9
$\begingroup$

Let $$\sum_{0 \leq i\leq n} a_ix^i$$

be a polynomial (real coefficients) with at least two real roots. Is there an algebraic way to show that for any two roots $k_1, k_2$ of this polynomial, the polynomial

$$\sum_{1 \leq i\leq n} i \cdot a_ix^{i-1} $$

admits at least one root $c$ satisfying $k_1 <c < k_2$?

Analytically, this is of course a consequence of Rolle's theorem.

Edit: "Algebra" is as broad as you want it to be. Elementary or abstract. The completeness of $\mathbb{R}$ is essential, so it won't be purely algebraic. I was mainly hoping for something without derivatives.

$\endgroup$
  • 1
    $\begingroup$ What would qualify as "purely algebraic proof" for you? Something only out of ring/ideal theory without any analysis at all? It may be hard, or perhaps even impossible, to accomplish. $\endgroup$ – DonAntonio Aug 24 '16 at 10:40
  • 1
    $\begingroup$ I also think that a "purey algebraic" proof would be impossible. Philosophically, my motivation is that algebra deals with equalities, while inequalities are the realm of analysis. $\endgroup$ – Daniel Robert-Nicoud Aug 24 '16 at 10:43
  • 1
    $\begingroup$ This isn't true for all rings, even for ordered fields (take the rationals). So my guess would be that this has to use the topology of the real numbers. Very good question though! $\endgroup$ – Nitrogen Aug 24 '16 at 10:46
  • $\begingroup$ It is true for the reals but not the rationals. The analytic completeness is difficult to define algebraically. $\endgroup$ – Arthur Aug 24 '16 at 10:47
  • $\begingroup$ @Arthur If the fundamental theorem of algebra can be proved algebraically, why can't this? $\endgroup$ – MathematicsStudent1122 Aug 24 '16 at 10:50
4
$\begingroup$

Probably not. Note that in $\mathbb Q(\sqrt{6})$, the polynomial $x^3 - 6x$ has three roots. However, its derivative has no roots. These would be $\pm \sqrt{2}$. Thus, whatever you mean by purely algebraic needs to be stronger than the theory of ordered fields.

If, however, you are willing to stretch your definition of "algebraic" somewhat, and add to the theory of ordered fields even the axiom "each positive number has a square root", and an axiom scheme containing for each polynomial of odd degree an axiom "this polynomial has a root", then you can prove any first order property of the reals. See for instance this Wikipedia page.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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