* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download background on constructible angles
Survey
Document related concepts
Mathematics of radio engineering wikipedia , lookup
Large numbers wikipedia , lookup
Mathematical proof wikipedia , lookup
Vincent's theorem wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
List of important publications in mathematics wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Factorization wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Four color theorem wikipedia , lookup
Elementary mathematics wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
History of trigonometry wikipedia , lookup
Transcript
Math 4000: Modern Algebra and Geometry I Spring 2014, Dr. Klipper Extra Examples for Week 15 Practice and Results with Constructible Angles In class, we defined an angle θ to be constructible when it is possible to construct two intersecting lines that form an angle of θ. (We allow ourselves to add or subtract multiples of 360◦ or 2π.) This isn’t the same definition as a constructible number ; indeed, π is not a constructible number, but π radians is clearly a constructible angle, since it’s formed by a straight line! Let’s say T is the set of constructible angles. We’ve seen that: • T is closed under sums and differences (as well as integer multiples): if θ ∈ T , then nθ ∈ T for n ∈ Z. • We can bisect angles in T : if θ ∈ T , then θ/2 ∈ T . • Trisection isn’t always possible: for instance, 60◦ ∈ T but 60◦ /3 6∈ T . • A regular n-gon is constructible iff θ = 2π/n (or 360◦ /n) is constructible. In this document, we’ll explore these ideas more, show a difficult example, and discover a surprising result about which degree values are constructible! Relationship to Sines and Cosines Let’s say F is the set of constructible numbers. We mentioned in class that a point (x, y) is constructible iff x, y ∈ F . When we apply that to points on the unit circle, where the angle θ yields the point (cos(θ), sin(θ)), we find that θ ∈ T if and only if cos(θ) and sin(θ) are constructible numbers. In fact, you only need to check one of these: cos(θ) ∈ F automatically implies sin(θ) ∈ F and vice versa. (Figure out why!) This allows us to make some nifty algebraic arguments for constructibility, as opposed to using geometric arguments! Theorem. If θ and φ are constructible angles (i.e. θ, φ ∈ T ), then so are θ + φ and nθ for any n ∈ N. Alternate proof. We will just show cos(θ + φ) and cos(nθ) are in F . Using the angle-addition formula for cosine, cos(θ + φ) = cos(θ) cos(φ) − sin(θ) sin(φ) By assumption, all four of these trigonometric values are in F , so their products and difference are in F . Hence, cos(θ + φ) ∈ F . (A similar identity shows that sin(θ + φ) ∈ F as well.) To get an identity for cos(nθ), we use DeMoivre’s Theorem and the Binomial Theorem to state that n X n n (cos(nθ) + i sin(nθ)) = (cos(θ) + i sin(θ)) = cos(θ)k sin(θ)n−k in−k k k=0 1 By taking the real part of that sum, we obtain a formula for cos(nθ) using a sum whose terms have products of integers, cos(θ), and sin(θ). All those factors are constructible numbers, so cos(nθ) ∈ F . (Similarly, the imaginary part shows that sin(nθ) ∈ F too.) Proving an Angle is Non-Constructible Questions like “is it possible to trisect an angle with compass and straightedge?” were very old mathematical questions, many of these questions dating back to the ancient Greeks and Euclid’s famous textbook Elements. With the advent of abstract algebra, a convenient technique finally emerged for showing something was non-constructible. We’ve seen in class that all elements of F have degrees over Q which are powers of 2. Thus, to show an angle θ is non-constructible, we should show that the degree of either cos(θ) or sin(θ) is not a power of 2. The textbook (and class) covers one example. We’ll provide another demonstration. The main technique, like in the proof on the previous page, is to relate cos(nθ) to cos(θ) by using an identity obtained from DeMoivre’s Theorem. Cutting arccos(5/7) into five parts: Let φ = arccos(5/7) and θ = φ/5. Certainly φ is constructible, because cos(φ) = 5/7 ∈ Q, and rational numbers are constructible. However, we will prove θ is non-constructible by showing that cos(θ) has degree 5 over Q. By using DeMoivre’s Theorem with n = 5, and a little algebra, you can eventually find that cos(5θ) = cos5 (θ) − 10 cos3 (θ) sin2 (θ) + 5 cos(θ) sin4 (θ) By using the identity sin2 (θ) = 1 − cos2 (θ) to substitute for sin2 (θ) (and sin4 (θ) becomes (1 − cos2 (θ))2 ), we expand and eventually obtain cos(5θ) = 16 cos5 (θ) − 20 cos3 (θ) + 5 cos(θ) Let’s say x = cos(θ) for clarity. Since 5θ = φ, cos(5θ) = cos(φ) = 5/7 as shown earlier. Therefore, 5 = 16x5 − 20x3 + 5x 7 Multiply the equation by 7, and then subtract 5, to obtain 112x5 − 140x3 + 35x − 5 = 0 This shows that cos(θ) is a root of a fifth-degree polynomial. This polynomial is irreducible thanks to Eisenstein’s Criterion with p = 5: note 112 is not divisible by 5, but the other coefficients are divisible by 5, and 5 is not a multiple of 52 . This proves cos(θ) has degree 5. Whole Degree Amounts Now, we turn to an interesting question: which degree amounts are constructible? Certainly, 1◦ cannot be constructible, or else all of its multiples would be constructible, and that would contradict the class proof that 20◦ is not constructible. In fact, no integer divisor of 20◦ could be constructible, so 2◦ and 4◦ are also non-constructible. However, it turns out that 3◦ is constructible. Here’s an outline of how we show this: 2 1. First, a regular hexagon is constructible, using six angles of 60◦ . In fact, we know 60◦ is constructible because its cosine is 1/2. 2. A regular pentagon is also constructible! One way to show this is to get an explicit formula for sin(2π/5) or cos(2π/5). (It’s a tricky task; see Exercise 5.2 #20 and its associated hints.) There are also some very elegant geometric constructions. I found a very nice applet at http://www.mathopenref.com/constinpentagon.html, and it seems that site has plenty of other geometric constructions as well. (However, the site does not prove why their final construction really produces a regular pentagon.) As a result, 360◦ /5 = 72◦ is constructible. 3. Since 72◦ and 60◦ are constructible, so is their difference 12◦ . 4. If we bisect 12◦ twice, we obtain 3◦ , so 3◦ is constructible. Therefore, the smallest n ∈ N for which n◦ is constructible is n = 3. This leads to the following theorem: Theorem. Let n ∈ N. The angle n◦ is constructible if and only if 3 | n. Proof. Divide n by 3 to get n = 3q + r with q, r ∈ Z and 0 ≤ r ≤ 2. Since 3◦ is constructible, so is its integer multiple (3q)◦ . Therefore, n◦ is constructible iff r◦ is constructible. However, 0◦ is constructible, whereas we showed earlier that 1◦ and 2◦ are not. Note, however, that this result only applies to integer degree amounts. Thus, it only helps us analyze polygons where the number of sides is a divisor of 360. We can’t, for example, use this result to study heptagons with 7 sides, because 360/7 isn’t a natural number. The textbook has some results that say Gauss figured out exactly which regular n-gons were constructible. Some of the results are surprising. For instance, it turns out that the regular 17-gon is constructible, whereas the regular 18-gon is not! 3