Algorithms in Real Algebraic Geometry, Second Edition by Saugata Basu, Richard Pollack, Marie-Francoise Roy,

By Saugata Basu, Richard Pollack, Marie-Francoise Roy,

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8: Since the union of a chain of proper cones is a proper cone, Zorn’s lemma implies the existence of a maximal proper cone C which contains C. It is then sufficient to show that C ∪ −C = F, and to define x ≤ y by y − x ∈ C. Suppose that −a∈C. 9, C[a] is a proper cone and thus, by the maximality of C, C = C[a] and thus a ∈ C. 7: a) ⇒ b) since in a real field F, F(2) is a proper cone. 8. n 2 2 c) ⇒ d) since in an ordered field, if x1 0 then i=1 xi ≥ x1 > 0. n 1, so 4 implies that 1 + i=1 x2i = 0 is d) ⇒ a), since in a field 0 impossible.

Proof: We can suppose without loss of generality that that 0 is not a root of A, that it that all the coefficients of A are positive. Then a p−1 a p−2 a0 , ap a p−1 a1 and a p−1 a p−2 a0 −x −x − x. ap a p−1 a1 Since a p > 0 and -a0 x<0, the coefficients of the polynomial (X − x) A = ap X p+1 + a p a p−1 − x Xp + ap + a1 a0 − x X − a0 x. a1 have exactly one sign variation. 35), is to interpret the even difference Var(Der(P ); a, b) − num(P ; (a, b]). This can be done through the notion of virtual roots.

F Q∈Q The sign condition σ is realizable if Reali(σ) is non-empty. 26. [Derivatives] Let P be a univariate polynomial of degree p in R[X]. We denote by Der(P ) the list P , P , , P (p). 27. [Basic Thom’s Lemma] Let P be a univariate polynomial of degree p and let σ be a sign condition on Der(P ) Then Reali(σ) is either empty, a point, or an open interval. Proof: The proof is by induction on the degree p of P . There is nothing to prove if p = 0. Suppose that the proposition has been proved for p − 1.

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