9
$\begingroup$

Why does the rational root theorem only work when the polynomial has integer coefficients?

$\endgroup$

closed as off-topic by Servaes, Lord Shark the Unknown, max_zorn, Chinnapparaj R, Brahadeesh Nov 29 '18 at 7:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Servaes, max_zorn, Chinnapparaj R, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ what would you prefer? $\endgroup$ – Will Jagy Jan 31 '14 at 3:12
  • $\begingroup$ 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. $\endgroup$ – coffeemath Jan 31 '14 at 3:13
  • $\begingroup$ 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. $\endgroup$ – Neil W Jan 31 '14 at 3:33
  • 1
    $\begingroup$ Cause divisibility only makes sense in integers $\endgroup$ – kingW3 Mar 2 '15 at 16:09
11
$\begingroup$

The Rational Root Test proof depends crucially on the polynomial coefficients being integers. Let's recall it. Suppose $f(x)\in\Bbb Z[x]\,$ has a rational root $\,a/b,\,$ wlog reduced, i.e. $\,\color{#0a0}{\gcd(a,b)=1}.$

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

Therefore $\,\ (\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,\ $ thus $\ 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, $\,a\mid b^nc\,\color{#0a0}{\Rightarrow}\,a\mid c.$

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 $\ b^n f(a/b) = f_na^n + ab (\ldots) + f_0 b^n\,$ 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 $\,\pm1$.

$\endgroup$
  • 4
    $\begingroup$ I love the color usage $\endgroup$ – qwr Jan 31 '14 at 4:13
  • $\begingroup$ See also this proof which trades off the induction in Euclid's Lemma for induction on degree of the polynomial. $\endgroup$ – Bill Dubuque Mar 20 '15 at 0:44
  • $\begingroup$ See also this proof, which simplifies the above proof by using modular fractions. $\endgroup$ – Bill Dubuque Jul 12 '16 at 20:03
3
$\begingroup$

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.

$\endgroup$

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