14 1. Trisecting Angles 1.5.3. Proposition. Every integer is a constructible number. The next result has only transitory interest. A far better result will be proved later with the help of this lemma. 1.5.4. Lemma. If (a, b) is a constructible point, then (b, a) is a constructible point. Proof. Let P = (a, b) be a constructible point. Also assume for the moment that both a and b are positive numbers, so that P lies in the first quadrant. We will show that (b, a) is the point of intersection of a constructible line and a constructible circle. The first step is to draw the constructible circle with center O and radius |OP|. Let X and Y be the points where this circle meets the x- and y-axes. Also draw the constructible line OP . See Figure 1.5.1. Figure 1.5.1 Now using Y as center, draw the constructible circle with radius |XP|. Let Q be the point where this circle meets the first one. Note that Q is a constructible point. We claim that Q = (b, a). To see this, draw the segments OQ, XP , and Y Q as in Figure 1.5.1. Note that all these lines are constructible, so all the points where they intersect are constructible. By the construction, |OP| = |OQ|, |OX| = |OY |, and |XP| = |Y Q|. Thus ΔOPX ΔOQY . Therefore the (perpendicular) distance from Q to the y-axis equals the distance from P to the x-axis. Again see Figure 1.5.1. But this latter distance is precisely b. This implies that the first coordinate of Q is b. Similarly the distance from Q to the x-axis equals the distance from P to the y-axis, which is the number a. Thus Q = (b, a), and the proof is complete when both a and b are positive. What happens if a and b are not both positive? Here we use Exercise 1 at the end of this section to complete the proof. The details are left to the reader.
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