Every equation $x^n + a_{n_1}x^{n-1} + \dots + a_0 = 0$ has a real root, if $n$ is odd.

Question

This is a theorem in Spivak's book "calculus"

Every equation $x^n + a_{n-1}x^{n-1} + \dots + a_0 = 0$ has a real root, if $n$ is odd.

However, I think his proof is overly complicated. Therefore, I would like to offer an alternative proof. It would be nice if someone can check if it correct.

Proof: Define $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^n + a_{n-1}x^{n-1} + \dots + a_0$

Then for $x \neq 0$, we have

$$f(x) = x^n(1+ a_{n-1}/x + \dots + a_0/x^n)$$

and we see that $f(x) \to + \infty$ when $x \to +\infty$ and that $f(x)\to - \infty$ when $x \to - \infty$.

Hence, there is $M> 0$ such that $f(M) > 0$ and $f(-M) < 0$.

By applying IVT on $f\vert_{[-M,M]}$, we deduce that there is $x \in [-M,M]$ such that $f(x) = 0$, and this proves the result.

Answer 1

Your proof using IVT is correct, in fact $x \in (-M, M)$.

Answer 2

Your proof is correct. Some algebraists prefer a purely algebraic proof. Because the Galois group $Gal(\mathbb{C}|\mathbb{R})$ has order $2$ and it operates on the set of zeros of the polynomial all orbits must have length one or two and since the sum of the orbit lengths must be odd there must be at least one orbit of length one that contains the real root. However You need the IVT to prove the completeness of $\mathbb{C}$ so that this proof just "seems to be purely algebraic".

Answer 3

Yes, your proof is correct, though I would state that the IVT applies only because $f$ is continuous on $[-M,M]$ .

(It seems overkill to prove that a polynomial is continuous on any interval, but I'm sure you could!)

I think your proof is a nice formalisation of how a high-school student might understand that a polynomial of odd degree must have a real root: "the graph goes down on the left and up on the right so must cross the $x$-axis".