Definition D1
p(x) = aₙxⁿ + ⋯ + a₀ (aₙ ≠ 0, n ≥ 1) is called a polynomial of degree n, and a c with p(c) = 0 is called a root.
Proof.
The degree n is the exponent of the leading term. Only non-constant polynomials are treated.convention
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Definition D2
Let ℚ(i) = { a + bi : a, b ∈ ℚ }.
Proof.
It is a field closed under the four operations — it satisfies all of the field axioms F.convention
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Lemma L1
x² − 2 has no root in ℚ(i).
Proof.
(1)Suppose there are a, b ∈ ℚ with (a + bi)² = 2.assumption, D2
(2)Expanding gives (a² − b²) + (2ab)i = 2 + 0·i.(1), A6, F, E
(3)Real and imaginary parts agree separately: a² − b² = 2, 2ab = 0.(2), A7
(4)From 2ab = 0, ab = 0 — therefore a = 0 or b = 0.(3), F, A8
(5)If b = 0, then a² = 2 — no such rational exists. Contradiction.(3), (4), A9, F, E
(6)If a = 0, then −b² = 2, that is, b² = −2 < 0 — squares are non-negative, a contradiction.(3), (4), A10, A11, F, E
(7)Both cases are blocked. There is no root.(5), (6)
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Theorem T1
If every non-constant polynomial has at least one root, then a degree-n polynomial has n roots (with multiplicity).
Proof.
(1)If there is a c₁ with p(c₁) = 0, then p(x) = (x − c₁)q(x) — q has degree n − 1.assumption, A4
(2)If q is also non-constant, it has a root c₂, and the degree drops by one again.(1), assumption, A4
(3)The degree falls by 1 each time and ends after n steps: p(x) = aₙ(x − c₁)⋯(x − cₙ).(2), D1
(4)Therefore it suffices to show 'there is at least one root'; the n roots then follow.(3)
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Theorem T2
A polynomial of odd degree with real coefficients has a real root.
Proof.
(1)Assume the leading coefficient is positive. If it is negative, replacing p by −p leaves the roots unchanged.convention, F
(2)The polynomial function p is continuous.A3
(3)The degree is odd, so for sufficiently large M, p(−M) < 0 < p(M).(1), A5, D1
(4)By the intermediate value theorem, there is a real c between −M and M with p(c) = 0.(2), (3), A2
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Statement T
Every non-constant polynomial with complex coefficients has a root in ℂ.
(1)A complete proof of this theorem exceeds the format of this board. Here it is set down as a statement, and what it rests on is made explicit.convention
(2)It cannot be proved from the field axioms alone — ℚ(i) satisfies all of them, yet x² − 2 has no root there.D2, L1
(3)In the odd-degree real-coefficient case, what produced the root was completeness (the intermediate value theorem). The general case rests on the same source.T2, A2
(4)What decides the existence of roots is not the algebra of the coefficients but the fact that the complex plane has no gaps.(2), (3), A1
(5)And one root yields n roots (T1). A degree-n equation has n roots in ℂ.(4), T1
Proof omitted.