16 1. Trisecting Angles Note that if we prove this when both a and b are positive, we have proved the whole claim. For example, if a 0 and b 0, then ab = −[(−a)b] and both −a and b are positive. (And −a K by Claim 1.) So we are allowed to assume that both a and b are positive. Let A = (a, 0) and B = (0,b). Draw the straight line from T = (0, 1) to A. Now construct the line that is parallel to TA and passes through B let C be the point where this line intersects the x-axis. See Figure 1.5.2. If C = (c, 0), then c K. It follows Figure 1.5.2 that ΔAOT ΔCOB (≈ means similar). (Why is this true?) Hence |OC| |OA| = |OB| |OT| . But |OT| = 1, |OA| = a, and |OB| = b. Solving the above equation we get that c = |OC| = ab. Claim 5. If b = 0, then a/b K. The proof of this is similar to the proof of the preceding claim and is left to the reader. (Exercise 2.) 1.5.7. Corollary. Every rational number is constructible. Proof. This an immediate consequence of Proposition 1.5.3 and the pre- ceding theorem. The reader likely knows that 2 / Q. The next result shows, amongst other things, that there are irrational numbers that are constructible (prob- ably suspected by the perspicacious reader). 1.5.8. Theorem. If a is a constructible number and a 0, then a is constructible. Proof. Assume that a K and a 0. By Theorem 1.5.6, 1 2 (1 + a) and 1 2 (1 a) are both constructible. Construct the circle with center the origin
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