This paper shows an elementary and direct proof of the Fundamental Theorem of Algebra, via Bolzano-Weierstrass Theorem on Minima and the Binomial Formula, that avoids: any root extraction other than the one used to define the modulus function over the complex plane, trigonometry, differentiation, integration, series, arguments by induction and arguments using epsilon's and delta's.
Deep Dive into The fundamental theorem of algebra: A most elementary proof.
This paper shows an elementary and direct proof of the Fundamental Theorem of Algebra, via Bolzano-Weierstrass Theorem on Minima and the Binomial Formula, that avoids: any root extraction other than the one used to define the modulus function over the complex plane, trigonometry, differentiation, integration, series, arguments by induction and arguments using epsilon’s and delta’s.
Besides the modulus function (derived from the inner product zw, with z, w ∈ C) and the 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:
• Polynomial continuity.
• 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 [4] proved for every k ∈ N \ {0}. The proof, via binomial formula, is a simplification of the one made by induction and given by Estermann. 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.
and, dividing the above inequality by r k > 0, we find the inequality 2Re
whose left side is a continuous function of r, r ∈ [0, +∞).
Thus, taking the limit as r → 0 we find,
(
and (since x < 0) we conclude that a ≤ 0. So, a = 0. Therefore, we get that ∓by ≥ 0. Hence, since y = 0, we conclude that b = 0. Hence α = 0 and then, P (0) = 0. Thus, the case k even is proved. The theorem is proved.
(1) The almost algebraic “Gauss’ Second Proof” (see [6]) 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.
∀r ∈ (0, 1). Hence,
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 proof of the theorem.
(3) It is worth to point out that this proof of the FTA easily implies an independent proof of the existence of a unique positive nth root, n ≥ 3, of any number a ≥ 0. In fact, given z ∈ C such that z n = a, we have that |z| n = a, with |z| ≥ 0. The uniqueness of such nth root is very trivial.
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