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Base Case $n=1$: if $ax+b=0$ for all values of $x$, then $a=b=0$

\begin{align} ax+b=0\\ (0)+b=0\\ b=0\\\\ ax+(0)=0\\ ax=0\\ a(1)=0\\ a=0 \end{align}

Inductive Step: if $(a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0)\Rightarrow(a_n=a_{n-1}=...=a_0=0)$ for all values of $x$, then $(a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1+a_0=0)\Rightarrow(a_{n+1}=a_n=...=a_0=0)$ is also true.

$$a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1+a_0=0$$ $$a_{n+1}(0)^{n+1}+a_n(0)^n+...+a_1(0)^1+a_0=0$$ $$\Rightarrow a_0=0$$

Now all that is left is to show that all the coefficients of $a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1=0$ are zero.

$$a_{n+1}x^{n+1}+a_{n}x^{n-1}+...+a_1x^1=0$$ $$x(a_{n+1}x^n+a_nx^{n-1}+...a_1)=0$$

Since I'm not interested in the $x=0$ possibility, set the right factor to equal $0$. Since that is in the form of what we are assuming to be true, we've shown that if the theorem works for $n$, it works for $n+1$.

$\therefore a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0\Rightarrow a_n=a_{n-1}=...a_1=a_0=0$, $\forall n\forall x$

Is this proof good enough? I'm a tiny bit unsure about where I just neglected the $x=0$ case. Is the inductive reasoning correct?

Lucien Jaccon
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    Not quite. With the way you have it written up, the induction hypothesis is that $\sum a_i x^i = 0 $ for ALL values of $x$. You've only shown it's true for all values of $x$ EXCEPT 0. So, to apply the IH, you need to show that $a_1 + a_2x + \ldots + a_{n+1} x^n = 0 $ for $ x = 0$. IE You need to show that $a_1 = 0 $, in order to prove that $a_1 = 0$, so your reasoning is circular. – Calvin Lin Feb 28 '24 at 23:09
  • In the title, you said $ \forall n$, but in the writeup, you said $ \forall x$. Please clarify which is intended. – Calvin Lin Feb 28 '24 at 23:10
  • Thank you! After that, the proof is complete? – Lucien Jaccon Feb 28 '24 at 23:10
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    To be clear, I'm pointing out that the current proof has a serious hole (circular reasoning) in it. $\quad$ If you can fix the hole, then the proof might be complete (assuming you introduce no other errors to it). – Calvin Lin Feb 28 '24 at 23:11
  • Wait, what is the circular reasoning? Should I have used $k$ instead of the generalized $n$ in the inductive step to make my intentions clearer? – Lucien Jaccon Feb 28 '24 at 23:14
  • As I said in my first comment, "you need to show that $a_1 = 0 $ in order to prove that $a_1 = 0 $". – Calvin Lin Feb 28 '24 at 23:15
  • How is that circular reasoning again? – Lucien Jaccon Feb 28 '24 at 23:16
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    I guess you consider polynomials with real coefficients and a real variable x . If so, you should include this information in your question. – Kritiker der Elche Feb 28 '24 at 23:50
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    Your statement if the problem is all over the place! What you mean is $\forall n ((\forall x(a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0))\Rightarrow a_n=a_{n-1}=...a_1=a_0=0)$. Or in words: If $a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0$ for all $x$, then $a_n=a_{n-1}=...a_1=a_0=0$. – TonyK Feb 29 '24 at 00:05
  • Thanks for telling me. Just to make sure, $a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0\Rightarrow a_n=a_{n-1}=...a_1=a_0=0$, $\forall n\forall x$ is never accepted as proper notation? – Lucien Jaccon Feb 29 '24 at 00:11
  • @Lucien: No, you should never put existential quantifiers at the end like that. It is too likely to result in ambiguity. – TonyK Feb 29 '24 at 13:08
  • Thanks for letting me know – Lucien Jaccon Feb 29 '24 at 13:51

1 Answers1

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Your proof has a gap. The equation $$x(a_{n+1}x^n+a_nx^{n-1}+...+a_1)=0 \tag{1}$$ means $x = 0$ or $a_{n+1}x^n+a_nx^{n-1}+\ldots+a_1 = 0$. This allows to conclude that $$p(x) = a_{n+1}x^n+a_nx^{n-1}+\ldots+a_1 = 0 \text{ for all } x \ne 0 . \tag{2}$$

Induction only works if we know that also $p(0) = 0$. The argument needed here is continuity. All polynomials are continuous, thus $$p(0) =\lim_{x \to 0} p(x) = 0 .$$

Paul Frost
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  • Yeah, but didn't I show that $a_1=0$ by substituting $x=0$? $p(x)$ is defined to equal zero for all values of $x$. – Lucien Jaccon Feb 28 '24 at 23:17
  • You are correct that the proof has a [very subtle!] gap and what the gap is, and that was the OP's question so +1. But I am not convinced that continuity can be used to fix the gap that is in the proof. – Mike Feb 28 '24 at 23:22
  • @LucienJaccon You cannot do that without an additonal argument. In the inductive step you correctly showed that $a_0 = 0$ and concluded that $(1)$ is true for all $x$. Howewer, this only proves $(2)$. But to apply the assumption you need to know it for all $x$. – Paul Frost Feb 28 '24 at 23:25
  • So all I have to show is for the $x=0$ example, but that is easy because I've already shown that $a_1=0$ for all $x$, meaning that the entire RHS is going to be $0$. What was the major circular reasoning that you were talking about? – Lucien Jaccon Feb 28 '24 at 23:28
  • @LucienJaccon Do you agree that $(1)$ does not imply $p(0) = 0$? – Paul Frost Feb 28 '24 at 23:35
  • Yeah but $p(x)$ is explicitly defined to be $0$ for all inputs, including $0$. – Lucien Jaccon Feb 28 '24 at 23:36
  • @LucienJaccon Of course $p(x)$ is defined for all $x$, but we only know that $p(x) = 0$ for $x \ne 0$. We need to prove that $p(0) = 0$ (whichs is equivalent to $a_1 = 0$). – Paul Frost Feb 28 '24 at 23:40
  • How do you know that the limit as $x$ goes to $0$ is $0$? I understand what you mean now, though. – Lucien Jaccon Feb 28 '24 at 23:47
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    @LucienJaccon Since $p(x) = 0$ for all $x \ne 0$, the limit is $0$ as $x \to 0$. – Paul Frost Feb 28 '24 at 23:56
  • Oooh, that's good. Thank you soooo much!! – Lucien Jaccon Feb 28 '24 at 23:57
  • I see it now Paul, thank you as well! – Mike Feb 29 '24 at 00:49