1.5. Constructible points and constructible numbers 15 We have already noticed that both coordinates of a constructible point are constructible numbers. What about the converse? The next result provides the answer and improves the preceding lemma. 1.5.5. Proposition. If a and b are constructible numbers, then (a, b) is a constructible point. Proof. Since a is a constructible number, there is a constructible point A such that a is one of its coordinates. By the preceding lemma, we may assume that A = (a, c) for some c. Similarly there is a constructible point B = (d, b). Construct the line a passing through A and perpendicular to the x-axis. This line meets the x-axis at the point (a, 0). Also construct the line b through B and perpendicular to the y-axis. This meets the y-axis at (0,b). Also the lines a and b intersect at the point (a, b), which must, therefore, be constructible. The next result is a linchpin in the proof that not every angle can be trisected. If you know the term, it says that the constructible numbers form a field, but we’ll get to that later. The underlying message and essence of the next theorem is that with constructible numbers we can do arithmetic to our hearts’ content. 1.5.6. Theorem. If a, b ∈ K, then a + b, a − b, and ab ∈ K. If b = 0, then a/b ∈ K. Proof. The proof will be accomplished in several small steps. Assume that a and b are as in the statement of the theorem. Claim 1. −a ∈ K. By Proposition 1.5.5, A = (a, 0) is a constructible point. Construct the circle with center O and radius |OA| = |a|. This intersects the x-axis at (−a, 0), so −a ∈ K. (Alternatively we could have used Exercise 1.) Claim 2. a + b ∈ K. Put A = (a, 0) and B = (b, 0), both constructible points. Draw the circle with center A and radius |OB| = |b|. This intersects the x-axis at (a+|b|, 0) and (a − |b|, 0). Depending on whether b 0 or b 0, one of these facts shows that a + b ∈ K. Claim 3. a − b ∈ K. Indeed, a − b = a + (−b), so Claim 3 follows from the first two claims. Claim 4. ab ∈ K.

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