26 1. Trisecting Angles Equation 1.7.4 will be referred to as the trisection equation. We might note that the question as to whether an angle can be trisected amounts to showing whether an angle with radians between 0 and π 2 radians can be trisected. Indeed, the π 2 angle can be trisected. So if we are given an angle with more than π 2 radians, we can subtract from it as many multiples of the right angle as are needed to have the resulting angle between 0 and π 2 radians. If this remainder angle can be trisected, the original one can also. Exercises 1. Show that a constructible angle having 3θ radians can be trisected if and only if sin θ is a constructible number. 2. Using the previous exercise, state and prove the result analogous to Theorem 1.7.3 where the sine function replaces cosine. 1.8. The π 3 angle cannot be trisected For an angle with π 3 radians, the associated number is cos π 3 = 1 2 . So the trisection equation becomes 1.8.1 x3 − 3x − 1 = 0. Theorem 1.7.3 says that to show that this angle cannot be trisected, we must show that this equation does not have a positive root in K, the field of constructible numbers. But by Theorem 1.6.6, this means we must show that (1.8.1) does not have a positive solution in any quadratic extension of Q. This is done by induction. The next lemma is the first step in the induction argument. 1.8.2. Lemma. The equation x3 − 3x − 1 = 0 has no solution in Q. Proof. Suppose, to the contrary, that (1.8.1) has a rational solution x = p/q, where p and q are integers with no non-trivial common divisor. Sub- stituting p/q for x in (1.8.1) and multiplying both sides by q3/p gives the equation p2 − 3q2 = q3 p . But the left side of this equation is an integer, so it must be that q3/p ∈ Z. That is, it must be that p divides q3. But the fact that p and q are relatively prime implies that p and q3 must be relatively prime. (Factor p and q as the product of prime numbers and see what this says. In other words, if a prime number divides both p and q3, then it must also divide q and thus be a common divisor of p and q.) This contradiction proves the lemma.

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