1 The Aiomatic Method Each theorem in a deductive sstem must be derived from previous theorems which must in turn be derived from previous theorems. Since this process must start somewhere we take a set of agreed upon notions as fact. These are called aioms or postulates. From these we ma deduce theorems. Eample: Blobs and rocks are objects that satisf the following aioms. P1 There eists at least two rocks. P2 For an two distinct rocks there eists a unique blob containing both. P3 Given an blob there is a rock not in it. P4 Given an blob B and a rock r not in B there eists a unique blob C containing r disjoint from B. Notice that there is no aiom above preventing rocks from containing blobs. Contains is actuall a relation, and does not necessaril mean live inside. We need some definitions before we proceed. Definition 1 If two blobs B and C contain the same rock r we sa that B and C intersect or meet in r. Definition 2 Two blobs B and C are disjoint if the have no rock in common. Definition 3 The unique blob containing the pair of rocks r and s will be denoted b rs. Note that if a blob contains rocks p, q and r then pq, qr and pr all denote the same blob. We are not claiming that blobs ma contain 3 rocks. We are now read to prove some results. We start with a simple one. Theorem 1 There are at least two disjoint blobs. Proof. Let p and q be two different rocks P1. Let B = pq be the blob containing p and q P2. Let r be a rock not in B and, consequentl, different from p and q P3. Let C be a blob disjoint from B containing r P4. B and C are two disjoint blobs. Can a blob have no rocks? Theorem 2 Ever blob contains at least one rock. Proof b contradiction. Assume there eists an empt blob B. Since B is empt an non-empt blob is disjoint from B. Let r and s be two distinct rocks P1. Then A = rs P2 is disjoint from B. Let t be a rock not in A P3 and C = rt P2. Blob C is also disjoint from B and C A. We now have two different blobs, A and C, disjoint from B that contain r. This contradicts P4. 1
Theorem 3 Two distinct blobs, each disjoint from a third blob are disjoint from each other. Proof b contradiction. Let A and B be two distinct blobs, both disjoint from a blob C. Suppose A and B are not disjoint. Let r be a rock common to both A and B. We have produced two different blobs that contain a rock not in C that are disjoint from C. This contradicts P4. Theorem 4 Ever blob contains at least two rocks. Proof b contradiction. Assume B is a blob containing eactl one rock r. Let s be a rock not in B P3, and C = rs P2. Let t be a rock not in C hence, not in B P3. Let A be a blob containing t that is disjoint from C P4. A must be disjoint from B as the onl rock of B is a rock of C. Blobs C and B are different and disjoint from A. But B and C are not disjoint from each other, a fact that contradicts Theorem 3. Definition 4 Two sets A and B, finite or infinite, have the same size if there is a one-to-one correspondence between the elements of set A and the elements of set B. Theorem 5 The number of blobs containing a given rock r is the same as the number of blobs containing an other rock s. Proof. Let r and s be two different rocks P1 and B = rs the unique blob containing r and s P2. Let t be a rock not in B P3 and C a blob containing t disjoint from B P4. Note that B is the unique blob containing r disjoint from C. Thus, each blob containing r, ecept B, will meet C in a rock P4 which is unique P2. Note that two different blobs cannot have two rocks in common. Thus there is a one-to-one correspondence, or bijection α, from blobs through r that meet C to rocks in C. Repeating the above argument with s we obtain a bijection β from blobs through s that meet C to rocks in C. A one-to-one correspondence between blobs through r and blobs through s is now easil obtained b defining fb = B and fa = β 1 αa, for blobs A B that contain r. Theorem 6 All blobs contain the same number of rocks, i.e., have the same size. Proof. Let B be a blob P1,P2 and r a rock not in B P3. Ecept for the unique blob containing r disjoint from B P4, ever blob containing r has a unique rock in common with B P2, Theorem 3. Thus, different blobs containing r determine different rocks of B and each rock of B determines a distinct blob containing r. Notice that the number of rocks in blob B is one less than the number of blobs containing the rock r. Similarl, for an other blob C let s be a rock not in C. The number of rocks in C is one less than the number of blobs containing s. Since the number of blobs containing r is the same as the number of blobs containing s Theorem 5 it follows that the number of rocks in B is the same as the number of rocks in C. If ou provide interpretations for undefined terms such as blob and rock that convert postulates into true statements then all of the derived theorems properl interpreted are also true. This is called a model. A popular model, for instance, is obtained b interpreting rocks as points and blobs as lines in the Euclidean plane. 2
Homework: Provide a model for the above aioms other than lines and points. Solution 1: Consider a tetrahedron. Let rocks be the vertices of the tetrahedron and blobs be the edges. Notice how all the theorems proved above continue to hold. Solution 2: Consider the same tetrahedron. Let rocks be the faces of the tetrahedron and blobs be the edges. Again, all the theorems proved above continue to hold. 2 Isometries Notation: R n is the set { 1,...,, n, i R} For, R n,, = 1 i n i i is the inner or dot product of and. E n is the pair R n,, i.e., the set R n together with a distance function also called norm or metric given b dp, Q = P Q where =,, R n. Sometimes we ma want to use other distance functions on R n = { 1,..., n, i R n } and define l n p as the pair R n, p : p = n i=1 i p 1/p, 0 < p < = ma 1 i n i Eample: the figure below shows various unit circles, i.e., the locus of points { R 2, p = 1}, using norms p = 1, 2,. Notice that for an R n, 2 1. Definition 5 An isometr of E 2 is an onto function T : E 2 E 2 such that dt, T = d,. Isometries satisf certain important properties: 1. Isometr are one-to-one functions, since if T = T then 0 = dt, T = d, which implies =. 2. Isometries have inverses which are also isometries. Since an isometr T is a bijection so is T 1. Then dt 1, T 1 = dt T 1, T T 1 = d,. 3. The composition of two isometries is an isometr. If S and T are isometries then dst, ST = dt T = d,. 3
Eample: Consider points A0, 2, B 3, 1 and C 3, 1 in E 2. We are interested in isometries T : E 2 E 2 that map the triangle ABC to itself. For simplicit we denote this triangle b. Suppose S and T are two such isometries that also satisf SA = T A, SB = T B, and SC = T C. Can S T? In other words, could there be a point in the such that T S? The answer to this important question is no but we postpone the proof until net class. Since a verte of must map to a verte of and an isometr of our set is full defined b its behavior on the vertices of, it follows that there are no more than 6 such isometries as there are onl 3! = 6 permutations of a set of 3 elements. It is, in fact, eas to come up with 6 such isometries, which must then necessaril constitute all the isometries that map to. Let [z] denote the unique isometr that satisfies T A =, T B =, T C = z. Our desired set of isometries includes the following it would be useful to draw a picture as ou are reading this list: [ABC]: the identit transformation, also denoted b I. [CAB]: a rotation b 2π/3 around the origin. [BCA]: a rotation b 2π/3 around the origin. [ACB]: a reflection about the line through A and the midpoint of BC. [CBA]: a reflection about the line through B and the midpoint of AC. [BAC]: a reflection about the line through C and the midpoint of AB. Since the composition of two isometries results in an isometr we can compute a multiplication table under the composition operator : I [CAB] [BCA] [ACB] [CBA] [BAC] I I [CAB] [BCA] [ACB] [CBA] [BAC] [CAB] [CAB] [BCA] I [BAC] [ACB] [CBA] [BCA] [BCA] I [CAB] [CBA] [BAC] [ACB] [ACB] [ACB] [CBA] [BAC] I [CAB] [BCA] [CBA] [CBA] [BAC] [ACB] [BCA] I [CAB] [BAC] [BAC] [ACB] [CBA] [CAB] [BCA] I As epected, the composition of two isometries ields an isometr and no new isometries are produced through this process. Definition 6 A group is a set G together with an operator : G G G, denoted b G,, that satisfies: 1. Associativit: a b c = a b c, for all a, b, c G 2. Identit: there eists e G such that a e = e a = a, for all a G. 4
3. Inverses: for all a G, there is a 1 G such that a a 1 = a 1 a = e. It is not difficult to verif that the set of isometries of E 2 constitutes a group under composition. We denote this group b I2. Definition 7 A subset H G of a group G, is a subgroup of G if 1 a b H, a, b H 2 a 1 H, a H Equivalentl, H is a subgroup of G if a b 1 H, a, b H. A quick glance through the above multiplication table for the isometries of shows that this set, under composition, is a subgroup of O2. The various isometries of can be easil computed b studing their effect on the unit vectors 1, 0 and 0, 1. For eample, [CAB] eerts a rotation of 2π/3 around the origin. Accordingl, [CAB]1, 0 = cos 2π/3, sin 2π/3 = 1/2, 3/2 and [CAB]0, 1 = sin 2π/3, cos 2π/3 = 3/2, 1/2. This can be compactl represented in matri form as: [CAB], = 1/2 3/2 3/2 1/2 = 1 2 3 3 Similarl, [BCA], = 1/2 3/2 3/2 1/2 = 1 2 3 3 + [ACB], = 1 0 0 1 = Consider now the line l 2 through B and the midpoint M = 3/2, 1/2 of AC. Since M = cos π/6, sin π/6, we have that M makes an angle π/6 with the -ais. Thus, when reflected about ell 2, the vector 1, 0 becomes cos 2π/6, sin 2π/6 = 1/2, 3/2. Similarl, 0, 1 forms an angle π/3 with ell 2. To reflect it we rotate it 2π/3 degrees obtaining cos π/6, sin π/6 = 3/2, 1/2. Thus [CBA], = 1/2 3/2 3/2 1/2 = 1 2 + 3 3 Similarl, [BAC], = 1/2 3/2 3/2 1/2 = 1 2 3 3 5
Reflection about an arbitrar line. We finish this eample with a discussion on how to reflect a point through an arbitrar line l that passes through a point p. The reflected point is denoted b. Let n be a unit normal to the line and θ, the angle between n and the vector p see the diagram below. Thus, n, p = p cos θ is the length of the projection of p onto a line orthogonal to l, i.e., the signed distance from to l. Thus, to project we must displace twice this distance in a direction opposite to n. In other words = 2 n, p n. n l θ p We now go back to the question of what it takes to uniquel define an isometr. Theorem 7 If T is an isometr that maps a triangle ABC to itself then T is unique, i.e., there is no other isometr S such that T A = SA, T B = SB, T C = SC and T S for some R 2. Proof. Assume T A = SA = a, T B = SB = b, T C = SC = c, where a, b, c are not collinear. If {A, B, C} we are done. Otherwise, since both T and S are isometries, dt A, T = da, = dsa, S = da, T = da, S = r a > 0 dt B, T = db, = dsb, S = db, T = db, S = r b > 0 dt C, T = dc, = dsc, S = dc, T = dc, S = r c > 0 This means that both T and S must lie on the boundar of each of the circles centered at a, b, and c with radii r a, r b and r c, respectivel. The circles centered at a and b intersect at most twice. If the intersect once we are done. Otherwise, since a, b, c are not collinear c cannot be equidistant from both intersections and onl one of them lies at distance r c from c. This intersection is T = S. Some isometr groups: I2: all isometries of E 2. O2: isometries of E 2 that fi the origin, i.e., O2 = {T I2, T 0, 0 = 0, 0}. This is also called the orthogonal group of E 2. SO2: all rotations about the origin. Note that SO2 is a subgroup of O2 which is a subgroup of I2. O2 is, in a sense, twice the size of SO2 as we can obtain two members of O from each member of T of SO b either appling a reflection or not before the rotation. 6
Theorem 8 Three Reflection Theorem. Let p be a point and α, β, γ, three different lines through p. Let R α, R β and R γ denote reflections through α, β and γ, respectivel. Then there eists a line δ through P such that the reflection R δ through δ satisfies: Proof. Do as homework. R δ = R α R β R γ Theorem 9 Ever isometr of E 2 can be written as the composition of at most three reflections. 3 Affine Transformations Definition 8 A collineation is a bijection i.e., both 1-1 and onto T : E 2 E 2 such that if points p, q, r are collinear then so are T p, T q, T r. Definition 9 An affine transformation is a map T : E 2 E 2 of the form T = A + b where A is a non-singular 2 2 matri and b is a vector in R 2. Definition 10 A map T : V W between two vector spaces is linear if 1 T u + v = T u + T v, for all u, v V. 2 T λu = λt u, for all scalars λ and u V. Eample: the isometries of the triangle ABC are all linear maps. The set of all 2D invertible linear maps is a group which we denote b GL2. Note that affine transformations are not, in general, linear as an affine transformation has both a linear and a translation component. Theorem 10 All collineations are affine transformations and vice versa. In other words, Definitions 8 and 9 are equivalent. How do isometries fit in all of this? All isometries are affine but the converse is not true. To see this it suffices to show that reflections are affine as ever isometr can be epressed as the composition of three reflections. Consider a reflection through a line l through p = p, p with unit normal n = n, n. For an point q =, T q = q 2 < q p, n n which has the form T q = Aq + b, T = 1 2n 2 2n n 2n n 1 2n 2 + p, n n Furthermore A is non-singular as det A = 1 2n 2 1 2n 2 4n 2 n 2 = 1 2n 2 + n 2 = 1. Theorem 11 Ever isometr is an affine transformation. Let AF 2 denote the set of affine transformations under E 2. 7 n
Theorem 12 AF 2, the set of affine transformations under E 2, is a group under composition. Proof. Let T = A + c and S = B + d be affine transformations. Then T S = T B + d = AB + d + c = AB + Ad + c = M + f. Since M 1 = B 1 A 1, T S is affine as well, so AF 2 is closed under composition. The identit matri is the identit of the group, and T 1 = A 1 c = A 1 A 1 c Definition 11 A direction [v] is the set of vectors proportional i.e., scalar multiples to a non-zero vector v. Thus, [v] = {tv : t R}, v 0. Definition 12 For an point p and direction [v] the line through p with direction [v] is the set l = { E n, p [v]}. Theorem 13 Let T be an affine transformation such that T = A + b and l = p + [v] a line in E 2. Then T l is also a line in E 2. Proof. A point on l has the form p+tv. Thus, T p+tv = Ap+tv+b = Ap+b+tAv = T p + tav T P + [Av]. In other words, T l is the line through T p with direction [Av]. Corollar 1 Let T = A + b be an affine transformation. A line p + [v] is a fied line of T iff the following two conditions hold, 1 Av = λv for some λ R \ {0}. 2 A IP + b [v]. 8