# Intermediate Value Theorem and polynomial of odd degree

As a consequence of Intermediate Value Theorem (IVT), it is illustrated in almost all calculus books and notes/courses that every polynomial of odd degree over real numbers have a real root.

I tried to see whether this result has been proved without the use of intermediate value property, and I did not find any link or question related to that on this and other sites.

Can one suggest me whether there is a proof of above mentioned consequence of IVT, which is done without using IVT?

I had seen that the fundamental theorem of algebra have been of interest to many people for proving just using algebraic techniques, but among all the known various proofs of FTA, epsilon amount of analysis is used.

So, my above question is of similar nature, but I did not see any comments on its proofs without IVT.

• Intuition: $\lim_{x\to -\infty} P(x) = -\infty$ while $\lim_{x\to +\infty} P(x) = +\infty$ ($\because$ odd degree) and since polynomials are continuous it should intersect $0$ somewhere. But it is IVT and very fundamental :) Dec 23, 2020 at 11:22
• I think that you have to say clearly what you mean by $\mathbb{R}$, the real numbers, and what you are going to assume about that field. If you assume the completeness axiom, then it is easy to see that $\sup\{x\in\mathbb{R}\mid f(x)<0\}$ is a root. [I am assuming the top coefficient of $f$ is $1$.) But this is just proving the IVT for the specil case. I believe that you can't avoid using Completeness, and that the most convenient stepping stone is the IVT. Dec 23, 2020 at 12:09
• One way to do this is to simply replace the citation of the intermediate value theorem (which would be one line in the usual proof) with the proof of the intermediate value theorem (which might amount to a page or so, if you allowed citation of the monotone sequence theorem; somewhat longer without that as long as you allow use of the completeness axiom; totally impossible otherwise). Would that be an acceptable answer to your question? Dec 23, 2020 at 15:00
• Also, your paragraph on the FTA is unclear, but if (as it seems) it is implying that there is no analysis in the "algebraic" proofs, that is incorrect: there is analysis in the algebraic proofs. Dec 23, 2020 at 15:02

Complex roots of a polynomial with real coefficients occur in pairs.

Hence?

https://en.m.wikipedia.org/wiki/Complex_conjugate_root_theorem

Every polynomial $$P$$ can be split (not sure of the word, the one in French is "scindé") in $$\mathbb{R}$$ : $$P = \prod (X^2-a_k X + b_k) \prod (X- \lambda_k)$$. As the degree of $$P$$ is odd, there is at least one factor $$(X-\lambda)$$, so there is at least one real root.

As for Vivid's proof, the proof of Descarte's Rule of Signs is based on IVT, so it is not really a different proof.

(Sorry if my words are incorrect, I'm French and I'm not used to writing maths in English.)

• I be very surprised if the proof that $P$ is split does not also use the IVT. Dec 23, 2020 at 15:06
• Actually no : it's an algebraic result. It comes from the fact that if $\gamma$ is a complex zero of $P\in \mathbb{R}[X]$, then $\overline{\gamma}$ is also a zero of $P$. (cf Peter Szilas's answer)
– math
Dec 23, 2020 at 15:13
• And how does one prove that there exists any complex zero at all? Dec 23, 2020 at 15:24
• It is the Fundamental theorem of algebra (cf en.wikipedia.org/wiki/Fundamental_theorem_of_algebra).
– math
Dec 23, 2020 at 15:51
• So to come to my point, the proof of the FTA uses 2-dimensional analytic principles which are generalizations of (and considerably harder to prove than) the 1-dimensional IVT. Dec 23, 2020 at 16:16

One could use Descarte's Rule of Signs. So, below is the idea:

Assume that we have $$P(x) = \sum_{k=0}^{n} a_kx^k$$

• If there is at least one sign change, then we are done. So we assume there is not.
• Assume there are terms of both odd and even exponents, then we analyze $$P(-x)$$. There were no sign changes in $$P(x)$$, therefore, we must have in $$P(-x)$$ since $$(-x)^{\text{odd}} = -x^{\text{odd}}$$ and $$(-x)^{\text{even}} = x^{\text{even}}$$.
• Assume there are only odd powers, then the polynomial is an odd function which means $$x=0$$ is a root.
• We cannot assume all exponents are even since $$P(x)$$ is of odd degree.