1.8. The π 3 angle cannot be trisected 27 One might note that there is a similarity between the proof of the pre- ceding lemma and the standard proof that √ 2 / Q. The next lemma will supply us with the means of performing the induction step in our argument. 1.8.3. Lemma. If F is a field and (1.8.1) has no solution in F , then (1.8.1) has no solution in any simple quadratic extension of F . Proof. It is easy to see by examining its graph that x3 − 3x − 1 = 0 has three real roots, a, b, c. Thus x3 − 3x − 1 = 0 = (x − a)(x − b)(x − c). Multiplying this out gives that 0 = x3 − 3x − 1 = x3 − (a + b + c)x2 + (ab + bc + ac)x − abc. Therefore 0 = a + b + c, −3 = ab + bc + ac, 1 = abc. It is only the first of these three equations that will be used. Suppose there is a solution of the trisection equation that lies in some quadratic extension of F . So there is a positive d in F such that d / F and one of the roots, say a, belongs to F ( √ d). So a = u + v √ d, where u, v ∈ F . Hence 0 = (u + v √ d)3 − 3(u + v √ d) − 1 = (u3 + 3uv2d − 3u − 1) + (3u2v + v3d − 3v) √ d. Therefore A = u3 + 3uv2d − 3u − 1 = 0 and B = 3u2v + v3d − 3v = 0 (Why?). Now a very fortunate thing happens: u − v d is also a root of the equation. Indeed, (u − v √ d)3 − 3(u − v √ d) − 1 = A − B √ d = 0. So we can assume that u − v √ d = b. But then the third root, c, satisfies c = −a − b = −2u ∈ F , contradicting the assumption that (1.8.1) has no root in F . Now for the denouement. By Lemma 1.8.2 there is no root of (1.8.1) in Q. Using the preceding lemma and induction we see that there is no root of (1.8.1) in any quadratic extension of Q. By Theorem 1.6.6 every constructible number lies in some quadratic extension of Q. Therefore there is no solution of (1.8.1) in K and the angle cannot be trisected.

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