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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.

MathematicsStudent1122
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    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. – DonAntonio Aug 24 '16 at 10:40
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    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. – Daniel Robert-Nicoud Aug 24 '16 at 10:43
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    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! – Nitrogen Aug 24 '16 at 10:46
  • It is true for the reals but not the rationals. The analytic completeness is difficult to define algebraically. – Arthur Aug 24 '16 at 10:47
  • @Arthur If the fundamental theorem of algebra can be proved algebraically, why can't this? – MathematicsStudent1122 Aug 24 '16 at 10:50
  • Let $f(x) = x(x-3)(x-6) = x^3-9x^2+18x$. Then $f(1) = 0$ and $f(2) = 0$. We have $f'(x) = 3x^2 - 18x + 18$, and its real roots, if they exist, are $3\pm\sqrt{3}$. Do you have a purely algebraic proof of the existence of a real square root of $3$? For that matter, do you have a purely algebraic proof that every positive real number has a real square root? – KCd Aug 24 '16 at 10:55
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    What purely algebraic proof of the Fundamental Theorem of Algebra do you have in mind? – KCd Aug 24 '16 at 10:58
  • @KCd sigh, I'll edit out "purely". Honestly, I just wanted a novel proof which doesn't use calculus. It's fine if it's not "pure". – MathematicsStudent1122 Aug 24 '16 at 11:00
  • @Daniel Robert-Nicoud, inequalities are not exclusive to the realm of analysis. The concept of an ordered field, for example, does not need analysis. Limits and infinite (nested) systems of inequalities belong to analysis. – KCd Aug 24 '16 at 11:02
  • You are not going to find a novel proof bypassing ideas from basic calculus/topology. Look at http://math.stackexchange.com/questions/25883/equivalence-of-rolles-theorem-the-mean-value-theorem-and-the-least-upper-boun to see how closely Rolle's theorem is connected to the least upper bound property of $\mathbf R$, an essentially "analytic" property. Granted, the reasoning there uses non-polynomial functions, but I think supplemented with the examples here you should see why it is quite dubious to think any "novel" proof of Rolle's theorem is out there. – KCd Aug 24 '16 at 11:11
  • @KCd Fair enough. Consider posting your comments as an answer; I would be happy to accept it. – MathematicsStudent1122 Aug 24 '16 at 11:12

1 Answers1

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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.

Mees de Vries
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