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Asked here before. But I had the idea to try and prove this using Rolle's Theorem and Induction. This is certainly not the easiest way to prove this, as seen in the link above. However, I am still new to analysis and I am trying to determine if what I have done actually works, I am a little hung up on how to prove non-distinct zeroes, but other than that I think what I have here is good. Here it is:

(Rolle's Theorem) If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$ then there is a point $c\in(a,b)$ so that $f'(c)=0$.

(Exercise 4.4) Every polynomial is a continuous function.

(Exercise 5.1) If $f(x)=x^n$, where $n\in\mathbb{N}$ then $f'(x)=nx^{n-1}$.

Proof. Let $P_n=a_0+a_1x+\ldots+a_nx^n$ be any $n$ degree polynomial where $n\in\mathbb{N}$ and $a_i\in\mathbb{R}$. By Exercises 4.4 and 5.1 and the arithmetic rules for limits, we know every polynomial is both continuous and differentiable on $\mathbb{R}$. By Exercise 5.1 and the arithmetic rules for limits, we know that the derivative of a polynomial is a polynomial. Therefore, for $n\geq k$, where $k\in\mathbb{N}$ the $k$th derivative of $P_n$, that is $P_n^{(k)}$, is continuous and differentiable on $\mathbb{R}$.

Suppose $P_n$ has $n+1$ distinct real zeroes $x_1<\ldots<x_{n+1}$. Then $P_n(x_1)=\ldots=P_n(x_{n+1})=0$. Since $P_n$ is continuous and differentiable for all $x\in\mathbb{R}$, $P_n$ is also continuous on $[x_i,x_{i+1}]$ and differentiable on $(x_i,x_{i+1})$ for each $i\in\{1,\ldots,n\}$. By Rolle's Theorem $\exists \ c_{1i}\in(x_i,x_{i+1})$ so that $P_n'(c_{11})=\ldots=P_n'(c_{1n})=0$. Then $P_n'$ has $n$ (that is $n+1-1$) distinct real zeroes where $c_{11}<\ldots<c_{1n}$.

Suppose $P_n^{(k)}$ has $n+1-k$ distinct real zeroes $c_{k1}<\ldots<c_{k(n+1-k)}$. Then $P_n^{(k)}(c_{k1})=\ldots=P_n^{(k)}(c_{k(n+1-k)})=0$. Since $P_n^{(k)}$ is continuous and differentiable on $\mathbb{R}$, we know it is continuous on $[c_{ki},c_{k(i+1)}]$ and differentiable on $(c_{ki},c_{k(i+1)})$ for each $i\in\{1,\ldots,n-k\}$. By Rolle's Theorem $\exists \ c_{(k+1)i}\in(c_{ki},c_{k(i+1)})$ such that $P_n^{(k+1)}(c_{(k+1)1})=\ldots=P_n^{(k)}(c_{(k+1)(n-k)})=0$. Then $P_n^{(k+1)}$ has $n-k$ (that is $(n+1)-(k+1)$) real zeroes. Thus by induction, if $P_n$ has $t$ distinct real zeroes, $P_n^{(k)}$ has $t-k$ distinct real zeroes.

Then we have that $P_n^{(n-1)}$ has $(n+1)-(n-1)=2$ distinct real zeroes $c_{(n-1)1}<c_{(n-1)2}$. So $P_n^{(n-1)}(c_{(n-1)1})=P_n^{(n-1)}(c_{(n-1)2})=0$. By Exercise 5.1 and arithmetic rules for limits we find that $P_n^{(n-1)}(x)=(n-1)!a_{n-1}+n!a_nx$. So it must be that $c_{(n-1)1}=c_{(n-1)2}$, a contradiction. Therefore, it must be that $P_n$ has at most $n$ real zeroes. $\square$

drfrankie
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    What do you mean by the "non-distinct zero" case? It only makes sense to define multiplicity of a root if you already know that any root of the polynomial can be factored, which is what I think you're trying to prove. – Ben Sep 06 '20 at 04:01
  • This theorem is true in any field, regardless of whether it is a complete, Archimedean ordered field (i.e. the real numbers). So it's hard to argue that proofs which depend on all those additional axioms are simpler than proofs which depend on the field axioms alone. – Lee Mosher Sep 08 '20 at 01:32

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I now have a much simpler way to do this still using Rolle’s Theorem:

Proof. Let $P_n$ be a non-zero degree $n$ polynomial. By Exercises 4.4 and 5.1 and the arithmetic rules for limits, we know every polynomial is both continuous and differentiable on $\mathbb{R}$. By Exercise 5.1 and the arithmetic rules for limits, $P_{n+1}'=P_n$. Since $P_0$ is non-zero, for some $a\in \mathbb{R}\backslash\{0\}$, $P_0(x)=a$ for all $x\in \mathbb{R}$. So there is no $x\in \mathbb{R} $ having the property $P_0(x)=0$, so $P_0$ has at most 0 zeroes. Suppose $P_k$ has at most $k$ real zeroes, and $P_{k+1}$ has more than $k+1$ real zeroes, say $k+2$ real zeroes $x_1<\ldots<x_{k+2}$. Since $P_{k+1}$ is continuous on $[x_i,x_{i+1}]$ and differentiable on $(x_i,x_{i+1})$ for each $i\in J=\{1,\ldots,k+1\}$, and $P_{k+1}(x_1)=\ldots=P_{k+1}(x_{k+2})=0$, by Rolle's Theorem $\exists \ c_i\in(x_i,x_{i+1})$ such that $P_{k+1}'(c_i)=0$ for each $i\in J$. So $P_{k+1}'$ has $k+1$ real zeroes, however $P_{k+1}'=P_k$ which has at most $k$ real zeroes, a contradiction. Therefore by induction $P_n$ can have at most $n$ real zeroes. $\square$

drfrankie
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