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4 Aug 2022
Q S A R. We have seen that Q and the constructible numbers, or the surds, S are subfields of R. The set A of Algebraic numbers is also a subfield of R. We want to do a partial proof of this by taking the following steps: given that s Element .A, start with a polynomial f(x), for simplicity assume f(x) = a_0 + a_1x + 02X^2 + a_3x^3, with a_i Element Q, that accepts s as a root, and construct another polynomial g(x) that can accept - s as a root. construct a polynomial h(x) that accepts 1/s as a root. (Bonus) For the next two cases we make work with simpler functions: let s and t be roots of polynomials f(x) = a_0 + a_1x + G2x^2 and g(x) = b_0 + b_1x + b_2x^2 and construct another h(x) that satisfies h(s + t) = 0, and another polynomial k(x) such that k(st) = 0.
Q S A R. We have seen that Q and the constructible numbers, or the surds, S are subfields of R. The set A of Algebraic numbers is also a subfield of R. We want to do a partial proof of this by taking the following steps: given that s Element .A, start with a polynomial f(x), for simplicity assume f(x) = a_0 + a_1x + 02X^2 + a_3x^3, with a_i Element Q, that accepts s as a root, and construct another polynomial g(x) that can accept - s as a root. construct a polynomial h(x) that accepts 1/s as a root. (Bonus) For the next two cases we make work with simpler functions: let s and t be roots of polynomials f(x) = a_0 + a_1x + G2x^2 and g(x) = b_0 + b_1x + b_2x^2 and construct another h(x) that satisfies h(s + t) = 0, and another polynomial k(x) such that k(st) = 0.
4 Aug 2022
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