关于恒零多项式系数全为零的数学归纳法证明合理性验证问询
Hey there! Let's break this down step by step—your proof is on the right track, but we can tweak a couple of things to make it rock-solid, especially around that $x=0$ concern you have.
Base Case ($n=1$)
We aim to show: If $ax + b = 0$ for all values of $x$, then $a = b = 0$.
Your reasoning here is perfect, formatted cleanly below:
$$
\begin{align}
ax + b &= 0 \
\text{Substitute } x=0: \quad 0 + b &= 0 \implies b = 0 \
\text{Substitute } b=0 \text{ and } x=1: \quad a(1) + 0 &= 0 \implies a = 0
\end{align}
$$
By using two distinct values of $x$ (since the equation holds for every $x$), you correctly force both coefficients to zero—this is exactly the right approach for the base case.
Inductive Step
First, a tiny easy-to-make typo to fix: your degree-$(n+1)$ polynomial should be $a_{n+1}x^{n+1} + a_nx^n + \dots + a_1x + a_0 = 0$ (you had $a_nx^{n-1}$ instead of $a_nx^n$). Let's formalize the corrected reasoning:
- Inductive Hypothesis: Assume that for any degree-$n$ polynomial $P(x) = a_nx^n + \dots + a_0$, if $P(x) = 0$ for all $x$, then all coefficients $a_n, \dots, a_0$ are zero.
- Goal: Prove that for any degree-$(n+1)$ polynomial $Q(x) = a_{n+1}x^{n+1} + a_nx^n + \dots + a_1x + a_0$, if $Q(x) = 0$ for all $x$, then all coefficients $a_{n+1}, \dots, a_0$ are zero.
Your core logic here is sound—let's clarify the $x=0$ piece you were unsure about:
$$
\begin{align}
Q(x) &= 0 \text{ for all } x \
\text{Plug in } x=0: \quad a_{n+1}(0)^{n+1} + a_n(0)^n + \dots + a_1(0) + a_0 &= 0 \
\implies a_0 &= 0
\end{align}
$$
We can now rewrite $Q(x)$ as:
$$
Q(x) = x\left(a_{n+1}x^n + a_nx^{n-1} + \dots + a_1\right) = x \cdot R(x)
$$
where $R(x)$ is a degree-$n$ polynomial.
Since $Q(x) = 0$ for every $x$, $x \cdot R(x) = 0$ no matter what $x$ we pick. For any $x \neq 0$, we can divide both sides by $x$ to get $R(x) = 0$. Here's the key point you might have missed: polynomials that vanish on infinitely many points must be the zero polynomial (they can't have infinitely many roots unless every coefficient is zero). Since there are infinitely many non-zero $x$ values, $R(x) = 0$ for all $x$—not just non-zero ones.
By our inductive hypothesis, this means all coefficients of $R(x)$ are zero: $a_{n+1} = a_n = \dots = a_1 = 0$. We already found $a_0 = 0$, so every coefficient of $Q(x)$ is zero.
This clears up your worry about neglecting $x=0$: we don't need to use $x=0$ for $R(x)$ because the infinite number of non-zero $x$ values already force $R(x)$ to be identically zero. Your original reasoning was almost there—you just needed to explicitly state that polynomial property to eliminate any doubt.
Final Takeaway
Your inductive reasoning is fundamentally correct! The only adjustments needed are:
- Fixing that small typo in the degree-$(n+1)$ polynomial's terms
- Adding the quick note about polynomials vanishing on infinitely many points being the zero polynomial to solidify the $R(x)$ step
Once you make those tweaks, your proof is rigorous and airtight. Great job questioning the edge case—it shows you're thinking critically, which is essential for strong mathematical proofs!
备注：内容来源于stack exchange，提问作者Lucien Jaccon
