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Page 1 of 150 Spring 2008 Math 330A Notes, Version 10.1 Reading Material for Foundations of Geometry I (by Mark Barsamian) 1. Relations..............................................................................................................................5 1.1. Cartesian Products ...................................................................................................5 1.2. Relations ..................................................................................................................8 1.3. Exercises ................................................................................................................16 2. Axiom Systems .................................................................................................................19 2.1. Introduction ............................................................................................................19 2.2. Definition of Axiom System ..................................................................................21 2.3. Primitive Relations and Primitive Terms ...............................................................22 2.4. Interpretations and Models ....................................................................................25 2.5. Properties of Axiom Systems.................................................................................28 2.6. Exercises ................................................................................................................36 3. Axiomatic Geometries .....................................................................................................39 3.1. Introduction ............................................................................................................39 3.2. Two Simple Finite Geometries ..............................................................................40 3.3. Terminology...........................................................................................................42 3.4. Fano’s and Young’s Finite Geometries .................................................................44 3.5. Incidence Geometry ...............................................................................................45 3.6. Models of Incidence Geometry..............................................................................46 3.7. Exercises ................................................................................................................49 4. Models of Incidence Geometry Whose Sets of Points are Infinite...............................51 4.1. A Model of Incidence Geometry Involving Straight-Line Drawings ....................51 4.2. A Model of Incidence Geometry Involving the Usual Analytic Geometry ...........52 4.3. A Model of Incidence Geometry Involving the Poincare Disk .............................52 4.4. Other Interpretations of Incidence Geometry ........................................................55 4.5. Exercises ................................................................................................................56 5. Incidence and Betweenness Geometry ...........................................................................61 5.1. Binary and Ternary Relations on a Set ..................................................................61 5.2. Introducing Incidence and Betweenness Geometry ...............................................62 Page 2 of 150 5.3. Line Segments and Rays ........................................................................................64 5.4. Plane Separation.....................................................................................................65 5.5. Line Separation ......................................................................................................67 5.6. Angles and Triangles .............................................................................................68 5.7. Exercises ................................................................................................................71 6. Neutral Geometry I ..........................................................................................................73 6.1. The need for a larger axiom system: Introducing Neutral Geometry ....................73 6.2. Line Segment Subtraction and the Ordering of Segments .....................................75 6.3. Triangle Congruence and its Role in the Neutral Geometry Axioms ....................76 6.4. Right Angles ..........................................................................................................82 6.5. Angle Addition and Subtraction, and Ordering of Angles ....................................83 6.6. The Alternate Interior Angle Theorem and Some Corollaries...............................85 6.7. Exercises ................................................................................................................87 7. Neutral Geometry II: Triangles ......................................................................................91 7.1. The Isosceles Triangle Theorem and Its Converse ................................................91 7.2. The Exterior Angle Theorem .................................................................................92 7.3. Comparison Theorems ...........................................................................................94 7.4. Midpoints and bisectors .........................................................................................95 7.5. Exercises ................................................................................................................96 8. Measure of Line Segments and Angles ..........................................................................99 8.1. Theorems Stating the Existence of Measurement Functions .................................99 8.2. Two length functions for Neutral Geometry........................................................100 8.3. An example of a curvy-looking Hline .................................................................102 8.4. Theorems about segment lengths and angle measures ........................................104 8.5. Exercises for Chapter 8 ........................................................................................106 9. Building Euclidean Geometry from Neutral Geometry .............................................111 9.1. Thirteen Statements that are Logically Equivalent in Neutral Geometry ............112 9.2. What’s up with the answer to THE BIG QUESTION?!? ....................................113 9.3. Proving Theorems about Equivalent Statements .................................................114 9.4. Finally, Euclidean Geometry ...............................................................................116 9.5. Exercises ..............................................................................................................117 Page 3 of 150 10. Euclidean Geometry I ....................................................................................................119 10.1. Some results from the previous chapter presented as theorems ..........................119 10.2. Circles, chords, and diameters and some Neutral Geometry Theorems ..............121 10.3. Some Euclidean Geometry Theorems about Circles and Triangles ....................122 10.4. Could Theorem 84 be a theorem of Neutral Geometry? ......................................123 10.5. Exercises ..............................................................................................................127 11. For Reference: Axioms, Definitions, and Theorems of Euclidean Geometry ..........135 11.1. The Axioms of Euclidean Geometry ...................................................................135 11.2. The Definitions of Euclidean Geometry ..............................................................136 11.3. The Theorems of Euclidean Geometry ................................................................143 Page 4 of 150 Page 5 of 150 1. Relations Consider the following three mathematical expressions: y x 2 (an equation that is also a function) x 2 y 2 1 (an equation that is not a function) x < y (an inequality) These are three very different kinds of expressions. But there is a mathematical framework that encompasses all three and even encompasses expressions like ―Line L1 is perpendicular to line L2‖ that don’t have anything to do with real numbers or real variables. All of these expressions are examples of what are called Relations. Relations play an important role in axiomatic geometry. They are on the list of topics for Math 306, but are not covered there in enough detail for our purposes. So we will start Math 330A by studying them. 1.1. Cartesian Products Central to the definition of relations is the terminology and notation of the Cartesian Product of two sets. This is a generalization of the so-called Cartesian plane—the set of order pairs (x,y) of real numbers. 1.1.1. Definition and examples The definition of Cartesian Product uses ordered pairs, but the pairs can be pairs of elements from any two sets, not just sets of real numbers. Definition 1 Cartesian Product Symbol: A B Spoken: The Cartesian Product of A and B. Usage: A and B are sets Meaning in words: A B is the set consisting of all ordered pairs a, b , where a is an element of A and b is an element of B. Meaning in symbols: A B a, b : a A and b B Example: Let A x, y and B 1, 2,3 a) Find A B . b) Find B A c) Answer the following true/false questions. If your answer is ―false‖, explain why. i) 2x A B true false ii) x2 A B true false iii) x 2 A B true false iv) x, 2 A B true false v) 2, x A B true false Notice that nothing in the definition of Cartesian Product requires that the two sets used in the product be different. The following example illustrates this. Page 6 of 150 Example: Let A x, y and B 1, 2,3 a) Find B B . b) Answer the following true/false questions. If your answer is ―false‖, explain why. i) 2x B B true false ii) x2 B B true false iii) 3 2 B B true false iv) 2, 2 B B true false v) 1,3 3,1 true false Our first two examples have involved finite sets. However, nothing in the definition of the Cartesian Product requires that the sets be finite, and in fact, you have all already used the Cartesian Product in a setting involving infinite sets. Example: Let A and let B . Then A B , which is commonly denoted as 2 . This is the set of ordered pairs of real numbers. Here is a proper definition. Definition 2 the Cartesian plane Symbol: 2 Spoken: ―r two‖, or ―the x, y plane‖, or ―the Cartesian plane‖. Meaning in symbols: Meaning in words: The set of ordered pairs of real numbers. Observations: 1) All real numbers are allowed, not just integers. So, pairs such as 5, 2.7 , , 10 , 2 , 0 , etc., are all elements of 2 . 2) Parentheses and commas are used, because 2 is a Cartesian product. So ― 10 ‖, ― ,10 ‖, etc., are not allowed. 3) Order is important: 5, 2.7 2.7,5 . 4) The definition of the Cartesian product makes no provisions for ―scalar multiplication‖. In the past, you may have seen computations such as 3 2,5 6,15 or 2,5 2, 5 . Computations such as these are perfectly valid, but they arise in contexts where one is dealing with ―vectors‖. Vectors may be typeset in a way that makes them look just like the elements of a Cartesian product, but be aware that a vector is something different. In a Cartesian product, there is no scalar multiplication. 5) Similarly, the definition of the Cartesian product makes no provisions for ―addition‖. In the past, you may have seen computations such as 2,5 1,8 3,13 . Again, computations such as this are perfectly valid, but they arise in contexts where one is dealing with ―vectors‖. In a Cartesian product, there is no addition. 1.1.2. Visualizing a Cartesian product Cartesian products involving finite sets can be visualized using tables. There are a number of conventions that can be followed. A common convention is that for a product such as A B , the elements of the set A correspond to the rows of the table; the elements of set B correspond to the Page 7 of 150 columns. Each cell of the table corresponds to a row, column pair. Elements of the Cartesian product are denoted by putting some sort of mark, such as an ―X‖, in a cell. Example: Let A x, y and B 1, 2,3 . Then A B is visualized as the table shown at right. Notice that the upper left corner gives the names of the sets used in constructing the product. Nowhere in the table is it written that the x and the y are from the set A, and that the 1,2,3 are from the set B. It doesn’t have to be written, because it is a convention. A B 2 3 1 2 3 x y A B The element x,3 A B could be displayed as shown at right. 1 x * y B A x y x y 1 On the other hand, B A is visualized as the table 2 3 B A The element 3, x B A is displayed as 1 2 3 * Sometimes, Cartesian products involving infinite sets can be visualized as well. You have done this for years, every time you drew a set of axes for the ―x,y plane‖. It is important to notice that some of the conventions in this case differ from the conventions used above, when illustrating finite Cartesian products with tables. Namely, in the case of the finite cartesian products, the elements of the left set are listed vertically, along the left edge of the table, while elements of the right set are listed horizontally, along the top of the table. In the x,y plane, however, elements of the left set are displayed on the horizontal axis, while elements of the right set are displayed on the vertical axis. Single elements in the ―x,y plane‖ are displayed as little dots, or little crosses. Remember that one of these little marks represents an ordered pair of real numbers. Example: y (right coordinate) 2 1.5, 2 2,1.5 1 1 2 3 x (left coordinate) Page 8 of 150 1.2. Relations Now that we are familiar with the Cartesian Product of sets, we are ready for the definition of Relations. 1.2.1. Definitions and Examples At the start of this chapter, it was mentioned that the concept of relations could include mathematical exparessions such as y x 2 and x 2 y 2 1 and x < y. Notice, however, that the initial definition of relation makes no mention at all of either an x or a y. And note that in the second definition, when an x and a y do appear, they do not stand for real numbers. Definition 3 Relation Words: R is a relation from A to B. Usage: A and B are sets. Meaning in words: R is a subset of A B . Meaning in symbols: R A B Definition 4 related to Symbol: xRy (Most often, other symbols besides R are used.) Spoken: x is related to y Usage: It assumed that R is a relation from A to B, where A and B are some sets. Meaning in words: the mathematical statement ― x, y is an element of the set R.‖ Meaning in symbols: ― x, y R ‖ Remark 1: Since it represents a mathematical statement, the symbol xRy can be true or false. Remark 2: It is not assumed that x A and y B . That is, there is nothing ―illegal‖ about writing the symbol down in some case where x A or y B . This will be elaborated in the examples. Example: Let A x, y and B 1, 2,3 . Above, we found that A B x,1 , x, 2 , x,3 , y,1 , y, 2 , y,3 . a) Let R x, 2 , y,1 . Then R is a relation from A to B. Observe that x, 2 R . In other words, we could write xR2, or ―x is related to 2‖, and it would be a true statement. b) Let R y,1 , y, 2 . Then R is a relation from A to B. Observe that the statement ―y is related to 2‖, abbreviated yR2, is true. However, the statement ―2 is related to y‖, abbreviated 2Ry, is false, because 2, y R . In fact, 2 A and y B , so there is no way that 2, y could possibly be an element of R. Even so, there is nothing ―illegal‖ about the expression 2Ry. The symbol represents a false statement, but it is not illegal. c) Let R . Is R a relation from A to B? If not, say why not. d) Let R A B x,1 , x, 2 , x,3 , y,1 , y, 2 , y,3 . Is R a relation from A to B? If not, say why not. Page 9 of 150 e) Let R x, 2 , 1, y . Is R a relation from A to B? If not, say why not. f) Let R x, 2 , x,3 , y,1 , y, 2 . Then R is a relation from A to B. i) Find all a such that aR2. ii) Find all a such that aR3. iii) Find all b such that yRb. 1.2.2. Visualizing relations Because a relation R from a set A to a set B is just a subset of the Cartesian product A B , any illustration of the set A B can just be ―filled in‖ to produce a picture of relation R. An alternate way to view a relation R from one finite set A to another finite set B is to use an arrow diagram. Elements of set A are denoted by dots in some arrangement to the left, and elements of set B are denoted by dots in some arrangement to the right. To indicate that aRb is true, one draws an arrow from the dot for element a to the dot for element b. 1 For example, the relation from Example (f) above can be illustrated with the table and arrow diagram shown at right. 1 2 3 x x * * y * * y 2 3 1.2.3. Relation on a set Observe that in the definition of relation, we used the symbols A and B, but there is no requirement that these two sets be different. That is, it could be that the set B is actually the same as the set A. In that case, we say that the relation is a ―relation on a set‖. The following definition make this more precise. Definition 5 Relation on a Set Words: R is a relation on A. Usage: A is a set. Meaning: R is a relation from A to A. Equivalent meaning in words: R is a subset of A A . Equivalent meaning in symbols: R A A Additional terminology: R is also called a binary relation on A. Examples: In the following examples, Let A 1, 2,3, 4,5, 6 . a) Let Ra 1,3 , 3,1 , 3,5 , 5,3 , 2, 4 , 4, 2 , 4, 6 , 6, 4 . Then Ra is a relation on the set A. b) Let Rb 1,3 , 3,5 , 1,5 , 2, 4 , 4, 6 , 2, 6 . Then Rb is a relation on the set A. 1,1 , 2, 2 , 3,3 , 4, 4 , 5,5 , 6, 6 , 1,3 , c) Let Rc Then Rc is a relation on the set A. 3,1 , 3,5 , 5,3 , 2, 4 , 4, 2 , 4, 6 , 6, 4 Page 10 of 150 So far, our examples of relations on a set have all been given by explicit lists of elements of sets. But relations on a set can also be described by a formula. Examples: In the following examples, Let A , the set of all real numbers. d) Let x Rd y means x y . The symbol Rd would mean ―the set of all pairs (x,y) such that x y .‖ That is, Rd x, y : x y . Then Rd is a relation on the set . because it is a subset of . It could be called the ―less than or equal to‖ relation. e) Let x Rey means xy = 0. The symbol Re would mean ―the set of all pairs (x,y) such that xy 0 .‖ That is, Re x, y : xy 0 . Then Re is a relation on the set . f) Let x Rf y means y 2 x 1. The symbol Rf would mean ―the set of all pairs (x,y) such that y 2 x 1.‖ That is, R f x, y : y 2x 1 . Then Rf is a relation on the set . 1.2.4. Visualizing a relation on a set A relation on a set is just a particular type of relation, so the same tools that used to visualize regular relations (tables and arrow diagrams) can be used to visualize a relation on a set. In the case of a relation on a finite set A, there is an additional visualization tool available, called the directed graph. A directed graph is a special kind of arrow diagram, in which elements of the set A are only shown once, rather than twice. Arrows are drawn as you might expect. Let’s revisit our previous examples Ra, Rb, and Rc from the previous section and make a table and a directed graph for each. Relation Table 1 2 3 4 5 6 1 5 * * 5 * * 1 2 3 4 5 6 2 Relation Rb 3 4 2 * 6 1 3 * 3 * 4 4 * 2 Relation Ra Directed Graph * * * 6 1 4 3 * * 5 2 * 5 6 6 1 Page 11 of 150 1 2 3 4 5 6 1 * 2 Relation Rc * * 3 * 4 5 6 3 4 * * * 5 * * * 2 * * * * 6 1 To visualize relations on the set of real numbers—an infinite set—we start with a picture of the underlying Cartesian plane and then simply darken all the (x,y) pairs in the plane that are elements of the relation. The resulting picture is called the graph of the relation. For examples, we can visualize the relations Rd, Re, and Rf above. Relation Graph y Rd x, y : x y y=x x y Re x, y : xy 0 x y y=2x+1 R f x, y : y 2x 1 x Page 12 of 150 1.2.5. Properties that a relation on a set may or may not have There are four common properties that a relation on a set may or may not have. The definitions follow. Notice that each definition is presented as a logical statement. Throughout the definitions, it is assumed that R is a relation on a set A. Definition 6 Reflexive Property Words: R is reflexive Meaning: a A, aRa Meaning in words: Every element of set A is related to itself. Definition 7 Symmetric Property Words: R is symmetric Meaning: a, b A, IF aRb THEN bRa Definition 8 Transitive Property Words: R is transitive Meaning: a, b, c A, IF aRb AND bRc THEN aRc Definition 9 Equivalence Relation Words: R is an equivalence relation Meaning: R is Reflexive and Symmetric and Transitive Each of the four properties above is a logical statement. Each may be true or false. If the statement of one of the properties is true for a certain relation, then we say that the given relation has that property. If the statement of one of the properties is false for a certain relation, then we say that the given relation does not have that property. Class Drill for Section Error! Reference source not found.: If the statement of one of the properties is false, then the negation of that statement will be true. Therefore, it is important to understand how to find the negations of each of the four statements above. Write the negations of each of the four statements here: Words Meaning R is reflexive: a A, aRa R is not reflexive: R is symmetric: a, b A, IF aRb THEN bRa R is not symmetric: R is transitive: a, b, c A, IF aRb AND bRc THEN aRc R is not transitive: R is an equivalence relation: R is Reflexive and Symmetric and Transitive Page 13 of 150 R is not an equivalence relation: Examples: a) Consider relation Ra described above. i) Is 3 related to itself? That is, is 3Ra3 true? No, so Ra is not reflexive. ii) Notice that Ra is symmetric. iii) Notice that 2Ra4 and 4Ra6, but 2Ra6 is not true. Therefore, Ra is not transitive. iv) Therefore, Ra is not an equivalence relation. b) Consider relation Rb described above. i) Notice that 3Rb3 is false, so Rb is not reflexive. ii) Notice that 1Rb3 is true, but 3Rb1 is not true. Therefore, Rb is not symmetric. iii) Notice that 1Rb3 and 3Rb5 are both true, and 1Rb5 is also true. And notice that 2Rb4 and 4Rb6 are both true, and 2Rb6 is also true. Therefore, Rb is transitive. iv) Therefore, Rb is not an equivalence relation. c) Consider relation Rc described above. Class Drill: i) Is Rc reflexive? ii) Is Rc symmetric? iii) Is Rc transitive? iv) Is Rc an equivalence relation? Notice that the sets and relations used in examples a), b), and c) above were very basic, and yet it was rather tedious to check for the reflexive, symmetric and transitive properties by looking only at the sets that define each relation. Using a picture to visualize the relation helps to make the checking of the reflexive and symmetric properties much easier. From studying the pictures for the three relations above, we can easily make the following general observations: When a relation is reflexive, All of the cells on the main diagonal of the table are filled in. Every dot in the directed graph has an arrow looping from the dot back to the dot. This happens in the pictures for relation Rc. When a relation is not reflexive, At least one of the cells on the main diagonal of the table is empty. At least one dot in the directed graph does not have an arrow looping around it. This happens in the pictures for relation Ra and Rb. When a relation is symmetric, The table is symmetric across the main diagonal. Every arrow in the diagram is a double arrow. This happens in the pictures for relation Ra and Rc. (The little loop arrows are effectively double arrows. Think about it.) When a relation is not symmetric, The table is not symmetric across the main diagonal. There is an arrow in the diagram that is not a double arrow. Page 14 of 150 This happens in the pictures for relation Rb . When a relation is transitive, The table is not much help in looking for transitivity. Every ―segmented path‖ in the arrows, there is a ―direct path‖ that goes from the same starting dot to the same ending dot.. This happens in the pictures for relation Rb . For instance, there is a segmented path that goes 1 3 5 . A direct path goes 1 5 . When a relation is not transitive, The table is not much help in looking for failures of transitivity, either. There is a ―segmented path‖ in the arrows, for which there is not a ―direct path‖ that goes from the same starting dot to the same ending dot.. This happens in the pictures for relation Ra and Rc. For instance, in the directed graph for Ra, there is a segmented path that goes 1 3 5 , but there is not a direct path that goes 1 5 . 1.2.6. Relations on the Set of Real Numbers In the drill on the following page, you will explore relations on the set of real numbers, . Remember that a ―relation on the set of real numbers‖ is just a fancy name for a ―subset of the cartesian plane, ‖. Note that just as the tables and directed graphs for a relation on a finite set could be used to check properties of the relation, the graph for a relation on the set of real numbers can be used to check properties of the relation: When a relation is reflexive, the entire line y = x is colored in. When a relation is not reflexive, at least one point on the line y = x is not colored in. When a relation is symmetric, the graph is symmetric across the line y = x. When a relation is not symmetric, the graph is not symmetric across the line y = x. The graph is not much help in looking for transitivity. The graph is not much help in looking for failures of transitivity, either. 1.2.7. Remark on Relations as Predicates Consider relation R5 from the previous section. The sentence ―2 R5 7‖ means ― 2 7 ‖. This is a true statement. The sentence ―7 R5 3‖ means ― 7 3 ‖.The sentence ―x R5 y‖ means ― x y ‖. This sentence is neither true nor false because we do not know the values of x and y. If we substitute in some actual values for x and y, the sentence becomes a sentence that is either true or false, but not both. In other words, the sentence ―x R5 y‖ is a predicate. But the sentence ― x , xR5 x ‖ means ― x , x x ‖. This is a true statement. And the sentence ― x, y , IF xR5 y THEN yR5 x ‖ means ― x, y , IF x y THEN y x ‖. This is a false statement. So the properties of the relations are statements. Page 15 of 150 R0 x R0 y means x – y = 2 R1 x R1 y means x < y R2 x R2 y means x y 1 R3 x R3 y means xy 0 R4 x R4 y means y x y 2 x 0 R5 x R5 y means x y R6 x R6 y means x y 1 R7 x R7 y means x 2 y 2 2 Transitive? Relation Symmetric? Reflexive? Class Drill for Section 1.2.6: For each of the following relations, draw a cartesian plane and sketch the points that are elements of the relation. Then decide if the relation is reflexive, symmetric, transitive. If a relation has one of the properties, then write ―yes‖ in the box. If a relation does not have one of the properties, write ―no‖ in the box and give a counterexample that shows that the relation does not have the property. That is, if you say that a relation is not reflexive, then you need to give an example of an x that shows that the relation is not reflexive. If you say that a relation is not symmetric, then you need to give an example of an x and a y that show that the relation is not symmetric. If you say that a relation is not transitive, then you need to give an example of an x, y, and z that shows that the relation is not transitive. no let x, y 8,6 no, let x, y, z 1,0,1 2 no let x 1 Big hint for R4. y x y 2 x 0 is logically equivalent to y x OR y 2 x Page 16 of 150 1.3. Exercises 1) Let A 4,5 and B 3, a, b a) b) c) d) Find A B . Find B A . Find A A . Answer the following true/false questions. If your answer is ―false‖, explain why. i) b4 A B true false ii) 4b A B true false iii) 5 a A B true false iv) 5, a A B true false v) a,5 A B vi) 5 3 A B vii) 15 A B viii) 4b A A ix) 4 5 A A x) 5,5 A A true true false true true true false false xi) 4,5 A A true false xii) 4,5 5, 4 true false true false false false 2) Let C be the set of celebrities and M be the set of months. Answer the following true/false questions. If your answer is ―false‖, explain why. a) Michael Jordan February M C true false b) September Woody Allen M C true false c) December,Tiger Woods M C true false 3) Consider the cartesian product 2 . What elements of the cartesian product 2 are denoted by the dots labelled a,b,c,d in the figure at right? (In your old jargon, you would have been asked to give the coordinates of each of the four points.) 4) Draw a picture to illustrate the cartesian product 2 , along with the following elements, a,b,c,d. (In your old jargon, you would have been asked to draw a set of axes, and then put in the following four points.) (a) 4, 2 (b) 3.5, 2 (c) 1, 3 (d) , 2 y 2 b a 1 x 1 2 d c Page 17 of 150 5) Let A 4,5 and B 3, d , e a) Let R 3, 4 , 4,3 . Is R a relation from A to B? If not, say why not. b) Let R 4,3 , 4, d , 5, d . Is R a relation from A to B? If not, say why not. c) Let R 4, e , 5,3 , 5, d , 5, e . Then R is a relation from A to B. i) Is 4 related to d ? ii) Is 5 related to d ? iii) Is d related to 5? iv) Is 3R5 true? v) Find all a such that aRd. vi) Find all a such that aRe. vii) Find all b such that 5Rb. 6) Consider the following relations on the set of real numbers, . Remember that a relation on the set of real numbers is just a fancy name for a subset of the cartesian plane, . Illustrate each relation by drawing it as a subset of the plane. Then decide if the relation is reflexive, symmetric, transitive. example Relation Reflexive? Symmetric? Transitive? x Ra y means 2 x y 1 Ra Rb x Rb y means x y Rc Rd Re x Rc y means x 2 y 2 1 x Rd y means xy 0 x Re y means xy 0 For the next two exercises, let A be the set of all lines in the Cartesian plane. 7) Define ―perpendicular lines‖ to mean two lines with the property that the products of their slopes is -1 OR one line is vertical and one line is horizontal. Define a relation on the set A as follows. For l1 , l2 A , ― l1 l2 ‖ means ― l1 is perpendicular to l2 ‖. Is this relation reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers. 8) In this exercise, you will explore three different definitions of ―parallel lines‖. Each definition can be recast as a relation. You will see that even though all three relations are spoken as ― l1 is parallel to l2 ‖, they are very different relations, with different properties. a) Define ―parallel lines‖ to mean two lines that have the same slope. Define a relation || on the set A as follows. For l1 , l2 A , ― l1 || l2 ‖ means ― l1 is parallel to l2 ‖. Is this relation reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers. b) Define ―parallel lines‖ to mean two lines that have the same slope or are both vertical. Define a relation || on the set A as follows. For l1 , l2 A , ― l1 || l2 ‖ means ― l1 is parallel to l2 ‖. Is this relation reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers. c) Define ―parallel lines‖ to mean two lines that do not intersect. Define a relation || on the set A as follows. For l1 , l2 A , ― l1 || l2 ‖ means ― l1 is parallel to l2 ‖. Is this relation reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers. Page 18 of 150 For the next four exercises, refer to the figure at right. l1 Let L l1 , l2 , l3 , l4 be the set of lines in the figure. Let P p1 , p2 , p3 , p4 , p5 , p6 be the set of dots in the figure. p2 p4 p3 p5 p6 p1 l2 l3 l4 p7 9) Answer the following true/false questions. If your answer is ―false‖, explain why. a) l2 l4 L L true false b) p5 , l1 L P p4 , l2 P L true d) p4 , l4 P L true c) true false false false true false p5 , l1 P L 10) Let R p2 , l1 , p4 , l2 , p4 , l4 , p6 , l4 . Then R is a relation from P to L. Notice that the e) relation could be defined in words by saying that the sentence pRl means ―the dot p touches the line l ‖. Answer the following questions. a) Is p4 Rl4 ? b) Is l4 related to p4 ? c) Is p4 related to l4 ? d) Is p4 related to l2 ? e) Is p2 related to l3 ? f) Is l2 Rl4 ? g) Is l2 Rl1 ? 11) Using relation R from the previous exercise, we could define a new relation S on the set L by saying that the statement laSlb means ―there exists a point p such that pRla and pRlb‖. In other words, we would say that two lines are related by relation S if there exists a point p that touches both of the lines. a) Is l2Sl4? b) Is l2Sl3? c) Is l1Sl1? d) Is l3Sl3? 12) Using relation R again, we could define another new relation T on the set P by saying that the statement paTpb means ―there exists a line l such that paRl and pbRl‖. In other words, we would say that two points are related by relation T if there is a line l that both points touch. a) Is p4Tp6? b) Is p6Tp4? c) Is p2Tp2? d) Is p3Tp3? Page 19 of 150 2. Axiom Systems 2.1. Introduction 2.1.1. Looking Back at Math 306 In Math 306, you studied conditional statements—statements of the form If Statement A is true then Statement B is true. For example, you proved this Theorem: If n is odd, then n2 is odd. You saw that the proof of such a conditional statement has the following form: Proof: 1. Statement A is true. (given) 2. Some statement (with some justification provided) 3. Some statement (with some justification provided) 4. Some statement (with some justification provided) 5. Statement B is true. (with some justification provided) End of Proof For example, the proof of the above theorem about odd integers is Proof: 1. 2. 3. 4. Suppose that n is odd. There exists some integer k such that n = 2k + 1. (by 1 and the definition of odd) n2 = (2k + 1)2 = (2k + 1)(2k + 1) = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 (arithmetic) (2k2 + 2k) is an integer (because the set of integers is closed under addition and multiplication) 5. n2 is odd. (by step 4 and the definition of odd) End of Proof In this proof, the only assumption is the given Statement A. The proof does not prove that Statement A is true; it only proves that if statement A is true, then statement B is also true. But in most of your proofs in Math 306, certainly in all of the proofs involving basic number theory, there were actually unspoken assumptions in addition to the explicitly given Statement A. The unspoken assumptions have to do with the basic properties of integerers. These basic properties of integers are not statements that can be proven. Rather, they are statements that are assumed to be true. Statements such as The set of integers is closed under addition and multiplication Page 20 of 150 or There is a special number called 1 with the following properties 1 is not equal to 0 for all integers n, n*1 = n It may seem silly that we are even discussing the above statements. They are so obviously true, and we’ve known that they are true since grade school math. We never think about the statements, even though we do integer arithmetic all the time. Well, the statements can not in fact be proven true (or false). We can’t imagine what we would do if they were false, though: all of the math we’ve lived with all of our lives would be out the window. So we just assume that all the basic properties of the integers are true and we go from there. But to be totally honest about all the given information in the Theorem about odd numbers, we would need to write it something like this: Theorem: If all of the axioms of the integers are true [they would have to be listed explicitly here] and n is odd, then n2 is also odd. And the proof would need to look something like this Proof: 1. Suppose that all of the axioms of the integers are true [they would have to be listed explicitly here] and n is odd 2. There exists some integer k such that n = 2k + 1. (by 1 and the definition of odd) 3. n2 = k2 + 4k + 1 (arithmetic) 4. k2 + 4k is an integer (because the set of integers is closed under addition and multiplication) 5. n2 is odd. (by 4 and the definition of odd) End of Proof But in Math 306, the emphasis is on learning how to build a proof. That is, you would have been concerned with steps 2 through 5 of the above proof. Since we all know how to work with the integers without having to refer to their underlying axioms, the axioms were not mentioned. 2.1.2. Looking Forward Having studied in Math 306 the building of proofs of basic number theory theorems, a setting where the underlying axioms could go unmentioned, we will study in Math 330A the building of proofs of theorems about geometry. In geometry, the underlying axioms are not obvious and cannot go unmentioned. In fact, Math 330A will primarily be about the axioms, themselves. We will start by studying very simple axioms systems, in order to learn how one builds proofs based on an explicit axiom system and to learn some of the basic terminology of axiom systems. Then, we will turn our attention to developing an axiom system to describe Euclidean geometry. Page 21 of 150 2.2. Definition of Axiom System We will use the term axiom system to mean a finite list of statements that are assumed to be true. The individual statements are the axioms. The word postulate is often used instead of axiom. Example 1 of an axiom system 1. Elvis is dead. 2. Chocolate is the best flavor of ice cream. 3. 5 = 7. Notice that the first statement is one that most people are used to thinking of as true. The second sentence is clearly a statement, but one would not have much luck trying to find general agreement as to whether it is true or false. But if we list it as an axiom, we are assuming it is true. The third statement seems to be problematic. If we insist that the normal rules of arithmetic must hold, then this statement could not possibly be true. There are two important issues here. The first is that if we are going to insist that the normal rules of arithmetic must hold, then that essentially means that our axiom system is actually larger than just the three statements listed: the axiom system would also include the axioms for arithmetic. The second issue is that if we do assume that the normal rules of arithmetic must hold, and yet we insist on putting this statement on the list of axioms, then we have a ―bad‖ axiom system in the sense that its statements contradict each other. We will return to this when we discuss consistency of axiom systems. So the idea is that regardless of whether or not we are used to thinking of some statement as true or false, when we put the statement on a list of axioms we are simply assuming that the statement is true. With that in mind, we could create a slightly different axiom system by modifying our first example. Example 2 of an axiom system 1. Elvis is alive. 2. Chocolate is the best flavor of ice cream. 3. 5 = 7. The statements of an axiom system are used in conjunction with the rules of inference to prove theorems. Used this way, the axioms are actually part of the hypotheses of each theorem proved. For example, suppose that we were using the axiom system from Example 2, and we were somehow able to use the rules of inference to prove the following theorem from the axioms. Theorem: If Bob is Blue then Ann is Red. Then what we really would have proven is the following statement: Theorem: If ((Elvis is alive) and (Chocolate is the best flavor of ice cream) and (5 = 7) and (Bob is Blue)) then Ann is Red. Page 22 of 150 2.3. Primitive Relations and Primitive Terms As you can see from the examples in the previous section, axiom systems may be comprised of statements that we are used to thinking of as true, or statements that we are used to thinking of as false, or some mixture of the two. More interestingly, an axiom system can be made up of statements whose truth we have no way of assessing. The easiest way to get such an axiom system is to build statements using words whose meaning has not been defined. In this course, we will be doing this in two ways. 2.3.1. Primitive Relations The first way of building statements whose meaning is undefined is to use nouns whose meaning is known in conjunction with transitive verbs whose meaning is not known. This can be described nicely in the jargon of relations. For instance, consider the set of integers. Introduce an undefined relation R on the set . That is, introduce a subset R , but don’t say what that subset is. Then the symbol 5R7 means that 5,7 R and would be spoken as ―5 is related to 7‖. This is a sentence with the noun 5 as the subject, the noun 7 as the direct object, and the words ―is related to‖ as the transitive verb. We have no idea idea what this sentence might mean, because the relation is undefined. In the context of axiom systems, an undefined relation is sometimes called a primitive relation. When one presents an axiom system that contains primitive relations—that is, undefined transitive verbs—it is important to introduce those primitive relations before listing the axioms. Here is an example of an axiom system consisting of sentences built using primitive relations in the manner described above. Axiom System: Axiom System #1 Primitive Relations: relation on the set , spoken ―x is related to y‖ Axioms: 1) 5 is related to 7 2) 5 is related to 8 3) For all real numbers x and y, if x is related to y, then y is related to x. 4) For all real numbers x, y, and z, if x is related to y and y is related to z, then x is related to z. With the symbols and terminology of relations that we learned in Chapter 1, we can easily abbreviate the presentation of this axiom system. Axiom System: Axiom System #1, abbreviated version Primitive Relations: relation R on the set , spoken ―x is related to y‖ Axioms: 1) 5R7 2) 5R8 3) Relation R is symmetric 4) Relation R is transitive Observe that each of the axioms is a statement whose truth we have no way of assessing, because the relation R is undefined. But we can prove the following theorem. Page 23 of 150 Theorem for axiom system #1: 7 is related to 8. Proof 1) 7 is related to 5 (by axioms 1 and 3) 2) 7 is related to 8 (by statement 1 and axioms 2 and 4) End of proof As mentioned in the previous section, the axioms could be stated explicitly as part of the statement of the theorem. (Then we would not really need to state the axiom system separately.) Theorem: If ((R is a relation on ) and (5R7) and (5R8) and (R is symmetric) and (R is transitive), then 7R8. 2.3.2. Primitive Terms The second way of building statements whose meaning is undefined is to use not only undefined transitive verbs, but also undefined nouns. A straightforward way to do this is to introduce sets A and B whose elements are undefined. For instance, let A be the set of akes and B be the set of bems, where ake and bem are undefined nouns. Introduce the following sentence: ―the ake is related to the bem‖. Note that this is a sentence with the undefined noun ake as the subject, the words ―is related to‖ as the transitive verb, and the undefined noun bem as the direct object. Of course we have no idea what this sentence might mean, because the nouns ake and bem are undefined. But we now have the following building blocks that can be used to build sentences. the undefined noun: ake the undefined noun: bem the undefined sentence: The ake is related to the bem. Since we don’t know the meaning of the sentence ―The ake is related to the bem‖, we have effectively introduced an undefined relation from set A to set B. We could call this undefined relation R. Using the standard notation for relations, we could write R A B . The sentence “The ake is related to the bem” would mean that ake, bem R and would be denoted by symbol akeRbem. In the context of axiom systems, an undefined noun is sometimes called an undefined term, or a primitive term, or an undefined object, or a primitive object. In presentations of axiom system that contains primitive terms, the primitive terms are customarily listed along with the primitive relations, before the axioms. Here is an example of an axiom system consisting of sentences built using primitive terms and primitive relations in the manner described above. Axiom System: Axiom System #2 Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei is related to the bem and akek is related to the bem. Page 24 of 150 3) For any bem, there is exactly one set of two akes, denoted akei , akek where i k , such that akei is related to the bem and akek is related to the bem. As with axiom system #1, each of these axioms in axiom system #2 is a statement whose truth we have no way of assessing, because the words ake and bem are undefined and the relation R is undefined. But we can prove the following theorem. Theorem #1 for axiom system #2: There are exactly 6 bems. Proof 1. By axiom #1, there are four akes. Therefore, it is possible to build six unique sets of two akes, denoted akei , akek where i k . 2. Axioms #2 and #3 tell us that there is bijective correspondence between the sets of two akes and the bems. 3. Therefore, there must be exactly 6 bems. End of proof As with the theorem that we proved in the previous section for Axiom System #1, we note that the theorem just presented could be written with all of the primitive terms, primitive relations, and axioms put into the hypothesis. The resulting theorem statement would be quite long. Theorem: If Blah Blah Blah then there are exactly 6 bems. In the exercises, you will prove the following: Theorem #2 for Axiom System #2: For every ake, there are exactly three bems that the ake is related to. Our examples of axiom systems with undefined terms and undefined relations seem rather absurd, because their axioms are meaningless. What purpose could such abstract collections of nonsense sentences possibly serve? Well, the idea is that we will use such abstract axiom systems to represent actual situations that are not so abstract. Then for any abstract theorem that we have been able to prove about the abstract axiom system, there will be a corresponding true statement that can be made about the actual situation that the axiom system is supposed to represent. This begs the question: why study the axiom system at all, if the end goal is to be able to prove statements that are about some actual situation? Why not just study the actual situation and prove the statements in that context? The answer to that is twofold. First, a given abstract axiom system can be recycled, used to represent many different actual situations. By simply proving theorems once, in the context of the axiom system, the theorems don’t need to be reproved in each actual context. Second, and more important, by proving theorems in the context of the abstract axiom system, we draw attention to the fact that the theorems are true by the simple fact of the axioms and the rules of logic, and nothing else. This will be very important to keep in mind when studying axiomatic geometry. Page 25 of 150 2.4. Interpretations and Models As mentioned above, an axiom system with undefined terms and undefined relations is often used to represent an actual situation. This idea of representation is made more precise in the following definition. You’ll notice that the representing sort of gets turned around: we think of the actual situation as a representation of the axiom system. Definition 10 Interpretation of an axiom system Suppose that an axiom system consists of the following four things an undefined object of one type, and a set A containing all of the objects of that type an undefined object of another type, and a set B containing all of the objects of that type an undefined relation R from set A to set B a list of axioms involving the primitive objects and the relation An interpretation of the axiom systems is the following three things a designation of an actual set A’ that will play the role of set A a designation of an actual set B’ that will play the role of set B a designation of an actual relation R’ from A’ to B’ that will play the role of the relation R As examples, for Axiom System #2 from the previous section we will investigate three different interpretations invented by Alice, Bob, and Carol. Recall Axiom System #2 included the following things an undefined term ake and a set A containing all the akes an undefined term bem and a set B containing all the bems a primitive relation R from set A to set B. a list of three axioms involving these undefined terms and the undefined relation Alice’s interpretation of Axiom System #2. Let A’ be the set of dots in the picture at right. Let set A’ play the role of set A. Let B’ be the set of segments in the picture at right. Let set B’ play the role of set B. Let relation R’ from A’ to B’ be defined by saying that the words ―the dot is related to the segment‖ mean ―the dot touches the segment‖. Let relation R’ play the role of the relation R. Bob’s interpretation of Axiom System #2. Let A’ be the set of dots in the picture at right. Let set A’ play the role of set A. Let B’ be the set of segments in the picture at right. Let set B’ play the role of set B. Let relation R’ from A’ to B’ be defined by saying that the words ―the dot is related to the segment‖ mean ―the dot touches the segment‖. Let relation R’ play the role of the relation R. Carol’s interpretation of Axiom System #2. Page 26 of 150 Let A’ be the set whose elements are the letters v, w, x, and y. That is, A v, w, x, y . Let set A’ play the role of set A. Let B’ be the set whose elements are the sets v, w , v, x , v, y , and w, x , w, y , and x, y . That is, B v, w , v, x , v, y , w, x , w, y , x, y . Let set B’ play the role of set B. Let relation R’ from A’ to B’ be defined by saying that the words ―the letter is related to the set‖ mean that ―the letter is an element of the set‖. Let relation R’ play the role of the relation R. Notice that Alice and Bob have slightly different interepretations of the axiom system. Is one better than the other? It turns out that we will consider one to be much better than the other. The criterion that we will use is to consider what happens when we translate the Axioms into statements about dots and segments. Using a find & replace feature in a word processor, we can simply replace every occurrence of ake with dot, every occurrence of bem with segment, and every occurrence of is related to with touches. Here are the resulting three statements. Statements: 1. There are four dots. These may be denoted dot1, dot2, dot3, and dot4. 2. For any set of two dots, denoted doti , dotk where i k , there is exactly one segment such that doti touches the segment and dotk touches the segment. 3. For any segment, there is exactly one set of two dots, denoted doti , dotk where i k , such that doti touches the segment and dotk touches the segment. We see that in Bob’s interpretation, all three of these statements are true. In Alice’s interpretation the first and third statements are true, but the second statement is false. What about Carol’s interpretation? We should consider what happens when we translate the Axioms into statements about letters and sets. We can simply replace every occurrence of ake with letter, every occurrence of bem with set, and every occurrence of is related to with is an element of. Here are the resulting three statements. Statements: 1. There are four letters. These may be denoted letter1, letter2, letter3, and letter4. 2. For any set of two letters, denoted letteri , letterk where i k , there is exactly one set such that letteri is an element of the set and letterk is an element of the set. 3. For any set, there is exactly one set of two letters, denoted letteri , letterk where i k , such that letteri is an element of the set and letterk is an element of the set. We see that in Carol’s interpretation, the translations of the three axioms are three statements that are all true. Let’s formalize these ideas with two definitions. Page 27 of 150 Definition 11 successful interpretation To say that an interpretation of an axiom system is successful means that when the undefined terms and undefined relations in the axioms are replaced with the corresponding terms and relations of the interpretation, the resulting statements are all true. Definition 12 model of an axiom system A model of an axiom system is an interpretation that is successful. So we would say that Bob’s and Carol’s interpretations are successful: they are models. Alice’s interpretation is unsuccessful: it is not a model. Notice also that Bob’s and Carol’s models are essentially the same in the following sense: one could describe a correspondence between the objects and relations of Bob’s model and the objects and relations of Carol’s model in a way that all corresponding relationships are preserved. Here is one such correspondence. objects in Bob’s model the lower left dot the lower right dot the upper right dot the upper left dot the segment on the bottom objects in Carol’s model the letter v the letter w the letter x the letter y the set v, w the segment that goes from lower left to upper right the set v, x the segment on the left side the set v, y the segment on the right side the set w, x the segment that goes from upper left to lower right the set w, y the segment across the top the set x, y relation in Bob’s model relation in Carol’s model the dot touches the segment the letter is an element of the set What did I mean above by the phrase ―...in a way that all corresponding relationships are preserved...‖? Notice that the following statement is true in Bob’s model. The lower right dot touches the segment on the right side. If we use the correspondence to translate the terms and relations from Bob’s model into terms and relations from Carol’s model, that statement becomes the following statement. The letter w is an element of the set w, x . This statement is true in Carol’s model. In a similar way, any true statement about relationships between dots and segments in Bob’s model will translate into a true statement about relationships between letters and sets in Carol’s model. Page 28 of 150 The notion of two models being essentially the same, in the sense described above, is formalized in the following definition. Definition 13 isomorphic models of an axiom system Two models of an axiom system are said to be isomorphic if it is possible to describe a correspondence between the objects and relations of one model and the objects and relations of the other model in a way that all corresponding relationships are preserved. It should be noted that it will not always be the case that two models for a given axiom system are isomorphic. We will return to this in the next section, when we discuss completeness. 2.5. Properties of Axiom Systems In this section, we will discuss three important properties that an axiom system may or may not have. They are consistency, completeness, and independence. 2.5.1. Consistency We will use the following definition of consistency. Definition 14 consistent axiom system An axiom system is said to be consistent if it is possible for all of the axioms to be true. The axiom system is said to be inconsistent if it is not possible for all of the axioms to be true. We will be interested in determining if a given axiom system is consistent or inconsistent. It is worthwhile to think now about how one would prove that an axiom system is consistent, or how one would prove that an axiom system is inconsistent. Suppose that one suspects that an axiom system is consistent and wants to prove that it is consistent. One proves that an axiom system is consistent by producing a model for the axiom system. For Axiom System #2, we have two models—Bob’s and Carol’s—so the axiom system is definitely consistent. Suppose that one suspects that an axiom system is inconsistent and wants to prove that it is inconsistent. One would be trying to prove that something is not possible. It is not obvious how that would be done. The key is found in the following rule of inference from Math 306. ~ pc Contradiction Rule p In this rule, the symbol c stands for a contradiction—a statement that is always false. We used this rule in to following way. To prove that statement p is true using the method of contradiction, we started by assuming that statement p was false. We then showed that we could reach a contradiction. Therefore, the statement p must be true. Suppose we replace the statement p with the statement ~q. Then the contradiction rule becomes Page 29 of 150 Contradiction Rule ~ ~ q c ~ q In other words, qc Contradiction Rule ~ q This version of the rule states that if one can demonstrate that statement q leads to a contradiction, then statement q must be false. There are other versions of this rule as well. Consider what happens if we use a statement q of the form statement1 statement2 . Contradiction Rule statement1 statement2 c ~ statement1 statement2 If we apply DeMorgan’s law to the conclusion of this version of the rule, we obtain Contradiction Rule statement1 statement2 c ~ statement1 ~ statement2 This version of the rule states that if one can demonstrate that an assumption that statement1 and statement2 are both true leads to a contradiction, then at least one of the statements must be false. Finally, consider what happens if we use a whole list of statements. Contradiction Rule statement1 statement2 statementk c ~ statement1 ~ statement2 ~ statementk This version of the rule states that if one can demonstrate that an assumption that a whole list of statements is true leads to a contradiction, then at least one of the statements must be false. Now return to the notion of an inconsistent axiom system. Recall in an inconsistent axiom system, it is impossible for all of the axioms to be true. In other words, at least one of the axioms must be false. We see that the Contradiction Rule could be used to prove that an axiom system is inconsistent. That is, if one can demonstrate that an assumption that a whole list of axioms is true leads to a contradiction, then at least one of the axioms must be false. We can create an example of an inconsistent axiom system by messing up Axiom system #2. We mess it up by appending a fourth axiom in a certain way. Page 30 of 150 Axiom System: Axiom System #3, an example of an inconsistent axiom system Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei is related to the bem and akek is related to the bem. 3) For any bem, there is exactly one set of two akes, denoted akei , akek where i k , such that akei is related to the bem and akek is related to the bem. 4) There are exactly five bems. Using Axiom System #3, we can prove the following two theorems. Theorem 1 for axiom system #3: There are exactly 6 bems. The proof is the exact same proof that was used to prove the identical theorem for axiom system #2. The proof only uses the first three axioms. Theorem 2 for axiom system #3: There are exactly 5 bems and there are exactly 6 bems. Proof: By Axiom #4 and Theorem 1 But this statement is a contradiction! So we have demonstrated that an assumption that the four axioms from Axiom system 3 are true leads to a contradiction. Therefore, least one of the axioms must be false. In other words, Axiom System #3 is inconsistent. Note that it is tempting to say that it must be axiom #4 that is false, because there was nothing wrong with the first three axioms before we threw in the fourth one. But in fact, there is not really anything wrong with the fourth axiom in particular. For for example, if one discards axiom #3, then it turns out that the remaining list of axioms, consisting of axiom #1, axiom #2, and axiom #4 is perfectly consistent. Here is such an axiom system, with the axioms re-numbered. You are asked to prove that this axiom system is consistent in Exercise #6. Axiom System: Axiom System #4, an example of an inconsistent axiom system Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei is related to the bem and akek is related to the bem. 3) There are exactly five bems. Page 31 of 150 So the problem is not with any one particular axiom. Rather, the problem is with the whole set of four. 2.5.2. Independence An axiom system that is not consistent could be thought of as one in which the axioms don’t agree; an axiom system that is consistent could be thought of as one in which there is no disagreement. In this sort of informal language, we could say that the idea of independence of an axiom system has to do with whether or not there is any redundancy in the list of axioms. The following definitions will make this precise. Definition 15 dependent and independent axioms An axiom is said to be dependent if it is possible to prove that the axiom is true as a consequence of the other axioms. An axiom is said to be independent if it is not possible to prove that it is true as a consequence of the other axioms. We will be interested in determining if a given axiom is dependent or independent. It is worthwhile to think now about how one would prove that an axiom is dependent, or how one would prove that an axiom is independent. Suppose that one suspects that a given axiom is dependent and wants to prove that it is dependent. To do that, one proves that the statement of the axiom must be true, with a proof that uses only the other axioms. That is, one stops assuming that the given axiom is a true statement and downgrades it to just an ordinary statement that might be true or false. If it is possible to prove the statement is true, using a proof that uses only the other axioms, then the given axiom is dependent. For an example of a dependent axiom, consider the following list of axioms that was constructed by appending an additional axiom to the list of axioms for Axiom System #2. Axiom System: Axiom System #5 Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei touches the bem and akek touches the bem. 3) For any bem, there is exactly one set of two akes, denoted akei , akek where i k , such that akei touches the bem and akek touches the bem. 4) There are exactly six bems. We recognize first three axioms. They are the axioms from Axiom System #2. And we also recognize the statement of axiom #4. It is the same statement as Theorem #1 for Axiom System #2. In other words, it can be proven that axiom #4 is true as a consequence of the first three axioms. So Axiom #4 is not independent; it is dependent. Page 32 of 150 Suppose that one suspects that a given axiom is independent and wants to prove that it is independent. To do that, one stops assuming that the given axiom is a true statement and downgrades it to just an ordinary statement that might be true or false. For the remaining, smaller axiom system, one must produce two models: one model in which all of the other axioms are true and the extra statement is also true, and a second model in which all of the other axioms are true and the extra statement is false. For an example of an independent axiom, consider axiom #3 from Axiom System #2. It is an independent axiom. To prove that it is independent, we treat it as an extra statement, instead of as an axiom, and we consider models of the remaining, smaller axiom system. Axiom System: Axiom System #6: This is just Axiom System #2 minus the third axiom Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei touches the bem and akek touches the bem. Extra Statement For any bem, there is exactly one set of two akes, denoted akei , akek , that used to be an where i k , such that akei touches the bem and akek touches the bem. axiom: Now consider two models for Axiom System #6 Bob’s model of Axiom System #6. Let A’ be the set of dots in the picture at right. Let set A’ play the role of set A. Let B’ be the set of segments in the picture at right. Let set B’ play the role of set B. Let relation R’ from A’ to B’ be defined by saying that the words ―the dot is related to the segment‖ mean ―the dot touches the segment‖. Let relation R’ play the role of the relation R. Dan’s model of Axiom System #6. Let A’ be the set of dots in the picture at right. Let set A’ play the role of set A. Let B’ be the set of segments in the picture at right. Let set B’ play the role of set B. Let relation R’ from A’ to B’ be defined by saying that the words ―the dot is related to the segment‖ mean ―the dot touches the segment‖. Let relation R’ play the role of the relation R. Consider the translation of the extra statement into the language of the models: Page 33 of 150 For any segment, there is exactly one set of two dots, denoted doti , dotk where i k , such that doti touches the segment and dotk touches the segment. We see that in Bob’s model of Axiom System #6, the extra statement is true, while in Dan’s model of Axiom System #6, the removed axiom is false. So based on these two examples, we can say that in Axiom System #2, the third axiom is independent. The following definition is self-explanatory. Definition 16 independent axiom system An axiom system is said to be independent if all of its axioms are independent. An axiom system is said to be not independent if one or more of its axioms are not independent. To prove that an axiom system is independent, one must prove that each one of its axioms is independent. That means that for each of the axioms, one must go through a process similar to the one that we went through above for Axiom #3 from Axiom System #2. This can be a huge task. To prove that an axiom system is not independent, one need only prove that one of its axioms is not independent. 2.5.3. Completeness Recall that in Section 2.4, we found that Bob’s and Carol’s models of Axiom System #2 were isomorphic models. It turns out that any two models for that axiom system are isomorphic. Such a claim can be rather hard—or impossible—to prove, but it is a very important claim. It essentially says that the axioms really ―nail down‖ every aspect of the behavior of any model. This is the idea of completeness. Definition 17 complete axiom system An axiom system is said to be complete if any two models of the axiom system are isomorphic. An axiom system is said to be not complete if there exist two models that are not isomorphic. It is natural to wonder why the word complete is used to describe this property. One might think of it this way. If an axiom system is complete, then it is like a complete set of specifications for a corresponding model. All models for the axiom system are essentially the same: they are isomorphic. If an axiom system is not complete, then one does not have a complete set of specifications for a corresponding model. The specifications are insufficient, some details are not nailed down. As a result, there can be models that differ from each other: models that are not isomorphic. For an example, consider axiom system #6. In Section 2.5.2, we found two models that are not isomorphic. (There are others as well.) In Bob’s model, there are six segments, but in Dan’s model, there is only one segment. Question for class: Why does the fact that Bob’s model has six segments and Dan’s model has only one segment necessarily mean that the two models are not isomorphic? Page 34 of 150 The discussion of Bob’s and Dan’s models for axiom system #6 can be generalized to the extent that it is possible to formulate an alternate wording of the definition of a complete axiom system. Remember that axiom system #6 has two axioms. Consider a feature that distinguished Bob’s mode from Dan’s model. One obvious feature is the number of segments. As observed above, in Bob’s model, there are six segments, but in Dan’s model, there is only one segment. Now consider the following statement: There are exactly six segments. This statement is an independent statement in axiom system #6. That is, there is a model for axiom system #6 in which the statement is true (Bob’s model) and there is a model for axiom system #6 in which the statement is false (Dan’s model). The fact that an independent statement can be made regarding the number of line segments indicates that axiom system #6 does not sufficiently specify the number of line segments. That is, axiom system #6 is incomplete. More generally, if it is possible to make a statement regarding the primitive objects in an axiom system that is an independent statement, then the axiom system is not complete (and vice-versa). Thus, an alternate way of wording the definition of a complete axiom system is as follows: Alternate definition of a complete axiom system: words: The axiom system is complete meaning: all models of the axiom system are isomorphic alternate wording: There are no independent statements regarding the primitive objects Alternate definition of an incomplete axiom system: words: The axiom system is incomplete meaning: there exist two models of the axiom system that are not isomorphic alternate wording: There is an independent statement regarding the primitive objects We have discussed the fact that the statement ―there are exactly six segments‖, or ―there are exactly six bems‖ is an independent statement in Axiom System #6, because the statement cannot be proven true on the basis of the axioms. If we wanted to, we could construct a new axiom system #7 by appending an axiom to Axiom System #6 in the following manner. Axiom System: Axiom System #7: This is Axiom System #6 with an added axiom Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei touches the bem and akek touches the bem. 3) There are exactly six bems. Page 35 of 150 Of course, Bob’s successful interpretation for Axiom System #6 would be a successful interpretation for Axiom System #7, as well. That is, Bob’s interpretation is a model for Axiom System #6 and also for Axiom System #7. But Dan’s successful interpretation for Axiom System #6 would not be a successful interpretation for Axiom System #7. That is, Dan’s interpretation is a model for Axiom System #6, but not for Axiom System #7. Keep in mind, that we could have appended a different axiom to Axiom System #6. Axiom System: Axiom System #8: This is Axiom System #6 with a different axiom added Primitive Terms: ake bem Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake is related to the bem‖ Axioms: 1) There are four akes. These may be denoted ake1, ake2, ake3, and ake4. 2) For any set of two akes, denoted akei , akek where i k , there is exactly one bem such that akei touches the bem and akek touches the bem. 3) There is exactly one bem. We see that Dan’s interpretation is a model for Axiom Systems #6 and #8. On the other hand, Bob’s interpretation is a model for Axiom System #6 but not for Axiom System #8. Page 36 of 150 2.6. Exercises The first four problems are about Axiom System #1. This axiom system was introduced in Section 2.3.1 and has an undefined relation. 1) Which of the following interpretations is successful? That is, which of these interpretations is a model? Explain. a. Interpret the words ―x is related to y‖ to mean ― xy 0 ‖. b. Interpret the words ―x is related to y‖ to mean ― xy 0 ‖. c. Interpret the words ―x is related to y‖ to mean ―x and y are both even or are both odd‖. Hint: One of the three is unsuccessful. The other two are successful. That is, they are models. 2) Consider the two models of Axiom System #1 that you found in exercise (1). For each model, determine whether the statement ―1 is related to -1‖ is true or false. 3) Based on your answer to exercise (2) is the statement ―1 is related to -1‖ an independent statement? Explain. 4) Based on your answer to exercise (3), is Axiom System #1 complete? Explain. 5) In Section 2.3.2, we discussed Axiom System #2, which had undefined terms and relations. Prove Theorem #2 for Axiom System #2. 6) In Section 2.5.1, we discussed Axiom System #4. Prove that this axiom system is consistent by demonstrating a model. (Hint: Produce a successful interpretation involving a picture of dots and segments.) The next few exercises refer to the following new axiom system. Axiom System: Axiom System #9 Primitive Terms: cet (pronounced “ket”) dag Primitive Relations: relation from the set of all cets to the set of all dags spoken ―the cet is related to the dag‖ Axioms: 1) There are exactly three cets. These may be denoted cet1, cet2, and cet3. 2) For any set of two cets, denoted ceti , cetk where i k , there is exactly one dag such that ceti is related to the dag and cetk is related to the dag. 3) For any set of two dags, denoted dagi , dagk where i k , there is at least one cet such that the cet is related to dagi and the cet is related to dagk. 4) For every dag, there is at least one cet that is not related to the dag. Page 37 of 150 7) Produce a model for Axiom System #9 that that uses dots and segments to correspond to cets and dags, and that uses the words ―the dot touches the segment‖ to correspond to the words ―the cet is related to the dag‖. That is, draw a picture that works. 8) Is Axiom System #9 consistent? Explain. 9) The goal is to prove that Axiom #1 is independent. In exercise #7, you produced a model involving dots and segments that demonstrates that it is possible for all four axioms to be true. Therefore, all you need to do is produce a model involving dots and segments that demonstrates that it is possible for Axioms #2, #3, and #4 to be true and Axiom #1 to be false. 10) The goal is to prove that Axiom #2 is independent. In exercise #7, you produced a model involving dots and segments that demonstrates that it is possible for all four axioms to be true. Therefore, all you need to do is produce a model involving dots and segments that demonstrates that it is possible for Axioms #1, #3, and #4 to be true and Axiom #2 to be false. 11) The goal is to prove that Axiom #3 is independent. In exercise #7, you produced a model involving dots and segments that demonstrates that it is possible for all four axioms to be true. Therefore, all you need to do is produce a model involving dots and segments that demonstrates that it is possible for Axioms #1, #2, and #4 to be true and Axiom #3 to be false. 12) The goal is to prove that Axiom #4 is independent. In exercise #7, you produced a model involving dots and segments that demonstrates that it is possible for all four axioms to be true. Therefore, all you need to do is produce a model involving dots and segments that demonstrates that it is possible for Axioms #1, #2, and #3 to be true and Axiom #4 to be false. 13) Is axiom system #9 independent? Explain. 14) Prove the following Theorem for Axiom System #9: Theorem 1 for Axiom System #9: For any set of two dags, denoted dagi , dagk where i k , there is exactly one cet such that the cet is related to dagi and the cet is related to dagk. Hint: Axiom #3 guarantees that there is at least one such cet. Suppose that there is more than one such cet and show that you can reach a contradiction. 15) Prove the following Theorem for Axiom System #9: Theorem #2 for Axiom System #9: There are exactly three dags. Page 38 of 150 Page 39 of 150 3. Axiomatic Geometries 3.1. Introduction For the remainder of the course, we will be studying axiomatic geometry. Before starting that study, we should be sure and understand the difference between analytic geometry and axiomatic geometry. 3.1.1. What is an analytic geometry? Very roughly speaking, an analytic geometry consists of two things: a set of points that is represented in some way by real numbers a means of measuring the distance between two points For example, in plane Euclidean analytic geometry, a point is represented by a pair x, y 2 . That is, a point is an ordered pair of real numbers. The distance between points P x1 , y1 and Q x2 , y2 is obtained by the formula d P, Q x2 x1 y2 y1 2 2 . In three dimensions, one adds a z-coordinate. In analytic geometry, objects are described as sets of points that satisfy certain equations. A line is the set of points x, y that satisfy an equation of the form ax + by = c; a circle is the set of points that satisfy an equation of the form x h y k r 2 , etc. Every aspect of the behavior of analytic geometric objects is completely dictated by rules about solutions of equations. For example, any two lines either don’t intersect, or they intersect exactly once, or they are the same line; there are no other possibilities. That this true is simply a fact about simultaneous solutions of a pair of linear equations in two variables ax by c dx ey f You will see that in axiomatic geometry, objects are defined in a very different way, and their behavior is governed in a very different manner. 2 2 3.1.2. What is an axiomatic geometry? Very roughly speaking, an axiomatic geometry is an axiom system with the following primitive (undefined) things. Primitive Terms: point line Primitive Relation: relation from the set of all points to the set of all lines, spoken the point lies on the line Remark: It is not a very confident definition that begins with the words ―…roughly speaking…‖. But in fact, one will not find general agreement about what constitutes an axiomatic geometry. My description above contains some of the essentials. Page 40 of 150 You’ll notice, of course, that the axiom system above is essentially the same sort of axiom system that we discussed in Chapter 2. The only difference is that we stick to the particular convention of using undefined terms called point and line, and an undefined relation spoken the point is on the line. It is natural to wonder why we bothered with the meaningless terms ake, bem, cet, and dag, when we could have used the more helpful terms point and line. The reason for starting with the meaningless terms was to stress the idea that the terms are always meaningless; they are not supposed to be helpful. When studying axiomatic geometry, it will be very important to keep in mind that even though you may think that you know what a point and a line are, you really don’t. The words are as meaningless as ake and bem. On the other hand, when studying a model of an axiomatic geometry, we will know the meaning of the objects and relations, but we will be careful to always give those objects and relations names other than point and line. For instance, we used the names dot and segment in our models that involved drawings. The word dot refers to an actual drawn spot on the page or chalkboard; it will be our interpretation of the word point, which is an undefined object. Because the objects and relations in axiomatic geometry are undefined things, their behavior will be undefined as well, unless we somehow dictate that behavior. That is the role of the axioms. Every aspect of the behavior of axiomatic geometric objects must be dictated by the axioms. For example, if we want lines to have the property that two lines either don’t intersect, or they intersect exactly once, or they are the same line, then that will have to be specified in the axioms. We will return to the notion of what makes axiomatic points and lines behave the way we ―normally‖ expect points and lines to behave in Section 3.5, when we study incidence geometry. 3.2. Two Simple Finite Geometries 3.2.1. Three Point Geometry A finite axiomatic geometry is one that has a finite number of points. Our first example has three. Axiom System: The Three Point Geometry Primitive Terms: point line Primitive Relation: relation from the set of all points to the set of all lines, spoken the point lies on the line Axioms: 1) There are exactly three points. 2) For any set of two points, there is exactly one line that both points lie on. 3) For any set of two lines, there is at least one point that lies on both lines. 4) For every line, there is at least one point that does not lie on the line. Notice that the Three-Point Geometry is almost the same as Axiom System #9, which was presented in the Exercises of Section 2.6. The differences are minor, just choices of names. In Axiom System #9, the primitive terms are ake and bem; in the Three-Point Geometry, the primitive terms are point and line. In Axiom System #9, the primitive relation is spoken the ake is related to the bem; In the Three-Point Geometry, the primitive relation is spoken the point lies on the line. Page 41 of 150 We will study the following theorems of Three-Point Geometry. Notice that Theorems #1 and #2 for Three Point Geometry are essentially the same as Theorems #1 and #2 for Axiom System #9. Three Point Geometry Theorem #1: For any two distinct lines, there is exactly one point that lies on both lines. You will prove this Theorem in the exercises. Three Point Geometry Theorem #2: There are exactly three lines. You will prove this Theorem in the exercises. Three Point Geometry Theorem #3: For any line, there is exactly one point that does not lie on the line. You will study a proof of this theorem in a class drill. Three Point Geometry Theorem #4: For any line, there are exactly two points that lie on the line. You will prove this theorem in the exercises. 3.2.2. Four Point Geometry Our next finite axiomatic geometry has four points. Axiom System: Four Point Geometry Primitive Terms: point line Primitive Relations: relation from the set of all points to the set of all lines, spoken the point lies on the line Axioms: 1) There are exactly four points. 2) For any set of two points, there is exactly one line that both points lie on. 3) Every line has exactly two points that lie on it. Notice that the Four Point Geometry is the same as Axiom System #2, presented in Section 2.3.2. The differences are minor, just the same differences in choices of names that we encountered before when we compared the Three Point Geometry to Axiom System #9. Following are two theorems of Four Point Geometry. Notice that they are almost the same as Theorems #1 and #2 for Axiom System #9. Four Point Geometry Theorem #1: There are exactly six lines. You will prove this Theorem in the exercises. Four Point Geometry Theorem #2: For every point, there are exactly three lines that pass through the point. You will prove this Theorem in the exercises. Page 42 of 150 3.3. Terminology 3.3.1. Is the primitive relation symmetric? Is the primitive relation symmetric? That is, if point P lies on line L, will it also be true that line L lies on point P? The answer must be no, and it has to do only with the conventions of the terminology of relations. The undefined relation spoken ―the point lies on the line‖ is a relation from the set of points to the set of lines. Therefore, one does not say ―the line lies on the point‖. So the relation cannot possibly be symmetric. Another way of explaining this is to say that the symmetry property of relations is something that is defined only for relations on a set. The ―lies on‖ relation is not a relation on a set; it is a relation from one set to a different set. So it cannot be symmetric and it cannot be turned around. But restricting ourselves to always having to say ―point P lies on line L” can lead to cumbersome and bland-sounding writing. For that reason alone, it is worthwhile to introduce some additional terminology. But we won’t introduce very much. We will introduce just one additional term that will enable us to change the word order when talking about points that lie on lines. Definition 18 passes through words: Line L passes through point P. meaning: Point P lies on line L. With this terminology, the axioms for the Four Point Geometry could be rephrased as follows 1. There are four points. 2. For any set of two points, there is exactly one line that passes through both points. 3. Every line passes through exactly two points. It is worth remarking that most introductory axiomatic geometry books are a bit sloppy regarding the order of words in expressions involving an undefined relation. One reason for this may be that most of those books do not use the terminology of relations, and so they do not need to stick to the conventions of that terminology. Here is a presentation of the Three Point Geometry copied verbatim from one such textbook. Axioms for Three-Point Geometry, copied from a Textbook 1) There exist exactly three distinct points in the geometry. 2) Each two distinct points are on exactly one line. 3) Not all the points of the geometry are on the same line. 4) Each two distinct lines are on at least one point. In your work, you will encounter a variety of wordings for the undefined relation. Here are four that we won’t use. Commonly used expressions that we will not use 1) Line L contains point P. 2) Line L is on point P. 3) Point P is incident upon line L. 4) Line L is incident upon point P. 3.3.2. New Definitions While we’re in the business of introducing new terminology, we may as well introduce some more. The next two definitions are self-explanatory. Page 43 of 150 Definition 19 intersecting lines words: Line L intersects line M. meaning: There exists a point P that lies on both lines. (at least one point) Definition 20 parallel lines words: Line L is parallel to line M. symbol: L M meaning: Line L does not intersect line M. Remember that you have encountered other definitions of parallel. In Exercise 8) of Section 1.3, you considered three definitions. The third definition presented in that exercise is the one that is presented here as the definition of parallel lines in axiomatic geometry. Discussion Question: Of the three definitions of parallel that were presented in Exercise 8) of Section 1.3, why do you suppose that the third one is the one that is being used here? Would there be any problem with using either of the other two definitions presented in that exercise? It’s worth noting that we have essentially introduced three new relations in this section. The words ―Line L passes through point P‖ indicate a relation from the set of all lines to the set of all points. The words ―Line L intersects line M‖ indicate a relation on the set of all lines. Similarly, the words ―Line L is parallel to line M‖ indicate another relation on the set of all lines. These relations are not primitive relations, because they have an actual meaning. Those meanings are given by the above definitions, and those definitions refer to primitive terms and relations and to previously defined words. We will refer to this kind of relation as a defined relation. It is customary to not list the definitions of the defined relations when presenting the axiom system. This is understandable, because often there are usually many defined relations, and to list them all would be very cumbersome. But it is unfortunate that the defined relations are not listed, because it makes the defined relations seem less important than the other components of an axiom system, and because the reader often does not remembe the definitions and must go looking for them. Here are two additional definitions, also straightforward. Definition 21 collinear points words: The set of points P1 , P2 , meaning: There exists a line L that passes through all the points. Definition 22 concurrent lines words: The set of lines L1 , L2 , , Pk is collinear. , Lk is concurrent. meaning: There exists a point P that lies on all the lines. It’s worth noting that these two definitions are not relations. Rather, they are simply statements that may or may not be true for a particular set of points or a particular set of lines. That is, they are properties that a set of points or a set of lines may or may not have. Page 44 of 150 3.3.3. The Big Question The following question is extremely important in axiomatic geometry. Given a line L and a point P that does not lie on L, how many lines M exist that pass through P and are parallel to L? It is so important that we will refer to it as The Big Question. The question is first raised in the current chapter, in discussions of finite geometries. But it will come up throughout the course. 3.4. Fano’s and Young’s Finite Geometries Let’s return to finite geometries. A more complicated finite geometry is the following. (Note the use of defined relations in the statements of the axioms.) Axiom System: Fano’s Geometry Primitive Terms: point line Primitive Relations: relation from the set of all points to the set of all lines, spoken the point lies on the line Axioms: 1) There exists at least one line. 2) Every line passes through exactly three points. 3) Not all points of the geometry lie on the same line. 4) For any set of two points, there is exactly one line that passes through both. 5) Any two lines intersect. In the exercises, you will prove the following things about Fano’s geometry. Fano’s Geometry Theorem #1: There are exactly seven points and seven lines. Fano’s Geometry Theorem #2: Any two lines intersect exactly once. Fano’s Geometry is a model of Incidence Geometry (defined below) Fano’s axioms are independent. (This one is hard!) By changing just the fifth axiom in Fano’s Geometry, we obtain Young’s Geometry Axiom System: Young’s Geometry Primitive Terms: point line Primitive Relations: relation from the set of all points to the set of all lines, spoken the point lies on the line Axioms: 1) There exists at least one line. 2) Every line passes through exactly three points. 3) Not all points of the geometry lie on the same line. 4) For any set of two points, there is exactly one line that passes through both. 5) For each line L, and for each point P that does not lie on L, there exists exactly one line M that passes through P and is parallel to L. In the exercises, you will prove the following things about Young’s geometry. Page 45 of 150 Young’s Geometry Theorem #1: There are exactly nine points and twelve lines. Young’s Geometry Theorem #2: Two lines parallel to a third line are parallel to each other. Young’s Geometry is a model of Incidence Geometry (defined below) Young’s axioms are independent. 3.5. Incidence Geometry 3.5.1. Incidence Relations and Axioms of Incidence You will notice that in each of the four finite geometries that we have encountered so far, the axioms can be classified into two types. One type of axiom is just about the primitive terms. Here are two examples. Four Point Geometry Axiom #1: There exist exactly four points. Fano’s Axiom #1: There exists at least one line. A second type of axiom is about the behavior of the primitive relation. Here are two examples. Three Point Geometry Axiom #3. For any set of two lines, there is at least one point that lies on both lines. Fano’s Axiom #3: Not all points of the geometry lie on the same line. In some early books on axiomatic geometry, the primitive relation was spoken ―the line is incident upon the point‖. Such a primitive relation could referred to as the incidence relation. Axioms such as the two above that described the behavior of the incidence relation were called axioms of incidence. Even though most books no longer use the words ―…is incident upon…‖ for the primitive relation, it is still fairly common for any axioms that describe the behavior of the primitive relation to be referred to as axioms of incidence. This can be confusing. 3.5.2. The Axiom System for Incidence Geometry To add to the confusion, the following axiom system is usually called Incidence Geometry. Axiom System: Incidence Geometry Primitive Terms: point line Primitive Relations: relation from the set of all points to the set of all lines, spoken the point lies on the line Axioms: IA1: There exist three distinct non-collinear points. (at least three) IA2: For every pair of distinct points, there exists exactly one line that passes through both points. IA3: Every line passes through at least two distinct points. Notice that the axioms for Incidence Geometry are less specific than any axiom system that we have seen so far. The number of points is not even specified. Even so, notice that axioms IA2 and IA3 do guarantee that the primitive points and lines in Incidence Geometry will have some of the ―normal‖ behavior that we associate with points and lines in analytic geometry, or in drawings that we have made all of our lives. For instance we are used to the fact that two lines can either be parallel, or intersect once, or be the same line. That is, distinct lines that are not parallel only intersect once. It was mentioned in Section 3.1.1 that in analytic geometry, lines behave this way as a consequence of behavior of Page 46 of 150 solutions of systems of linear equations. In Section3.1.2, it was mentioned that in axiomatic geometry, every aspect of the behavior of points and lines will need to be specified by the axioms. Well, the axioms of Incidence Geometry do in fact guarantee that lines have the particular behavior we are discussing. The first of the following five theorems articulate this fact. You will study the proofs of these theorems in a class drill. Theorem 1 In Incidence Geometry, if L and M are distinct lines that are not parallel, then there is exactly one point that both lines pass through. Theorem 2 In Incidence Geometry, there exist three lines that are not concurrent Theorem 3 In Incidence Geometry, given any line L, there exists a point not lying on L. Theorem 4 In Incidence Geometry, given any point P, there exists a line that does not pass through P. Theorem 5 In Incidence Geometry, given any point P, there exist two lines that pass through P. 3.6. Models of Incidence Geometry 3.6.1. Abstract versus Concrete Models; Relative versus Absolute Consistency It is interesting to consider the question of whether or not the Axiom System for Incidence Geometry is Consistent. That is, whether or not it is possible to find a model. Remember that a model is a successful interpretation. A very simple interpretation of Incidence Geometry is possible using the Three Point Geometry: objects in Incidence Geometry objects in the Three Point Geometry points points lines lines relation in Incidence Geometry relation in Three Point Geometry the point lies on the line the point lies on the line To determine whether or not the interpretation is successful, we use the interpretation to translate the axioms of Incidence Geometry into statements in Three Point Geometry, and then consider whether or not the resulting statements are true in Three Point Geometry. axioms of Incidence True? statements in Three Point Geometry Geometry IA1) There exist three Statement 1: There exist three distinct true because of distinct non-collinear non-collinear points. (at least three) Three-Point Axiom 1 points. (at least three) IA2) For every pair of Statement 2: For every pair of distinct distinct points, there exists true because of Three points, there exists exactly one line exactly one line that passes Point Axiom 2 that passes through both points. through both points. IA3) Every line passes true because of Three Statement 3: Every line passes through at least two Point Geometry through at least two distinct points. distinct points. Theorem 4 Page 47 of 150 The table above demonstrates that Three Point Geometry can provide a successful interpretation of Incidence Geometry. That is, Three Point Geometry can be a model of Incidence Geometry. This demonstrates that Incidence Geometry is consistent. Sort of… The concept of consistency seemed to be about demonstrating that the words of the axiom system could be interpreted as actual, concrete things. It is a little unsatisfying that we have only demonstrated that the words of Incidence Geometry can be interpreted as other words from another axiom system. That’s a little bit like paying back an I.O.U. with another I.O.U. It would be more satisfying if we had an interpretation involving actual, concrete things. Here’s one that uses a picture: objects in Incidence Geometry objects in the picture at right points dots lines line segments relation in Incidence Geometry relation in the picture at right the point lies on the line the dot touches the line segment To determine whether or not the interpretation is successful, we use the interpretation to translate the axioms of Incidence Geometry into statements about the picture, and then consider whether or not the resulting statements about the picture are true. axioms of Incidence Geometry statements about the picture True? IA1) There exist three distinct nonStatement 1: There exist three distinct nontrue collinear points. (at least three) collinear dots. (at least three) IA2) For every pair of distinct Statement 2: For every pair of distinct dots, points, there exists exactly one line true there exists exactly one line segment that that passes through both points. passes through both dots. IA3) Every line passes through at Statement 3: Every line segment passes true least two distinct points. through at least two distinct dots. There is terminology that applies to the above discussion Definition 23 Abstract Model and Concrete Model An abstract model of an axiom system is a model that is, itself, another axiom system. A concrete model of an axiom system is a model that uses actual objects and relations. The Three Point Geometry is an example of an abstract model for Incidence Geometry. The picture with three dots and three line segments is an example of a concrete model. Definition 24 Relative Consistency and Absolute Consistency An axiom system is called relatively consistent if an abstract model has been demonstrated. An axiom system is called absolutely consistent if a concrete model has been demonstrated. It is possible for an axiom system to be both relatively consistent and absolutely consistent. The fact that Three Point Geometry is a model for Incidence Geometry merely proves that Incidence Page 48 of 150 Geometry is Relatively Consistent.We would say that Incidence Geometry is absolutely consistent because of the model involving the picture with three dots. In the exercises, you will prove that the Four Point Geometry, Fano’s Geometry, and Young’s Geometry are also models of incidence geometry. Page 49 of 150 3.7. Exercises The first six exercises deal with the Three Point Geometry that was introduced in Section 3.2.1. In the reading, it was pointed out that Three Point Geometry is just Axiom System #9 from the Exercises of Section 2.6, but with some of the wording changed. In those exercises, you proved a bunch of facts about Axiom System #9. All proofs of facts about Axiom System #9 can be recycled, with their wording changed, into proofs of corresponding facts about the Three Point Geometry. The first three exercises are of that type. 1) Prove Three Point Geometry Theorem #1. 2) Prove Three Point Geometry Theorem #2. 3) Are the Three Point Geometry axioms independent? Explain. The remaining three exercises about Three Point Geometry are not simply adaptations of facts from Axiom System #9. They are new. 4) Prove Three Point Geometry Theorem #4. 5) In the Three Point Geometry, do parallel lines exist? Explain. 6) In the Three Point Geometry, what is the answer to The Big Question? Explain. The next six exercises deal with the Four Point Geometry that was introduced in Section 3.2.2. In the reading, it was pointed out that Four Point Geometry is basically just Axiom System #2 from Section 2.3.2, but with some of the wording changed. All proofs of facts about Axiom System #2 can be recycled, with their wording changed, into proofs of corresponding facts about the Four Point Geometry. 7) Prove Four Point Geometry Theorem #1. (Hint: translate the proof of Theorem #1 of Axiom System #2.) 8) Prove Four Point Geometry Theorem #2. (Hint: translate the proof of Theorem #2 of Axiom System #2.) 9) Are the Four Point Geometry axioms independent? Explain. (Hint: see your class notes from Thursday, April 10. There, I discussed the corresponding question about Axiom System #2.) The remaining three exercises about Four Point Geometry are not simply adaptations of facts from Axiom System #2. They are new. 10) Prove that the Four Point Geometry is a model of Incidence Geometry. 11) Prove that in the Four Point Geometry, parallel lines exist. 12) In the Four Point Geometry, what is the answer to The Big Question? Explain. Page 50 of 150 The next nine exercises explore Fano’s Geometry, introduced in Section 3.4. 13) Give a model for Fano’s geometry that interprets points as dots and interprets lines as curves. 14) Give a model for Fano’s geometry that interprets points as letters and interprets lines as sets. 15) Are Fano’s axioms consistent? Explain. 16) Prove Fano’s Geometry Theorem #1: There are exactly seven points and seven lines. 17) Prove Fano’s Geometry Theorem #2: Any two lines intersect exactly once. 18) Prove that Fano’s Geometry is a model of Incidence Geometry. 19) Prove that Fano’s axioms are independent. (All the steps are easy except for one killer.) 20) Are there parallel lines in Fano’s Geometry? Explain. 21) In Fano’s Geometry, what is the answer to The Big Question? Explain. The next eight exercises explore Young’s Geometry, introduced in Section 3.4. 22) Give a model for Young’s geometry that interprets points as letters and interprets lines as sets. 23) Are Young’s axioms consistent? Explain. 24) Prove Young’s Geometry Theorem #1: There are exactly nine points and twelve lines. 25) Prove Young’s Geometry Theorem #2: Two lines parallel to a third line are parallel to each other. Hint: Do a proof by contradiction, involving axiom #5. 26) Prove that Young’s Geometry is a model of Incidence Geometry. 27) Prove that Young’s axioms are independent. 28) Are there parallel lines in Young’s Geometry? Explain. 29) In Young’s Geometry, what is the answer to The Big Question? Explain. Recall that incidence geometry was presented in Section 3.5.2. 30) Explain why each of the pictures below could not be an incidence geometry. picture (a) picture (b) picture (c) picture (d) Page 51 of 150 4. Models of Incidence Geometry Whose Sets of Points are Infinite In the axiom systems for Three Point Geometry, Four Point Geometry, Fano’s Geometry, and Young’s Geometry, there were axioms that specified a certain number of points. Nothing in the axioms for Incidence Geometry specifies the number of points and, as a result, there are models of Incidence Geometry with many different numbers of points. As varied as they may be, however, each of the models that we studied in the previous chapter had a finite number of points. It is natural to wonder if Incidence Geometry has an infinite model—one whose set of points is an infinite set. 4.1. A Model of Incidence Geometry Involving Straight-Line Drawings Consider the following interpretation of Incidence Geometry as ―straight-line drawings‖. objects in Incidence Geometry points lines relation in Incidence Geometry the point lies on the line objects in straight-line drawings drawn dots (drawn with a pencil or computer) drawn lines (drawn with a ruler or computer) relation in straight-line drawings the drawn dot touches the drawn line Is this interpretation successful? That is, is it a model? As always, we consider what happens when we translate the axioms of Incidence Geometry into statements about straight-line drawings. axioms of Incidence Geometry IA1) There exist three distinct noncollinear points. (at least three) IA2) For every pair of distinct points, there exists exactly one line that passes through both points. IA3) Every line passes through at least two distinct points. statements about straight-line drawings Statement 1: There exist three distinct non collinear drawn dots. (at least three) Statement 2: For every pair of distinct drawn dots, there exists exactly one drawn line that touches both drawn dots. Statement 3: Every drawn line passes through at least two distinct drawn dots. True? yes no yes There are problems with this interpretation. Whether drawing on paper with a pencil or pen, or on a chalkboard with chalk, or on a computer screen, a dot will have some breadth, and a line will have some width as well as length. In extreme cases, the breadth of a point or the width of a line will cause violations of the Incidence Axioms. For example, in the picture below, drawn line L is meant to touch drawn dots A, B, and C. Line M is meant to touch drawn dots A and D. But because of the thickness of the dots and lines, line M also touches dot B. So there are two lines that touch dots A and B. Thus, Statement 2 is false. line M D A line L C B Of course the solution to the problem with the above drawing would be to simply reduce the line width, maybe from 1.5 to 1, and reduce the dot size. But even then, a smaller drawing could that would still cause Statement 2 to be false. Page 52 of 150 As an alternative, we could continue to use actual drawn dots and drawn lines, but think of them as crude versions of idealized imaginary drawings. In those imaginary drawings, dots would have no breadth, and lines would have no width. And while we’re at it, in our imaginary drawings, the flat sheet of paper would continue on forever in all directions. These imaginary, idealized straight-line drawings would be a model of incidence geometry. And note that the set of all points in such a drawing is an infinite set. 4.2. A Model of Incidence Geometry Involving the Usual Analytic Geometry Consider the following interpretation of Incidence Geometry as the usual Analytic Geometry of the (x,y) plane. objects in Incidence Geometry objects in Analytic Geometry points (x,y) pairs where x and y are real numbers sets of the form L x, y : ax by c , where a, b, and lines c are real number constants and x and y are real variables. relation in Incidence Geometry relation in Analytic Geometry the pair (d,e) is an element of the set x, y : ax by c the point lies on the line . Is this interpretation successful? That is, is it a model? As always, we consider what happens when we translate the axioms of Incidence Geometry into statements about Analytic Geometry. axioms of Incidence Geometry IA1) There exist three distinct noncollinear points. (at least three) IA2) For every pair of distinct points, there exists exactly one line that passes through both points. statements about Analytic Geometry Statement 1: There exist three distinct pairs, P = (x1,y1), Q = (x2,y2), R = (x3,y3) such that there is no one set L of the form L x, y : ax by c such that P L and Q L and R L . Statement 2: For every pair of distinct pairs P = (x1,y1) and Q = (x2,y2), there exists exactly one set L of the form L x, y : ax by c such that True? yes yes P L and Q L . Statement 3: For every set L of the form L x, y : ax by c , there exist at least two IA3) Every line passes through at yes least two distinct points. distinct pairs P = (x1,y1) and Q = (x2,y2), such that P L and Q L . The three statements are true in Analytic Geometry. (We know the statements are true because of the behavior of solutions of linear equations in two unknowns x and y.) So the usual Analytic Geometry of the (x,y) plane is a model of Incidence Geometry. And notice that it is a model whose set of points is an infinite set. 4.3. A Model of Incidence Geometry Involving the Poincare Disk Our third interpretation of Incidence Geometry is another Analytic Geometry. That is, points will be represented in some way by real numbers and lines will be certain kinds of sets of points. Page 53 of 150 Before introducing the interpretation, we need to introduce some terms pertaining to the unit circle and circles that intersect it in a certain way. The first of these is called the Hpoint. y (0,1) Definition 25 Hpoint word: Hpoint meaning: an x, y pair in the interior of the unit circle (1,0) (-1,0) x The interior of the unit circle is the shaded region shown in the figure at right. The boundary pairs, including the four that are labeled, are NOT part of the interior, so they are not Hpoints. (0,-1) We need to discuss something called ―orthogonal circles‖. To do that, we must first introduce the idea of the angle of intersection of curvy objects. You are used to the idea of the angles created by two intersecting lines: One simply considers the angle created by a pair of rays that emanate from the intersection point and that lie on the two lines. This same idea can be generalized to the intersection of curvy objects. Given two intersecting objects, there are two straight lines that are tangent to the two objects at their point of intersection. We say that the two curvy objects are perpendicular (at the place where they intercept) if the tangent lines are perpendicular. Now consider two intersecting circles. The circles may intersect in just one point. But notice that if the circles intersect at more than one point, then the angles created at the two intersection points will be congruent, in the sense that when the two angles are placed on top of each other, they match. (To see this in the pictures below, draw a line that connects the centers of two circles that are touching. Fold the paper along the line.) In particular, if the two tangent lines at one intersection point are perpendicular, then the two tangent lines at the other intersection point are also perpendicular. Page 54 of 150 two intersection points with two intersection points with perpendicular tangent lines non-perpendicular tangent lines We will say that a circle is orthogonal to the unit circle if the tangent lines at the point of intersection are perpendicular. Such circles will play a key role in our definition of Hlines. one intersection point Definition 26 Hlines word: Hline meaning: Either of the following particular types of sets of Hpoints A ―straight-looking‖ Hline is the set of Hpoints that lie on a diameter of the unit circle. A ―curved-looking‖ Hline is the set of Hpoints that lie on a circle that is orthogonal to the unit circle. Note that because an Hline must be made up of Hpoints, and all Hpoints lie in the interior of the unit circle, an Hline (of either flavor) therefore does not include the two ―ends‖. We will refer to these as the ―missing ends‖. To indicate that the ends are missing, we will put open circles or arrowheads on the ends of our Hlines. (My typesetting program won’t do open circles, so I’ll use arrowheads.) Shown at right are some examples of Hlines. Two important observation about Hlines The only ones that are straight are the ones that go through the center of the circle. The ones that curve always curve away from the center of the circle. We are now ready for our new interpretation of incidence geometry. It is called the Poincare disk interpretation. . objects in Incidence Geometry objects in the Poincare disk points Hpoints lines Hlines relation in Incidence Geometry relation in the Poincare disk the point lies on the line the Hpoint is an element of the Hline Is this interpretation successful? That is, is it a model? As always, we consider what happens when we translate the axioms of Incidence Geometry into statements about the Poincare Disk. axioms of Incidence Geometry IA1) There exist three distinct noncollinear points. (at least three) statements about the Poincare Disk True? Statement 1: There exist three distinct Hpoints, P, Q, and R such that there is no one Hline L such yes that P L and Q L and R L . Page 55 of 150 IA2) For every pair of distinct points, there exists exactly one line that passes through both points. IA3) Every line passes through at least two distinct points. Statement 2: For every pair of distinct Hpoints, P and Q, there is exactly one Hline L such that P L and Q L . Statement 3: For every Hline L, there exist at least two distinct Hpoints P = (x1,y1) and Q = (x2,y2), such that P L and Q L . yes yes It should come as no surprise that Statements 1 and 3 are true. But what about Statement 2? It turns out that given two Hpoints A and B, there is exactly one Hline containing both points. (We would use the symbol AB to denote the Hline, being sure to remember that the meaning of this symbol in the context of the Poincare disk is different from the meaning of the same symbol in the context of the usual Analytic Geometry of the plane.) This can be shown by some rather messy mathematics involving lines and circles. In class, in the exercises, and in the computer lab, you will explore the Poincare disk by drawing examples of Hpoints and Hlines. There, you will hopefully convince yourself that the Poincare disk interpretation is actually a model for incidence geometry. Your explorations won’t constitute a proof, but they are enough for our purposes. 4.4. Other Interpretations of Incidence Geometry In a class drill, we will investigate a variety of interpretations of Incidence Geometry. Some of the interpretations will involve the sphere. These, we will investigate using the ―Lenart Sphere‖. For each interpretation, the goal will be to determine whether or not the interpretation is successful. That is, to determine whether or not the interpretation is a model. Page 56 of 150 4.5. Exercises Recall that the Poincare disk interpretation of incidence geometry was presented in Section 4.3. [1] For each picture, draw Hline PQ . The center of the circle is marked with a small cross for reference. Q Q P P Q P P Q Page 57 of 150 [2] On the labeled circle below Draw Hline AB . Label this Hline L. Draw Hline PQ . Label this Hline M. Draw Hline PR . Label this Hline N. Observe that Hpoint P is not on Hline L, and that both Hlines M and N pass through P and do not intersect Hline L. [3] On the labeled circle below, draw Hlines AB , BC , and CA . [4] On the labeled circle below, draw Hlines AB , BC , and CA . [5] On the labeled circle below, draw Hlines AB , BC , CA , and CD . C R P Q B A A B Exercise [2] Exercise [3] C B A B D A C Exercise [4] Exercise [5] Page 58 of 150 [6] Below are eight interpretations of Incidence Geometry. For each interpretation, do the following: i) Determine whether or not the interpretation is a model. Explain your answers. ii) If the interpretation is a model, are there any parallel lines in the model? Explain. iii) If the interpretation is a model, what is the answer to THE BIG QUESTION? Interpretation (a): ordered pairs and straight lines in the interior of the unit circle objects in Incidence Geometry objects in Interpretation (a) point an (x,y) pair in interior of unit circle the set L consisting of the portion of an ordinary line that line lies in the interior of the unit circle. relation in Incidence Geometry relation in Interpretation (a) The point lies on the line. The pair (x,y) is an element of the set L. Interpretation (b): ordered pairs and straight lines through the origin objects in Incidence Geometry objects in Interpretation (b) point an (x,y) pair in the plane line an ordinary line L through the origin in the plane. relation in Incidence Geometry relation in Interpretation (b) The point lies on the line. The pair (x,y) is an element of the set L. Interpretation (c): lines and planes in three-space objects in Incidence Geometry point line relation in Incidence Geometry The point lies on the line. objects in Interpretation (c) an ordinary line M in three-space. an ordinary plane T in three-space. relation in Interpretation (c) The set M is a subset of the set T. Interpretation (d): lines through the origin and planes through the origin in three-space objects in Incidence Geometry objects in Interpretation (d) point an ordinary line M through the origin in three-space. line an ordinary plane T through the origin in three-space. relation in Incidence Geometry relation in Interpretation (c) The point lies on the line. The set M is a subset of the set T. Interpretation (e): Ordered pairs and circles in the plane objects in Incidence Geometry objects in Interpretation (e) point an (x,y) pair in the plane an ordinary circle C in the plane line (just the circle, not the interior) relation in Incidence Geometry relation in Interpretation (e) The point lies on the line. The pair (x,y) is an element of the set C. Page 59 of 150 Interpretation (f): Ordered pairs and circles centered at the origin in the plane objects in Incidence Geometry objects in Interpretation (f) point an (x,y) pair in the plane an ordinary circle C in the plane, centered at the origin line (just the circle, not the interior) relation in Incidence Geometry relation in Interpretation (f) The point lies on the line. The pair (x,y) is an element of the set C. The next two interpretations involve the sphere. By sphere, we mean the surface of a ball of radius 1. That is, The ball is the set B x, y, z : x 2 y 2 z 2 1 . The sphere is the set S x, y, z : x 2 y 2 z 2 1 . Interpretation (g): points and great circles on the sphere objects in Incidence Geometry objects in Interpretation (g) an ordinary point on the sphere point that is, a triple (x,y,z) such that x 2 y 2 z 2 1 a great circle C on the sphere line (that is, a circle of radius 1 on the sphere) relation in Incidence Geometry relation in Interpretation (g) The point lies on the line. The triple (x,y,z) is an element of the set C. Interpretation (h): sets of antipodal points and great circles on the sphere objects in Incidence Geometry objects in Interpretation (h) a set consisting of two antipodal points on the sphere point (that is, a set P x, y, z , x, y, z where (x,y,z) is a point on the sphere) a great circle C on the sphere line (that is, a circle of radius 1 on the sphere) relation in Incidence Geometry relation in Interpretation (h) The point lies on the line. The set P is a subset of the set C. Page 60 of 150 Page 61 of 150 5. Incidence and Betweenness Geometry We shall make it our goal to build an axiom system that prescribes the behavior of points and lines in a way that accurately represents the ―straight line‖ drawings that we have been making all our lives. Remember, though, that in the language of axiom systems, we turn the idea of representation around, and say that we want to the ―straight line‖ interpretation to be a model of our axiom system. And because our ―straight line‖ drawings and the usual Analytic Geometry of the x-y plane behave the same way, we would expect that the usual Analytic Geometry would be a model. But we want those to be the only models of our axiom system. Or rather, we want any other models of the axiom system to be isomorphic to the Analytic Geometry Model. That is, we would like our axiom system to be complete. Complete Axiom System our goal: model Straight Line Drawings model isomorphic Ordinary Analytic Geometry of the x-y plane In the previous chapter, we saw that the ―straight line‖ interpretation was a model of Incidence Geometry. But we studied a number of other models as well, and we saw that they varied widely in their properties. As a result, we were able to conclude that the axioms of Incidence Geometry are not complete. We need to consider what axioms to add to list of Incidence Geometry Axioms in order to make a complete axiom system. We will do this in stages. Notice that many of the models of Incidence Geometry were finite geometries. Our ―straight line‖ drawings clearly have an infinite number of points. In this chapter, we will take the first step towards our goal of building a complete axiom set by adding a set of axioms that insure that any model will have an infinite number of points. (The axioms specify other kinds of behavior, as well.) Much of the material is follows Marvin Greenberg’s book, Euclidean and NonEuclidean Geometry. The axioms that we will add are called The Betweenness Axioms. Before we can introduce them, however, we need to quickly learn about a new kind of relation. 5.1. Binary and Ternary Relations on a Set In Section 1.2.3, you were introduced to the idea of a relation on a set. You learned that to say that R is a relation on a set A means that R is a subset of the Cartesian product A A . Such a relation is more properly called a binary relation, because of the two ―factors‖ of A in the Cartesian product. (Notice that this terminology was introduced in Definition 5.) So for instance, the ―less than‖ symbol, <, represents a binary relation on the set of real numbers. When one uses the symbol, it goes between two real numbers, such as 5 < 7. A binary relation of more interest to us in our study of geometry is the ―parallel‖ relation on the set of lines, studied in Section 1.3 Page 62 of 150 Exercise 8).The idea of a relation on a set A generalizes to other numbers of ―factors‖ of A, as well. In particular, we will be interested in what are called ternary relations. Definition 27 Ternary Relation on a Set Words: R is a ternary relation on A. Usage: A is a set. Meaning: R is a subset of A A A . Equivalent meaning in symbols: R A A A For example, you are undoubtedly familiar with the following ternary relation on the set of real numbers. Definition 28 The Betweenness Relation on the set of real numbers Words: ―x is between y and z.‖ Usage: x, y, and z are real numbers. Meaning: ―x < y < z or z < y < x.‖ Remark: This is a ternary relation on the set of real numbers. Warning: This is NOT the same as betweenness for points, discussed in the next section. When we introduce the batch of axioms that will be appended to the list of Incidence Axioms, we will introduce a new ternary relation that will also be called The Betweenness Relation, but it will be a ternary relation on the set of points and it will be an undefined relation. 5.2. Introducing Incidence and Betweenness Geometry As mentioned earlier, our first ―upgrade‖ of Incidence Geometry will be to add a batch of axioms that will insure, among other things, that any model will have an infinite number of points. In addition to the undefined relation “the point lies on the line”, this axiom system has another undefined relation, spoken “point B is between point A and point C”. In the jargon of relations, we would say that this is a ternary relation on the set of points. Notice that the first batch of axioms is The Incidence Axioms, which dictate the behavior of the primitive relation ―the point lies on the line‖. The second batch of axioms are called The Betweenness Axioms; they dictate the behavior of the primitive relation “point B is between point A and point C”. This axiom system does not have a name in the literature, but we will refer to it as Incidence and Betweenness Geometry. It is presented on the next page. Notice that in addition to the primitive terms and relations, the Incidence and Betweenness axioms also use the defined relation ―passes through‖ and the defined property ―collinear‖, both of which were introduced in the previous chapter. But there are new defined terms as well, in axiom BA4: ―same side‖ and ―opposite side‖. You might wonder why I presented the axiom system without first explaining what these new words mean. The reason is that the meaning of the new words can only be understood after proving some theorems involving the earlier axioms. In the next section, we will study those theorems and others of Incidence and Betweenness Geometry, defining new terminology along the way. Page 63 of 150 Axiom System: Incidence and Betweenness Geometry (abbreviated I&B Geometry) Primitive Terms: point line Primitive Relations: relation from the set of points to the set of lines, spoken ―the point lies on the line.‖ ternary relation on the set of points, spoken ―point B is between point A and point C‖, denoted A*B*C. Defined stuff: ―passes through‖, ―intersect‖, ―parallel‖, ―concurrent‖, ―collinear‖, ―same side‖, ―opposite side‖ Axioms: Incidence Axioms IA1: There exist three distinct non-collinear points. (at least three) IA2: For every pair of distinct points, there exists exactly one line that passes through both points. IA3: Every line passes through at least two distinct points. Betweenness Axioms BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A. BA2: If A, B, and C are three distinct points lying on the same line, then exactly one of the points is between the other two. BA3: If B and D are distinct points, and L is the unique line that passes through both points, then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E. BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then (i) If A and B are on the same side of L and B and C are on the same side of L, then A and C are on the same side of L. (ii) If A and B are on opposite sides of L and B and C are on opposite sides of L, then A and C are on the same side of L. It should be noted that any theorem of Incidence Geometry will also be a theorem of Incidence and Betweenness Geometry, because the proofs will still work. So for instance, we know that in Incidence and Betweenness Geometry, if two distinct lines intersect, they only intersect in one point. Because of this, the theorem numbering that began in Section 3.5 will be continued as we continue to build on the axioms of Incidence Geometry in this section and throughout the rest of this book. Much of what we will discuss about Incidence & Betweenness Geometry will require the definition of additional terminology. But there are some small theorems that we can discuss that do not require any words that we don’t already know. Here is one, our first example of a theorem for Incidence and Betweenness geometry. (Note that the theorem number continues the numbering that began in Section 3.5.) Theorem 6 In I&B Geometry, if A*B*C and A*C*D then A, B, C, D are distinct and collinear. (You will justify the steps in a proof of this theorem in a class drill.) Page 64 of 150 But as I said, most of what we will discuss in this chapter will require new terms. In the next section, we introduce line segments and rays, and prove some theorems about their basic properties. 5.3. Line Segments and Rays Our first three definitions make reference only to concepts from Incidence Geometry. Definition 29 symbol for a line symbol: AB meaning: the (unique) line that passes through points A and B It is worth reminding ourselves of the extent to which points and lines are undefined in our geometry. In many definitions of geometries, including some of the models that we study in our course, a line will be defined as a particular kind of set of points. But that is not the case in our geometry. Lines are not sets of points; lines are undefined things. But it will often make sense to refer to the set of points that lie on a line, and we should have a symbol for that. Here it is. Definition 30 the set of points that lie on a line symbol: AB meaning: the set of points that lie on line AB Again, in our geometry, a line is not the same thing as the set of points that lie on the line. While we’re on the subject of things that are sets of points, we should introduce the definition of the plane. Definition 31 the plane meaning: the set of all points In this course, we are studying ―plane geometry‖. Our models are the straight-line drawings that we have been making all of our lives, and the usual analytic geometry of the x-y plane. Both of these models are 2-dimensional things. The axiom system that we develop will be for plane geometry. In an axiom system plane geometry, points and lines are undefined, and the plane is defined to be the set of all points. This is in contrast to what we could call ―space geometry‖, analogous to 3-dimensional sculptures. In space geometry, points, lines, and planes are all undefined, and space is defined to be the set of all points. We are not studying the geometry of space. So to reiterate, in our geometry points and lines are undefined, and the plane is defined to be the set of all points. Our next three definitions refer to the Betweenness relation; they could not have been introduced in the previous chapter. The terms they introduce are meant to be analogous to the identical terms from the world of straight line drawings. Definition 32 line segment, endpoints of a line segment symbol: AB Page 65 of 150 spoken: ―line segment A, B‖, or ―segment A, B‖ usage: A and B are points. meaning: the set AB A C : A * C * B B additional terminology: points A and B are called endpoints of segment AB . Definition 33 ray, endpoint of a ray symbol: AB spoken: ―ray A, B‖ usage: A and B are points. meaning: the set AB A C : A * C * B B D : A * B * D additional terminology: Point A is called the endpoint of ray AB . We say that ray AB emanates from point A. Definition 34 opposite rays words: BA and BC are opposite rays meaning: A*B*C Here are a few simple theorems that refer to some of these new terms. Theorem 7 In I&B Geometry, every ray has an opposite ray. The proof of this theorem is short and gives a taste of the combined use of axioms and definitions. See if you can justify the steps: Proof of Theorem 7 1) Let QR be a ray. 2) There exists a points P such that P*Q*R (by ______________________________) 3) Rays QP and QR are opposite rays. (by _________________________________) End of Proof Theorem 8 In I&B Geometry, for any two distinct points A and B, AB BA AB . (The proof of this theorem is not particularly hard, but it is not particularly illuminating, either. So we will omit the proof and assume the theorem as given.) Theorem 9 In I&B Geometry, for any two distinct points A and B, AB BA AB . (The proof of this theorem is hard and not particularly illuminating. So we will omit the proof and assume the theorem as given.) 5.4. Plane Separation Keep in mind that with the axioms, we are trying to prescribe behavior for our points and lines (undefined terms) that matches behavior that we have observed in our ―straight line‖ drawings. One thing that we observe in our drawings is that a drawn line splits the plane of the paper into two halves. This is referred to as plane separation. The role of Axiom BA4 in I&B geometry is Page 66 of 150 to insure that our undefined points and lines behave the same way. Our next definitions introduce new terms that are used in the Betweeness Axiom BA4. If we wanted to present the axioms and definitions in logical, chronological order, we would put these definitions after axiom BA3, before axiom BA4. Definition 35 same side words: ―A and B are on the same side of L.‖ usage: A and B are points and L is a line that does not pass through either point. meaning: either A B or ( A B and line segment AB does not intersect line L.) Definition 36 opposite side words: ―A and B are on the opposite side of L.‖ usage: A and B are points and L is a line that does not pass through either point. meaning: A B and line segment AB does intersect line L. With these most recent definitions, we are equipped to understand the language of Axiom BA4. (Go back and read that axiom now.) In Greenberg’s book, the following theorem is called a corollary of Axiom BA4. That word can mean a number of different things. In this case, it could be taken to mean that the theorem follows from Axiom BA4 with a very small proof. One simple proof of the theorem uses the technique of proof by contradiction. Theorem 10 In I&B Geometry, if A and B are on opposite sides of line L, and B and C are on the same side of L, then A and C are on opposite sides of L. (You will justify the steps in a proof of this theorem in the exercises.) Notice that Axiom BA4 does not seem to say anything about halves of the plane. The beauty of the axiom, however, is that it doesn’t have to: the concept of half planes and their behavior can be left to definitions and theorems, as follows. Definition 37 half plane words: half-plane bounded by L, containing point A. symbol: symbol: HA usage: L is a line and A is a point not on L. meaning: the set of points that are on the same side of L as point A. Theorem 11 In I&B Geometry, every line L partitions the plane into three sets: (1) the set of points that lie on L, (2) a half plane, and (3) a second half-plane. In other words, every point of the plane either lies on L or is an element of one (not both) of the half planes. It is difficult to write a proof of Theorem 11 that shows enough detail to be rigorous but not so many that it becomes difficult to read. A proof that I have written uses the method of proof by division into cases. You will justify the steps of the proof in a class drill. Page 67 of 150 5.5. Line Separation Just as we are familiar with the way a drawn line splits the plane in two, we are also familiar with the way a dot on a drawn line splits the line in two. This is referred to as line separation. The Incidence & Betweenness axioms can be used to prove that our undefined points and lines have this same behavior. But by now, you are certainly getting the idea that it can be very tedious to prove even simple behavior of points and lines using only the axioms. It turns out that the proof of the line separation property is rather difficult. Its proof refers to two preliminary theorems whose proofs are also rather hard. I will state the two preliminary theorems here, along with the Line Separation theorem. You will prove only the middle one. Theorem 12 In I&B Geometry, if A*B*C and A*C*D, then B*C*D and A*B*D. (This theorem will not be proved in this course. We will assume it as given.) Theorem 13 In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D. (You will justify the proof of this theorem in the exercises.) Theorem 14 (Line Separation) In I&B Geometry, if A, B, C, and D are collinear points such that A*B*C, and D B , then either D AB or D AC , but not both. (This theorem will not be proved in this course. We will assume it as given.) Another obvious behavior of drawn points and lines is that if a line goes into one side of a triangle, it must come out through another side. We have not formally introduced triangles yet, but even so, this behavior of points and lines can be addressed, in what is called Pasch’s Theorem. The proof of the theorem uses the plane separation property, Theorem 11.) Theorem 15 (Pasch’s Theorem) In I&B Geometry, if A, B,and C are non-collinear points and line L intersects segment AB at a point between A and B, then L also intersects either segment AC or segment BC . Furthermore, if C does not lie on L, then L does not intersect both segment AC and segment BC .. (You will justify a proof of this theorem in a class drill.) Following are two more theorems that demonstrate that our undefined points and lines behave in the same way that our drawn points and lines behave. Theorem 16 In I&B Geometry, if A*B*C then AC AB BC and B is the only point common to segments AB and BC . (This theorem will not be proved in this course. We will assume it as given.) Theorem 17 In I&B Geometry, if A*B*C then B BA BC and AB AC . (This theorem will not be proved in this course. We will assume it as given.) Theorem 18 In I&B Geometry, if point A lies on line L and point B does not lie on line L, then every point of ray AB except point A is on the same side of L as B. (You will justify the steps in a proof of this theorem in the exercises.) Page 68 of 150 5.6. Angles and Triangles With the notion of rays and line segments, we are able to define angles and triangles. In this section, we will define them and explore some of their simpler behavior. Definition 38 angle symbol: ABC usage: A, B, and C are non-collinear points meaning: BA BC additional terminology: point B is called the vertex of ABC , rays BA and BC are called the sides. observations: because BA BC BC BA , the symbols ABC and CBA represent the same angle. Notice that an angle is a set of points, because it is defined as the union of two rays and rays are sets of points. Furthermore, note that as sets of points, BA BC BC BA , so the symbols ABC and CBA represent the same angle. Definition 39 supplementary angles words: supplementary angles meaning: two angles that share a common side and whose other sides are opposite rays. We won’t use the idea of supplementary angles in this chapter. I put the definition here to emphasize that the notion of supplementary angles does not depend on any concept of angle congruence or angle measure. This is worth restating explicitly: the definition of supplementary angles does not define them as two angles whose measures add up to 180 ! We don’t get to the notion of angle congruence until the next chapter, and we won’t get to a notion of angle measure until a later chapter. Definition 40 interior of an angle words: the interior of ABC meaning: the set of points P such that (P is on the same side of line BA as point C) and (P is on the same side of line BC as point A). Our first theorem about the behavior of angles states something that is fairly easy to visualize and not difficult to prove. Theorem 19 In I&B Geometry, given BAC and point D lying on BC , point D is in the interior of BAC if and only if B*D*C. (You will justify the steps in a proof of this theorem in a class drill.) Page 69 of 150 The next theorem about the behavior of angles also states something that is fairly easy to visualize. But the proof of the theorem is messy and not particularly illuminating, so we will omit the proof Theorem 20 In I&B Geometry, given angle BAC ; point D in the interior of BAC , and point E such that C*A*E, the following three statements are all true: (1) Every point on ray AD except A is in the interior of angle BAC . (2) No point on the ray opposite to AD is in the interior of angle BAC . (3) Point B is in the interior of angle DAE . (We won’t prove this theorem in this class, but will take it as a given.) Remember that in Incidence and Betweenness Geometry, the betweenness relation for points is a primitive (undefined) relation. What is interesting is that there is also a betweenness relation for rays and that it is not a primitive relation: it is defined in terms of other stuff. The definition follows. Definition 41 ray between two other rays words: ray BD is between BA and BC . meaning: Point D is in the interior of ABC . The next theorem about the behavior of angles is famous enough that it shows up in some elementary Geometry books. Theorem 21 (The Crossbar Theorem) In I&B Geometry, if ray AD is between rays AB and AC then ray AD intersects segment BC . (You will justify the proof of this theorem in the exercises.) Our final three definitions for this section have to do with triangles. Definition 42 triangle symbol: ABC spoken: triangle A, B, C usage: A, B, and C are non-collinear meaning: the set AB BC CA additional terminology: o The points A, B, C are called the vertices of the triangle. o The segments AB, BC, CA are called the sides of the triangle. o Side BC is said to be opposite vertex A. Similarly for the other sides. o The angles ABC , BCA , and CAB are called the angles of the triangle. o The angle ABC is called angle B when there is no chance of this causing confusion. Similarly for the other angles. Definition 43 interior of a triangle words: the interior of ABC Page 70 of 150 meaning: the set of points P such that (P is on the same side of line BA as point C) and (P is on the same side of line BC as point A) and (P is on the same side of line AC as point B) Definition 44 exterior of a triangle meaning: the set of points Q that are neither an element of the triangle itself, nor of the interior of the triangle. Notice that a triangle and the interior of a triangle and the exterior of a triangle are all defined as sets of points. The final three theorems of the section have to do with basic facts about the intersection of rays and lines with triangles. All three theorems are easy to visualize. Theorem 22 In I&B Geometry, if ray r emanates from an exterior point of triangle ABC and intersects side AB in a point between A and B, then ray r also intersects side AC or BC . (We won’t prove this theorem in this class, but will take it as a given.) Theorem 23 In I&B Geometry, if a ray emanates from an interior point of a triangle, then it intersects one of the sides, and if it does not pass through a vertex, it intersects only one side. (We won’t prove this theorem in this class, but will take it as a given.) Theorem 24 In I&B Geometry, a line cannot be contained in the interior of a triangle. (You will prove this theorem in the exercises.) Page 71 of 150 5.7. Exercises [1] Recall the statement of Theorem 10: In I&B Geometry, if A and B are on opposite sides of line L, and B and C are on the same side of L, then A and C are on opposite sides of L. Justify the steps in the follwing proof: Proof. 1. Let A, B, and C be any points such that A and B are on opposite sides of line L, and B and C are on the same side of L. 2. Suppose that A and C are not on opposite sides of L. (_______________________) 3. None of the points A, B, or C lie on line L. (___________________________) 4. Points A and C are on the same side of L. (___________________________________) 5. Points A and B are on the same side of L. (___________________________________) 6. But (____________________) contradicts (__________________) Therefore, our assumption must be wrong. Conclude that A and C must be on opposite sides of L. End of Proof [2] Recall the statement of Theorem 13: In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D Justify the steps in the following proof Step 1: Show that A*C*D. (Note that we will create a line through C) 1. Suppose that A*B*C and B*C*D. 2. A, B, C, and D are distinct collinear points. (by ________________) Let L be the line that passes through all four points. 3. There exists a point E not on the line L. (by ____________________) 4. point C is an element of segment BD (because___________) and point C also lies on line EC and. So segment BD intersects line EC . 5. Points B and D are on opposite sides of line EC . (by ______________________) 6. Point C is the only point of intersection of line EC and line L. (by _________________) 7. Point C is not an element of segment AB , because of the fact that C ≠ A and C ≠ B (by _______) and and the fact that C is not between A & B. (by _______________________) 8. Segment AB does not intersect line EC . (by ______________) 9. Points A and B are on the same side of line EC . (by ___________________________) 10. Points A and D are on opposite sides of line EC . (by ____________________________) 11. Segment AD intersects line EC at a point between A and D. (by ___________________) 12. The point of intersection of segment AD and line EC must be point C. (by _________) 13. A*C*D. (by ____________________________) Step 2: Use a previous theorem to show that A*B*D 14. A*B*D (by ____________________________) End of Proof Page 72 of 150 [3] Recall the statement of Theorem 18: In I&B Geometry, if point A lies on line L and point B does not lie on line L, then every point of ray AB except point A is on the same side of L as B. Justify the steps in the following proof: 1. Suppose that point A lies on line L and point B does not lie on line L, and let P be any point such that P AB and P A . 2. Either A*P*B or P = B or A*B*P. (by ________________________________) 3. Assume that P is on the opposite side of line L from point B. (________________) 4. Segment PB intersects L at a point between P and B. (by ________________________) 5. Point A is an intersection point of line L and ray AB . (by given information) 6. Line L and line AB can’t have any other intersection points. (by ___________________) 7. Point A must be the intersection point of segment PB and line L. (by 4 and 6 and fact that points P and B lie on line AB .) 8. Point A is between points P and B. That is, P*A*B. (by __________________________) 9. Step 8 contradicts step (___________). So our assumption in step 3 was wrong. Point P cannot be on the opposite side of line L from point B. Therefore, P must be on the same side of L as B. End of Proof [4] Recall the statement of Theorem 21 (The Crossbar Theorem): In I&B Geometry, if ray AD is between rays AB and AC then ray AD intersects segment BC . Justify the steps in the following proof: 1. Suppose ray AD is between rays AB and AC . 2. Let E be a point such that C*A*E (we know such a point exists by _____________) 3. Segment CE . intersects AD at point A, which is between C and E (by __________). 4. Points E and C are on opposite sides of AD . (by ____________________________) 5. Point B is in the interior of DAE (by Theorem _________) 6. Point B is on the same side of AD as E is. (by definition of __________________) 7. Point B is on the opposite side of AD from point C. (by 4, 6, and Theorem ______) 8. Segment BC intersects AD at a point between points B and C. (by ______________) Let G be the point of intersection of segment BC and line AD . So B*G*C. 9. Point G is in the interior of CAB . (by step ____ and Theorem _______) 10. Point G is on ray AD , and not on the ―opposite ray‖. (by Theorem _______) 11. Ray AD intersects segment BC . (by steps _____ and _______) End of Proof [5] Prove Theorem 24: In I&B Geometry, a line cannot be contained in the interior of a triangle. Hint: Do a proof by contradiction. Suppose that a line L is contained in the interior of a triangle. Show that it is possible to reach a contradiction of an earlier theorem. Page 73 of 150 6. Neutral Geometry I 6.1. The need for a larger axiom system: Introducing Neutral Geometry The Incidence and Betweenness Axioms discussed in previous chapters guarantee that our undefined points and lines will have some of the behavior that we are accustomed to in our drawings of points and straight lines. But there are many important aspects of the behavior of our drawings that are not addressed in those axioms. One thing that we can observe about pairs of drawings is whether or not one drawing can be slid on top of another drawing in a way that the two drawings ―fit‖. If they do fit, the two drawings are called congruent. Using fancy jargon, we could say that congruence is a binary relation on the set of drawings. We would like to have a comparable notion of congruence for our primitive points and lines, and for triangles made from them. For that, we will introduce congruence relations into our axiom system. Recall that in the previous chapter, we introduced the notion of betweenness of points as an undefined relation on the set of points, and we included a set of betweenness axioms that specified how that undefined relation behaved. But later, we were able to introduce a notion of betweenness of rays that was described by a definition. The behavior of that defined relation was described in a few theorems. In this chapter, we will see the congruence of line segments and the congruence of angles introduced as undefined relations that in our axiom system, and there will be a number of congruence axioms that will specify how these undefined relations behave. But we will be able to make an actual definition for congruence of triangles. One aspect of the behavior of the congruence of triangles will be given by an axiom, but all other aspects will be described by theorems. In addition to the undefined congruence relations and the congruence axioms, we will also add two axioms under the heading Axioms of Continuity. These two axioms clearly describe behavior that we are familiar with in drawings. It probably won’t be obvious to the reader why these statements need to be included as axioms—that is, why they can’t simply be proven as theorems that are a consequence of the other axioms—but it is a fact that they are independent of the other axioms. So, if we want these two statements to be true in our abstract geometry, we need to include them as axioms. With the resulting collection of axioms—the incidence, betweenness, congruence, and continuity—we have specified an enormous amount about the behavior of our abstract geometry. What is most fascinating about this collection of axioms, however, is what they do not specify. We will see that our ―straight line‖ drawings are not the only model of this geometry: the drawings of Hpoints and Hlines in the Poincare disk are also a model. It turns out that these are the only two kinds of models of our axiom system that exist, but they are two non-isomorphic models, all the same. In other words, this axiom system is not complete. The axiom system is often referred to as Neutral Geometry. It is presented on the following page. Page 74 of 150 Axiom System: Neutral Geometry Primitive Terms: point, line Primitive Relations: relation from the set of points to the set of lines, spoken ―the point lies on the line.‖ ternary relation on the set of points, spoken ―point B is between point A and point C‖, denoted A*B*C. binary relation on the set of line segments, spoken ―segment AB is congruent to segment CD‖, denoted AB CD . binary relation on the set of angles, spoken ―angle A, B, C is congruent to angle D, E, F‖, denoted ABC DEF . Incidence Axioms IA1: There exist three distinct non-collinear points. (at least three) IA2: For every pair of distinct points, there is exactly one line that passes through both points. IA3: Every line passes through at least two distinct points. Betweenness Axioms BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A. BA2: If A, B, and C are three distinct points lying on the same line, then exactly one of the points is between the other two. BA3: If B and D are distinct points, and L is the unique line that passes through both points, then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E. BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then (i) If A and B are on the same side of L and B and C are on the same side of L, then A and C are on the same side of L. (ii) If A and B are on opposite sides of L and B and C are on opposite sides of L, then A and C are on the same side of L. Congruence Axioms CA1: The congruence relation on the set of line segments is an equivalence relation CA2: (segment construction axiom) For any segment AB and ray PQ , there exists a unique point R on PQ such that PR AB . CA3: (segment addition axiom) If A*B*C, A'*B'*C', AB AB , and BC BC , then AC AC . CA4: The congruence relation on the set of angles is an equivalence relation CA5: (angle construction axiom) Given an angle BAC , distinct points A' and B', and a choice of one of the two half-planes bounded by the line AB , there exists a unique ray AC such that C' lies in the chosen half plane and BAC BAC . CA6: (SAS axiom) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Axioms of Continuity Archimedes Axiom: Given any segment CD and ray AB , there exists some positive integer n and a set of points P0 , P1 , , Pn on AB such that P0 A ; for all k, Pk Pk 1 CD ; and either Pn B or ( A * Pn1 * B and A * B * Pn ). Circular Continuity Axiom: If a circle intersects both the interior and exterior of another circle, then the two circles intersect in exactly two points. Page 75 of 150 6.2. Line Segment Subtraction and the Ordering of Segments Notice that numbers have played very little role in our geometry so far. In particular, we have not introduced the notion of the length of a line segment. In Incidence & Betweenness Geometry, we found that the axioms of betweenness can be used to prove many things about points and lines that we might have thought could only be proved by referring to length. In Neutral Geometry, we have not said that congruent segments are ones that have the same length. Indeed, we have been very clear about the fact that line segment congruence is an undefined relation. But as was the case with Incidence & Betweenness Geometry, we will find that it is possible to do things in Neutral Geometry that we might have thought impossible without the notion of length. We are able to do this because of the Segment Construction Axiom, CA2, and the Segment Addition Axiom, CA3. These axioms are statements that would automatically be true in a geometry with a notion of length. But because we have no notion of length in our geometry, the statements must be made axioms if we want them to be true. What is remarkable, though, is that there are many other statements about segments that would automatically be true in a geometry with a notion of length, but we don’t have to make them all axioms in order for them to be true. It turns out that with just axioms CA1, CA2, and CA3, we can prove many other statements as theorems. For example, the following theorem is a statement very similar to the segment addition axiom, CA3. But this statement does not need to be included in the list of axioms. (That’s why it is here, listed as a theorem.) The proof is fairly simple. You will provide the justifications in a homework exercise. Theorem 25 (segment subtraction) In Neutral Geometry, if A*B*C and D*E*F and AB DE and AC DF , then BC EF . The following theorem proves aanother claim that would be obvious for drawn line segments, but which is a little tricky to prove for abstract segments. You will justify the steps in the proof in a class drill. Theorem 26 In Neutral Geometry, if AC DF and A*B*C, then there exists a unique point E such that D*E*F and AB DE . The following definition is very important because introduces a way to compare two line segments without a having a notion of length. Definition 45 the order relation on the set of line segments Symbol: AB CD Spoken: “Segment A,B is less than segment C,D.” Meaning: There exists a point E between C and D such that AB CE . Remark: The order relation is a binary relation on the set of line segments. It turns out that the order relation for line segments has behavior that is entirely analogous to the behavior of line segment length, even though the order relation does not use the notion of length. Page 76 of 150 That behavior is spelled out in the following theorem. You will justify the steps in a proof of this theorem in a homework exercise. Theorem 27 Facts about the order relation on the set of line segments Given: Neutral Geometry; line segments AB , CD , and EF . Claim (a) (trichotomy) Exactly one of the following is true: AB CD , AB CD , or CD AB . (b) If AB CD and CD EF , then AB EF . (c) (transitivity)If AB CD and CD EF , then AB EF . 6.3. Triangle Congruence and its Role in the Neutral Geometry Axioms As mentioned in Section 6.1, congruence of line segments and congruence of angles are undefined relations in Neutral Geometry, but congruence of triangles is a defined relation. The definition of triangle congruence—indeed the definition of any term or relation—is as important in the axiom system as the primitive terms, primitive relations, and axioms. But the definitions are traditionally left out of the presentation of the axiom system, because they would take up an enormous amount of space. In this section, our goal is simply to present the definition of the congruence relation on the set of triangles, and to understand the significance of the one Neutral Geometry axiom that pertains to triangle congruence. We will focus on the terminology and will not prove any theorems. Even so, the section is quite lengthy. A remark: We won’t prove any theorems about triangles until the next chapter. The current chapter deals only with theorems about lines, rays, line segments, and angles. It is natural to wonder, then, why the current section on triangles is even included in this chapter. The reason is that some of the theorems about lines, rays, line segments, and angles that we prove in this chapter use triangles and triangle congruence. 6.3.1. Correspondences between parts of triangles The term correspondence is used in almost any discussion of triangle congruence, and in almost any theorem about triangle congruence. So we start our presentation of the concept of triangle congruence by defining what we mean by a correspondence. Definition 46 ―function‖, ―domain‖, ―codomain‖, ―image‖, ―machine diagram‖; ―correspondence‖ Symbol: f : A B Spoken: “ f is a function that maps A to B ” Usage: A and B are sets. Set A is called the domain and set B is called the codomain. Meaning: f is a machine that takes an element of set A as input and produces an element of set B as output. More notation: If an element a A is used as the input to the function f , then the symbol f (a ) is used to denote the corresponding output. The output f (a ) is called the image of a under the map f . Page 77 of 150 Machine Diagram: a f input f a output Domain: Codomain: the set A the set B Additional notation: If f is both one-to-one and onto (that is, if f is a bijection), then the symbol f : A B will be used. In this case, f is called a correspondence between the sets A and B. Correspondences play a key role in the concept of triangle similarity and congruence, and they will also play a key role in the concept of polygon similarity and congruence, so we should do a few examples to get more familiar with them. Examples 1) Let f : be the cubing function, f x x3 . Then f is one-to-one and onto, so we could say that f is a correspondence, and we would write f : . 2) Let S1 A, B, C , D, E and S 2 L, M , N , O, P . Define a function f : S1 S2 by this picture: S1 S2 A L f B M C N D O E P Then we would say that f is a correspondence, and we would write f : S1 S2 . 3) For the same example as above, we could display the correspondence more concisely: A N BP CL DM EO This takes up much less space, and is faster to write, than the picture. However, notice that this way of displaying the correspondence still uses a lot of space. 4) There is an even more concise way to display the correspondence from the above example. To understand the notation, though, we should first recall some conventions about brackets and parentheses. When displaying sets (where order is not important), curly brackets are used. For example, S1 A, B, C , D, E C , A, E , D, B . But when displaying an ordered list, parentheses are used instead. So whereas the symbol A, B denotes the same set as the symbol B, A , the symbol A, B denotes a different ordered pair from the symbol B, A . With that notation in mind, we will use the symbol below to denote the function f described Page 78 of 150 in the previous examples. A, B, C, D, E N , P, L, M , O The parentheses indicate that the order of the elements is important, and the double arrow symbol indicates that there is a correspondence between the lists. Notice that this way of displaying the function is not as clear as the one in the previous example, but it takes up much less space. Definition 47 Correspondence between vertices of two triangles Words: “f is a correspondence between the vertices of triangles ABC and DEF .” Meaning: f is a one-to-one, onto function with domain A, B, C and codomain D, E , F . Examples of correspondences between the vertices of 1) A, B, C D, E, F 2) 3) 4) ABC and DEF . A, B, C D, F , E B, A, C D, E, F B, C, A D, F , E Notice that the third and fourth examples are actually the same. Each could be illustrated by the figure shown at right. A B C f D E F If a correspondence between the vertices of two triangles has been given, then there is an automatic correspondence between any other geometric items that are defined purely in terms of those vertices. For example, suppose that we are given the following correspondence B, A, C D, E, F between the vertices of ABC and DEF . For clarity, we can display the correspondence vertically. There is a correspondence between the sides of triangle ABC and the sides of DEF , and a correspondence between the angles of triangle ABC and the angles of DEF , since those items are defined only in terms of the vertices. BD the given correspondence between AE vertices of ABC and vertices of DEF CF BA DE AC EF the automatic correspondence between CB FD BAC DEF parts of ABC and parts of DEF ACB EFD CBA FDE Based on the ideas of this discussion, we make the following definition. Definition 48 Corresponding parts of two triangles Words: Corresponding parts of triangles ABC and DEF . Usage: A correspondence between the vertices of ABC and DEF has been given. Page 79 of 150 Meaning: As discussed above, there is an automatic correspondence between the sides of triangles ABC and the sides of DEF , and also between the angle of triangles ABC and the angles of DEF . Suppose the correspondence between vertices were B, A, C D, E, F . Corresponding parts would be pairs such as the pair of sides, AC EF , or the pair of angles, ACB EFD . 6.3.2. Triangle congruence In Neutral Geometry, line segment congruence is an undefined binary relation on the set of line segments, and angle congruence is an undefined binary relation on the set of angles. But triangle congruence is a defined binary relation on the set of triangles. Here is the definition. Definition 49 Triangle congruence Symbol: ABC DEF Words: “ ABC is congruent to DEF .” Meaning: “There is a correspondence between the vertices of the two triangles such that corresponding parts of the triangles are congruent.” Remark: Triangle congruence is a defined binary relation on the set of triangles. Additional terminology: If a correspondence between vertices of two triangles has the property that corresponding parts are congruent, then the correspondence is called a congruence. Remark: Many students remember the sentence ―Corresponding parts of congruent triangles are congruent‖ from their high school geometry course. The acronym is, of course, ―CPCTC‖. We see now that ―CPCTC‖ is really a summary of the definition of triangle congruence. That is, to say that two triangles are congruent is the same as saying that corresponding parts of those two triangles are congruent. This is worth restating: CPCTC is not an axiom and it is not a theorem; it is merely shorthand for the definition of triangle congruence. It is important to discuss notation at this point, because there is an abuse of notation common to many geometry books. In most books, the symbol ABC DEF is used to mean not only that triangle ABC is congruent to DEF , but also that the correspondence between the vertices is given by A, B, C D, E, F . That is, in most books, the order of the vertices in the correspondence is assumed to be the same as the order of the vertices in the symbol ABC DEF . This can cause problems. To see why, notice C that given any three non-collinear points, A, B, and C, the A B symbols ABC and ACB represent the same triangle. That is because a triangle is defined to be the union of three line segments, and the three line segments that make up ABC are exactly the same as the three line segments that make up ACB . Now consider the correspondence A, B, C A, B, C of vertices of triangles ABC and ACB . Below is a list of the resulting correspondence of parts. A A the given correspondence between BB vertices of ABC and vertices of ACB . C C Page 80 of 150 AB AB BC BC the automatic correspondence between CA CA ABC ABC parts of ABC and parts of ACB . BCA BCA CAB CAB Clearly, each of the pairs of corresponding parts are congruent to each other. Therefore, we would say that the correspondence A, B, C A, B, C is a congruence, and triangle ABC is congruent to ACB . Now consider the correspondence A, B, C A, C, B of vertices of triangles ABC and ACB . Below is a list of the resulting correspondence of parts. A A the given correspondence between B C vertices of ABC and vertices of ACB . CB AB AC BC CB the automatic correspondence between CA BA ABC ACB parts of ABC and parts of ACB . BCA CBA CAB BAC Clearly, the pairs of corresponding parts are not congruent to each other. So the correspondence A, B, C A, C, B is not a congruence. So we see that triangle ABC is congruent to ACB , but that the correspondence that should be used is A, B, C A, B, C , not A, B, C A, C, B . In most books, the symbol ABC ACB is used to mean ―triangle ABC is congruent to ACB , and the correspondence used is A, B, C A, C, B .‖ This is bad, because although the triangles ABC and ACB are congruent, the correspondence that should be used is A, B, C A, B, C , not A, B, C A, C, B . It would be useful to have a symbol that not only indicates congruence but also tells which correspondence was used. But there is not such a symbol, and so we have to make do with the symbols available. When the particular choice of correspondence is an issue, we should write things like ― ABC ACB using the correspondence A, B, C A, B, C ‖, instead of merely writing ― ABC ACB ‖. 6.3.3. Introducing CA6, the Side Angle Side Congruence Axiom As mentioned earlier, we have a notion of congruence for drawn line segments and drawn angles, having to do with sliding one drawn segment or angle on top of another and seeing if they ―fit‖. The primitive relations of line segment congruence and angle congruence in the axiom Page 81 of 150 system of Neutral Geometry are meant to behave in the same way as our drawings. Congruence axioms CA1 through CA5 prescribe much of that behavior. But we can also draw triangles, of course, and we have a notion of congruence for drawn triangles, again having to do with sliding one drawn triangle on top of another and seeing if they ―fit‖. And in our drawings, we know that if enough parts of one drawing ―fit‖ on top of the corresponding parts of another drawing, then all of the other parts will fit, as well. This can be said more precisely. To determine whether or not two drawn triangles fit on top of each other, one would officially have to verify that every pair of corresponding line segments fit on top of each other and also that every pair of corresponding angles fit on top of each other. That is total of six fits that must be checked. But we know that with drawings, one does not really need to check all six fits. Here are four examples of methods that can be used to verify that two drawn triangles are congruent by checking only three fits instead of six: If two sides and the included angle of the first triangle fit on top of the corresponding parts of the second triangle, then all the remaining corresponding parts always fit, as well. If two angles and the included side of the first triangle fit on top of the corresponding parts of the second triangle, then all the remaining corresponding parts always fit, as well. If two angles and some non-included side of the first triangle fit on top of the corresponding parts of the second triangle, then all the remaining corresponding parts always fit, as well. If all three sides the first triangle fit on top of the corresponding parts of the second triangle, then all the remaining corresponding parts always fit, as well. We would like our undefined relations of line segment congruence and angle congruence and our defined relation of triangle congruence to have this same sort of behavior. But if we want them to have that behavior, we must specify it in the Neutral Geometry axioms. One might think that it would be necessary to include four axioms, to guarantee that the four kinds of behavior that we observe in drawn triangles will also be observed in abstract triangles. But the amazing thing is that we don’t need to include four axioms. We can include just one axiom, about just one kind of behavior that we want abstract triangles to have, and then we can prove theorems that show that triangles will also have the other three kinds of desired behavior. Here is the axiom that will be included, along with the three theorems that will be proven later. CA6 (SAS axiom): If there is a correspondence between parts of two triangles such that two sides and the included angle of one triangle are congruent to the corresponding parts of the other triangle, then all remaining corresponding parts are congruent as well, so the triangles are congruent. The ASA Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and the included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Page 82 of 150 The AAS Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and a non-included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. The SSS Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a correspondence between parts of two triangles such that the three sides of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Two of these three theorems, the ASA and SSS congruence theorems, will be proven in this chapter. The third, the AAS congruence theorem, will be proven in the next chapter. 6.4. Right Angles In Section 6.2, we saw that even though we have no notion of line segment length, we are able to have an order relation on the set of line segments. In this section, we will do something comparable with angles. Remember that in our axiom system, we do not have a notion of measure of angles. Even so, we will see that it is possible to define supplemental angles and right angles and to define an order relation on the set of angles. Remember that the term supplementary angle was introduced in Definition 39, in Section 5.6. What was interesting about that definition was that it does not say that supplementary angles are angles whose measures add up to 180. It is worth noting that every angle has a supplementary angle. You will prove this fact in the exercises. In the exercises, you will justify the steps in the proof of the following theorem. Theorem 28 In Neutral Geometry, supplements of congruent angles are congruent. That is, if ABC and CBD are supplementary, and EFG and GFH are supplementary, and ABC EFG , then CBD GFH . Do any of you know why vertical angles are called vertical angles? Maybe we should have prize for the best back story. Regardless of the reason for the name, the definition should be familiar, as is the theorem that follows. You will justify a proof of the theorem in the exercises. Definition 50 vertical angles words: vertical angles Meaning: Two angles that share a vertex and whose sides are opposite rays. That is, two angles that can be labeled ABC and DBE where A*B*D and C*B*E. Theorem 29 In Neutral Geometry, vertical angles are congruent. As mentioned earlier, it is possible to define right angles without any reference to angle measure. This is worth re-stating more explicitly: Right angles are NOT defined to be angles whose Page 83 of 150 measure is 90 ! (We do not even have a notion of angle measure.) The trick is to define right angles in terms of supplementary angles, as follows. Definition 51 right angle words: right angle Meaning: An angle that is congruent to its supplementary angle There are a bunch of obvious sounding statements about right angles that we will have to prove. Some are hard to prove; some are easy. The following one is easy to prove. You will justify a proof of it in the exercises. Theorem 30 In Neutral Geometry, any angle congruent to a right angle is also a right angle. Now that we have definition of right angles, we can define perpendicular lines. But before doing that, we should describe how intersecting lines create angles. Suppose that L and M are distinct lines that intersect at point P. By the Incidence & Betweenness axioms, there are points A and B on line L such that A*P*B, and there are points C and D on line M such that C*P*D. Notice that APC and BPD are vertical angles, and that APD and BPC are vertical angles. So when two distinct lines intersect, two pairs of vertical angles are created. Definition 52 perpendicular lines symbol: L M spoken: L and M are perpendicular Meaning: L and M are distinct, non-parallel lines that create a pair of vertical angles that are right angles. The next two theorems are extremely important in Neutral Geometry. You will justify the steps of the proofs in class drills. Some of the justifications are simple, amounting to a citation of a single axiom, theorem or definition. But some of the justifications will be more complicated. Theorem 31 In Neutral Geometry, for every line L and every point P not on L there exists a line through P perpendicular to L. Theorem 32 In Neutral Geometry, for every line L and every point P on L there exists a unique line through P perpendicular to L. 6.5. Angle Addition and Subtraction, and Ordering of Angles Recall that the notion of line segment addition is specified by an axiom (CA3). But the notions of segment subtraction and ordering of segments were introduced in theorems. In this section, we will see that notions of angle addition, angle subtraction and ordering of angles can all be introduced in theorems; no axioms are necessary. The proofs of the theorems are not terribly hard, but they would not enhance our understanding very much. (They are in the same spirit of mind-numbing attention to detail that we found in the proofs of the theorems about segment subtraction and ordering of line segments.) So we will not discuss the proofs of these theorems, and will just assume them as given. Page 84 of 150 Theorem 33 (angle addition) Given: Neutral Geometry; angle ABC with point D in the interior; angle point H in the interior; DBA HFE ; DBC HFG Claim: ABC EFG EFG with Just as the principle of line segment subtraction followed from the principle of line segment addition, the principle of angle subtraction follows from the principle of angle addition. Theorem 34 (Angle Subtraction). Given: Neutral Geometry; angle ABC with point D in the interior; angle point H in the interior; DBA HFE ; ABC EFG Claim: DBC HFG EFG with The following theorem about rays and angles is analogous to Theorem 26, which was about segments. Theorem 35 (Theorem about Rays and Angles) In Neutral Geometry, if ABC EFG and ray BD is between rays BA and BC , then there is a unique ray FH between rays FE and FG such that ABD EFH . Notice that we have not introduced the notion of the measure of an angle and we have not said that congruent angles are ones that have the same measure. Indeed, we have been very clear about the fact that angle congruence is an undefined relation. But it is possible to compare two angles without any notion of angle measure. We will do this by introducing a new relation on the set of angles. Definition 53 the order relation on the set of angles Symbol: ABC DEF Spoken: “Angle ABC is less than angle DEF .” Meaning: There exists a ray EG between ED and EF such that ABC GEF . Remark: The order relation is a binary relation on the set of angles. The following theorem says that the order relation for line segments has behavior that is entirely analogous to the notion of angle measure, even the order relation does not use the notion of measure. You will prove this theorem in the exercises. Theorem 36 (facts about the order relation on the set of angles) Given: Neutral Geometry; angles P , Q , R . Claim (a) (trichotomy): Exactly one of the following is true: P Q , (b): If P Q and Q R then P R . (c): (transitivity): If P Q and Q R then P R . P Q , or Q P. Page 85 of 150 Now that we have the order relation on the set of angles, it is possible to prove one important remaining theorem about right angles. Theorem 37 (Euclid’s 4th postulate). All right angles are congruent to each other. 6.6. The Alternate Interior Angle Theorem and Some Corollaries In this final section of the chapter, we will study a major theorem called the Alternate Interior Angle Theorem. It is important because it enables us to prove a number of important theorems, including one that gives a partial answer to THE BIG QUESTION. In order to understand the wording of the Alternate Interior Angle Theorem, we need two definitions. Definition 54 ―transversal‖ Words: “Line T is transversal to lines L and M.” Meaning: “T intersects L and M in distinct points.” Definition 55 ―alternate interior angles‖, ―corresponding angles‖ Usage: Lines L, M, and transversal T are given. Labeled points: Let B be the H T intersection of lines T and L, and let E D be the intersection of lines T and M. M E F (By definition of transversal, B and E A B C L are not the same point.) By the G betweenness axioms, there exist points A and C on line L such that A*B*C, points D and F on line M such that D*E*F, and points G and H on line T such that G*B*E and B*E*H. Without loss of generality, we may assume that points D and F are labeled such that it is point D that is on the same side of line BE as point A. Meaning: ABE and FEB is a pair of alternate interior angles. CBE and DEB is a pair of alternate interior angles. ABG and DEG is a pair of corresponding angles. ABH and DEH is a pair of corresponding angles. CBG and FEG is a pair of corresponding angles. CBH and FEH is a pair of corresponding angles. Theorem 38 The Alternate Interior Angle Theorem Given: Neutral Geometry, lines L and M and a transversal T Claim: If a pair of alternate interior angles is congruent, then lines L and M are parallel. It should be pointed out that the name of the Alternate Interior Angle Theorem follows the usual unspoken custom for theorem names. That is, the name of the theorem refers to something that is in the hypotheses of the theorem, not the conclusion. Page 86 of 150 The Alternate Interior Angle Theorem is a major theorem in two senses: its proof is a little tricky, and it can be used to build fairly simple proofs of a bunch of other theorems. Those other theorems could be considered corollaries of this theorem. You will be asked to justify the steps of the proof of the Alternate Interior Angle Theorem in the exercises. The corollaries are listed below; you will be asked to prove them in the exercises. Theorem 39 The Corresponding Angle Theorem Given: Neutral Geometry, lines L and M and a transversal T Claim: If a pair of corresponding angles is congruent, then lines L and M are parallel. Theorem 40 In Neutral Geometry, two distinct lines perpendicular to the same line are parallel. Remember that Theorem 31 tells us that for any line L and any point P not on L, there exists a line that passes through P and is perpendicular to L. To say ―there exists a line‖ means ―at least one line‖. Nothing in the statement or proof of that theorem said anything about the possibility of there being more than one such line. The following theorem states that there cannot be more than one line. Theorem 41 Uniqueness of the perpendicular from a point to a line Given: Neutral Geometry, line L and a point P not on L Claim: There is not more than one line that passes through P and is perpendicular to L. (Remark: We know that there exists at least one perpendicular by Theorem 31.) Notice that none of the Neutral Geometry axioms say anything about THE BIG QUESTION. Even so, the following theorem proves that the answer is ―at least one line‖. This makes it one of the most important theorems of Neutral Geometry. You will prove it in a class drill. Theorem 42 Existence of parallel lines (The answer to THE BIG QUESTION) Given: Neutral Geometry, line L and a point P not on L Claim: There exists at least one line that passes through P and is parallel to L. Page 87 of 150 6.7. Exercises [1] Recall the statement of Theorem 25: (Segment Subtraction) In Neutral Geometry, if A*B*C and D*E*F and AB DE and AC DF , then BC EF . Justify the steps in the following proof. Make a drawing in the indicated step. Proof 1. Assume that BC EF . (justify) 2. There is a point G on EF such that BC EG . (justify) (drawing) 3. G F . (justify) 4. AC DG . (justify) 5. DG DF . (justify) 6. G F . (justify) 7. G F and G F . (justify) 8. BC EF (justify) End of proof [2] Recall the statement of Theorem 27: (Facts about the order relation on the set of line segments) Given: Neutral Geometry; line segments AB , CD , and EF . Claim (a) (trichotomy) Exactly one of the following is true: AB CD , AB CD , or CD AB . (b) If AB CD and CD EF , then AB EF . (c) (transitivity)If AB CD and CD EF , then AB EF . Justify the steps in the following proof. Make drawings where indicated: Proof of Claim (a) 1) There exists a unique point G on ray CD such that CG AB . (by axiom _____) 2) Exactly one of the following statements is true about point G:C*G*D or C*D*G or G=D. (by definition of ________________) Case I 3) If C*G*D then AB CD . (by definition of ___________________________) Case II 4) If C*D*G, then there exists a unique point H such that A*H*B and AH CD . (by step ________ and Theorem ______________) 5) In this case, CD AB . (by definition of ____________________) Case III 6) If G = D then AB CD . End of Proof of Claim (a) Proof of Claim (b) 1) Suppose that AB CD and CD EF . Page 88 of 150 2) There exists a point G such that C*G*D and CG AB . (by definition of __________) 3) There exists a point H such that E*H*F and EH CG . (by Theorem ______) 4) EH AB (by steps ____ and ____ and axiom _____) 5) AB EF . (by steps ___ and ___ and the definition of _________________________) End of proof of Claim (b) Proof of Claim (c) 1) Suppose that AB CD and CD EF . 2) There exists a point G such that C*G*D and CG AB . (by definition of __________) 3) There exists a point H such that E*H*F and EH CD . (by definition of __________) 4) There exists a point I such that E*I*H and EI CG . (by steps ___ and ___ and Theorem ____) 5) EI AB (by steps ___ and ___ and axiom ____) 6) E*I*F (by steps ___ and ___ and Theorem ______) 7) AB EF .(by steps ___ and ___, and the definition of _________________________) End of proof of Claim (c) [3] Recall the statement of Theorem 28: In Neutral Geometry, supplements of congruent angles are congruent. That is, if ABC and CBD are supplementary, and EFG and GFH are supplementary, and ABC EFG , then CBD GFH . Justify the steps in the following proof. Make a drawing where indicated. 1. We can assume that points E, G, and H have the property that AB EF , CB GF , and DB HF . Otherwise, we can find three points on those rays that do have this property, and rename those three points as E, G, and H. (_________________________________) (draw a picture to illustrate) 2. ABC EFG . (_________________________________________________________) 3. AC EG , and A E . (________________________________________________) 4. AD EH . (_____________________________________________________________) 5. CAD GEH . (_________________________________________________________) 6. CD GH and D H . (________________________________________________) 7. CBD GFH . (_________________________________________________________) 8. CBD GFH (________________________________________________________) End of proof [4] The term supplementary angle was introduced in Definition 39, in Section 5.6. Prove that every angle has a supplementary angle. [5] Recall the statement of Theorem 28:In Neutral Geometry, supplements of congruent angles are congruent. That is, if ABC and CBD are supplementary, and EFG and GFH are supplementary, and ABC EFG , then CBD GFH . Page 89 of 150 Justify the steps in the following proof. Draw a picture where indicated. Proof 1. Suppose that if ABC and CBD are supplementary, and EFG and GFH are supplementary, and ABC EFG . 2. We can assume that points E, G, and H have the property that AB EF , CB GF , and DB HF . Otherwise, we can find three points on those rays that do have this property, and rename those three points as E, G, and H. (justify) (draw a picture to illustrate) 3. ABC EFG . (____________________________________________________) 4. AC EG , and A E . (___________________________________________) 5. AD EH . (________________________________________________________) 6. CAD GEH . (____________________________________________________) 7. CD GH and D H . (___________________________________________) 8. CBD GFH . (____________________________________________________) 9. CBD GFH (___________________________________________________) End of proof [6] Prove Theorem 29, which says that in Neutral Geometry, vertical angles are congruent. Solution: [7] Recall the statement of Theorem 30: In Neutral Geometry, any angle congruent to a right angle is also a right angle. Justify the steps in the following proof: Make a drawing where indicated. 1) Suppose that angles ABC and EFG are given; that ABC EFG ; and that EFG is a right angle. 2) There exists a point D is a point such that A*B*D. (by axiom _____________) 3) Angle ABC , is supplementary to angle CBD . (by definition of ______________) 4) There exists a point H is a point such that E*F*H. (by ____________________) 5) Angle EFG , is supplementary to angle GFH . (by _____________________) (make a drawing) 6) EFG GFH (by steps ____ and ____ and definition of _________________) 7) CBD GFH (by steps ____ and ____ and Theorem ) 8) ABC CBD (by steps ____ and ____ and ____ and axiom ____) 9) Angle ABC is a right angle. (by steps ____ and ____ and the definition of _______) End of Proof [8] Prove Theorem 39 (The Corresponding Angle Theorem) Given: Neutral Geometry, lines L and M and a transversal T Claim: If a pair of corresponding angles is congruent, then lines L and M are parallel. Hint: Use Theorem 38 in your proof. [9] Prove Theorem 40: In Neutral Geometry, two distinct lines perpendicular to the same line are parallel. Hint: Use Theorem 38 or Theorem 39 in your proof. [10] Recall the statement of Theorem 41 (uniqueness of the perpendicular from a point to a line): Page 90 of 150 Given: Neutral Geometry, line L and a point P not on L Claim: There is not more than one line that passes through P and is perpendicular to L. Justify the steps in the following proof: 1. Suppose that L is a line and P is a point not on L. 2. Assume that there are two different lines that pass through P and are perpendicular to L. Call the lines M and N. 3. Lines M and N are not parallel. (because ____________________________________) 4. Lines M and N are parallel. (by Theorem ____________________________________) 5. Steps 4 and 5 are a contradiction. Therefore, our assumption in step 3 was wrong. There cannot be two different lines that pass through P and are perpendicular to L. That means there must only be the one line, M. End of Proof Page 91 of 150 7. Neutral Geometry II: Triangles In this section, we will find that many important theorems that are commonly thought of as Euclidean Geometry theorems can actually be proven to be true in Neutral Geometry. 7.1. The Isosceles Triangle Theorem and Its Converse In Section 6.3.3, it was mentioned that there were four important statements that we would like to be true in Neutral Geometry: SAS congruence, ASA congruence, SSS congruence, and AAS congruence. As mentioned in that section, the statement of SAS congruence is included in the list of axioms, as CA6, while the other three statements will be proven as theorems. Here is the first of those three theorems. You will justify the steps of this proof in a homework exercise. Theorem 43 (Angle Side Angle Congruence) (ASA). In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and the included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. The next theorem is called the Isosceles Triangle Theorem. We should officially define the term isosceles before stating the theorem. Definition 56 isosceles triangle Words: isosceles triangle Meaning: two sides of the triangle are congruent to each other (at least two) Additional Terminology: The angles opposite the two congruent sides are called the base angles. Here is the theorem: Theorem 44 (The Isosceles Triangle Theorem) Given: Neutral Geometry, ABC Claim: If AB AC then B C Alternate wordings for the statement of the Isosceles Triangle Theorem: If two sides are congruent, then the two opposite angles are congruent. If the triangle is isoscelese, then the base angles are congruent. Notice that in the Isosceles Triangle Theorem, the fact that the triangle is isosceles is part of the hypothesis. So the name is in keeping with the naming convention that we have discussed. The following theorem is the Converse of the Isosceles Triangle Theorem. It is worth noting that it is the converse, not the original, that has the words ―isosceles triangle‖ as part of the conclusion. Theorem 45 (Converse of the Isosceles Triangle Theorem) Page 92 of 150 Given: Neutral Geometry, ABC Claim: If B C then AB AC (If two angles are congruent, then the two opposite sides are congruent.) Proofs of both the Isosceles Triangle Theorem and its converse will be studied in a class drill. As a summary of the statements of the two theorems, it might be useful to put them together in a table: Isosceles Triangle Theorem: If two sides are congruent, then the two opposite angles are congruent Converse of the Isosceles Triangle Theorem: If two angles are congruent, then the two opposite sides are congruent. In other words, in triangles, congruent angles are always opposite congruent sides. This summary is easy to remember, and has a convenient acronym, which we will present as the title of a theorem. Theorem 46 The CACS Theorem: In Neutral Geometry triangles, Congruent Angles are always opposite Congruent Sides. Again, Theorem 46 is not really a new theorem, but rather just a combination of Theorem 44 and Theorem 45 with a name that is perhaps easier to remember. Here is a theorem that is presented only because it is useful in the proofs of some other more important theorems. In other words, it could be called a lemma. The proof is not difficult, but it is also not particularly illuminating, and so we will omit the proof and assume the theorem as given. Theorem 47 The Triangle Construction Theorem Given: Neutral Geometry, ABC , DE AB , and a point G not on DE . Claim: There exists a unique point F in half plane HG such that ABC DEF . The Triangle Construction Theorem just presented and the Isosceles Triangle Theorem can be used to prove the Side Side Side Congruence Theorem. Theorem 48 (The Side Side Side Congruence Theorem) (SSS) In Neutral Geometry, if there is a correspondence between parts of two triangles such that the three sides of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. The proof of the Side Side Side Congruence Theorem takes awhile to digest because there are a number of cases to consider. You will study the proof in a homework exercise. 7.2. The Exterior Angle Theorem The next major theorem that we will study is the Exterior Angle Theorem. In order to understand the wording of the theorem, we need to define some terms. Page 93 of 150 Definition 57 ―exterior angle‖, ―interior angle‖, ―remote interior angles‖ Words: An exterior angle of a triangle Meaning: An angle that is supplemental to one of the angles of the triangle Additional terminology: The angles of the triangle are also called “interior angles”. Given an exterior angle, there will be an interior angle that is its supplement, and two other interior angles that are not its supplement. Those other two interior angles are called remote interior angles. Theorem 49 The Exterior Angle Theorem In Neutral Geometry, each of the remote interior angles is less than the exterior angle. The proof of the Exterior Angle Theorem is rather tricky: it uses the method division into cases and the method of contradiction. You will study it in a class drill. The Exterior Angle Theorem plays an important role in the proof of the Angle Angle Side Congruence Theorem. Remember that the statement of this theorem is the last of the four congruence conditions (SAS, ASA, SSS, AAS) that were discussed in Section 6.3.3. Like the proof of the Exterior Angle Theorem, the proof of the Angle Angle Side Theorem uses both the method division into cases and the method of contradiction. You will study it in a class drill. Theorem 50 The Angle Angle Side Congruence Theorem (AAS) In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and a non-included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. In addition to the four congruence conditions introduced Section 6.3.3, there is another famous congruence theorem that applies only to right triangles. First, some terminology. Definition 58 ―hypotenuse‖ and ―legs‖ of a right triangle Meaning: The hypotenuse of a right triangle is the side that is opposite the right angle. The other two sides are called legs. Here is the theorem: Theorem 51 The Hypotenuse Leg Congruence Theorem In Neutral Geometry, if the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. You will study a proof of the Hypotenuse Leg Congruence Theorem in the homework. Page 94 of 150 7.3. Comparison Theorems In this section, we will turn our attention to theorems that are not about pairs of objects that are congruent, but rather are about pairs objects in which one of the objects can be judged less than the other. Since we are comparing objects, we could call these comparison theorems. Theorem 52 In Neutral Geometry, in triangle ABC , if AC BC then B A. It turns out that the converse of the statement of the above theorem is also a theorem: Theorem 53 In Neutral Geometry, in triangle ABC , if B A then AC BC . Proofs of both these theorems will be studied in a class drill. As a summary of the statements of the two theorems, it might be useful to put them together in a table, just as we did when studying the Isoscelese Triangle Theorem and its converse: Theorem 52: If one side of a triangle is smaller than another side, then the two opposite angles have the same ordering. Theorem 53: If one angle of a triangle is smaller than another angle, then the two opposite sides have the same ordering. In other words, in triangles, smaller angles are always opposite smaller sides. This summary is easy to remember, and has a convenient acronym, which we will present as the title of a theorem. Theorem 54 The SASS Theorem: In Neutral Geometry triangles, Smaller Angles are always opposite Smaller Sides. Note that Theorem 54 is not a new theorem, but rather just a combining of Theorem 52 and Theorem 53. It actually helps to take the combining of old theorems further. One can combine Theorem 46 (the CACS Theorem), and Theorem 54 (the SASS Theorem) into a single theorem: Theorem 55 The CACS and SASS Theorem: In triangles of Neutral Geometry, Congruent Angles are always opposite Congruent Sides, and Smaller angles are always opposite Smaller Sides. In summary, Theorem 55 is a combination of the two theorems: Theorem 46 (the CACS Theorem), and Theorem 54 (the SASS Theorem). But these in turn are combinations of four theorems: Theorem 44, Theorem 45, Theorem 52 and Theorem 53. We end this section with the Hinge Theorem. The name refers to our imagining two triangle legs of fixed length forming an included angle that is not fixed but rather is a hinge that can be opened varying amounts. The idea is that when the angle of the hinge is reduced, the ends of the legs will be brought closer together. We are imagining drawings, where we have the notion of line segment length and angle measure. In our axiomatic world of Neutral Geometry, the Page 95 of 150 equivalent is to consider two triangles that have two corresponding sides congruent and included angles not congruent. In this world, remember that line segment congruence is undefined. The ―less than‖ relation for angles is defined, but its definition does not refer to angle measure. Rather, it refers to earlier definitions that ultimately refer to betweenness and angle congruence, both undefined concepts. Theorem 56 The Hinge Theorem: Given: Neutral Geometry, ABC , DEF , AB DE , AC DF Claim: The following are equivalent (1) A D . (2) BC EF . The proof of the Hinge theorem is difficult and would is not particularly illuminating, so we will omit the proof and assume the Hinge Theorem as given. 7.4. Midpoints and bisectors We end the chapter with a short presentation of three theorems about midpoints and bisectors. We most likely will not discuss this section in class, but you will study the proofs in homework exercises. Definition 59 ―midpoint‖ of a line segment Words: C is a midpoint of segment A, B Meaning: A*C*B and CA CB . Theorem 57 Existence of Midpoint of a Line Segment In Neutral Geometry, every segment has exactly one midpoint. Definition 60 ―bisector‖ of a line segment, ―perpendicular bisector‖ of a line segment Words: Line L is a bisector of segment AB . Meaning: L is distinct from line AB and passes through the midpoint of segment AB . Additional Terminology: If L is perpendicular to line AB and is also a bisector of segment AB , then L is said to be a perpendicular bisector of segment AB . Theorem 58 In Neutral Geometry, every line segment has exactly one perpendicular bisector. Definition 61 ―bisector‖ of an angle Words: a bisector of angle ABC Meaning: a ray BD between rays BA and BC such that Theorem 59 DBA DBC . In Neutral Geometry, every angle has exactly one bisector. Page 96 of 150 7.5. Exercises [1] Recall the statement of Theorem 43 (ASA congruence): In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and the included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Justify the steps in the following proof. Draw a picture where indicated. 1. Suppose that ABC and DEF are triangles such that A D , C F , and AC DF . 2. There is a unique point G on DE such that DG AB . (by axiom _________________) 3. ABC DGF . (by steps ___ and ___ and axiom _____) (Draw a picture.) 4. C DFG . (by step ___ and definition of __________________________________) 5. Ray FE must be the same ray as ray FG . (by axiom ________________) 6. Point E must be the same point as point G. (There can be only one point of intersection of lines DE and FE , by Theorem _____) 7. ABC DEF . (by steps ___ and ___ and ___ and axiom ______________) End of proof [2] Recall the statement of Theorem 48 (The Side-Side-Side Congruence Theorem) (SSS): In Neutral Geometry, if there is a correspondence between parts of two triangles such that the three sides of one triangle are congruent to the corresponding sides of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Justify the steps in the following proof. Draw pictures where indicated. We will consider four cases. In the first three cases, triangles ABC and DEF are ―back-toback‖. In the fourth case, they are not. case 1: A = D, C = F, and the points B and E are on opposite sides of AC , and points A and C are on opposite sides of BE . case 2: A = D, C = F, and the points B and E are on opposite sides of AC , and point A is between B and E. That is, B*A*E. case 3: A = D, C = F, and the points B and E are on opposite sides of AC , and points A and C are on the same sides of BE . case 4: All other possible configurations. Proof of Case 1 1) Suppose that triangles ABC and DEF have a correspondence between their parts such that the three sides of one triangle are congruent to the corresponding sides of the other triangle, and suppose further that A = D, C = F, and the points B and E are on opposite sides of BE , and points A and C are on opposite sides of BE . (Draw a picture.) 2) AB DE (by step ______) 3) ABE AEB (by step ___ and Theorem ______________________________) 4) BC EF (by step ___) 5) CBE CEB (by step ___ and Theorem ______________________________) 6) ABC DEF (by steps ___ and ___ and Theorem ______________________) 7) ABC DEF (by steps ___ and ___ and ___ and axiom ___________________) Page 97 of 150 End of proof of case 1 Proof of Case 2 1) Suppose that triangles ABC and DEF have a correspondence between their parts such that the three sides of one triangle are congruent to the corresponding sides of the other triangle, and suppose further that A = D, C = F, and the points B and E are on opposite sides of BE , and point A is between B and E. That is, B*A*E. (Draw a picture.) 2) BC EF (by step ___) 3) CBE CEB (by step ___ and Theorem ___________________________) 4) BA ED (by step ___) 5) ABC DEF (by steps ___ and ___ and ___ and axiom ____________________) End of proof of case 2 Proof of Case 3 1) Suppose that triangles ABC and DEF have a correspondence between their parts such that the three sides of one triangle are congruent to the corresponding sides of the other triangle, and suppose further that A = D, C = F, and the points B and E are on opposite sides of BE , and points A and C are on the same sides of BE . (Draw a picture.) 2) AB DE (by step ___) 3) ABE AEB (by step ___ and Theorem _______________________________) 4) BC EF (by step ___) 5) CBE CEB (by step ___ and Theorem _______________________________) 6) ABC DEF (by steps ___ and ___ and Theorem _______________________) 7) ABC DEF (by steps ___ and ___ and ___ and axiom _____________________) End of proof of case 3 Proof of Case 4 Suppose that triangles ABC and DEF have a correspondence between their parts such that the three sides of one triangle are congruent to the corresponding sides of the other triangle. One can use Theorem 47, the Triangle Construction Theorem, to build a new triangle GHI that is congruent to DEF and that is ―back-to-back‖ with ABC . Then, case 1, 2, or 3 (with the letters D,E,F replaced by G,H,I) would describe the configuration of triangles ABC and GHI . The results of those cases could be used to conclude that ABC GHI . Transitivity would then say that ABC DEF . End of proof of case 4 [3] Recall the statement of Theorem 51, the Hypotenuse-Leg Congruence Theorem In Neutral Geometry, if the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Justify the steps in the following proof. (Draw pictures where indicated.) 1. Let ABC and DEF be triangles such that angles A and D are right angles, BC EF , and BA ED . Construct a special point G that creates a segment congruent to segment DF . 2. There exists a point H such that H*A*C. (by axiom BA3) 3. There exists a point G on ray AH such that AG DF . (by axiom CA2, the segment construction axiom) (Draw a picture.) Page 98 of 150 Show that the resulting angle is congruent to EDF . 4. G*A*C (a brief argument of the type we did in Chapter 5 would show this) 5. Angles BAG and BAG are supplemental. (by step 5 and definition of supplemental) 6. BAG BAC (by steps 1 and 7 and definition of right angle) 7. BAC EDF (by Theorem 37 ) 8. BAG EDF (by steps 7 and 8 and axiom CA4) Show that the resulting triangle is congruent to DEF . 9. ABG DEF . (steps 1 and 8 and 3 and axiom CA6 (SAS)) Show that the resulting triangle is congruent to ABC . 10. BG EF . (by step 9 and definition of triangle congruence, that is, CPCTC) 11. BG BC . (by steps 1 and 10 and axiom CA1) 12. G C . (by step 11 and Theorem 44, the Isosceles Triangle Theorem) 13. ABG ABC . (by steps 6 and 12 and 11 and Theorem 50 (AAS)) 14. DEF ABC . (by steps 9 and 13 and the fact that triangle congruence is an equivalence relation (which follows from the fact that line segment congruence is an equivalence relation, by CA1, and angle congruence is an equivalence relation, by CA4)) End of Proof [4] Recall the statement of Theorem 57 (Existence of Midpoint of a Line Segment) In Neutral Geometry, every segment has exactly one midpoint. Justify the steps in the following proof. Draw pictures where indicated. 1. Let AB be a line segment. 2. There exists a point C that is not on line AB . (by Theorem ____________________) 3. There exists a unique ray BX such that point X is on the opposite side of line AB from point C and such that XBA CAB . (by axiom ___________________________) 4. There exists a unique point D on ray BX such that BD AC . (by axiom ___________) 5. Points D and X are on the same side of line AB , (by step ___ and Theorem _________) 6. Points D and C are on opposite sides of AB , (by steps ___ and ___ and Theorem _____) 7. Line AB intersects segment CD . (by step ___ and definition of ___________________) We can call the intersection point E. 8. Point E is between A and B. (This is a very tricky step to justify. We will assume it as given.) (Draw a picture.) 9. AEC BED . (by Theorem _________________________________________) 10. AEC BED . (by steps ___ and ___ and ___ and Theorem ____________________) 11. AE BE . (by step ___ and the definition of _________________________________) 12. Point E is the midpoint of segment AB . (by steps ___ and ___ and definition of ______________________) End of Proof [5] Prove Theorem 58: In Neutral Geometry, every line segment has exactly one perpendicular bisector. [6] Prove Theorem 59: In Neutral Geometry, every angle has exactly one bisector. Page 99 of 150 8. Measure of Line Segments and Angles 8.1. Theorems Stating the Existence of Measurement Functions So far, real numbers have played almost no role in our axiomatic geometry. We have used the non-negative integers to count things, but have not used rational numbers and certainly not used real numbers. This is in sharp contrast to the style of axiomatic geometry found in high school textbooks. There, real numbers play a central role: they are mentioned explicitly in the axiom system. The following two theorems show two ways that real numbers can be introduced into our geometry: as line segment length and angle measure. We will not prove these theorems in our class. Theorem 60 Existence of a Length Function for Line Segments Given: Neutral Geometry and some reference line segment AB . Claim: There exists a unique function length : line segments properties. with the following 1. length AB 1 . 2. The length function is onto. That is, for every positive real number L, there exists a line segment CD such that length CD L . length CD length EF if and only if. CD EF . length CE length CD length DE if and only if C*D*E. 3. length CD length EF if and only if CD EF . 4. 5. Theorem 61 Existence of a Measure Function for Angles Given: Neutral Geometry Claim: There exists a unique function degrees : angles 0,180 with the following properties. 1. degrees A 90 if and only if A is a right angle. 2. The degrees function is onto. That is, for every real number 0 m 180 , there exists an angle A such that degrees A m . 3. degrees A degrees B if and only if A B. 4. degrees A degrees B if and only if A B. 5. If ray AC is between rays AB and AD then degrees BAD degrees BAC degrees CAD . 6. If angle A is supplementary to angle B , then degrees A degrees B 180 . Page 100 of 150 Notice that both the length function and the degrees function make use of a reference input. In the case of the length function, a choice of a line segment AB whose length will be declared to be 1 is central to the definition. In the case of the degrees function, it is declared that the measure of a right angle will be 90. The length definition obviously allows an amount of choice. That is, if one uses a different reference segment AB (more precisely, if one uses a different reference segment that is not congruent to the first), then the resulting distance function will be different, as well. There is choice in the definition of angle measure, as well, but not in the choice of the reference input. The reference angle is always a right angle. But there is choice in the positive real number that is assigned to be the measure of the right angle. In the definition of the degrees function, the number 90 is used. Correspondingly, the degree measures of supplementary angles add up to 180, and the codomain of the degrees function is the set (0,180). But we could have used some number other than 90 for the measure of a right angle. When a number other than 90 is used, the name of the angle measurement function is changed, as well. A common choice of measure for right angles is the number . When that number is used, the name of the angle measurement 2 function is changed to radians, the measures of supplementary angles add up to and the codomain of the degrees function is the set 0, . A less common choice of measure for right angles is the number 100. When that number is used, the name of the angle measurement function is changed to gradians, the measures of supplementary angles add up to 200, and the codomain of the degrees function is the set (0,200) (You have probably noticed that most calculators allow you to choose between degrees, radians, or gradians.) 8.2. Two length functions for Neutral Geometry In this section, we will explore two length functions for Neutral Geometry. One will be familiar to you: it comes from the standard distance formula that you have been using since high school. But the other will be new. Definition 62 the length function for line segments in straight-line drawings (Euclidean) Meaning: the function lengthE : Euclideanline segments defined by the following formula lengthE AB x1 x2 y1 y2 2 2 , where segment AB is a Euclidean line segment with endpoints A x1 , y1 and B x2 , y2 . (This implies that x1, y1, x2,and y2 are real numbers.) Definition 63 the length function for line segments in the Hyperbolic plane Meaning: the function lengthH : Hyperbolicline segments defined by the following formula Page 101 of 150 lengthE AR lengthE AS , lengthH AB ln lengthE BR length BS E where segment AB is a Hyperbolic line segment with endpoints that are the Hpoints A and B, and R and S are the ―missing endpoints‖ of the line AB . Two typical configurations of points A, B, R, and S are shown in the figures below. R R S A A B B S Notice that the Hyperbolic line segment AB can be a curvy thing that does does not look the same as the Euclidean line segment AB . Also note that when computing the Hyperbolic length of the Hyperbolic segment AB , the Euclidean lengths of four Euclidean line segments are used. S Hyperbolic line segment AB is the curvy arc. It does not look the same as Euclidean line segment AB. Euclidean line segments AR, AS, BR, and BS are the straight-looking things. A B R You might be a little disturbed by the fact that in one of the figures above, the ordering of points on the line is R-A-B-S, while in the other, the ordering is S-A-B-R. Shouldn’t it matter what names we give to the missing endpoints, at least when using the function dH? The answer is no, it doesn’t matter. To see why, consider the effect interchanging the names of the missing endpoints R & S. That would mean that we should interchange the symbols R & S in the formula for the distance between A & B. But rules of logarithms can be used to show that if u, v, x, and y are positive numbers, then u v v ln ln u . x y y x Page 102 of 150 (You will prove this in the exercises.) Therefore, interchanging the symbols C & D in the formula would have no effect on the outcome. For an example of the use of the distance function lengthH, let A 0, 1 and let B , 3 . 4 4 Both of these points qualify as Hpoints, and they lie on the Hline consisting of the portion of the y-axis that lies inside the unit circle. The missing endpoints of this line are R 0,1 and S 0, 1 . To compute the hyperbolic distance between A and B, we first compute the Euclidean distances needed in the formula for lengthH. The results are lengthE AR 5 , 4 lengthE AS 3 , lengthE BR 1 , and lengthE BS 7 . Plugging these into the formula 4 4 4 for lengthH, we obtain lengthE AR 54 lengthE AS 34 57 35 lengthH AB ln ln 1 ln ln 2.46 1 3 3 lengthE BR 47 4 lengthE BS Notice that the hyperbolic length of line segment AB is not the same as the Euclidean length. The Euclidean length of the segment is just lengthE AB 1 . 8.3. An example of a curvy-looking Hline When Hpoints A and B happen to lie on a diameter of the unit circle, the Hline AB is easy to determine because it is straight: it will be described by the standard equation for a straight line containing A and B. Because this straight line is a diameter of the unit circle, we know that it must go through the origin. So the form of the equation will be y = mx for a non-vertical diameter or x = 0 for the one diameter that is vertical. Determination of the missing endpoints R and S is not terribly difficult: one would simply find the intersection of the straight non-vertical line y = mx or the straight vertical line x = 0 with the unit circle. That is, one would simply solve the pair of equations y mx 2 2 x y 1 in the case of a non-vertical diameter. In the case of a vertical diameter, of course, the missing endpoints are the points 0,1 and 0, 1 . In general, it can be very difficult to find the equation for a curvy-looking Hline. Given Hpoints A and B, one must find the equation for the one circle that contains A and B and is orthogonal to the unit circle. This involves solving some messy equations. Then, one must find the coordinates of the two points of intersection of the orthogonal circle and the unit circle. This also involves solving messy equations. It is very useful, therefore, to have at least one example of a curvylooking Hline whose equation and missing endpoints are not so hard to determine. Page 103 of 150 One simple example of an orthogonal circle is simply the unit circle moved to the right. Recall that the unit circle contains twelve famous points as shown in the figure below. (0,1) 1 3 1 3 , , 2 2 2 2 2 2 , 2 2 3 1 , 2 2 2 2 , 2 2 3 1 , 2 2 (-1,0) (1,0) 3 1 , 2 2 2 2 , 2 2 1 3 , 2 2 A circle of radius 1 centered at the point (0,-1) 3 1 , 2 2 2 2 , 2 2 1 3 , 2 2 2, 0 will have twelve famous points obtained by 2 to the x-coordinates of the points above. The result is shown in the figure below. 2,1 1 1 3 3 2, 2, 2 2 2 2 2 2 3 2 2 , , 2 2 2 2 3 3 1 1 2, 2, 2 2 2 2 adding 1 2, 0 3 1 2, 2 2 2 2 , 2 2 1 3 2, 2 2 1 2, 0 3 1 2, 2 2 2, 1 3 2 , 2 1 2, 2 2 2 3 2 Page 104 of 150 2 2 2 2 This new circle intersects the unit circle at points and , , . Furthermore, the 2 2 2 2 two circles are orthogonal, as shown in the figure below. So the set of points on the second circle that lie in the interior of the unit circle qualifies as an Hline. That is, the set of points x, y that satisfy the following two equations: x 2 x2 y 2 1 2 y2 1 The point x, y lies on the circle of radius 1 centered at 2, 0 . The point x, y lies in the interior of the unit circle. Three of the twelve famous points on the right circle above qualify as Hpoints. They are are 3 1 3 1 A 2, , B 1 2, 0 , C 2, . 2 2 2 2 Since these three points all lie on the same Hline, they are collinear Hpoints. (Of course, they are not collinear as points of Euclidean Geometry.) The two points of intersection of the two circles are the missing endpoints of the Hline. They are 1 1 1 1 R , , , S . 2 2 2 2 In the exercises, you will compute the hyperbolic lengths of Hyperbolic segments AB ands AC . 8.4. Theorems about segment lengths and angle measures In this section, we will discuss five theorems of Neutral Geometry that use the concept of line segment length or angle measure. Three of the theorems have fairly easy proofs, but two have hard proofs. Page 105 of 150 The first of these theorems might seem a little strange to you. It says that in Neutral Geometry, the sum of the measures of any two angles of a triangle is less than 180. This might seem obvious: we all know that the sum of the measures of all three angles of a triangle is exactly 180, so therefore the sum of any two of them must be less than 180, right? But we have not proven that all triangle angle sums equal 180. What’s more, the statement that all triangle angle sums equal 180 cannot be proven in Neutral Geometry, because it is not true! So, since we don’t yet know anything about triangle angle sums, and since we will eventually know less than we thought we knew, the following theorem is actually interesting. The proof of this theorem is not hard; you will justify it in the exercises. Theorem 62 In Neutral Geometry, the sum of the measures of any two angles of a triangle is less than 180. The Triangle Inequality is well known, and also has a fairly simple proof that you will justify in the exercises. Theorem 63 (The Triangle Inequality) In Neutral Geometry, the length of any side of a triangle is less than the sum of the lengths of the two other sides. As mentioned earlier, you are all used to the fact that the angle sum of any triangle is exactly 180 degrees, but that fact cannot be proven in Neutral Geometry. In fact, it is not even true. All that can be proven is the following very famous theorem. Theorem 64 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any triangle is less than or equal to 180. The proof of Theorem 64 is difficult; we will accept the theorem as given. But the following theorem is not very hard to prove. You will justify a proof of it in the exercises. Theorem 65 In Neutral Geometry, the sum of the degree measures of any two angles of a triangle is less than or equal to the degree measure of their remote exterior angle. Page 106 of 150 8.5. Exercises for Chapter 8 u v v [1] In the reading, you were told that ln ln u . Prove this. x y y x [2] Let l be the Hline consisting of the portion of the x-axis that lies inside the unit circle. The ―missing endpoints‖ of this line are R 1,0 and S 1,0 . (They are not Hpoints.) Let A, B, C, and D be the Hpoints A 0,0 , B 1 , 0 , C 1 , 0 , and D 3 ,0 . Find 4 2 4 lengthH AB and lengthH CD . (Be sure to use R & S as the missing endpoints!) [3] In Section 8.3, an example of a curvylooking Hline was presented. It has missing endpoints R and S, and famous points A, B, and C as shown in the figure at right. Compute AC . lengthH AB and lengthH 2 2 S , 2 2 3 1 C 2, 2 2 B 1 2, 0 3 1 A 2, 2 2 2 2 R , 2 2 [4] Let A 0,0 , and let B x,0 where x is some unknown real number such that 0 x 1. Compute lengthH AB . For consistency, label the missing endpoints R 1,0 and S 1,0 . (b) Your answer to part (a) will be a symbolic expression that includes the absolute value symbol. In the next problem, it will be necessary to have this expression simplified further, with the absolute value symbol eliminated. Determine how to simplify the expression in a way that the absolute value symbol is eliminated. (Hint: to eliminate the absolute value symbol, you will have to determine whether the thing inside it is positive or negative. You can do that by exploiting the fact that 0 x 1. Because of 1 x that, you know that 1 1 x 2 and 0 1 x 1 . In the fraction , you are dividing a 1 x 1 x number that is greater than 1 by a positive number that is less than 1. So the fraction will 1 x number greater than 1. What does that tell you about the logarithm of that fraction? More specifically, will the logarithm be positive, negative or zero? With that information, decide what to do with the absolute value symbol.) Page 107 of 150 [5] The goal of this exercise is to prove that the lengthH function is onto. To do that, you must prove that given any unknown positive number L, there exists a segment AB such that lengthH AB L . There are lots of such segments. The easiest one to find is the one that has point A at the origin, that is A 0,0 , and point B somewhere on the positive x-axis, inside the unit circle. That is, point B of the form B x, 0 where 0 x 1. Given an unknown L, find the value of x such that lengthH AB L . (Hint: Think of the result of the previous problem as a function that computes length as a function of x. Call the resulting length L. Then you have an equation involving x and L, and that equation is solved for L in terms of x. Simply solve that equation for x in terms of L.) [6] Recall the statement of Theorem 62: In Neutral Geometry, the sum of the measures of any two angles of a triangle is less than 180. Justify the steps in the following proof. Draw a picture where indicated. 1. Let two angles of a triangle be given. Without loss of generality, we can assume that the triangle vertices are labeled such that the two given angles are A and B of ABC . 2. There exists a point D such that A*B*D. (by axiom _______) (Draw a picture.) 3. Angle DBC is an exterior angle of ABC , and A is one of its remote interior angles. (by steps ___ and ___ and definition _________________________________________) 4. A DBC . (by step ___ and Theorem _____________________________________) 5. degrees A degrees DBC . (by step ___ and Theorem _____________________) 6. degrees A degrees B degrees DBC degrees B . (by step ___ and algebra) 7. Angles DBC and B are supplemental. (by steps ___ and ___ and definition of ______________________________________________________) 8. degrees DBC degrees B 180 . (by step ___ and Theorem _________________) 9. degrees A degrees B 180 . (by steps ___ and ___ and algebra) End of proof Page 108 of 150 [7] Recall the statement of Theorem 63: (The Triangle Inequality) In Neutral Geometry, the length of any side of a triangle is less than the sum of the lengths of the two other sides. Justify the steps in the following proof. Draw a picture where indicated. 1. Let a side of a triangle be given. Without loss of generality, we can assume that the triangle vertices are labeled such that the given side is side AC of ABC . 2. There exists a point D such that A*B*D and BC BD . (Axiom ____ tells us that a point D exists such that A*B*D. Axiom _________________________________________ tells us that a special point exists on ray BD that will create a segment congruent to segment BC . With no loss of generality, we can just rename that special point as point D.) (Draw a picture.) 3. length BC length BD . (by step ___ and Theorem ___________________________ 4. length AB length BC length AB length BD . (by step ___ and algebra) 5. length AB length BD length AD . (by step ___ and Theorem _______________) 6. length AB length BC length AD . (by steps ___ and ___) 7. ADC DCB . (by step ___ anTheorem ____________________________________) 8. Ray CB is between rays CA and CD . (Accept this as given.) 9. DCB DCA .(by definition ___________________________________________) 10. ADC DCA . (by steps ___ and ___ and Theorem ___________________________ _______________________________________________________________________) 11. AC AD . (by Theorem __________________________________________________) 13. length AC length AB length BC . (by steps ___ and ___ and algebra) 12. length AC length AD . (by step ___ and Theorem __________________________) End of proof [8] Recall the statement of Theorem 65: In Neutral Geometry, the sum of the degree measures of any two angles of a triangle is less than or equal to the degree measure of their remote exterior angle. Justify the steps in the following Proof. Draw a picture where indicated. 1. Let two angles and a remote exterior angle be given. Without loss of generality, we can assume that the triangle vertices are labeled so that the given interior angles are A and B of ABC , with remote exterior angle BCD , where D is a point such that A*C*D. (Draw a picture.) 2. Angles C and BCD are supplementary. (by step ___ and definition ____________ ______________________________________________________________________) 3. degrees C degrees BCD 180 . (by step ___ and Theorem ________________ _______________________________________________________________________) 4. degrees A degrees B degrees C 180 . (by Theorem 64, The SaccheriLegendre Theorem) 5. degrees A degrees B degrees BCD . (by steps ___ and ___ and algebra) End of proof Page 109 of 150 [10] Three questions about proof structure (a) What is the statement that is proven by a proof with the following structure? Proof 1) In Neutral Geometry, suppose that triangle ABC has Property X. 2) 3) 4) 5) ABC has Property P. End of Proof (b) What is the statement that is proven by a proof with the following structure? Proof (Indirect proof by method of contraposition.) 1) In Neutral Geometry, suppose that triangle ABC does not Property X. 2) 3) 4) ABC does not have Property P. End of Proof (c) What is the statement that is proven by a proof with the following structure? Proof Part 1 1)In Neutral Geometry, suppose that triangle ABC has Property Y. 2) 3) 4) 5) ABC has Property M. Part 2 6) In Neutral Geometry, suppose that triangle ABC does not Property Y. 7) 8) 9) ABC does not have Property M. End of Proof Page 110 of 150 Page 111 of 150 9. Building Euclidean Geometry from Neutral Geometry At the beginning of this course, we were introduced to the notion of an axiom system. We also learned about three properties that an axiom system may or may not have: consistency, independence, and completeness. In Chapters 2 and 3, we studied simple axiom systems and their properties. Then in Chapter 5, we embarked on a program to build an axiom system that would describe the straight line drawings that we have been making for much of our lives. (The axiom system would also describe the usual analytic geometry of the x-y plane, in which the graphs of lines look straight.). One goal was to have an axiom system that is independent—has no redundancy. Another goal was to have an axiom system that is complete. That is, we wanted the axiom system to thoroughly specify the geometry, so that there would be no other model besides our straight-line drawings. Our first step towards this goal was actually already taken back in Chapter 4, when we studied Incidence Geometry. We observed that the incidence axioms insured that abstract points and lines in our axiomatic geometry have some of the ―normal‖ behavior that we expect of points and lines. But we observed that it was not very difficult for a geometry to satisfy all of the incidence axioms. As a result, there are many different models. (The axiom system for incidence geometry is not complete.) Since then, we have studied Incidence and Betweenness Geometry (in Chapter 5) and Neutral Geometry (in Chapters 6, 7, and 8). During that time, we have studied only two models: straightline drawings and drawings in the Poincare Disk. But we never really proved that these two models were actually models. It would have been more correct to just call them interpretations instead of calling them models. And we never considered the possibility that there might be other models as well. But the fact is that the straight-line interpretation and the Poincare Disk interpretation are both models of Neutral Geometry. And the two models are not the same: From studying the two types of drawings in the computer lab, we have seen many instances where the two kinds of drawings have important differences beyond the obvious difference that they simply look different. For example, in Computer Project 3, we studied the following two statements: Statement S: If a pair of alternate interior angles is congruent, then lines L and M are parallel Converse of S: If lines L and M are parallel, then a pair of alternate interior angles is congruent. We observed that Statement S was true in straight-line drawings and in the Poincare Disk. (That makes sense, because the statement is actually a theorem of Neutral Geometry.) However, the Converse of S was true in straight-line drawings but not in the Poincare Disk. Based on this observation, we would conclude that the straight-line drawings and the Poincare Disk drawings are non-isomorphic models of Neutral Geometry. We did not consider the possibility of other models of Neutral Geometry besides these two. But the fact is that these are the only two models. More correctly, these are the only two types of Page 112 of 150 models. There are other models of Neutral geometry that look different, but they all behave like either the straight-line model or the Poincare Disk model. Most of you had probably not seen Poincare Disk drawings before this course. They look very different from straight-line drawings and, as we have observed, they have some important differences in terms of the behavior of objects. But the similarities between the straight-line drawings and the Poincare Disk drawings are more remarkable. In spite of the fact that the drawings look very different, there are many instances in which the objects in the drawings behave in identical ways. The extent to which this is true may best be illustrated by noting that our list of Neutral Geometry theorems ends with Theorem 65. Every one of those theorems describes something that is true in both the straight line drawing model and the Poincare Disk model. It turns out that many important facts about lines and triangles that we might have previously associated with straight-line drawings are also true in Poincare Disk drawings. So the Neutral Geometry axiom system is very very rich but in the end is not complete: it does not thoroughly specify the world of straight-line drawings and, as a result, there are two nonisomorphic models. In this chapter, we turn our attention to the question of what axioms must be added to the Neutral Geometry in order to turn it into a complete axiom system that does fully specify the world of straight-line drawings. 9.1. Thirteen Statements that are Logically Equivalent in Neutral Geometry What is remarkable is that it is possible to turn Neutral Geometry into a complete axiom system by adding just one more axiom. And what is equally remarkable is that there is a huge list of candidate statements that will work as the one additional axiom. The reason is that a long list of statements can be proven to be logically equivalent to each other: Theorem 66 Thirteen Statements that are Logically Equivalent in Neutral Geometry Given: The axioms for Neutral Geometry Claim: The following statements are logically equivalent: 1) Euclid’s Fifth Postulate: ―That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.‖ 2) A reworded version of SMSG Postulate #16, which we will call EPP: The Euclidean Parallel Postulate: For any line L and for any point P not on L, there is at most one line that passes through P and is parallel to L. 3) Playfair’s Postulate: For every line L and every point P not on L, there is a unique line that passes through P and is parallel to L. 4) ―If a line intersects one of two parallel lines, then it intersects the other.‖ 5) ―If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.‖ 6) ―If two parallel lines are cut by a transversal, then any pair of alternate interior angles created is congruent.‖ (This statement is the converse of Theorem 38, The Alternate Interior Angle Theorem.) 7) ―If two parallel lines are intersected by a transversal, then any pair of corresponding angles formed is congruent.‖ (This statement is the converse of the statement of Theorem 39,The Corresponding Angle Theorem,) Page 113 of 150 8) ―If two parallel lines are intersected by a transversal, then the measures of pairs of interior angles on the same side of the transversal add up to 180.‖ 9) ―Every triangle has angle sum exactly 180.‖ 10) ―Given any triangle PQR and any line segment AB , there exists a triangle ABC having AB as one of its sides such that ABC is similar to PQR but not congruent to PQR .‖ 11) ―The opposite sides of a parallelogram are congruent.‖ 12) ―The opposite angles of a parallelogram are congruent.‖ 13) ―The diagonals of a parallelogram bisect each other.‖ Because these statements are equivalent, if one of them is true, then all thirteen of them are true, and if one of them is false, then all thirteen of them are false. That means that if we make any one of these thirteen statements an axiom (that is, we declare it to be true), and we add it to the set of axioms describing Neutral Geometry, then regardless of which statement we have chosen to make an axiom, the resulting larger axiom set will describe the same thing. That thing is called Euclidean Geometry. The resulting larger axiom set will be complete: its only model will be straight-line drawings (and the usual analytic geometry of the xy plane). On the other hand, if we make the negation of any one of these thirteen statements an axiom (that is, we declare the negation of the statement to be true), and we add it to the set of axioms describing Neutral Geometry, then regardless of which statement we have chosen to make an axiom, the resulting larger axiom set will describe the same thing. That thing is called Hyperbolic Geometry.The resulting larger axiom set will be complete: its only model will be Poincare Disk drawings (and the analytic geometry of the Hyperbolic plane). This ―completion‖ of Euclidean geometry has a fascinating history. For roughly 2000 years, from the time of Euclid (300bc) until the 1700’s, mathematicians thought that the Neutral Geometry axioms were complete. This belief was fostered by the fact that there was only one known model for Neutral Geometry: straight-line drawings. These mathematicians, including Euclid, himself, thought that Euclid’s Fifth Postulate was not needed in the list of axioms. They sought to prove that the statement of the Fifth Postulate was always true as a consequence of the other axioms. That is, they sought to prove that it was reudundant, or not independent. But such a proof was elusive. Eventually, mathematicians began to wonder if perhaps the statement of the fifth postulate might be independent, after all. To demonstrate that the statement was independent, a model would have to be found that satisfied all of the Neutral Geometry axioms, but in which the statement of the Fifth postulate was false. Such a model was discovered in the 1800’s. 9.2. What’s up with the answer to THE BIG QUESTION?!? Notice that there is a difference between statements 2 and 3 on the list statement 2: For any line L and for any point P not on L, there is at most one line that passes through P and is parallel to L. Page 114 of 150 statement 3: For every line L and every point P not on L, there is a unique line that passes through P and is parallel to L. Which of these statements is correct? And why does the theorem say that they are equivalent when they clearly say different things? Well, first of all remember that correctness is not the issue here. If we put a statement on the list of axioms, we are simplay playing a game where we pretend that the statement is true. But what about the issue that the two statements seem to say different things? The answer to that is that even though the two statements seem to say different things, in the context of Neutral Geometry they actually are equivalent. To see why, consider what various statements say about the answer to THE BIG QUESTION. First, recall the statement of Theorem 42: Theorem 42 (Existence of parallel lines) (The answer to THE BIG QUESTION): Given: Neutral Geometry, line L and a point P not on L Claim: There exists at least one line that passes through P and is parallel to L. Now, let n be the number of lines that pass through P and are parallel to L. Notice that Theorem 42 says that n 1. Notice that Statement 2 above says that n 1. Notice that Statement 3 above says that n = 1. If we add statement 2 to the list of Neutral Geometry axioms, then Theorem 42 will still be true. Since both Theorem 42 and Statement 2 will have to be true, that will require that n 1 and n 1 . The only way for this to be true is if n = 1. So even though Statement 2 does not say that the number of parallel lines is exactly 1, the end result will still be that the number of lines is exactly 1. 9.3. Proving Theorems about Equivalent Statements We won’t prove all of Theorem 66, but we will discuss proof strategies for theorems of this type and we then prove parts of the theorem. Recall that we have encountered a few different methods of proving the conditional statement If P then Q. A direct proof has the following form. Proof 1) Suppose that P is true 2) 3) 4) Q is true. (by some justification) End of proof Alternately, we could prove the contrapositive statement, Page 115 of 150 If not Q then not P. Such a proof would have the following form Proof 1) Suppose that Q is false 2) 3) 4) P is false. (by some justification) End of proof We could use an indirect proof by the method of contradiction. We have used this method to prove the original conditional statement: Proof 1) Suppose that P is true 2)Assume that Q is false. (assumption) 3) 4) 5)Reach some contradiction. The assumption must be wrong. Therefore, Q must be true. End of proof But we could also use the method of contradiction to prove the contrapositive statement: Proof 1) Suppose that Q is false 2)Assume that P is true. (assumption) 3) 4) 5)Reach some contradiction. The assumption must be wrong. Therefore, P must be false. End of proof In summary, if we wish to prove that P Q, we may do it by proving that P Q or by proving that ~Q ~P. And we may choose to use the method of contradiction or not. Now, suppose that we wish to prove that P is logically equivalent to Q. To say that P and Q are logically equivalent means that P Q and Q P. In light of the discussion above, we could use the following four proof strategies: Strategy 1: Part 1: Prove that P Q. Part 2: Prove that Q P. Strategy 2: Part 1: Prove that ~Q ~P. Part 2: Prove that Q P. Strategy 3: Part 1: Prove that P Q. Page 116 of 150 Part 2: Prove that ~P ~Q. Strategy 4: Part 1: Prove that ~Q ~P. Part 2: Prove that ~P ~Q. Four possible strategies!! It’s actually worse than that. Remember that we can also choose whether or not to use the method of contradiction for any of the parts. So there are many many proof strategies to choose from. Now, what if we want to prove that three statements are logically equivalent? The available strategies multiply like bunnies. Here are just a few: (1) (2) (1) (2) (1) (2) (3) (3) (3) Strategy 2 Strategy 1 Strategy 3 I have left out the multitude of choices available such as whether to prove a conditional statement or its contrapositive, or whether or not to use the method of proof by contradiction. For each of the arrows in the diagrams above, those choices are still available. What you should notice is that there are new choices to be made. Notice that Strategy 1 would involve a proof of the equivalence of statements (2) and (3), but it would not involve a proof of the equivalence of statements (1) and (3). This is different from Strategy 2. Notice that Strategy 3 is very different from the first two strategies. What is cool about Strategy 3 is that it would involve a total of three proofs, whereas the other two strategies would involve four proofs. Theorem 66 is about the logical equivalence of thirteen statements! Using the method of Strategy 1 or 2 in the above diagram would involve twenty four proofs. Using the method of Strategy 3 would involve thirteen proofs. So it seems obvious that we ought to use Strategey 3. But the decision is not so simple. For one thing, some statements can be much harder to prove than others. So the fact that Strategy 3 involves only thirteen proofs does not mean that it would be the simplest strategy. Also, realize that when using Strategy 3, no equivalences are proven until the very end, when the circle of thirteen proofs is completed. At that time, all thirteen statements will have been proven to be complete. We don’t have the time for thirteen proofs. But we would like to get a taste of the equivalence proofs. For that reason, we will study the following proofs this week: (3) (6) (9) 9.4. Finally, Euclidean Geometry With Theorem 66, we come to the end of our study of Neutral Geometry. We will add one of the thirteen equivalent statements to the list of Neutral Geometry axioms, and the resulting axiom system will be called Euclidean Geometry. It doesn’t matter which of the thirteen statements we choose to add, but the most popular one to add is Statement (2), the one that we are call EPP. Page 117 of 150 9.5. Exercises [1] Negations of statements from Theorem 66. (a) Write Statement (2) and its negation. (b) Write Statement (4) and its negation. (c) Write Statement (6) and its negation. (d) Write Statement (8) and its negation. (e) Write Statement (9) and its negation. (f) Write Statement (10) and its negation. (g) Write Statement (11) and its negation. (h) Write Statement (13) and its negation. [2] Some missing parts of the proof of Theorem 66. (a) Prove that (11) (12). (b) Prove that (13) (11). [3] Supply one of the missing parts of the proof of Theorem 66. (Not one of the parts that we proved in class and not one of the parts from exercise [2]. Page 118 of 150 Page 119 of 150 10.Euclidean Geometry I In the previous chapter we studied Theorem 66, which presented thirteen statements that are equivalent in Neutral Geometry. We ended the chapter by adding one of the statements to the Neutral Geometry axiom list to create a longer axiom list. This new, longer axiom list describes Euclidean Geometry. The statement that we added to the axiom list is Statement (2), the one that we call EPP. By adding that statement to the axiom list, we declared it to be true. (Remember that there is no way to prove that Statement (2) is true, because Statement (2) is an independent statement in Neutral Geometry. That is, there are models of Neutral Geometry where Statement (2) is true, and there are models of Neutral Geometry where Statement (2) is false.) Because Statement (2) is true in Euclidean Geometry and Euclidean Geometry is a kind of Neutral Geometry, Theorem 66 tells us that all of the other twelve statements are also true in Euclidean Geometry. They are all true, but they need to be proven true, as theorems. That is, there would be twelve theorems, each proving that one of those remaining twelve statements is true in Euclidean Geometry. In the previous chapter, we did not thoroughly present proofs of all of the equivalences mentioned in Theorem 66. If we had thoroughly proved all the equivalences, then those proofs could now serve as proofs of the twelve theorems. So we have not yet done the job of proving that all of those twelve statements are true. We start this chapter by presenting some of those twelve statements as theorems, with explicit proofs. Following that, we will introduce circles and prove some theorems about them. Then we will go to Coldstone. 10.1. Some results from the previous chapter presented as theorems As discussed in class, a full proof of Theorem 66 would be quite long, because we would have to prove equivalences. We only proved one of the equivalences in class. (We proved (2) (4).) In the current section, we will not prove equivalences. We will just prove that a bunch of the thirteen statements are true as a consequence of statement (2) being true. Here, the statements will be presented as theorems, and they will be added to our list of Euclidean Geometry Theorems. The first theorem that we will prove is the statement that is number (4) on the list of thirteen statements. We discussed a proof of it in class. You will justify a proof of it in the exercises. Theorem 67 In Euclidean Geometry, if a line intersects one of two parallel lines, then it also intersects the other. Remember the naming convention for theorems: They are named for the situation described in their hypotheses. So the Alternate Interior Angle Theorem, Theorem 38, tells us something about congruent alternate interior angles: it tells us that they create parallel lines. It is important to remember that the Alternate Interior Angle Theorem can never be used to prove that alternate interior angles are congruent. Page 120 of 150 Finally now we come to a theorem that can be used to prove that alternate interior angles are congruent. It is the Converse of the Alternate Interior Angle Theorem. It was the statement that was number (6) on the list of thirteen. You will justify a proof of this theorem in the exercises. Theorem 68 In Euclidean Geometry, if two parallel lines are cut by a transversal, then any pair of alternate interior angles created is congruent.‖ (This statement is the converse of the statement of Theorem 38, The Alternate Interior Angle Theorem.) The next two theorems could really be called corollaries of Theorem 68, because their proofs are very short if that theorem is used. The statements are the ones that were numbered (7) and (8) on the list of thirteen. The first one is very easy to prove, and will not be assigned on homework or discussed in class. But the second one is a little harder to prove. You will justify a proof of it in the exercises. Theorem 69 In Euclidean Geometry, if two parallel lines are intersected by a transversal, then any pair of corresponding angles formed is congruent.‖ (This statement is the converse of the statement of Theorem 39,The Corresponding Angle Theorem.) Theorem 70 In Euclidean Geometry, if two parallel lines are intersected by a transversal, then the measures of pairs of interior angles on the same side of the transversal add up to 180. This next theorem was statement (5) on the list of thirteen statements. Its proof is also very short, but it uses both Theorem 67 and Theorem 68. You will prove it in the exercises. Theorem 71 In Euclidean Geometry, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. Statement (9) on the list of thirteen statements says that every triangle has angle sum exactly 180. I proved this statement in class. The proof is not difficult, but it is worth reviewing. You will justify a proof of the statement in the exercises. Theorem 72 In Euclidean Geometry, every triangle has angle sum exactly 180. The three statements numbered (11), (12), and (13) on the list of thirteen are presented here as the folloing three theorems. In Chapter 9 Exercise [2], you were asked to verify that (13) (11) and (11) (12). Here, though, the statements are listed as theorems. The ordering of the theorems means that Theorem 73 will have to be proved without any reference to the subsequent two thereoms. That means that the proof of Theorem 73 will be very different than the proof of (13) (11) that you did in Chapter 9 Exercise [2]. You will justify a proof of Theorem 73 in the homework. Theorem 73 In Euclidean Geometry, the opposite sides of a parallelogram are congruent. Theorem 74 In Euclidean Geometry, the opposite angles of a parallelogram are congruent. Theorem 75 In Euclidean Geometry, the diagonals of a parallelogram bisect each other. Page 121 of 150 10.2. Circles, chords, and diameters and some Neutral Geometry Theorems In this section, we will define circles and some related things, and we will prove a few theorems about them. We start with the definition of the circle. It is noteable that this is the first new definition since the introduction of Theorem 60, which was the theorem about existence of a length function for line segments. Now that we have the notion of length, all of our definitions and theorems can use it. However, it is useful to observe that many definitions and theorems can be formulated without referring to the notion of length. This is useful because it reminds us that the notion of length is not essential to Euclidean geometry. And it is useful because it would allow us to postpone the introduction of length until later, if we wanted to. And most importantly, it is useful to know about both versions because not all books use the same version. Definition 64 Circle Version of the definition that does not use the notion of distance o Symbol: Circle A, AB . o Words: Circle centered at A, with radial segment AB . o Meaning:The set of points P such that segment AP is congruent to segment AB . o Meaning in symbols: P : AP AB . Version of the definition that does use the notion of length o Symbol: Circle A, r . o Words: Circle centered at A, with radius r. o Meaning:The set of points P that are a distance r from point A. o Meaning in symbols: P : length AP r . The theorems about circles that we prove in this section will be about the relationship between circles and certain segments and lines. The segments and lines that we will be interetested in are presented in the following definition. Definition 65 Segments and Lines related to circles. Given a circle, o A chord is a line segment whose endpoints are points on a circle. o A diameter is a chord that also contains the center of the circle. o A secant is a line that passes through exactly two points of the circle. o A tangent is a line that passes through exactly one point of the circle. The theorems in this section can be proven using only the Neutral Geometry axioms. Because of that, I present them here as a Neutral Geometry Theorems. Page 122 of 150 The first two theorems have fairly easy proofs. The theorems are presented here mainly because they enable us to eliminate some of the clutter from later proofs. (Most books would call these Lemmas.) We discussed them in class on Monday June 2 and Tuesday June 3. Theorem 76 In Neutral Geometry, the following are equivalent: (1) Point P lies on the perpendicular bisector of segment AB . (2) Point P is equidistant from A and B. That is, AP BP and length AP length BP Theorem 77 In Neutral Geometry, all diameters of a given circle are congruent. The next four theorems about circles are more difficult to prove. You will justify their proofs in class and/or in the exercises. Theorem 78 In Neutral Geometry, diameters of a circle are the longest chords. Given a circle, if segment CD is a chord but not a diameter, and segment AB is a diameter, then segment CD AB . That is, length CD length AB . Theorem 79 In Neutral Geometry, the following are equivalent: (1) Diameter AB is perpendicular to non-diameter chord CD . (2) Diameter AB bisects non-diameter chord CD . Theorem 80 In Neutral Geometry, the perpendicular bisector of a chord passes through the center of the circle. Theorem 81 In Neutral Geometry, the following are equivalent: (1) Line L is perpendicular to segment AP at point P. (2) Line L is tangent to Circle A, AP at point P. 10.3. Some Euclidean Geometry Theorems about Circles and Triangles You might be a little bit puzzled by the fact that each of the theorems that we have so far proven about circles has been a theorem of Neutral Geometry. That tells us that certain aspects of the behavior of circles is the same in Euclidean and Hyperbolic Geometries. That should not be a big surprise. We previously found that many aspects of the behavior of triangles are the same in Euclidean and Hyperbolic Geometries. We will will now set out to state and prove a famous statement that is true about circles in Euclidean Geometry but is not true about circles in Hyperbolic Geometry. Before presenting the big theorem, it is useful to identify two smaller theorem that can serve as lemmas. These theorems are not famous, but they turn out to be the key to the proof of some famous theorems. Theorem 82 (really a Lemma) In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is perpendicular to line AB , and line M is perpendicular to line BC , then lines L and M intersect. Page 123 of 150 Theorem 83 (really a Lemma) In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is the perpendicular bisector of segment AB , and line M is the perpendicular bisector of segment BC , and line N is the perpendicular bisector of segment CA , then lines L, M, and N are concurrent. Theorem 84 In Euclidean Geometry, If A, B, and C are non-collinear points, then there exists exactly one circle that contains all three points. You will justify proofs of Theorem 82 and Theorem 83 and Theorem 84 in class drills. Theorem 84 is often presented in a different way, in terms of triangles. The term ―circumscribed‖ is often used. Here is a definition. Definition 66 Circumscribe Words: Triangle Meaning: Points A, B, and C lie on the circle. ABC is circumscribed by circle Circle P, PQ . With this terminology, we can present the following theorem, which is really just a re-packaging of the previous theorem: Theorem 85 In Euclidean Geometry, every triangle can be circumscribed by exactly one circle. That is for every triangle ABC , there exists exactly one circle Circle P, PQ such that points A, B, and C lie on the circle. 10.4. Could Theorem 84 be a theorem of Neutral Geometry? As an introduction, consider the following statement and a proof: Statement S: If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two triangles are congruent. Proof Let a1 , b1 , and h1 be the lengths of the two legs and hypotenuse of the first triangle, and let a2 , b2 , and h2 be the lengths of the two legs and hypotenuse of the second triangle. We are told that the hypotenuses are congruent, so h1 and h2 must actually be the same number, that we can just call h. We are also told that one leg of the first triangle is congruent to one leg of the second triangle. We can assume that it is the sides labeled a1 and a2 that are congruent, so a1 and a2 must actually be the same number, that we can just call a. The Pythagorean Theorem of Euclidean Geometry tells us that the three sides of a right triangle must satisfy the equation a 2 b 2 h 2 , where a and b are the lengths of the two legs and h is the length of the hypotenuse. Applying the Pythagorean Theorem to the first triangle, we have a 2 b12 h 2 . Solving this equation for b1, we obtain the equation b1 h 2 a 2 . Page 124 of 150 Applying the Pythagorean Theorem to the second triangle, we have a 2 b2 2 h 2 . Solving this equation for b2, we obtain the equation b2 h 2 a 2 . We see that b1 and b2 must actually be the same number. So the second legs of the two triangles are also congruent. Therefore, the triangles are congruent by Side-Side-Side. End of Proof The proof is perfectly valid, and it shows that Statement S is true in Euclidean Geometry. But it is a famous fact that the Pythagorean Theorem is not valid in Hyperbolic Geometry. Based on that fact, one might reasonably suspect that Statement S is not true in Hyperbolic Geometry, and so it cannot be a theorem of Neutral Geometry. But notice that S is the statement of Theorem 51, the Hypotenuse-Leg Congruence Theorem of Neutral Geometry. We have already proven that statement S is true in Neutral Geometry. What’s wrong? Nothing is wrong. But this example illustrates the idea that one should not assume that because a proof of some statement cites a theorem of Euclidean Geometry, the statement is only true in Euclidean Geometry. Maybe there is some other proof possible that only uses the axioms and theorems of Neutral Geometry. In light of this discussion, notice that Theorem 82 is listed as a theorem of Euclidean Geometry. You might think that the reason for that is that one of the steps in the proof is justified by a theorem of Euclidean Geometry. But that is not the reason. Just because you find a proof of Theorem 82 that uses a theorem of Euclidean geometry, you should not assume that Theorem 82 has to be listed as a Euclidean geometry theorem. You should wonder if maybe the statement of Theorem 82 is actually true in Neutral Geometry, and I was just not clever enough to come up with a proof that used only the Neutral Geometry axioms and theorems. We should consider the statement of Theorem 82 and try to determine if it is also true or false in Hyperbolic Geometry. It is useful to write down the statement of the theorem with explicit quantifiers, and then also write down the negation of the theorem, with explicit quantifiers. Statement S: For all points A, B, and C, and for all lines L and M, if A, B, and C are noncollinear , and line L is perpendicular to line AB , and line M is perpendicular to line BC , then lines L and M intersect. Statement ~S: There exist points A, B, and C, and lines L and M such that A, B, and C are noncollinear, and line L is perpendicular to line AB , and line M is perpendicular to line BC , and lines L and M do not intersect. Notice that statement S is a universal statement. To prove that statement S is true, we would need to come up with a general proof. If we want to prove that statement S is false, then we would need to do it by proving that the negation of S is true. But the negation, ~S, is an existential statement. To prove that ~S is true, we would need to come up with an example. Page 125 of 150 I will present an example that can prove that statement ~S is true. Consider the picture at right showing three circles all of radius 1. The bottom left circle is centered at the (0,0), the top circle is centered at 0, 2 , and the circle on the right is centered at 2, 0 . The top and right circles are orthogonal to the bottom left circle. So the top and right circle can be used as a source of Hlines that contain some known points. line L The picture at right shows the standard unit circle, along with the two Hlines obtained from the top and right circles of the previous picture. We can call the lines L and M. One known Hpoint on each line is labelled. Those are the points P and Q. P 0, 1 2 1 1 missing endpoint , 2 2 line M Q 1 2, 0 B = (0,0) Notice that Hlines L and M share a missing endpoint. Remember, though, that the missing endpoint is a point on the unit circle, not in the interior. So the missing endpoint is not an Hpoint. That means that Hlines L and M do not intersect. Notice also that Hline L is perpendicular to the y-axis and Hline M is perpendicular to the x-axis. Page 126 of 150 By axioms BA3 and CA2, we know that there is some point A such that A*P*B and such that AP BP . So point P is the midpoint of segment AB and Hline L is the perpendicular bisector of segment AB . Similarly there is some point C such that C*Q and such that CQ BQ . So point Q is the midpoint of segment CB and Hline M is the perpendicular bisector of segment CB . The points A, B, C, P, Q are shown at right. line L point A point P missing endpoint line M point C point Q point B This picture is a remarkable thing. It was created without the aid of the NonEuclid program, using only some translations of the unit circle and some references to the axioms of Neutral Geometry. It shows that there exist Hpoints A, B, and C, and Hlines L and M such that A, B, and C are non-collinear, and line L is perpendicular to line AB , and line M is perpendicular to line BC , and lines L and M do not intersect. So the picture serves as an example that shows that Statement ~S is true in Hyperbolic Geometry. In other words, it illustrates the fact that Statement S is not true in Hyperbolic Geometry. This tells us that Theorem 82 cannot be a theorem of Neutral Geometry. It is only true in Euclidean geometry. Notice that the three non-collinear Hpoints A, B, and C define a triangle. So the same picture (with those three points connected by line segments) could serve as an example of a triangle in Hyperbolic Geometry that cannot be circumscribed! Page 127 of 150 10.5. Exercises The first six exercises have to do with proofs of theorems that were originally on the list of thirteen equivalent statements. Now, instead of trying to prove that the statements are equivalent to each other, we are trying to prove that each is true asa theorem of Euclidean Geometry. [1] Recall the statement of Theorem 67: In Euclidean Geometry, if a line intersects one of two parallel lines, then it also intersects the other. Justify the steps in the following proof (we discussed this proof in class). Make a drawing. 1. Suppose that lines L and M are parallel and line T intersects line M. Denote by P the point of intersection of lines T and M, and observe that line M is a line that passes through P and is parallel to L, and also observe that line T is a line that passes through P. (Make a drawing) 2. There is at most one line that passes through P and are parallel to L. (by _____________ ___________________________________________, so line T cannot be parallel to L. 3. Line T intersects line L. (by ________________________________________________ ______________________________________________________________________) End of proof Page 128 of 150 [2] Recall the statement of Theorem 68: In Euclidean Geometry, if two parallel lines are cut by a transversal, then any pair of alternate interior angles created is congruent.‖ (This statement is the converse of the statement of Theorem 38, The Alternate Interior Angle Theorem.) Justify the steps in the following proof. Make a drawing where indicated. 1. Suppose that L and M are parallel lines cut by a transversal T. 2. There exist points A, B, C on line L and points D, E, F on line M such that point B is the point of intersection of lines L and T, and point E is the point of intersection of lines M and T, and A*B*C, and D*E*F, and such that points A and D are on the same side of line T. (This step can be justified by labelling the intersection points and by invoking the betweenness axioms.) (Make a drawing) With this labelling scheme, angles ABE and FEB are alternate interior angles. (Our goal is to show that these two angles are congruent.) 3. There exists a ray EG such that points G and A are on opposite sides of line T and such that GEB ABE . (by axiom ________________________) (Add to your drawing) 4. Line EG is parallel to L. (by step ___ and Theorem ___________________________ _____________________________________________________________________) 5. Line EG must be the same line as line M. (because ___________________________ _____________________________________________________________________ _____________________________________________________________________) 6. Rays EF and EG are the same ray. (By step 2, 3, and 5, we know that E, F, and G are collinear and that points F and G are on the same side of line T. Theorem __________ _____________________________________________________________________ can be used to justify that the rays are the same.) So angles GEB and FEB are the same angle. 7. FEB ABE (by seps ___ and ___) End of proof Page 129 of 150 [3] Recall the statement of Theorem 70: In Euclidean Geometry, if two parallel lines are intersected by a transversal, then the measures of pairs of interior angles on the same side of the transversal add up to 180. Justify the steps in the following proof. Make a drawing where indicated. 1. Suppose that L and M are parallel lines cut by a transversal T. 2. There exist points A, B, C on line L and points D, E, F on line M such that point B is the point of intersection of lines L and T, and point E is the point of intersection of lines M and T, and A*B*C, and D*E*F, and such that points A and D are on the same side of line T. (This step can be justified by labelling the intersection points and by invoking the betweenness axioms.) (Make a drawing) With this labelling scheme, angles ABE and DEB are a pair of angles on the same side of the transversal. (Our goal is to show that the measures of these two angles add up to 180.) 3. Angles DEB and FEB are supplementary angles. (by step ___ and definition of ________________________________________________) 4. degrees DEB degrees FEB 180 (by step ___ and Theorem _____________ ____________________________________________________________________) 5. Angles ABE and FEB are alternate interior angles. (by _____________________ _____________________________________________________________________) 6. ABE FEB (by steps ___ and ___ and Theorem ___________________________ ______________________________________________________________________) 7. degrees ABE degrees FEB (by step ___ and Theorem ___________________ ______________________________________________________________________) 8. degrees DEB degrees AEB 180 (by steps ___ and ___ and algebra) End of proof [4] Prove Theorem 71: In Euclidean Geometry, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. Hint: the proof can be done in just three lines if you use two recent theorems. Page 130 of 150 [5] Recall the statement of Theorem 72: In Euclidean Geometry, every triangle has angle sum exactly 180. Justify the steps in the following proof. (We discussed the proof in class). Make a drawing 1. Let ABC be given. (make a drawing, and try to make the triangle lopsided) 2. There exists a line L that passes through C and is parallel to line AB . (by Theorem ____ ____________________________________________________) (add to your drawing) 3. There exist points D and E on line L such that D*C*E and such that points D and A are on the same side of line BC . (This step can be justified by betweenness axioms and theorems.) (add to your drawing) 4. Angles DCB and BCE are supplementary. (by steps ___ and ___ and definition of ______________________________.) 5. degrees DCB degrees BCE 180 (by step ___ and Theorem _______________ _______________________________________________________________________) 6. degrees DCB degrees DCA degrees ACB (by Theorem ______________) 7. degrees DCA degrees ACB degrees BCE 180 (by steps ___ and ___ and algebra) 1. Angles DCA and CAB are alternate interior angles. (by ______________________ ______________________________________________________________________) 2. DCA CAB (by steps ___ and ___ and Theorem ___________________________ _______________________________________________________________________) 3. degrees DCA degrees CAB (by step ___ and Theorem ____) 4. Angles ECB and CBA are alternate interior angles. (by ______________________ ______________________________________________________________________) 5. ECB CBA (by steps ___ and ___ and Theorem ___________________________ ______________________________________________________________________) 6. degrees ECB degrees CBA (by step ___ and Theorem ______________) 7. degrees CAB degrees ACB degrees CBA 180 (by steps ___ and ___ and ___ and algebra) End of proof Page 131 of 150 [6] Recall the statement of Theorem 73: In Euclidean Geometry, the opposite sides of a parallelogram are congruent. Justify the steps in the following proof. Make a drawing where indicated. And supply the second half of the proof. 1. Let ABCD be a parallelogram. (so AB CD and AD BC .) (Make a drawing) Part 1: Prove that DA BC by using diagonal AC . 2. Angles DCA and BAC are alternate interior angles. (by _________________ ________________________________________________) (add to your drawing) 3. DCA BAC (by steps ___ and ___ and Theorem _______________________ __________________________________________________________________) 4. Angles DAC and BCA are alternate interior angles. (by ___________________ ___________________________________________________________________) 5. DAC BCA (by steps ___ and ___ and Theorem ________________________) 6. AC AC (by _______________________________________________________) 7. DAC BCA (by steps ___ and ___ and ___ and __________________________) 8. DA BC (by step _ and the definition of __________________________________) Part 2: Prove that AB CD by using steps analogous to steps 2 – 8, but using diagonal BD . End of Proof Page 132 of 150 The next three exercises deal with justifications of proofs of Neutral Geometry Theorems about Circles. [7] Recall the statement of Theorem 78: In Neutral Geometry, diameters of a circle are the longest chords. Given a circle, if segment CD is a chord but not a diameter, and segment AB is a diameter, then segment CD AB . That is, length CD length AB . Justify the steps in the following proof. Make a drawing where indicated. 1. Suppose that segment CD is a chord but not a diameter, and segment AB is a diameter. 2. Let point F be the center of the circle. We can denote the circle by Circle F , FC . Let point G be the second intersection of line CF with the circle. Then segment CG is a diameter. (Make a drawing showing the circle and chord CD and diameter CG .) Part 1: show that CD CG . 3. FD FG (by _________________________________________) (add to your drawing) 4. FGD FDG (by ____________________________________________________) 5. Observe that point F is in the interior of angle ADG , so that ray DF is between rays DC and DG . 6. FDG CDG (by step ___ and definition of ________________________________) 7. FGD CDG (by steps ___ and ___ and Theorem ___________________________) 8. segment CD CG . (by step ___ and Theorem _________________________________ _______________________________________________________________________) Part 2: show that CD AB . 9. CG AB (by steps ___ and ___ and Theorem _________________________________ _______________________________________________________________________) 10. CD AB (by steps ___ and ___ and Theorem __________________________________ _______________________________________________________________________) End of proof Page 133 of 150 [8] Recall the statement of Theorem 79: In Neutral Geometry, the following are equivalent: (1) Diameter AB is perpendicular to non-diameter chord CD . (2) Diameter AB bisects non-diameter chord CD . Justify the steps in the following proof. Make drawings where indicated. Part 1: Prove that (2) (1) 1. Suppose that diameter AB bisects non-diameter chord CD . Let E be the center of the circle and let F be the point of intersection of AB and CD . (Make a drawing) 2. CF DF (by definition of ____________________) (add to your drawing) 3. EF EF (by ___________________________________________________) 4. EC ED (by ____________________________________________________) 5. CFE DFE (by steps ___ and ___ and ___ and Theorem _________________ __________________________________________________________________) 6. CFE DFE (by step ___ and the definition of _________________________ __________________________________________________________________) 7. Angles CFE and DFE are supplementary. (by ________________________) 8. Angles CFE and DFE are right angles. (by steps ___ and ___ and definition of ___ ______________________________________________________________________) 9. AB is perpendicular to CD . (by step ___ and definition of perpendicular) Part 2: Prove that (1) (2) 1. Suppose that diameter AB is perpendicular to non-diameter chord CD . Let E be the center of the circle and let F be the point of intersection of AB and CD . (We must show that F is the midpoint of segment CD .) (make a drawing) 2. Assume that F is not the midpoint of segment CD . (assumption) 3. Segment CD has a midpoint that we can label G. (by Theorem _________________) 4. Line AG bisects segment CD . (by step ___ and definition of _________________) 5. Line AG contains a diameter HJ that also bisects segment CD . (by steps 4 and 3 and definition of diameter) 6. Diameter HJ is perpendicular to segment CD . (by step 5 and Part 1 of this proof) 7. Line AG is perpendicular to line CD . (by steps 5 and 6) 8. Line AB is perpendicular to line CD . (by step 1) 9. Lines AG and AB are parallel. (by step ___ and Theorem _______________________ ______________________________________________________________________) 10. Step 9 contradicts the fact that lines AG and AB intersect at point A. Therefore, our assumption in step 2 was wrong. It cannot be that F is not the midpoint of segment CD . Therefore, F is the midpoint of segment CD . End of proof Page 134 of 150 Recall the statement of Theorem 80:In Neutral Geometry, the perpendicular bisector of a chord passes through the center of the circle. Justify the steps in the following proof. Make a drawing where indicated. 1. Suppose that segment CD is a chord of Circle A, AB . 2. Segment CD has a perpendicular bisector that we can call line L. (by Theorem ______ ______________________________________________________________) We must show that line L passes through the center of the circle. 3. Line L is perpendicular to segment CD at a point E that is the midpoint of segment CD . (by definition of perpendicular bisector) 4. Either segment CD is a diameter chord or segment CD is a non-diameter chord. Case 1: Segment CD is a diameter chord. 5. Suppose that segment CD is a diameter chord. (make a drawing) 6. Then the midpoint E of segment CD is actually the center of the circle, so line L passes through the center of the circle. Case 2: Segment CD is a non-diameter chord 7. Suppose that segment CD is a non-diameter chord. (make a new drawing) 8. Line AE intersects the circle at two points that we can call F and G, creating diameter segment FG . (add to your drawing) 9. Diameter FG bisects non-diameter chord CD . (because diameter FG is part of line AE that passes through point E.) 10. Diameter FG is perpendicular to non-diameter chord CD . (by step ___ and Theorem __________________________________________________________________) 11. Line AE is a perpendicular bisector of segment CD . (by steps ___ and ____ and definition of _____________________________________________) 12. Line L and line AE must be the same line. (because segment CD has only one perpendicular bisector, by Theorem ____________________________) So line L passes through the center of the circle. End of proof Page 135 of 150 11.For Reference: Axioms, Definitions, and Theorems of Euclidean Geometry 11.1. The Axioms of Euclidean Geometry Primitive Relations: incidence relation, written ―the point lies on the line.‖ betweenness relation for points, written ―B is between A and C‖, denoted A*B*C. congruence of segments, written ―segment AB is congruent to segment CD‖, AB CD . congruence of angles, written ―angleABC is congruent to angleDEF‖, ABC DEF Incidence Axioms IA1: There exist three distinct non-collinear points. (at least three) IA2: For every pair of distinct points, there is exactly one line that passes through both points. IA3: Every line passes through at least two distinct points. Betweenness Axioms BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A. BA2: If A, B, and C are three distinct points lying on the same line, then exactly one of the points is between the other two. BA3: If B and D are distinct points, and L is the unique line that passes through both points, then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E. BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then (i) If A and B are on the same side of L and B and C are on the same side of L, then A and C are on the same side of L. (ii) If A and B are on opposite sides of L and B and C are on opposite sides of L, then A and C are on the same side of L. Congruence Axioms CA1: The congruence relation on the set of line segments is an equivalence relation CA2: (segment construction axiom) For any segment AB and ray PQ , there exists a unique point R on PQ such that PR AB . CA3: (segment addition) If A*B*C, A'*B'*C', AB AB , and BC BC , then AC AC . CA4: The congruence relation on the set of angles is an equivalence relation CA5: (angle construction axiom) Given an angle BAC , distinct points A' and B', and a choice of one of the two half-planes bounded by the line AB , there exists a unique ray AC such that C' lies in the chosen half plane and BAC BAC . CA6: (SAS axiom) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Axioms of Continuity Archimedes Axiom: Given any segment CD and ray AB , there exists some positive integer n and a set of points P0 , P1 , , Pn on AB such that P0 A ; for all k, Pk Pk 1 CD ; and either Pn B or ( A * Pn1 * B and A * B * Pn ). Circular Continuity Axiom: If a circle intersects both the interior and exterior of another circle, then the two circles intersect in exactly two points. The Euclidean Parallel postulate (EPP): For any line L and for any point P not on L, there is at most one line that passes through P and is parallel to L. Page 136 of 150 11.2. The Definitions of Euclidean Geometry Definition 18 passes through words: Line L passes through point P. meaning: Point P lies on line L. Definition 19 intersecting lines words: Line L intersects line M. meaning: There exists a point P that lies on both lines. (at least one point) Definition 20 parallel lines words: Line L is parallel to line M. symbol: L M meaning: Line L does not intersect line M. Definition 21 collinear points words: The set of points P1 , P2 , meaning: There exists a line L that passes through all the points. Definition 22 concurrent lines words: The set of lines L1 , L2 , , Pk is collinear. , Lk is concurrent. meaning: There exists a point P that lies on all the lines. Definition 23 Abstract Model and Concrete Model An abstract model of an axiom system is a model that is, itself, another axiom system. A concrete model of an axiom system is a model that uses actual objects and relations. Definition 24 Relative Consistency and Absolute Consistency An axiom system is called relatively consistent if an abstract model has been demonstrated. An axiom system is called absolutely consistent if a concrete model has been demonstrated. Definition 25 Hpoint word: Hpoint meaning: an (x,y) pair in the interior of the unit circle Definition 26 Hline word: Hline meaning: Either of the following particular types of sets of Hpoints A ―straight-looking‖ Hline is the set of Hpoints that lie on a diameter of the unit circle. A ―curved-looking‖ Hline is the set of Hpoints that lie on a circle that is orthogonal to the unit circle. Definition 27 Ternary Relation on a Set Page 137 of 150 Words: R is a ternary relation on A. Usage: A is a set. Meaning: R is a subset of A A A . Equivalent meaning in symbols: R A A A Definition 28 The betweenness relation on the set of real numbers Words: ―x is between y and z.‖ Usage: x, y, and z are real numbers. Meaning: ―x < y <z or z <y <x.‖ Remark: This is a ternary relation on the set of real numbers. Warning: This is NOT the same as betweenness for points, discussed in the next section. Definition 29 symbol for a line symbol: AB meaning: the (unique) line that passes through points A and B Definition 30 the set of points that lie on a line symbol: AB meaning: the set of points that lie on line AB Definition 31 the plane meaning: the set of all points Definition 32 line segment, endpoints of a line segment symbol: AB spoken: ―line segment A, B‖, or ―segment A, B‖ usage: A and B are points. meaning: the set AB A C : A * C * B B additional terminology: points A and B are called endpoints of segment AB . Definition 33 ray, endpoints of a ray symbol: AB spoken: ―ray A, B‖ usage: A and B are points. meaning: the set AB A C : A * C * B B D : A * B * D additional terminology: Point A is called the endpoint of ray AB . We say that ray AB emanates from point A. Definition 34 opposite rays words: BA and BC are opposite rays meaning: A*B*C Definition 35 same side Page 138 of 150 words: ―A and B are on the same side of L.‖ usage: A and B are points and L is a line that does not pass through either point. meaning: either A B or ( A B and line segment AB does not intersect line L.) Definition 36 opposite side words: ―A and B are on the opposite side of L.‖ usage: A and B are points and L is a line that does not pass through either point. meaning: A B and line segment AB does intersect line L. Definition 37 half plane words: half-plane bounded by L, containing point A. meaning: the set whose elements are some point A not on L, along with all the other points that are on the same side of L as point A symbol: HA Definition 38 angle symbol: ABC usage: A, B, and C are non-collinear points meaning: BA BC additional terminology: point B is called the vertex of ABC , rays BA and BC are called the sides. observations: because BA BC BC BA , the symbols ABC and CBA represent the same angle. Definition 39 supplementary angles words: supplementary angles meaning: two angles that share a common side and whose other sides are opposite rays. Definition 40 interior of an angle words: the interior of ABC meaning: the set of points P such that (P is on the same side of line BA as point C) and (P is on the same side of line BC as point A). Definition 41 ray between two other rays words: ray BD is between BA and BC . meaning: Point D is in the interior of ABC . Definition 42 triangle symbol: ABC spoken: triangle A, B, C usage: A, B, and C are non-collinear meaning: the set AB BC CA additional terminology: Page 139 of 150 o o o o o The points A, B, C are called the vertices of the triangle. The segments AB, BC, CA are called the sides of the triangle. Side BC is said to be opposite vertex A. Similarly for the other sides. The angles ABC , BCA , and CAB are called the angles of the triangle. The angle ABC is called angle B when there is no chance of this causing confusion. Similarly for the other angles. Definition 43 interior of a triangle words: the interior of ABC meaning: the set of points P such that (P is on the same side of line BA as point C) and (P is on the same side of line BC as point A) and (P is on the same side of line AC as point B) Definition 44 exterior of a triangle meaning: the set of points Q that are neither an element of the triangle, itself, nor of the interior of the triangle. Definition 45 the order relation on the set of line segments Symbol: AB CD Spoken: “Segment A,B is less than segment C,D.” Meaning: There exists a point E between C and D such that AB CE . Remark: The order relation is a binary relation on the set of line segments. Definition 46 ―function‖, ―domain‖, ―codomain‖, ―image‖, ―machine diagram‖; ―correspondence‖ Symbol: f : A B Spoken: “ f is a function that maps A to B ” Usage: A and B are sets. Set A is called the domain and set B is called the codomain. Meaning: f is a machine that takes an element of set A as input and produces an element of set B as output. More notation: If an element a A is used as the input to the function f , then the symbol f (a ) is used to denote the corresponding output. The output f (a ) is called the image of a under the map f . Machine Diagram: a input f f a output Domain: Codomain: the set A the set B Additional notation: If f is both one-to-one and onto (that is, if f is a bijection), then the symbol f : A B will be used. In this case, f is called a correspondence between the sets A and B. Page 140 of 150 Definition 47 Correspondence between the vertices of two triangles Words: “f is a correspondence between the vertices of triangles ABC and DEF .” Meaning: f is a one-to-one, onto function with domain A, B, C and codomain D, E , F . Definition 48 Corresponding parts of two triangles Words: Corresponding parts of triangles ABC and DEF . Usage: A correspondence between the vertices of ABC and DEF has been given. Meaning: As discussed above, there is an automatic correspondence between the sides of triangles ABC and the sides of DEF , and also between the angle of triangles ABC and the angles of DEF . Suppose the correspondence between vertices were B, A, C D, E, F . Corresponding parts would be pairs such as the pair of sides, AC EF , or the pair of angles, ACB EFD . Definition 49 Triangle Congruence Symbol: ABC DEF Words: “ ABC is congruent to DEF .” Meaning: “There is a correspondence between the vertices of the two triangles such that corresponding parts of the triangles are congruent.” Remark: Triangle congruence is a defined binary relation on the set of triangles. Additional terminology: If a correspondence between vertices of two triangles has the property that corresponding parts are congruent, then the correspondence is called a congruence. Definition 50 vertical angles words: vertical angles Meaning: Two angles that share a vertex and whose sides are opposite rays. That is, two angles that can be labeled ABC and DBE where A*B*D and C*B*E. Definition 51 right angle words: right angle Meaning: An angle that is congruent to its supplementary angle Definition 52 perpendicular lines symbol: L M spoken: L and M are perpendicular Meaning: L and M are distinct, non-parallel lines that create a pair of vertical angles that are right angles. Definition 53 the order relation on the set of angles Symbol: ABC DEF Spoken: “Angle ABC is less than angle DEF .” Meaning: There exists a ray EG between ED and EF such that ABC GEF . Page 141 of 150 Remark: The order relation is a binary relation on the set of angles. Definition 54 transversal Words: “Line T is transversal to lines L and M.” Meaning: “T intersects L and M in distinct points.” Definition 55 transversal―alternate interior angles‖; ―corresponding angles‖ Usage: Lines L, M, and transversal T are given. Labeled points: Let B be the H T intersection of lines T and L, and let E D be the intersection of lines T and M. M E F (By definition of transversal, B and E A B C L are not the same point.) By the G betweenness axioms, there exist points A and C on line L such that A*B*C, points D and F on line M such that D*E*F, and points G and H on line T such that G*B*E and B*E*H. Without loss of generality, we may assume that points D and F are labeled such that it is point D that is on the same side of line BE as point A. Meaning: ABE and FEB is a pair of alternate interior angles. CBE and DEB is a pair of alternate interior angles. ABG and DEG is a pair of corresponding angles. ABH and DEH is a pair of corresponding angles. CBG and FEG is a pair of corresponding angles. CBH and FEH is a pair of corresponding angles. Definition 56 ―isosceles triangle‖ Words: isosceles triangle Meaning: two sides of the triangle are congruent to each other (at least two) Additional Terminology: The angles opposite the two congruent sides are called the base angles. Definition 57 ―exterior angle‖, ―interior angle‖, ―remote interior angles‖ Words: An exterior angle of a triangle Meaning: An angle that is supplemental to one of the angles of the triangle Additional terminology: The angles of the triangle are also called “interior angles”. Given an exterior angle, there will be an interior angle that is its supplement, and two other interior angles that are not its supplement. Those other two interior angles are called remote interior angles. Definition 58 ―hypotenuse‖ and ―legs‖ of a right triangle Meaning: The hypotenuse of a right triangle is the side that is opposite the right angle. The other two sides are called legs. Definition 59 ―midpoint‖ of a line segment Page 142 of 150 Words: C is a midpoint of segment A, B Meaning: A*C*B and CA CB . Definition 60 ―bisector‖ of a line segment, ―perpendicular bisector‖ of a line segment Words: Line L is a bisector of segment AB . Meaning: L is distinct from line AB and passes through the midpoint of segment AB . Additional Terminology: If L is perpendicular to line AB and is also a bisector of segment AB , then L is said to be a perpendicular bisector of segment AB . Definition 61 ―bisector‖ of an angle Words: a bisector of angle ABC Meaning: a ray BD between rays BA and BC such that DBA DBC . Definition 62 the length function for line segments in straight-line drawings (Euclidean) Meaning: the function lengthE : Euclideanline segments defined by the following formula lengthE AB x1 x2 y1 y2 2 2 , where segment AB is a Euclidean line segment with endpoints A x1 , y1 and B x2 , y2 . (This implies that x1, y1, x2,and y2 are real numbers.) Definition 63 the length function for line segments in the Hyperbolic plane Meaning: the function lengthH : Hyperbolicline segments defined by the following formula lengthE AR lengthE AS , lengthH AB ln lengthE BR length BS E where segment AB is a Hyperbolic line segment with endpoints that are the Hpoints A and B, and R and S are the ―missing endpoints‖ of the line AB . Definition 64: Circle Version of the definition that does not use the notion of distance o Symbol: Circle A, AB . o Words: Circle centered at A, with radial segment AB . o Meaning:The set of points P such that segment AP is congruent to segment AB . o Meaning in symbols: P : AP AB . Version of the definition that does use the notion of length o Symbol: Circle A, r . o Words: Circle centered at A, with radius r. Page 143 of 150 o Meaning:The set of points P that are a distance r from point A. o Meaning in symbols: P : length AP r . Definition 65: Segments and lines related to a circle Given a circle, o A chord is a line segment whose endpoints are points on a circle. o A diameter is a chord that also contains the center of the circle. o A secant is a line that passes through exactly two points of the circle. o A tangent is a line that passes through exactly one point of the circle. Definition 66: Circumscribe Words: Triangle Meaning: Points A, B, and C lie on the circle. ABC is circumscribed by circle Circle P, PQ . 11.3. The Theorems of Euclidean Geometry Theorem 1: In Incidence Geometry, if L and M are distinct lines that are not parallel, then there is exactly one point that both lines pass through. Theorem 2: In Incidence Geometry, there exist three lines that are not concurrent. Theorem 3: In Incidence Geometry, given any line L, there exists a point not lying on L. Theorem 4: In Incidence Geometry, given any point P, there exists a line that does not pass through P. Theorem 5: In Incidence Geometry, given any point P, there exist two lines that pass through P. Theorem 6: In I&B Geometry, if A*B*C and A*C*D then A, B, C, D are distinct and collinear. Theorem 7: In I&B Geometry, every ray has an opposite ray. Theorem 8: In I&B Geometry, for any two distinct points A and B, AB BA AB . Theorem 9: In I&B Geometry, for any two distinct points A and B, AB BA AB . Theorem 10: In I&B Geometry, if A and B are on opposite sides of line L, and B and C are on the same side of L, then A and C are on opposite sides of L. Theorem 11: In I&B Geometry, every line L partitions the plane into three sets: (1) the set of points that lie on L, (2) a half plane, and (3) a second half-plane. In other words, every point of the plane either lies on L or is an element of one (not both) of the half planes. Page 144 of 150 Theorem 12: In I&B Geometry, if A*B*C and A*C*D, then B*C*D and A*B*D. Theorem 13: In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D. Theorem 14: (Line Separation) In I&B Geometry, if A, B, C, and D are collinear points such that A*B*C, and D B , then either D AB or D AC , but not both. Theorem 15: (Pasch’s Theorem) In I&B Geometry, if A, B,and C are non-collinear points and line L intersects segment AB at a point between A and B, then L also intersects either segment AC or segment BC . Furthermore, if C does not lie on L, then L does not intersect both segment AC and segment BC . Theorem 16: In I&B Geometry, if A*B*C then AC AB BC and B is the only point common to segments AB and BC . Theorem 17: In I&B Geometry, if A*B*C then B BA BC and AB AC . Theorem 18: In I&B Geometry, if point A lies on line L and point B does not lie on line L, then every point of ray AB except point A is on the same side of L as B. Theorem 19: In I&B Geometry, given BAC and point D lying on BC , point D is in the interior of BAC if and only if B*D*C. Theorem 20: In I&B Geometry, given angle BAC ; point D in the interior of BAC , and point E such that C*A*E, the following three statements are all true: (1) Every point on ray AD except A is in the interior of angle BAC . (2) No point on the ray opposite to AD is in the interior of angle BAC . (3) Point B is in the interior of angle DAE . Theorem 21: (The Crossbar Theorem) In I&B Geometry, if ray AD is between rays AB and AC then ray AD intersects segment BC . Theorem 22: In I&B Geometry, if ray r emanates from an exterior point of triangle ABC and intersects side AB in a point between A and B, then ray r also intersects side AC or BC . Theorem 23: In I&B Geometry, if a ray emanates from an interior point of a triangle, then it intersects one of the sides, and if it does not pass through a vertex, it intersects only one side. Theorem 24: In I&B Geometry, a line cannot be contained in the interior of a triangle. Theorem 25 (segment subtraction): In Neutral Geometry, if A*B*C and D*E*F and AB DE and AC DF , then BC EF . Page 145 of 150 Theorem 26: In Neutral Geometry, if AC DF and A*B*C, then there exists a unique point E such that D*E*F and AB DE . Theorem 27: Facts about the order relation on the set of line segments Given: Neutral Geometry; line segments AB , CD , and EF . Claim (a) (trichotomy) Exactly one of the following is true: AB CD , AB CD , or CD AB . (b) If AB CD and CD EF , then AB EF . (c) (transitivity)If AB CD and CD EF , then AB EF . Theorem 28: In Neutral Geometry, supplements of congruent angles are congruent. That is, if ABC and CBD are supplementary, and EFG and GFH are supplementary, and ABC EFG , then CBD GFH . Theorem 29: In Neutral Geometry, vertical angles are congruent. Theorem 30: In Neutral Geometry, any angle congruent to a right angle is also a right angle. Theorem 31: In Neutral Geometry, for every line L and every point P not on L there exists a line through P perpendicular to L. Theorem 32: In Neutral Geometry, for every line L and every point P on L there exists a unique line through P perpendicular to L. Theorem 33 (angle addition): Given: Neutral Geometry; angle ABC with point D in the interior; angle point H in the interior; DBA HFE ; DBC HFG Claim: ABC EFG Theorem 34 (angle subtraction): Given: Neutral Geometry; angle ABC with point D in the interior; angle point H in the interior; DBA HFE ; ABC EFG Claim: DBC HFG EFG with EFG with Theorem 35 (Theorem about Rays and Angles): In Neutral Geometry, if ABC EFG and ray BD is between rays BA and BC , then there is a unique ray FH between rays FE and FG such that ABD EFH . Theorem 36 (facts about the order relation on the set of angles): Given: Neutral Geometry; angles P , Q , R . Claim (a) (trichotomy): Exactly one of the following is true: P Q , (b): If P Q and Q R then P R . (c): (transitivity): If P Q and Q R then P R . P Q , or Theorem 37 (Euclid’s 4th postulate): All right angles are congruent to each other. Q P. Page 146 of 150 Theorem 38 (The Alternate Interior Angle Theorem): Given: Neutral Geometry, lines L and M and a transversal T Claim: If a pair of alternate interior angles is congruent, then lines L and M are parallel. Theorem 39 (The Corresponding Angle Theorem): Given: Neutral Geometry, lines L and M and a transversal T Claim: If a pair of corresponding angles is congruent, then lines L and M are parallel. Theorem 40: In Neutral Geometry, two distinct lines perpendicular to the same line are parallel. Theorem 41 (Uniqueness of the perpendicular from a point to a line): Given: Neutral Geometry, line L and a point P not on L Claim: There is not more than one line that passes through P and is perpendicular to L. Theorem 42 (Existence of parallel lines) (The answer to THE BIG QUESTION): Given: Neutral Geometry, line L and a point P not on L Claim: There exists at least one line that passes through P and is parallel to L. Theorem 43 (ASA congruence): In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and the included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Theorem 44 (The Isosceles Triangle Theorem): Given: Neutral Geometry, ABC Claim: If AB AC then B C . (If two sides are congruent, then the two opposite angles are congruent.) Theorem 45 (The Converse of the Isosceles Triangle Theorem) Given: Neutral Geometry, ABC Claim: If B C then AB AC (If two angles are congruent, then the two opposite sides are congruent.) Theorem 46 (The CACS Theorem): In Neutral Geometry triangles, Congruent Angles are always opposite Congruent Sides. Theorem 47 (The Triangle Construction Theorem): Given: Neutral Geometry, ABC , DE AB , and a point G not on DE . Claim: There exists a unique point F in half plane HG such that ABC DEF . Theorem 48 (The Side-Side-Side Congruence Theorem) (SSS): In Neutral Geometry, if there is a correspondence between parts of two triangles such that the three sides of one triangle are congruent to the corresponding sides of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Page 147 of 150 Theorem 49 (The Exterior Angle Theorem): In Neutral Geometry, each of the remote interior angles is less than the exterior angle. Theorem 50 (The Angle Angle Side Angle Congruenc Theorem) (AAS): In Neutral Geometry, if there is a correspondence between parts of two triangles such that two angles and a non-included side of one triangle are congruent to the corresponding parts of the other triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Theorem 51 (The Hypotenuse-Leg Congruence Theorem): In Neutral Geometry, if the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then all the remaining corresponding parts are congruent as well, so the triangles are congruent. Theorem 52: In Neutral Geometry triangle ABC , if AC BC then Theorem 53: In Neutral Geometry triangle ABC , if B B A. A then AC BC . Theorem 54 (The SASS Theorem): In Neutral Geometry triangles, Smaller Angles are always opposite Smaller Sides. (Not a new theorem, but rather just a combining of Theorem 52 and Theorem 53.) Theorem 55: The CACS and SASS Theorem: In triangles of Neutral Geometry, Congruent Angles are always opposite Congruent Sides, and Smaller angles are always opposite Smaller Sides. (Not a new theorem, but rather just a combination of the two theorems: Theorem 46 (the CACS Theorem), and Theorem 54 (the SASS Theorem). But these in turn are combinations of four theorems: Theorem 44, Theorem 45, Theorem 52 and Theorem 53.) Theorem 56, The Hinge Theorem: Given: Neutral Geometry, ABC , DEF , AB DE , AC DF Claim: The following are equivalent (1) A D . (2) BC EF . Theorem 57: In Neutral Geometry, every segment has exactly one midpoint. Theorem 58: In Neutral Geometry, every line segment has exactly one perpendicular bisector. Theorem 59: In Neutral Geometry, every angle has exactly one bisector. Theorem 60 Existence of a Length Function for Line Segments. Given: Neutral Geometry and some reference line segment AB . Claim: There exists a unique function length : line segments properties. with the following Page 148 of 150 1. length AB 1 . 2. The length function is onto. That is, for every positive real number L, there exists a line segment CD such that length CD L . length CD length EF if and only if. CD EF . length CE length CD length DE if and only if C*D*E. 3. length CD length EF if and only if CD EF . 4. 5. Theorem 61 Existence of a Measure Function for Angles Given: Neutral Geometry Claim: There exists a unique function degrees : angles 0,180 with the following properties. 1. degrees A 90 if and only if A is a right angle. 2. The degrees function is onto. That is, for every real number 0 m 180 , there exists an angle A such that degrees A m . 3. degrees A degrees B if and only if A B. 4. degrees A degrees B if and only if A B. 5. If ray AC is between rays AB and AD then degrees BAD degrees BAC degrees CAD . 6. If angle A is supplementary to angle B , then degrees A degrees B 180 . Theorem 62 In Neutral Geometry, the sum of the measures of any two angles of a triangle is less than 180. Theorem 63 (The Triangle Inequality) In Neutral Geometry, the length of any side of a triangle is less than the sum of the lengths of the two other sides. Theorem 64 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any triangle is less than or equal to 180. Theorem 65 In Neutral Geometry, the sum of the degree measures of any two angles of a triangle is less than or equal to the degree measure of their remote exterior angle. Theorem 66: Thirteen Statements that are Logically Equivalent in Neutral Geometry Given: The axioms for Neutral Geometry Claim: The following statements are logically equivalent: 1) Euclid’s Fifth Postulate: ―That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.‖ Page 149 of 150 2) A reworded version of SMSG Postulate #16, which we will call EPP: The Euclidean Parallel Postulate: For any line L and for any point P not on L, there is at most one line that passes through P and is parallel to L. 3) Playfair’s Postulate: For every line L and every point P not on L, there is a unique line that passes through P and is parallel to L. 4) ―If a line intersects one of two parallel lines, then it intersects the other.‖ 5) ―If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.‖ 6) ―If two parallel lines are cut by a transversal, then any pair of alternate interior angles created is congruent.‖ (This statement is the converse of Theorem 38, The Alternate Interior Angle Theorem.) 7) ―If two parallel lines are intersected by a transversal, then any pair of corresponding angles formed is congruent.‖ (This statement is the converse of the statement of Theorem 39,The Corresponding Angle Theorem,) 8) ―If two parallel lines are intersected by a transversal, then the measures of pairs of interior angles on the same side of the transversal add up to 180.‖ 9) ―Every triangle has angle sum exactly 180.‖ 10) ―Given any triangle PQR and any line segment AB , there exists a triangle ABC having AB as one of its sides such that ABC is similar to PQR but not congruent to PQR .‖ 11) ―The opposite sides of a parallelogram are congruent.‖ 12) ―The opposite angles of a parallelogram are congruent.‖ 13) ―The diagonals of a parallelogram bisect each other.‖ Theorem 67: In Euclidean Geometry, if a line intersects one of two parallel lines, then it also intersects the other. Theorem 68: In Euclidean Geometry, if two parallel lines are cut by a transversal, then any pair of alternate interior angles created is congruent.‖ (This statement is the converse of the statement of Theorem 38, The Alternate Interior Angle Theorem.) Theorem 69: In Euclidean Geometry, if two parallel lines are intersected by a transversal, then any pair of corresponding angles formed is congruent.‖ (This statement is the converse of the statement of Theorem 39,The Corresponding Angle Theorem.) Theorem 70: In Euclidean Geometry, if two parallel lines are intersected by a transversal, then the measures of pairs of interior angles on the same side of the transversal add up to 180. Theorem 71: In Euclidean Geometry, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. Theorem 72: In Euclidean Geometry, every triangle has angle sum exactly 180. Theorem 73: In Euclidean Geometry, the opposite sides of a parallelogram are congruent. Theorem 74: In Euclidean Geometry, the opposite angles of a parallelogram are congruent. Page 150 of 150 Theorem 75: In Euclidean Geometry, the diagonals of a parallelogram bisect each other. Theorem 76: In Neutral Geometry, the following are equivalent: (1) Point P lies on the perpendicular bisector of segment AB . (2) Point P is equidistant from A and B. That is, AP BP and length AP length BP Theorem 77: In Neutral Geometry, all diameters of a given circle are congruent. Theorem 78: In Neutral Geometry, diameters of a circle are the longest chords. Given a circle, if segment CD is a chord but not a diameter, and segment AB is a diameter, then segment CD AB . That is, length CD length AB . Theorem 79: In Neutral Geometry, the following are equivalent: (1) Diameter AB is perpendicular to non-diameter chord CD . (2) Diameter AB bisects non-diameter chord CD . Theorem 80: In Neutral Geometry, the perpendicular bisector of a chord passes through the center of the circle. Theorem 81: In Neutral Geometry, the following are equivalent: (1) Line L is perpendicular to segment AP at point P. (2) Line L is tangent to Circle A, AP at point P. Theorem 82: In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is perpendicular to line AB , and line M is perpendicular to line BC , then lines L and M intersect. Theorem 83: In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is the perpendicular bisector of segment AB , and line M is the perpendicular bisector of segment BC , and line N is the perpendicular bisector of segment CA , then lines L, M, and N are concurrent. Theorem 84: In Euclidean Geometry, If A, B, and C are non-collinear points, then there exists exactly one circle that contains all three points. Theorem 85: In Euclidean Geometry, every triangle can be circumscribed by exactly one circle. That is for every triangle ABC , there exists exactly one circle that contains points A, B, and C.