Is my proof for the existence of roots of an odd-degree polynomial correct?

Question

$\color{crimson}{\text{Problem}}$

Show that if $f:\mathbb{R}\to\mathbb{R}$ be a polynomial of odd degree with real coefficients then it has at least one real root.

$\color{green}{\text{Proof}}$

Let $$f(x)=a_{2n+1}x^{2n+1}+a_{2n}x^{2n}+\ldots+a_1x+a_0$$ be a polynomial where $a_i\in\mathbb{R}$ for all $i\in \{1,2,\ldots,2n+1\}$ and $a_{2n+1}\ne 0$. Without loss of generality we can also assume that $a_0>0$. Then, $$\boxed{f(0)>0}$$

Case 1

Let $a_i\ne 0$ for all $i\in \{1,2,\ldots,2n+1\}$. Then we rewrite $f(x)$ as,

$$f(x)=x^{2n}\left(a_{2n+1}x+a_{2n}\right)+\ldots+(a_1x+a_0)$$Now observe that the sign of the terms of the form $x^{2i}\left(a_{2i+1}x+a_{2i}\right)$ depends only on the sign of $\left(a_{2i+1}x+a_{2i}\right)$. So, if we want to show that $f(x)<0$ for some $x\in\mathbb{R}$, it suffices to show that each terms of the form $\left(a_{2i+1}x+a_{2i}\right)<0$. This arguments shows that $f(x_0)<0$ for any $x_0\in \mathbb{R}$ such that, $$x_0<\min\left(-\dfrac{a_{2n}}{a_{2n+1}},-\dfrac{a_{2n-2}}{a_{2n-1}},\ldots,-\dfrac{a_{0}}{a_{1}}\right)$$

Case 2

If on the other hand $a_{2i+1}=0$ for some $i$'s (if $a_{2i}$'s are $0$, it won't affect the argument of Case 1) then we define the polynomial $g:\mathbb{R}\to\mathbb{R}$ defined by, $$g(x)=b_{2n+1}x^{2n+1}+b_{2n}x^{2n}+\ldots+b_1x+b_0$$ where, $$b_{2i+1}\begin{cases}=-1 & \text{if}\ a_{2i+1}=0\\=a_{2i+1} & \text{if}\ a_{2i+1}\ne0\end{cases}$$Then, to the polynomial $g(x)$, the argument of Case 1 applies.

Now for this $g(x)$ we choose $x_0$ is such a manner that $x_0<0$. Then $g(x_0)<0$ and also $g(x_0)>f(x_0)$. Consequently $f(x_0)<0$

What is left now is to apply Bolzano's Theorem to conclude the existence of a root.

Answer 1

It might not be possible to find an $x$ such that all the $a_{2i+1}x + a_{2i}$ are negative. The interval you suggested, $x < \min -\frac{a_{2i}}{a_{2i+1}}$, is not valid because $a_{2i+1}x < -a_{2i}$ does not imply $x < -a_{2i}/a_{2i+1}$. The direction of the inequality changes when you divide by a negative number. So, sometimes, your argument will need $x$ to be bigger than some things and smaller than others. This may not always happen: as in André's example $x^3 - x^2 - x + 2$, you would want $x - 1 < 0$ and $-x + 2 < 0$, i.e. $x < 1$ and $x > 2$. No such $x$ exists!

Answer 2

You are really proving the simpler result that an odd degree polynomial attains both positive and negative values. This is true in more general situations where the polynomial might not have a root (over $\mathbb{Q}$, for example, or for any subset of $\mathbb{R}$ that contains unboundedly large and small values).

For large values of $|x|$, the sign of the polynomial is the same as the sign of the highest-degree term. This is what leads to the positive and negative values and is the method used in every published proof of the theorem (that I have ever seen).

The complication with using the highest two terms, $x^{2n}(px+q)$, is that the $q$ part can work against the $px$ part, and to show that this doesn't ultimately cause problems amounts to showing that the highest term by itself dominates everything. Which is the traditional proof.

Answer 3

For a rigorous proof, you can use Least Upper Bound principle of real numbers (which defines of real numbers- the smallest set containing rational numbers and satisfying the least upper bound property) and the continuity of polynomials.

The roots of any monic polynomial are bounded (Some bounds can be found here). So above the positive boound und f(x)>0 and below the negativ bound f(x)<0. These give sufficint conditions to apply least upper bound principle and deduce that the set {$x:f(x)<0$ has a supremum, say $X$. Continuity has not been used till this point. $f(X)<0$ and $f(X)>0$ can be discounted using continuity arguments, which completes the proof.