This paper presents an elementary and direct proof of the Fundamental Theorem of Algebra, via Bolzano-Weierstrass Theorem on Minima, that avoids: every root extraction, angle, non-algebraic functions, differentiation, integration, series and arguments by induction.
Deep Dive into The Fundamental Theorem of Algebra: From The Four Basic Operations.
This paper presents an elementary and direct proof of the Fundamental Theorem of Algebra, via Bolzano-Weierstrass Theorem on Minima, that avoids: every root extraction, angle, non-algebraic functions, differentiation, integration, series and arguments by induction.
some of these proofs include a proof by induction of the existence of every nth root, n ∈ N, of every complex number (see [8], [12], [15] and [16]).
It is well known that all norms over C are equivalent and that C, equipped with any norm, is complete. In what follows it is considered the norm
Moreover, in what follows it will be needed the well known Binomial Formula (z + w) n = n j=0 n j z j w n-j , z ∈ C, w ∈ C, n ∈ N, n j = n! j!(n-j)! and 0! = 1. It is assumed, without proof, only:
• Bolzano-Weierstrass Theorem: Any continuous function f : D → R, D a bounded and closed disc, has a minimum on D.
Right below, we show, for the case k even, k ≥ 2, a pair of inequalities that Estermann [5] proved for every k ∈ N \ {0}. The proof, via binomial formula, simplifies Estermann’s proof, which uses root extraction and also induction. The case k odd can be proved similarly, if one wishes.
Lemma (Estermann).
Proof. Since k = 2m and 2k = 4m, for some m ∈ N, applying the formulas
, we end the proof by noticing that for every j ∈ N, 1 ≤ j ≤ k -1, we have
Theorem. Let P be a complex polynomial, with degree(P ) = n ≥ 1.
Then, P has a zero.
we have
Hence, P (z)P (z) → ∞ as |z| 1 → ∞ and, by continuity, P P has a global minimum at some z 0 ∈ C. We can clearly assume that z 0 = 0. Therefore,
(1) P (z)P (z)-P (0)P (0) ≥ 0 , ∀ z ∈ C , and P (z) = P (0) + z k Q(z), for some k ∈ {1, …, n}, where Q is a polynomial and Q(0) = 0. Substituting this equation, at z = rζ, where r ≥ 0 and ζ is arbitrary in C, in inequality (1), we arrive at
and, cancelling r k > 0, we find the inequality
whose left side is a continuous function of r, r ∈ [0, +∞). Thus, taking the limit as r → 0 we find,
( (2) The almost algebraic “Gauss’ Second Proof” (see [7]) of the FTA uses only that “every real polynomial of odd degree has a real zero” and the existence of a positive square root of every positive real number.
Nevertheless, this proof by Gauss is not elementary.
So, we conclude that -2Re P (0)ζ k Q(0) ≤ 0, with ζ arbitrary in C. Now, obviously, the proof continues as in the theorem proof. The uniqueness of a positive nth root of c is trivial.
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