Why Rational Root Theorem requires integer coefficient polynomials? [closed]

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

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

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• what would you prefer? – Will Jagy Jan 31 '14 at 3:12
• 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 3:13
• 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 3:33
• Cause divisibility only makes sense in integers – kingW3 Mar 2 '15 at 16:09

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 $$\,\pm1$$.

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

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.