree $\:\!q\in\Bbb Z\:\!$ then $\:\!q\mid f_0,f_1,\ldots,f_{n-1}\Rightarrow q\mid a,\, $ $\,q\mid f_1,f_2,\ldots f_n\Rightarrow q\mid b$.

Proof $\ \ 0 = f(a/b)\ \Rightarrow\ 0 = b^n f(a/b)\ =\, f_n\, a^n\! + f_{n-1}\, a^{n-1}b+\cdots+f_1\, ab^{n-1}\! + f_0\, b^n$

Thus $\,\ (\overbrace{f_{n}\, a^{n-1}+f_{n-1}\,a^{n-2}b+\cdots+f_1\, b^{n-1}}^{\large{\rm an\ integer,\ since}\,\ \color{#c00}{f_i\ {\rm are\ integers}}})\,a\,=\, -f_0\, b^n,\ $ hence $\ a\mid b^n f_0\,\color{#0a0}{\Rightarrow}\, a\mid f_0,\, $ since $\,\color{#0a0}{\gcd(a,b)=1},\,\ a\mid bc\,\Rightarrow\,a\mid c,\,$ by Euclid's Lemma, so, by induction, $\,a\mid b^nc\,\color{#0a0}{\Rightarrow}\,a\mid c.\,$ Further if $\,q\mid f_0,f_1,\ldots,f_{n-1}\,$ then by the first equation above also $\,q\mid f_na^n.\,$But $\,q\nmid f_n$ (else $\,q\mid{\rm all}\ f_i)$ so by Euclid $\,q\mid a^n,\,$ so $\,q\mid a\,$ by $\,q\,$ squarefree. Similarly follows the dual claim that $\,q\mid b$.

Notice how the above proof depends crucially on the polynomial coefficients $\,\color{#c00}{f_i\,\ \rm being\ integers},\,$ which implies that the overbraced term is an integer and, hence, that $\,a\mid b^n f_0.\,$ Exactly the same applies to the reversed case, which deduces, symmetrically that $\, b\mid a^n f_n\,\Rightarrow\,b\mid f_n\ $ [or use $\ b^n f(a/b) = f_n\,a^n + ab (\ldots) + f_0\, b^n\,$ for $\,(\ldots) \in \Bbb Z\,$]

Besides identifying where the proof breaks down, there are obvious counterexamples, e.g. $\,x-a/b\,$ has a root $\,a/b\,$ that need not be an integer. Less trivial are quadratic examples

$\quad (x-a/b)\,(x-b/a)\, =\, x^2-(a/b+b/a)\,x + 1\,$ has a root $\,a/b\,$ that need not be $\,\pm1$.

Note also the hypothesis that the rational root is $\color{#0a0}{\rm reduced}$ (in lowest or least terms) is necessary, else e.g. $\, x = 4/6\, [= 2/3]\,$ is a root of $\,3\,x-2\,$ but $\, 6\nmid 3,\, 4\nmid 2.$

The above proof requires only that gcds exist, so it works over any GCD domain, e.g. any UFD (see here for more on this general case).

The Rational Root Test can also viewed as a special case of Gauss's Lemma (for polynomials).

We can view the divisibility bounds in RRT as instances of the method of simpler multiples.

Below is a typical application of the "further" lower bounds on $\,a\,$ and $\,b.$

Corollary $ $ If squarefree $\,q\in\Bbb Z,\,$ all $\,a_i\in \Bbb Q\,$ and $\,e_i\,$ are elementary symmetric polynomials in $\,a_i,\,$ i.e. $\,(x\!-\!a_1)(x\!-\!a_2)(x\!-\!a_3) = x^3-e_1\:\! x^2 + e_2\:\! x - e_3,\,$ then all $\,\color{#0a0}{q^i e_{i+1}\in\Bbb Z}\,\Rightarrow\,$ all $\,a_i\in\Bbb Z.\,$

Proof $\,\ (x-qa_1)(x-qa_2)(x-qa_3) = x^3-\color{#c00}q\:\!\color{#0a0}{e_1}\:\!x^2 + \color{#c00}q\:\!\color{#0a0}{q e_2}\:\! x + \color{#c00}q\:\!\color{#0a0}{q^2 e_3},\, $ all $\rm\color{#0a0}{greens}\in\Bbb Z$
thus by the theorem $\,qa_i\in\Bbb Z\,$ and further $\,\color{#c00}q\mid qa_i\,$ i.e. $\,a_i = (q a_i)/q\in \Bbb Z$.

The proof clearly generalizes to any number $\,k\,$ of $\,a_i\,$ (case $\,k=2=q\,$ is this question)

Bill Dubuque
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If the coefficients are rational, you can multiply the polynomial by the least common denominator to get a second polynomial in integer coefficients that has the same zeroes. Rational root theorem applies.

If the coefficients are irrational numbers, all bets are off.

Tom
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