Geometry Without Axioms

Geometry without axioms Notes I used the notation AB or (AB) for a straight line, ‚ABƒ for a segment of straight line and (ABƒ for a half of line. The notation A-B-C means that B is between A and C. The notation ‚ACƒB indicates the shortest way from A to C through point B. Introduction (I skip it.) Definitions1 The definition of a straight line: a straight line is the group of (successive) points in which, whatever two points we would take, the shortest way between them is made up of points that belong to the group.2 The definition of a perpendicular line: a perpendicular line to a given straight line is the group of successive (or coplanar) points for which the shortest way to the straight line ends on the same point on the straight line. The definition of parallel straight lines: two straight lines are parallel if they are coplanar and they don’t intersect. The definition of plane: a plane is the group of the points of the straight lines which intersect at least two out of three given straight lines which intersect each other.3 The definition of geometrical space4: the geometrical space is the group of all the points whose existence is possible, non-contradictory. Theorems Theorem 1 (the theorem or the principle of the existence of geometrical entities) If it’s possible for a point to exist (its existence is not contradictory) then it exists in the geometrical space.5 1

The concept of point is taken as prime, undifined concept. It reduces to the classical definition of a straight line: “two points determine a straight line”. 3 If a straight line intersects two out of three straight lines in the same point it is included in plane if by one of its points we can draw a straight line which satislets the conditions to be included in plane (in this case the straight line has at least two points in common with the three intersectin straight lines, that is, at least two points in plane, which is required to be included in that plane). 4 I used the expresion «geometrical space» to make it distincti from a real physical space which could have in principle less than all possible points, less than all possible dimentions. 5 The extended principle is this: it the existence of a geometrical entity (made up of more points) is noncontradictory then it exists in the geometrical space.

2

Demonstration The theorem is deduced from the definition of geometrical space.

Theorem 2 In a plane, a straight line intersects at least one of two concurrent straight lines. Demonstration If in a plane a straight line did not intersect both of two concurrent straight lines, then, even if it intersected a third straight line which intersects the two concurrent straight lines, it would intersect one line at the most out of the three intersecting lines, which contradicts the definition of plane. Theorem 3 In a plane, the perpendicular line to a straight line is a straight line. Demonstration First, we will show that two points (A and B) of the perpendicular line which are on two different sides of the straight line (a) and the point (C) on the straight line in which the shortest way from any of the points of the perpendicular line to the straight line end are collinear. Let us suppose A, B and C, are not collinear. Then, since A and B are on two different sides of the straight line a, the shortest way from A to B intersects line a. Point C cannot be the point of intersection because in this case A, B and C would be collinear. Therefore, there is (at least6) a point D on line a, different from C, so that A, B and D are collinear and A, B and C are not collinear. Therefore, we have ‚ABƒD 6 ‚ABƒC (1). We also have: ‚ABƒC ˆ ‚ABƒ ‡ ‚BCƒ and ‚ABƒD ˆ ‚ADƒ ‡ ‚DCƒ. But since ‚ACƒ is the shortest way from A to line a ‚ACƒ ≤ ‚ADƒ.7 The same way, ‚BCƒ ≤ ‚BDƒ. We will have: ‚ACƒ ‡ ‚BCƒ ≤ ‚ADƒ ‡ ‚BDƒ. Therefore, ‚ABƒC ≤ ‚ABƒD (2). Relations (1) and (2) are contradictory. Therefore, the premise is wrong and A, B and C are collinear (3). Now we will show that A, B and C are collinear when A and B are on the same side of line a. Let us consider the same construction as above with the exception that A and B are on the same side. We will take a point D on the other side of line a. According The construction of line, plane, tridimensional space, perpendicular and parallel lines and geometrical figures is based on this principle (see Appendix B). It eliminates the need for axioms to postulate the existence of line or plane, for example. 6 Since we have not proved yet that two points determine only one straight line, it would be possible that there would be more D’s. 7 The equality is ‚ACƒ ≤ ‚ADƒ and not ‚ACƒ 6 ‚ADƒ because we have not proved yet that the shortest way from a point to a straight line is unique.

2 to conclusion (3) A, C and D are collinear – A is on line CD (4) and also B, C and D are collinear – B is on line CD (5). From conclusions (4) and (5) we draw the conclusion that A, B, C and D are collinear – A and B are on the same line CD (6). In conclusion, all the points of a perpendicular line, either that they are on different sides (1) or on the same side (6), are collinear. Therefore, a perpendicular line is a straight line.

Theorem 4 There is only one straight line which passes through two points. Demonstration Since there are two distinct points there is one or more ways to get from one to another. At least one of them is the shortest. Let us take two points, A and B, and let us suppose that there are two distinct straight lines which pass through them: (AB)1 and (AB)2. We will have: ‚ABƒ1 ˆ ‚ABƒ2 (1). Let us construct through B a straight line a so that (AB)1 ⊥ a (2). From (1) and (2) we draw the conclusion that (AB)2 ⊥ a too (3). Relations (2) and (3) contradict theorem 3, therefore the initial premise is false. Theorem 5 In a plane, from an exterior point of a line, there is only one perpendicular line to that line. Demonstration Let us consider a line a and a exterior point A. Let us suppose that from it there are two distinct lines drawn so that they are perpendicular to line a. They will intersect it in points B and C. We take on AB a point D so that B-A-D. Since AB is distinct from AC, D is not collinear with A and C (according to theorem 4). Therefore, there is a way from D to C shorter than the way through point A. Let F be a point collinear with D and C and D-F-C. We have: ‚DCƒA Ž ‚DCƒF (1). ‚DCƒA ˆ ‚DAƒ ‡ ‚ACƒ (2). But AB ⊥ BC and AC ⊥ BC. Therefore, ‚ABƒ ˆ ‚ACƒ (3). From equalities (2) and (3) we have: ‚DCƒA ˆ ‚DAƒ ‡ ‚ABƒ ˆ ‚DBƒ (4). From (1) and (4) we have: ‚DCƒF 6 ‚DBƒ. But DB ⊥ BC, threrefore ‚DBƒ is the shortest way from D to BC. The last two conclusions contradict each other, therefore, the initial premise is false. Theorem 6 There is only one parallel line from a point exterior to a given line. Demonstration First, we will prove the existence of parallel lines. Let us take a line a and two points on it, B and C. In the same plane we draw two perpendicular lines, b and c, on line a in point B and C, respectively. Lines b and c cannot intersect each

other because that would contradict theorem 5. Therefore, they need to be parallel. Therefore, through a point (C) exterior to a given line (b) we can draw at least one parallel line (c) to that line (b ⊥ c). The existence of parallel lines can also be deduced from the principle of existence of geometrical entities (theorem 1) because their existence is not contradictory. The demonstration of the uniqueness of a parallel line from an exterior point of a given line can be directly deduced from theorem 2.

Theorem 7 The intersection of two planes is a unique line. Demonstration Let us suppose that two planes intersect each other in a point. Let us consider a plane α and a point A in which it intersects with a second plane β. Through A we draw two lines (a and b) which intersect plane α in A and which belong to plane β. We take on line a a point B and a point E on the other side and on line b a point C on the same side with E and a point G on the same side with B. Line BC will intersect plane α in a point different from A which we will call D. The same way, EG will intersect it in a point F. Points A, D and E need to be collinear because otherwise we will have two planes instead of β – one formed by lines BA, AD and DB, and another one formed from lines BA, AF and FB. But, there is just one plane – plane β, threrefore A, D and F need to be collinear. Therefore, the intersection of two planes is at least on line. Let us suppose that it is two lines. Let us suppose that the lines intersect each other (the case in which they don’t intersect can be reduced to the case in which they intersect by drawing a third line which intersects the two parallel lines). We take a point on each line so that the points are distinct. In order that two distinct planes would pass through the two lines it should be the case that at least two distinct lines would pass through the two points considered. This conclusion contradicts theorem 4. Theorem 8 From a point exterior to a plane we can draw only one perpendicular line to that plane. 8

We construct the perpendicular line c to the line b based on theorem 3. Otherwise, based on the principle that the laws of geometry are the same in any plane and for all the lines (either we can draw a perpendicular line from an exterior point either we cannot) we draw the conclusion that there is no perpendicular lines and then that there is no plane. We can also construct c based on theorem 1.

3 Demonstration The point is exterior to the plane, therefore, it is at a distance from the plane. Therefore, there is at least one shortest way from it to the plane. Let us suppose that there are two shortest ways, therefore two perpendicular lines to the plane from the same point. It implies that there are two coplanar concurrent perpendicular lines on the line of intersection between the initial plane and the plane formed by the two concurrent lines. This conclusion contradicts theorem 5.

Theorem 9 If a ⊥ b then b ⊥ a too. Demonstration 1 Let A be the intersection point. Let us suppose that the perpendicular line a to line b is not the line on which b is perpendicular but it is another line c so that b ⊥ c in point A. Let α be a plane on which b is perpendicular in point A – b ⊥ α in point A (plane α will include line a). Let β be the plane formed by the perpendicular lines on b in point A – β ⊥ b in point A (plane β will include line c).9 Let d be a line of the intersection of the of the plane β with plane α. It will have the properties of both planes: b ⊥ d and d ⊥ b. This conclusion contradicts the initial premise, therefore the theorem is true.10 Demonstration 2 The theorem can also be demonstrated if we define the concept of angle. Let angle be the degree by which two half of lines which have the same origin are away from each other so that using the concept of angle we could uniquely determine or describe the relation between two lines. Let us define the angle formed by a half of line and its complementary half of line to be 1800. Now let us suppose that the relation between a perpendicular 9

Even if those lines didn’t form a plane, they will form a continuous surface formed by lines which intersect each other in A and are perpendicular on b. Since this countinuous surface intersect line b it will also intersect plane α which passes through line b. Since both the continuous surface and plane α are formed by straight lines which passes through A their intersection needs to be a straight line (d) which passes through A (and is also perpendicular on b). 10 The demonstration is based on the principle that the laws of geometry are the same in any plane. This principle is implied any time we suppose that a demonstration of a theorem for a plane is true for any plane. The principle is based on the fact that the lines or planes that we start with can be any other lines or planes so that the conclusions are true for any other plane. In this case the principle is either either . The last one proved to be correct.

line and the line on which it is perpendicular is not described by the half of this angle, that is by 900. In this case, from a point of a given line we could draw two distinct perpendicular lines since the angle between them is not zero. This conclusion contradicts theorem 3, therefore, the relation between a perpendicular line a and the line b on which a is perpendicular is described by an angle of 900 (we will call it a right angle) (1). Line b needs to be unique (2). Otherwise there would be more than one angle that describes the relation between the perpendicular line a and line b and that contradicts conclusion (1). If the angle between b and a is different than a right angle b is not unique anymore (the double of an angle which is not a right angle is not 1800). In this case we can draw another line, b', so that a ⊥ b’ but this contradicts conclusion (2). Therefore, the angle between b and a is also a right angle (3). From conclusions (1) and (3) we can directly derive the validity of the theorem. We can arrive to the same conclusion another way. The process of rotation of a half of a line in order to form an angle needs to be symmetric in time or if we change the direction or rotation (geometry needs to be independent of movement or time direction, see Appendix 1). Therefore, if we have two halves of lines ‚AB) and ‚AC) then angle BAC is equal to angle CAB. If one is a right angle then the other is a right angle too. Therefore, if a ⊥ b then b ⊥ a.

Theorem 10 If two lines are parallel they are perpendicular on the same line. Demonstration Let us suppose that two parallel lines are not perpendicular on the same line. We take two parallel lines a and b which intersect a third line c in A and B respectively so that a ⊥ AB. If b is not perpendicular on AB then there will be another line d which will be perpendicular on AB in point B. Since a does not intersect line b, based on theorem 2, a will intersect line d. This means that two perpendicular lines to the same line intersect each other which contradicts theorem 5. Therefore, if two lines are parallel they are perpendicular on the same line. Theorem 11 If in a plane we draw two perpendicular lines to a line a and through two points of the perpendicular that are at the same distance from line a and are on the same side of it we draw a second line b, the perpendicular lines on line a are perpendicular on line b too.

4 Demonstration We draw the figure mentioned in the theorem. Let A and B be the points on line a from which we draw the perpendicular lines and C and D the points on the perpendicular lines which are at the same distance from line a on the same side so that the two perpendicular lines are AC and BD. Let us suppose that AC is not perpendicular on b. Then there is another line EC so that EC ⊥ CD. Let E be the point of intersection with line a. AC and BD, being perpendicular on the same line, don’t intersect each other. EC intersects AC. Threrefore, based on theorem 2, EC intersects line BD. Let F be the point of intersection. ‚ECƒ ≥ ‚ACƒ because otherwise AC would not be perpendicular on AB. But ‚ACƒ ˆ ‚BDƒ (mentioned in theorem), threrefore ‚CEƒ ≥ ‚BDƒ (1). The same way, ‚EFƒ ≥ ‚FBƒ (2) because otherwise FB would not be perpendicular on AB. ‚FCƒ ˆ ‚FEƒ ‡ ‚ECƒ (3) and ‚FDƒ ˆ ‚FBƒ ‡ ‚BDƒ (4). From relations (1) – (4) we get: ‚FCƒ ≥ ‚FDƒ (5). But, ‚FDƒ ≥ ‚FCƒ (6) because otherwise FC would not be perpendicular on CD. From relations (5) and (6) we get that ‚FCƒ ˆ ‚FDƒ. From this relation and relations (1) – (4) we get: ‚CEƒ ˆ ‚ASƒ. Then, since AS ⊥ AB, CE needs to be perpendicular on AB too. But that contradicts theorem 5. Threrefore, initial premise is false and the above theorem is true.

Theorem 12 The distance between two parallel lines remains constant. Demonstration From theorem 9 and 11 we derive the following conclusion: a line which passes through two points which are at the same distance from a given line is parallel with that given line (otherwise contradicts theorem 5). In other words, if we have a line a and a point A exterior to it through which we draw a line b parallel to line a, then a point B at the same distance to line a as point A (and on the same side) is situated on line b parallel to line a. Threrefore, theorem is true. Theorem 13 A perpendicular to a given line is perpendicular on any line parallel with that given line. Demonstration The demonstration can be derived from theorem 12 together with theorem 3 or 5. Theorem 14 If ab and bc then ac.11

11

The Lobacevskian geometry constradicts this theorem also.

Demonstration The demonstration can be derived from theorem 13 and theorem 5.

Commentaries about geometries in relation with:

possible

1. The theorem of the perpendicular line to a plane (theorem 5) From this theorem we derive the conclusion that four distinct dimensions are not just impossible to imagine but, even more, their existence is geometrically impossible (no matter of the physical nature of the dimensions when the geometrical model is applied to a physical model). A forth distinct dimension requires that it should be perpendicular on the other three. But, since from an exterior point of a plane there is only one perpendicular line to that plane, the existence of a forth perpendicular line on the first three lines perpendicular on each other is impossible. The existence of a forth dimension requires that from an exterior point of a plane it should be possible to draw at least two perpendicular lines to that plane, which should also be perpendicular on each other (which implies that we can draw an infinity of distinct perpendicular lines to that plane from the same point). In order that the fourth dimension would form, together with the other dimensions, one space (in order to be a dimension of the same space) it needs to be possible to relate it to the other dimensions. It needs to be perpendicular on them (or it should be possible to be “converted” so that it is perpendicular). Otherwise we cannot say that it is exterior to the space formed by the others, independent from them. This fourth dimension would form together with one of the other dimensions a plane which is perpendicular on the space formed by the three initial dimensions. In other words, in this extra plane there is an infinity of perpendicular lines to a plane formed by two of the three initial dimensions. This conclusion contradicts theorem 8. We can talk about “curved” geometries as geometries on bi-dimensional curved surfaces (which are particular figures of the flat threedimensional space), but we cannot talk about geometries on three-dimensional curved surfaces (which are particular figures of the flat 4dimensional space, which is geometrically impossible). The theorem of the uniqueness of the perpendicular line to a plane from an exterior point of the plane (theorem 8) disproves the physical models which are based on the existence of more

5 than three geometrical dimensions, as the theory general relativity (it implies four distinct dimensions, three spatial and a temporal one) or string theories which are based on 10 or even 26 dimensions. This theorem is closely linked with theorem 5 (which is used for its demonstration) and, through it, by the parallel theorem (theorem 6). The theory of relativity is based on a non-Euclidean space which contradicts the parallel theorem.

2. The theorem of the parallels (theorem 6) The definition of space used in this essay implies that three lines which intersect each other determine a plane and, in this respect is equivalent with the proposition that three non-collinear points determine a plane (the three non-collinear points determine three lines which intersect each other). But, the definition used here is precise and excludes any curved plane and makes the “parallel postulate” as easy to prove as possible.If from a point exterior to a line we can draw two parallel lines to that line then either one of them is not in the same plane with the given line, either the plane is not flat. In fact, these possibilities are equivalent. If we “condense” the two planes (formed by the given line with each of the two lines parallel to it) by the requirement of coplanarity of parallel lines then the unique plane resulted is not flat anymore. Thus, the postulate that from an exterior point of a line we can draw more than one parallel lines to that line “hides” an extra dimension or plan in the curvature of that dimension or plan. As we shall see in the next section, this curved plane is inadequate. If we introduce a curvature of the space we, in effect, introduce a new dimension. Without this extra dimension there would be no room for curvature. The extra dimension offers the room in which the old space can be curved. Let us take an example and suppose we have an uni-dimensional space. Does it make sense to talk about its curvature? By the very fact that it has a one dimension this dimension is necessarily straight. By the fact that the way determined by this one dimension is unique (because there are no other dimensions there are no other ways) it is implicitly straight. The concept of “curvature” has no meaning in lack of the plurality of possible ways. “Straight” is linked with “unique” or “minimal” and “curved” is linked with “plural”. “Curved” means not-straight; it is a modification of what is straight. Since there are a lot of ways in which we can modify what is straight “curved” implies the plurality of possible ways. In order to talk about more possible ways we need to add an extra dimension so that we have two. The “corrupted” concept of plane of the nonEuclidean geometries implies, of course, a

“corrupted” concept of straight line. Let us consider a sphere (corresponding to a Riemannian space) and two “lines” on it. In how many points do they intersect? In two! This conclusion contradicts the theorem of the uniqueness of the line determined by two points (theorem 4).12

3. The definition of the plane The classic definition of the plane is made through the axiom which states that: . This phrasing is deficient because it is imprecise. We can ask the question: “How do the three points determine a plane?>, <Which is the relation which relates to the plane determined by them?>. There is no direct relation between the three necolinear points and the plane! This <determining> or relation needs to be expressed using the elements or relations already established and defined. In the process of building the system of concepts of a given field we need to start from prime undefined concepts or relations.13 It is preferably to start from as less undefined concepts as possible because by the fact that they are not defined they are not as definite and precise as the defined concepts. Each undefined concept introduces therefore, some degree of imprecision in our system. These undefined prime concepts need to be elementary concepts, irreducible concepts. If a concept is complex, that is, it can be reduced to other concepts and relations between them, then it should be a defined concept. If it is taken as a undefined concept and thus, its sub-concepts and sub-relations are not determined this also introduces an indetermination it the system which is to be built. Any new complex concept is to be defined only on already defined concepts and relations, if possible. Introducing a new undefined concept makes the system less precise. When we defined the concept of “straight line” we used the elementary concept of “point” and an undefined relation of “shortest way.” The (intuitive) concept of “plane” is a complex concept. It is based on more elementary concepts like “line” and “point” and relations between them. Therefore, it needs to be defined. The concept of “line” and “intersection” (which is the property of a point to belong to two or more lines) are enough to define the concept of “plane”. They are enough to establish a relation 12

Saccheri and Lambert used this argumentation to prove the invalidity of the “obtuse angle hypothesis” (the riemannian geometry). 13 It’s evident that we cannot define all the concepts (each concept is defined based on other concepts which are to be defined and so on indefinitely). Therefore, we need to start from some prime undefined terms, based only on an intuitive or knowledge of them.

6 between the three non-collinear points and the plane determined by them (see the definition given in this essay). The concept of “plane”, as any concept, needs to have a unity and an order of the wholeness (that is needed for the oneness of the concept). The concept of “point” is too elementary to base the definition of plane only on it. Therefore, we will base the definition on the concept of “line” (which already implies an order, a unity in the group of all possible points). Therefore, the plane is composed of lines. But if the plane does not consist of all the lines (in which case it is identical with the geometrical space) there should be a relation so that a line’s satisfying it implies the line’s inclusion in that plane. The only possible relation is that given in the definition of plane (or an equivalent one). If a line intersects only one line of the three that is not enough (the plane would be identical with the space) and if it needs to intersect all three lines that is too much (it excludes the parallel lines). If a line intersects two out of the three intersecting lines which determine the plane then the line has two points in plane (the points of intersection with two of the three lines) and, based on theorem 4, it is included in plane (because the plane is composed of lines and there is a unique line that links the two points that line needs to belong to the plane). A. N. Whitehead, co-author together with B. Russell of the famous book Principia Mathematica, being primarily a philosopher and logician, was more preocupated than geometers to define as precisely as possible the concepts to be used. In his book Process and Reality, he gives the same definition for plane as the one given here. He states that this definition applies to any geometry, including the non-Euclidean ones.14 Although, sadly, he did not examined the geometrical implications of the precise definition of plane. The problem of geometry was not related to the postulate of parallels but to a mere definition.15

Appendices Appendix A: A demonstrations of the parallel 14

N. A. Whitehead, Process and Reality, New York, The Free Press, 1969, pag. 388. Whitehead also gives another definition which, as he himself says, is equivalent with one given here. 15 Some mathematicians have considered modifying definitions to solve the parallel problem but they dealed with the definition of parallel lines. Proclus (sec. V), for example, sugested redefining the concept of a line parallel to a given line as the locus (or the group) of points which are situated at a given constant distance from the given line. But he couldn’t prove that the group of points make a stright line.

axiom which does not require the new definition of plane Note: We will use the symbol Σ for the sum of the angles of a geometrical figure and σ (that we will call geometrical shift) for the difference between the sum of the angles of that figure in Euclidean geometry and the sum of the angles of the figure in a given case of the Lobacevskian geometry. We have the relation: σ ˆ (n – 2)π – Σ (where n is the number of segments that the figure has). It is easy to see that the geometry shift of a figure is the sum of the geometrical shifts of the figures that the initial figure is made up of. For example, the geometrical shift of the quadrangle ABCD is the sum of the geometrical shifts of the triangles ABD and CBD. We can prove the validity of the parallel axiom without the new definition of plane. We start from the premise that the sum of the angles in a triangle is less than π (the axiom of the Lobacevskian geometry) is true. First we will show that any triangle needs to have the sum of its angles greater than a given value (we will take π/2). We take a triangle (which can be any triangle) ∆ABC, where AB is the biggest segment of the triangle (let’s say its length is l). We draw a line a on which we take two points, D and E, so that DE Ž l. We draw two perpendicular lines, b and c, on a in points D and E. We take a point F on the line c so that ‚FEƒ 6 l and we draw the perpendicular line from it to line b which intersects it in point G. We draw the line GE and we take point H on FE so that GH ⊥ FE. The sum of the angles 6EGD and 6GED is less than π/2 (according to our initial premise, the sum of the angles in a triangle needs to be less than π). Therefore, the sum of the angles 6FGE and 6FEG is greater than π/2 (angles 6FGD and 6FED are right angles) (1). In Lobacevskian geometry we have ‚FGƒ Ž ‚DEƒ and, therefore, ‚FGƒ Ž ‚FEƒ.16 The circle with the center in F and the diameter twice as ‚FGƒ will intersect ‚FE) in a point I so that 16

‚FGƒ Ž ‚GHƒ and so on up to ‚DEƒ. Also, we can move line b far enough from c (or c far eough from b) so that ‚FGƒ Ž ‚FEƒ. Anyway, if ‚FGƒ 6 ‚FEƒ then FG is the shortest segment of the triangle FGE and therefore, angle 6FEG is the smallest angle. If it is greater than π/4 then we use triangle FGE for our demonstration instead of triangle JGE (triangle FGE has segment GE greater than l and both of its angles 6FGE and 6FEG are greater than π/4). If angle 6FEG is smaller than π/4 then angle 6GED is greater than π/4 and we take triangle DEG for our demonstration (it has segment DE equal with l and both of its angles 6GDE and 6GED greater than π/4).

7 F–E–I (‚FGƒ ≡ ‚FIƒ). Let’s consider a point J movable on line c (J–F–E). Let K be the point of intersection of the circle with the center in point J and the diameter twice as ‚JGƒ (point K is corresponding to point I for the circle centered in F) – ‚JGƒ ≡ ‚JKƒ. The farther away from point F we take point J the closer to point H point K gets (point K will reach point H when point J is infinitely far away from point H). Therefore, there is a point J so that point K is identical with point E (‚JGƒ ≡ ‚JEƒ).17 Therefore, for that point J we have: 6JGE ≡ 6GEJ (2). But, 6JGE ‡ 6GEJ Ž π/2 (see conclusion (1)) (3). From conclusions (2) and (3) we draw the further conclusion that 6JGE ˆ 6GEJ Ž π/4 (4). Now we draw the initial triangle ABC in the interior of the isosceles triangle JGE so that point A will be identical with point G (or E) and segment AB and GE will be on the same line (line a) and point C will be on the same side of line a as point J. If triangle ABC will be completely inside triangle JGE then σABC 6 σJGE and ΣABC Ž ΣJGE (see note at the beginning of this appendice). If triangle ABC is not completely inside triangle JGE then it is necessary that at least one of the angles 6CAB or 6ABC are greater than 6JGE or 6JEG (which are both greater than π/4 – see conclusion (4)). Therefore, at least one of these two angles are greater than π/4. But segment ‚ABƒ is the longest segment of the triangle ABC, therefore, angle 6ACB is the greatest angle of the triangle. Therefore, 6ACB also needs to be greater than π/4 and ΣABC Ž π/2. In conclusion in any case the sum of the angles of the triangle ABC is greater than π/2. Triangle ABC can be any triangle, therefore the sum of any triangle needs to be greater than π/2, that is, the geometrical shift needs to be smaller than π/2 (5). The second part of the demonstration is to prove that if the sum of the angles of a triangle is less than π then it can decrease indefinitely towards zero (therefore it can be smaller than π/2). Let’s consider an equilateral triangle, ABC. All its angles have the same value, α1. So far it is required that π/2 6 ΣABC 6 π (and σABC 6 π/2, let’s 17

If we take a point J so that J–F–E then point K (‚GJƒ ( ‚KJƒ) will be between I and J: ‚IJƒ ˆ ‚IFƒ ‡ ‚FJƒ ˆ ‚GFƒ ‡ ‚FJƒ Ž ‚GJƒ ˆ ‚KJƒ. Now, if we take a point J2 we will get a point K2 between K1 and J and so on. Each K will get closer and closer to point H and K∞ will be identical with point H (for an J∞ infinitely far away from point H which implies that ‚GJ∞ƒ ≡ ‚HJ∞ƒ). There is a point J for each point between I and H (for each K, so that ‚GJƒ ≡ ‚JKƒ) therefore, there will be a point J for point E so that ‚GJƒ ≡ ‚JEƒ).

call it σ1) and therefore, π/6 6 α 6 π/3. We draw other triangles identical with triangle ABC so that one of the corners of all triangles are in the same point and that triangles are adjacent to each other. We draw n triangles so that 2π – α1 6 n·α1 ≤ 2π (6) (in Euclidean geometry α1 ˆ π/3 and n ˆ 6 and the total figure is a hexagon). If n·α1 6 2π then we draw the segment that links the exterior corners of the two triangles which are not adjacent with other two triangles. The total figure will consist of n triangles identical with triangle ABC and we will have the following relation for the geometrical shift: σt ≥ n·σ1 (7) (where σt is the geometrical shift for the total figure). The total figure has at the most (n ‡ 1) segments. Therefore, we can construct out of it ‚(n ‡ 1) – 2ƒ ˆ (n – 1) triangles. There will be at least one triangle among them (let us call it ∆ DEF) which has a geometrical shift σ2 so that σ2 ≥ σt / (n – 1) (8). From relatioins (7) and (8) we get the following relation: σ2 ≥ n·σ1 / (n – 1) which is equivalent with n ≥ σ2 / (σ2 – σ1) (9). Also, σ1 ˆ π – ΣABC ˆ π – 3· α1 (10). From relations (6), (9) and (10) we get: σ2 ≥ 6π·σ1 / (5π – σ1) (11). Now we consider triangle DEF. Let us say that DE is the longest segment. We take a point G on DE so that FG ⊥ DE. Let us say that FE is the shortest segment. We take a point H on line DE (D–E–H) so that triangles DGE and HGE are identical. In triangle FDH ‚FDƒ ≡ ‚FHƒ and DH is the longest segment. Now we take a point I on line FG so that ‚IDƒ ≡ ‚IHƒ ≡ ‚DHƒ. Triangle IDH is an equilateral triangle. Point F needs to be between I and G because ‚IHƒ Ž ‚FHƒ and ‚IDƒ Ž ‚FDƒ. Therefore, triangle FDH is included in triangle IDH. Therefore, σIDH Ž σFDH ≥ σ2. Thus we have constructed an equilateral triangle that has the geometrical shift greater than σ2. Now it starts a cycle. We do with triangle IDH what we did with triangle ABC and we will get a third equilateral triangle with a bigger geometrical shift, σ3, and so on. The relation between the geometrical shifts is relation (11): σ2 ≥ 6π·σ1 / (5π – σ1). According to this relation the geometrical shift grows closer and closer to π (which means that the sum of the angles decreases towards zero). For example, if we start with a geometrical shift σ1 ˆ π/3 we will get after four cycles: σ5 ˆ 0,51π Ž π/2. This conclusion contradicts the conclusion of the first part, conclusion (5), which states that in any triangle the geometrical shift needs to be smaller than π/2. Consequently, the initial premise – the axiom of Lobacevskian geometry – is an invalid axiom.