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I am having difficulty understanding the intuition behind the proof for the following theorem:

Suppose $a_0,a_1,\dots, a_m \in \mathbb F.$ If $$a_0 + a_1z^1+ \cdots + a_mz^m = 0$$ for every $z \in \mathbb F$, then $a_0 =a_1 =\dots = a_m = 0$

The proof I am referring to below is from Sheldon Axler's Linear Algebra Done Right.

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Question 1 : Doesn't the choice of $z$ only prove the theorem for $z \ge1$?

Question 2 : Why should we let $z$ be represented in terms of the coefficients? Normally in a function, we wouldn't define the independent variable ($z$) in terms of the parameters(the coefficients) would we?

Question 3 : How is $(|a_0|+ |a_1|+ \cdots|a_{m-1}|)z^{m-1} < |a_m z^m|$ for possible cases in which coefficients, in addition to $a_m$, do not equal to $0$?

Side Yes/No Question: Can this theorem be proven with induction by taking the derivative and showing the constant term to be zero on each step also?

Thanks in Advance!

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Bunny
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    Hint Q1: this is a proof by contradiction, so it's enough to find one $z$ for which $P(z) \ne 0$, which is precisely what the proof does. – dxiv Nov 16 '16 at 01:46
  • Hint Q2: that is a choice, a tricky choice. – MattG88 Nov 16 '16 at 01:50
  • @dxiv Because the original statement is of the form $P \Longrightarrow Q $ by assuming $\lnot Q$ we prove $\lnot P$, will it be a contradiction as we will have $P \land \lnot P$ ?? I think my reasoning is wrong here however – Bunny Nov 16 '16 at 02:06
  • The way it's phrased it actually proves the contrapositive, which is equivalent to the original implication. Lookup proof by contrapositive. I called it proof by contradiction in my previous comment since the two are closely related, to the point where it's essentially a simple matter of rephrasing between the two. – dxiv Nov 16 '16 at 02:15
  • To the side question: Yes, but this requires you setting up the whole machinery of derivatives beforehand. A better proof along these lines would use finite differences instead of derivatives. – darij grinberg Nov 17 '16 at 03:03

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