Hilbert’s Axioms PDF

Hilbert’s Axioms March 26, 2013 1 Flaws in Euclid The description of “a point between two points, line separating the plane into two sides, a segment is congruent to another segment, and an angle is congruent to another angle,” are only demonstrated in E...

Hilbert’s Axioms March 26, 2013 1 Flaws in Euclid The description of “a point between two points, line separating the plane into two sides, a segment is congruent to another segment, and an angle is congruent to another angle,” are only demonstrated in Euclid’s Elements. 2 Axioms of Betweenness Points on line are not unrelated. We assume that there is a ternary relation among points, named as “point B is between point A and point C,” abbreviated as A∗B∗C Given distinct collinear points A, B, C, D. We use A∗B∗C ∗D to denote the following simultaneous relations of betweenness A ∗ B ∗ C, A ∗ B ∗ D, A ∗ C ∗ D, B ∗ C ∗ D. (1) Betweenness Axiom 1 (BA1) (Collinearity and symmetrization). If A ∗ B ∗ C, then A, B, C are three distinct points all lying on the same line, and C ∗ B ∗ A. Betweenness Axiom 2 (BA2) (Extension). Given two distinct points B and D on a line l. There exist points A, C, E lying on line l such that A ∗ B ∗ D, B ∗ C ∗ D, and B ∗ D ∗ E; see Figure 1. A B C D E Figure 1: Betweenness Axiom 2 Betweenness Axiom 3 (BA3) (Uniqueness). Let A, B, C br three distinct points on a line. Then one and only one of the three points is between the other two. Definition 1 (Line, segment, and ray). The line determined by two distinct points A and B is denote by AB. 1 We also use AB to denote the set of all points incident with the line determined by points A and B. A segment with endpoints A and B, denoted AB, is the set of points A, B, and all points between A and B. A ray emanating from a point A to another point B, denoted r(A, B), is the set of all points on AB and all points C such that A ∗ B ∗ C. An open ray emanating from a point A to another point B is the set r̊(A, B) := r(A, B) r {A}. Proposition 2.1. For any two distinct points A and B, AB = r(A, B) ∩ r(B, A), AB = r(A, B) ∪ r(B, A). Proof. Note that AB ⊆ r(A, B) ∩ r(B, A) by definition of segment and ray. For each point P ∈ r(A, B) ∩ r(B, A), we have P ∈ r(A, B) and P ∈ r(B, A). Suppose P 6∈ AB. By definition of ray, we have A ∗ B ∗ P by P ∈ r(A, B) and P ∗ A ∗ B by P ∈ r(B, A). Then A, B, P are three distinct collinear points by BA1. This is contradictory to BA3 that there is only one point of the three A, B, P between the other two. It is clear that r(A, B) ∪ r(B, A) ⊆ AB. For each P ∈ AB, if P ∈ AB, it is clear that P ∈ r(A, B) ∪ r(B, A). Assume P 6∈ AB, then A, B, P are three distinct points by BA1, and one of them is between the other two by BA3. Since P is not between A and B, we have either A is between B and P or B is between A and P. In the formal case, we have P ∈ r(B, A); in the latter case, we have P ∈ r(A, B). Hence P ∈ r(A, B) ∪ r(B, A). Definition 2 (Same side and opposite side). Two points A, B not on a line l are said to be on the same side of l if A = B or the segment AB does not meet l. Two points A, B not on a line l are said to be on opposite sides of l if AB does not meet l. Betweenness Axiom 4 (BA4) (Plane separation). Let A, B, C be three distinct points not on a line l. (i) If A, B are on the same side of l and B, C are on the same side of l, then A, C are on the same side of l. (ii) If A, B are on opposite sides of l and B, C are on opposite sides of l, then A, C are on the same side of l. The relation of being on the same side of a fixed line l is an equivalence relation on the set of points not on the line l, since it is reflexive, symmetric, and transitive by definition and Betweenness Axiom 4(i). Each equivalence class is called an open half-plane bounded by l. For each point P not on l, we denoted by H̊(l, P ) the open half-plane that contains P. The set H(l, P ) := H̊(l, P ) ∪ l is called a half-plane (or closed half-plane) bounded by l. Corollary 2.2. For each line l there are exact two half-planes bounded by l. (iii) If A, B are on opposite sides of l and B, C are on the same side of l, then A, C are on opposite sides of l. 2 Proof. Let A, B be two points on opposite sides of a line l. We have two distinct half-planes H(l, A) and H(l, B). Given an arbitrary point C not on l. If A, C are on the same side of l, then H(l, C) = H(l, A). If A, C are on opposite sides, then B, C are on the same side of l by Betweenness Axiom 4(ii). Thus H(l, C) = H(l, B). Therefore there are at most two half-planes bounded by l. Given a point B on l and a point D not on l. By Betweenness Axiom 2 there exist points A, C, E such that A ∗ B ∗ D, B ∗ C ∗ D and B ∗ D ∗ E. Then A, D are on opposite sides of l. So there are at least two half-planes bounded by l. Proposition 2.3 (Linearity rules). Let A, B, C, D be distinct points on a line l. Then (a) A ∗ B ∗ C, A ∗ C ∗ D ⇒ A ∗ B ∗ C ∗ D. (b) B ∗ C ∗ D, A ∗ B ∗ D ⇒ A ∗ B ∗ C ∗ D. (c) A ∗ B ∗ C, B ∗ C ∗ D ⇒ A ∗ B ∗ C ∗ D. Proof. (a) Pick a point E outside l and make line EC; see Figure 2. Then C is the unique A B C D E Figure 2: Betweenness and separation axioms imply linearity intersection of l and EC. The points A, B must be on the same side of line EC. (Otherwise, AB meets EC at C; we then have A∗C ∗B, which contradicts A∗B ∗C.) Since A∗C ∗D, then A, D are on opposite sides of EC. Hence B, D are on opposite sides of EC by Corollary 2.2, i.e., BD meets EC at C. We then obtain B ∗ C ∗ D. Draw line EB; the point B is the unique intersection of l and EB; see Figure 2. Since A ∗ B ∗ C, then A, C are on opposite sides of EB. Since B ∗ C ∗ D, we must have C, D on the same side of EB. (Otherwise B would be between C and D, contradicting to B ∗ C ∗ D.) Thus A, D are on opposite sides of EB by Corollary 2.2, i.e., AD meets EB at B between A and D. We then obtain A ∗ B ∗ D. (b) is similar to (a) by reversing the order. (c) Note that A, B, C are distinct and B, C, D are distinct. If A = D, then B ∗ C ∗ D becomes B ∗ C ∗ A, which is contradictory to A ∗ B ∗ C. So A, B, C, D are distinct. Pick a point E outside l and draw the line EC. Since B ∗C ∗D, then B, D are on opposite sides of EC by definition. Likewise, A ∗ B ∗ C implies that A, B are on the same side of EC. (Otherwise, A, B are on opposite sides of EC, i.e., AB meet EC at C; so A ∗ C ∗ B, contradicting to A ∗ B ∗ C.) It follows from Corollary 2.2 that A, D are on opposite sides of EC. Hence AD meets EC at C between A and D, i.e., A ∗ C ∗ D. Definition 3 (Strict total order). A binary relation ≺ on a set X is called a strict total order if (TO1) Irreflexivity: x 6≺ x for all x ∈ X; (TO2) Transitivity: if x ≺ y and y ≺ z then x ≺ z; (TO3) Completeness: either x ≺ y or y ≺ x but not both for all x, y ∈ X with x 6= y. For a strict total order on X, the relation ¹, defined on X by x ¹ y if x = y or ≺ y, is called a total order. For an order relation, we also write x ≺ y and x ¹ y as y  x and y º x respectively. The set X with a total order is said to be totally ordered. 3 Proposition 2.4 (Strict total order of line). For each line l with two distinct points A, B there exists a unique total order on l such that A ≺ B and if C ∗ D ∗ E then either C≺D≺E or E ≺ D ≺ C but not both. Proof. Define A ≺ B. For each point P of l other than A, B, we define (1) P ≺ A and P ≺ B if P ∗ A ∗ B, (2) A ≺ P and P ≺ B if A ∗ P ∗ B, (3) A ≺ P and B ≺ P if A ∗ B ∗ P. For any two distinct points P, Q other than A and other than B, we define P ≺ Q if one of the following holds: (I) P ∗ Q ∗ A ∗ B, (II) P ∗ A ∗ Q ∗ B, (III) P ∗ A ∗ B ∗ Q, (IV) A ∗ P ∗ Q ∗ B, (V) A ∗ P ∗ B ∗ Q, (VI) A ∗ B ∗ P ∗ Q. We claim that ≺ is a strict total order on l. It is clear that ≺ satisfies irreflexive and completeness. For transitivity, let P ≺ Q and Q ≺ R, we claim P ≺ R. If {P, Q, R} ∩ {A, B} 6= ∅, we clearly have P ≺ R by definition of ≺. If {P, Q, R} ∩ {A, B} = ∅, we verify the six cases. Case I. P ∗ Q ∗ A ∗ B. (I.1) Q ∗ R ∗ A ∗ B: Since P ∗ Q ∗ A and Q ∗ R ∗ A, then P ∗ Q ∗ R ∗ A by Proposition 2.3(b). Since P ∗ R ∗ A and R ∗ A ∗ B, then P ∗ R ∗ A ∗ B by Proposition 2.3(c). Hence P ≺ R by definition. (I.2) Q ∗ A ∗ R ∗ B: Since P ∗ Q ∗ A and Q ∗ A ∗ R, then P ∗ Q ∗ A ∗ R by Proposition 2.3(c). Since P ∗ A ∗ R and A ∗ R ∗ B, then P ∗ A ∗ R ∗ B by Proposition 2.3(c). Hence P ≺ R by definition. (I.3) Q ∗ A ∗ B ∗ R: Since P ∗ A ∗ B and A ∗ B ∗ R, then P ∗ A ∗ B ∗ R by Proposition 2.3(c). By definition P ≺ R. Case II. P ∗ A ∗ Q ∗ B. (II.1) A∗Q∗R ∗B: Since P ∗A∗B and A∗R ∗B, then P ∗A∗R ∗B by Proposition 2.3(b). By definition P ≺ R. (II.2) A∗Q∗B ∗R: Since P ∗A∗B and A∗B ∗R, then P ∗A∗B ∗R by Proposition 2.3(c). By definition P ≺ R. Case III. P ∗ A ∗ B ∗ Q. (III.1) A∗B ∗Q∗R: Since P ∗A∗B and A∗B ∗R, then P ∗A∗B ∗R by Proposition 2.3(c). By definition P ≺ R. Case IV. A ∗ P ∗ Q ∗ B. (IV.1) A∗Q∗R∗B: Since A∗P ∗Q and A∗Q∗R, then A∗P ∗Q∗R by Proposition 2.3(a). Since P ∗ Q ∗ B and Q ∗ R ∗ B, then P ∗ Q ∗ R ∗ B by Proposition 2.3(b). We then have A ∗ P ∗ R and P ∗ R ∗ B. Thus A ∗ P ∗ R ∗ B by Proposition 2.3(c). By definition P ≺ R. (IV.2) A∗Q∗B ∗R: Since A∗P ∗B and A∗B ∗R, then A∗P ∗B ∗R by Proposition 2.3(a). By definition P ≺ R. Case V. A ∗ P ∗ B ∗ Q. (V.1) A∗B ∗Q∗R: Since A∗P ∗B and A∗B ∗R, then A∗P ∗B ∗R by Proposition 2.3(a). By definition P ≺ R. Case VI. A ∗ B ∗ P ∗ Q. (VI.1) A∗B ∗Q∗R: Since B ∗P ∗Q and B ∗Q∗R, then B ∗P ∗Q∗R by Proposition 2.3(a). Since B ∗P ∗R and A∗B ∗R, then A∗B ∗P ∗R by Proposition 2.3. By definition P ≺ R. 4 Proposition 2.5 (Line separation). Let A, B, O be three distinct points such that A∗O∗B. Then r(O, A) ∩ r(O, B) = {O}, r(O, A) ∪ r(O, B) = AB. If P ∈ AB, then either P ∈ r(O, A) or P ∈ r(O, B). The rays r(O, A) and r(O, B) are said to be opposite each other. Proof. Let ≺ be the strict total order on the line l such that A ≺ B. By definition of the total order ¹, the rays r(O, A), r(O, B), and the segment AB, we have r(O, A) = {P ∈ l : P ¹ O}, r(O, B) = {P ∈ l : O ¹ P }, AB = {P ∈ l : A ¹ P ¹ B}. Then r(O, A) ∩ r(O, B) = {O} and r(O, A) ∪ r(O, B) = AB by the total ordering property of ≺. Corollary 2.6 (Line separation). Let l, m be two distinct lines intersecting at a point O. Let ≺ be a strict total order on l. Then the two sets r̊(O, −) := {P ∈ l : P ≺ O}, r̊(O, +) := {P ∈ l : O ≺ P } are on opposite sides of m. We also call them on opposite sides of O on l. Proof. Let A, B be two distinct points on l. If A, B ∈ r̊(O, −), i.e., A ≺ O, B ≺ O, then for all P between A and B, we have either A ≺ P ≺ B or B ≺ P ≺ A. In either case we have P ≺ O by transitivity. So AB is contained in r̊(O, −). Clearly, AB does not meet m (since O is the unique intersection of l and m). Hence A, B are on the same side of m by definition. Likewise, if A, B ∈ r̊(O, +), i.e., O ≺ A, O ≺ B, then A, B are on the same side of m. If A ≺ O ≺ B or B ≺ O ≺ A, then in either case AB meets m at O between A and B; so A, B are on opposite sides of m by definition. Theorem 2.7 (Pasch’s Theorem). Let A, B, C be distinct points of not collinear. Let l be a line meeting AB at a point D between A and B. Then one and only one of the three holds: (i) l meets AC at a point between A and C, (ii) l meets BC at a point between B and C, (iii) l meets both AC and BC at a point C. Intuitively, this theorem says that if a line “goes into” a triangle through one side then it must “come out” through another side. B P Q l A C Figure 3: A line passes through a triangle Proof. The points A, B are on the opposite sides of the line l. If C is on l, then l does not meet AC between A and C, otherwise l = AC; and l does not meet BC between B and C. If C is not on l, then either A, C are on the same side of l, or B, C are on the same side of l, butt not both. In the formal case, then B, C are the opposite sides of l. Thus l meets BC at a point between meets B and C, and is disjoint from AC. In the latter case, l meets AC at a point between A and C, and is disjoint from BC. 5 Definition 4 (Interior of angle). Given points A, O, B not collinear. The interior of an angle ∠AOB, denoted ˚ ∠AOB, is the set of points P such that P, A are on the same side of line OB, and P, B are on the same side of line OA, in other words, ˚ ∠AOB := H̊(OB, A) ∩ H̊(OA, B); see Figure 4. We also define ∠AOB := H(OB, A) ∩ H(OA, B). It is convenient to consider a closed half-plane as a flat angle. B B P P O O A A Figure 4: Interior of an angle Proposition 2.8 (Between-Cross Lemma). Given an angle ∠AOB and a point P on AB. Then P belongs to ˚ ∠AOB if and only if A ∗ P ∗ B. Proof. “⇒”: The point P belongs to ˚ ∠AOB. By definition P, B are on the same side of line OA. Suppose P ∗ A ∗ B. Then P, B are opposite sides of OA, since P B meets OA at A between P and B. This is a contradiction. Likewise, A ∗ B ∗ P leads to a similar contradiction. Then we must have A ∗ P ∗ B by trichotomy of betweenness. “⇐”: We have A ∗ P ∗ B. Note that line AB meets line OB at the unique point B. Then AP does not meet OB. So A, P are on the same side of OB. Likewise, points B, P are on the same side of OB. Hence by definition P belongs to ˚ ∠AOB. Proposition 2.9. Let P be a point in ˚ ∠AOB. Then (a) The open ray r̊(O, P ) is contained in ˚ ∠AOB. (b) The opposite ray to r(O, P ) is disjoint from ˚ ∠AOB. See the left of Figure 5. 0 ˚ (c) If B ∗ O ∗ B , then A belongs to ∠P OB. 0 B B P C O P P’ O A A B’ Figure 5: Property of interior of an angle and Crossbar Theorem Proof. (a) Let Q be a point on the open ray r̊(O, P ). It is clear that P Q is disjoint from OA (since the intersection of the two lines P Q, OA are the unique point O). This means that P, Q are on the same side of OA by definition. Since P ∈ ˚ ∠AOB, i.e., P, B are on the same side of OA, then B, Q are on the same side of OA. Likewise, A, Q are on the same side of OB. Thus Q is an interior point of ∠AOB. 6  (b) Let P 0 be a point on the opposite ray of r(O, P ); see the left of Figure 5. Then P, P 0 are on opposite sides of OB. Since A, P are the same side of OB, then A, P 0 are on the opposite sides of OB. Thus P 0 is not an interior point of ∠AOB by definition. (c) Note that P, A are on the same side of OB 0 (since OB 0 = OB and P ∈ ˚ ∠AOB). We claim that A, B 0 are on the same side of OP. If so, we have A ∈ ˚ ∠P OB 0 by definition. Suppose that A, B 0 are on opposite sides of OP , i.e., OP intersects AB 0 at C between A and B 0. Then A ∗ C ∗ B 0 and C ∈ ˚ ∠AOB 0 by Proposition 2.8. Since C ∈ OP and C 6= O, we have either C ∈ r̊(O, P ) or C ∈ r̊(O, P 0 ). If C ∈ r̊(O, P ), then P ∈ r̊(O, C), which is contained in ˚ ∠AOB 0 by part (a). Thus P, B 0 are on the same side of OA (since P ∈ ˚ 0 ∠AOB ). Since P, B are the same side of OA, we see 0 that B, B are on the same side of OA. This is a contradiction. If C ∈ r̊(O, P 0 ), then P 0 ∈ r̊(O, C), which is contained in ˚∠AOB 0 by part (a). Thus A, P 0 0 0 are on the same side of OB (= OB ) by definition. Since P , P are on opposite sides of OB, we see that A, P are on the opposite sides of OB. This is a contradiction. Definition 5 (Between rays). A ray r(O, P ) is between two non-opposite rays r(O, A) and r(O, B) if P is in the interior of ∠AOB (independent of the choice of P on the ray r(O, P )). Proposition 2.10 (Crossbar Theorem). If a ray r(O, P ) is between two rays r(O, A) and r(O, B), then r(O, P ) intersects AB at C between A and B. See the right of Figure 5. The interior of ∠AOB is a disjoint union of interiors ˚ ∠AOP , ˚ ∠BOP , and open ray r̊(O, P ). Proof. Note that B, B 0 are on opposite sides of OP , and B 0 , A are on the same side of OP ; see the left of Figure 5. Then A, B are on opposite sides of OP. Thus OP intersects AB. Since the ray r(O, P 0 ) (opposite to the ray r(O, P )) is disjoint from the interior of ∠AOB, and since the open segment (A, B) is contained in the interior ˚ ∠AOB, then the open ray r̊(O, P ) must intersect AB at C between A and B; see the right of Figure 5. Definition 6 (Interior of triangle). The interior of a triangle ∆ABC is the intersection ˚ of interiors of its three angles, denoted ∆ABC. The boundary of ∆ABC is the union of the three sides, i.e., ∂∆ABC := AB ∪ AC ∪ BC. We also use ∆ABC to denote the union of the interior and the boundary of ∆ABC. Proposition 2.11. Given a triangle ∆ABC and O ∈ ∆ABC.˚ Let l = AB, m = AC, n = BC. Then ˚ (a) ∆ABC = H̊(l, O) ∩ H̊(m, O) ∩ H̊(n, O). (b) Any ray r(O, P ) meets the boundary of ∆ABC at a unique point Q. Proof. (a) Trivial by ˚ ∠ABC = H̊(l, O) ∩ H̊(n, O) and other two interiors of angles. (b) Let l = OP. The line OA meet BC at D between B and C. We then have A ∗ O ∗ D, the open ray r̊(D, O) is contained in H̊(n, O), and its opposite half-line is contained in the opposite side of n. So (A, D) := AD r {A, D} is contained in the interior of ∆ABC. Case 1. OP = OA. Then Q = A if A ∗ P ∗ O; Q = D if A ∗ O ∗ P. See the left of Figure 6. Case 2. OP 6= OA. The line OA separates the triangle ∆ABC into two triangles ∆ABD and ∆ACD. Since OP meets AD at O between A and D, then OP meets the boundary of ∆ABD at a unique point E and the boundary of ∆ACD at a unique point F. Moreover, E ∈ AB ∪ BD and F ∈ AC ∪ CD. If r(O, P ) = r(O, E), then Q = E. If r(O, P ) = r(O, F ), then Q = F. 7  C C F P O P D O D A A E B B Figure 6: Ray emanating from the interior of a triangle 3 Axioms of Segment Congruence Segments are not unrelated. We assume that there is a binary relation between segments, named as “segment AB is congruent to segment CD,” abbreviated as AB ∼ = CD. Congruence Axiom 1 (CA1). Given two distinct points A, B, and a ray r emanating from a point A0. There exists exactly one point B 0 on r such that B 0 6= A0 and AB ∼ = A0 B 0. 0 0 Moreover, if r = r(A, B), then B = B; if r = r(B, A), then B = A. A B B’ A’ Figure 7: Congruence of segments Congruence Axiom 2. (CA2) If AB ∼ = CD and CD ∼ = EF , then AB ∼ = EF. Proposition 3.1. (1) AB ∼ = AB, AB ∼ = BA. (2) If AB ∼ = CD, then CD ∼ = AB. Proof. (1) It follows from the latter part of CA1. (2) Let CD ∼ = AB 0 , where B 0 is a point on the ray r(A, B). Then AB ∼ = AB 0 by transitivity. Hence B 0 = B by CA1. We then have ∼ CD = AB. Congruence Axiom 3 (Segment addition) (CA3). If A ∗ B ∗ C, A0 ∗ B 0 ∗ C 0 and AB ∼ = A0 B 0 , BC ∼ = B 0 C 0 , then AC ∼ = A0 C 0. Proposition 3.2 (Segment subtraction). Given A ∗ B ∗ C and A0 ∗ B 0 ∗ C 0. If AB ∼ = A0 B 0 and AC ∼ = A0 C 0 , then BC ∼ = B0C 0. Proof. Let BC ∼= B 0 P , where P is a point on the ray r(A0 , B 0 ). Then AC ∼ = A0 P by CA2. Since AC ∼= A C , then A P ∼ 0 0 0 = A C by CA2. Thus P = C by CA1. So BC ∼ 0 0 0 = B0C 0. Proposition 3.3 (Betweenness preserving by congruence of segments). Given AC ∼ = A0 C 0 and A ∗ B ∗ C. Then there exists a unique point B 0 between A0 and C 0 such that AB ∼ = A0 B 0 and BC ∼ = B0C 0. Proof. Let AB ∼= A0 B 0 , where B 0 is the unique point on the ray r(A0 , C 0 ). Let BC ∼= B0P , where P is the unique point such that A ∗ B ∗ P. Since AB ∼ 0 0 = A B and BC ∼ 0 0 0 = B P , then ∼ 0 ∼ 0 0 0 AC = A P by CA3. Since AC = A C , then P = C by CA2. So A ∗ B ∗ C. 0 0 0 8 Proposition 3.4 (Congruence of lines). For any two lines l and l0 , there exists a one- to-one correspondence f : l → l0 such that AB ∼ = f (A)f (B) for distinct points A, B ∈ l and if A ∗ B ∗ C then f (A) ∗ f (B) ∗ f (C). Proof. Pick two points O ∈ l and O0 ∈ m. We have open rays r̊(O, −),r̊(O, +) of l and open rays r̊(O0 , −),r̊(O0 , +) of l0. Define f (O) = O0. For each P ∈ r̊(O, −), there exists a unique point P 0 ∈ r̊(O0 , −) such that OP ∼ = O0 P 0 ; define f (P ) = P 0. For each Q ∈ r̊(O, +), there exists a unique point Q0 ∈ r̊(O0 , +) such that OQ ∼ = O0 Q0 ; define f (Q) = Q0. We then 0 0 0 have a map f : l → l. Likewise we have a map f : l → l defined in similar fashion. Then f 0 ◦ f : l → l and f ◦ f 0 : l0 → l0 are identity maps. So f and f 0 are bijections. Given distinct points A, B ∈ l. If A ∗ B ∗ O or B ∗ A ∗ O, then either A, B ∈ r̊(O, −) or A, B, ∈ r̊(O, +); thus AB ∼ = f (A)f (B) by segment subtraction. If A ∗ O ∗ B, then ∼ AB = f (A)f (B) by segment addition. If A ∗ B ∗ C on l, then there exists a unique point B 00 be between f (A) and f (C) such that AB ∼ = f (A)B 00 and B 00 f (C) by the congruence of preserving betweenness. Since AB ∼ = f (A)f (B), we must have f (B) = B 00. Hence f (A) ∗ f (B) ∗ f (C). Definition 7 (Linear order of segments). For segments AB, CD, if there exists a point E between C and D such that AB ∼= CE, we write AB < CD or CD > AB. Theorem 3.5 (Strict total order of segments). For two segments AB and CD, one and only one of the three holds: AB < CD, AB ∼ = CD, AB > CD (trichotomy). Moreover, ∼ (a) If AB = CD and CD < EF , then AB < EF. (b) If AB < CD and CD ∼ = EF , then AB < EF. (c) If AB < CD and CD < EF , then AB < EF. Proof. Given segments AB and CD. Let AB ∼ = CE, where E is the unique point on the ray r(C, D). We have one and only one of the three: C ∗ E ∗ D, E = D, C ∗ D ∗ E. These are exactly the three cases: AB < CD, AB ∼ = CD, AB > CD. (a) Let P be a point such that E ∗ P ∗ F and CD ∼ = EP. Then AB ∼= EP by CA2. Thus AB < EF by definition. (b) Let P be a point such that C ∗ P ∗ D and AB ∼ = CP by definition. Then there exists ∼ a point Q such that E ∗ Q ∗ F and CP = EQ by Proposition 3.3 (congruence of preserving betweenness). Then AB ∼ = EQ by CA2. Thus AB < EF by definition. (c) Let P be such that AB ∼ = CP and C ∗ P ∗ D. Let R be such that E ∗ R ∗ F and CD ∼= ER. Then there exists a point Q such that E ∗Q∗R and CP ∼ = EQ. Thus AB ∼ = EQ, E ∗ Q ∗ R ∗ F , and of course E ∗ Q ∗ F. Therefore AB < EF. 4 Axioms of Angle and Triangle Congruence Angles are not unrelated. We assume that there is a binary relation between angles, named as “angle ∠ABC is congruent to angle ∠DEF ,” abbreviated as ∠ABC ∼ = ∠DEF. Congruence Axiom 4 (CA4). Given an angle ∠AOB and a ray r(O0 , A0 ), where the rays r(O, A), r(O, B) are not opposite. There exists a unique ray r(O0 , B 0 ) on each side of the line O0 A0 such that ∠A0 O0 B 0 ∼ = ∠AOB. Moreover, if r(O0 , A0 ) = r(O, A) and the side of OA is the half-plane H̊(OA, B), then r(O, B 0 ) = r(O, B). If r(O0 , A0 ) = r(O, B) and the side of OB is the half-plane H̊(OB, A), then r(O, B 0 ) = r(O, A). 9 Congruence Axiom 5 (CA5). If ∠A ∼ = ∠B and ∠B ∼ = ∠C, then ∠A ∼ = ∠C. It is easy to see that angle congruence is reflexive, symmetric, and transitive. So angle congruence is an equivalence relation on angles. Definition 8 (Congruence of triangles). A triangle is a collection of three non-collinear points A, B, C together with three segments AB, AC, BC (called sides), and three angles ∠ABC, ∠ACB, ∠BAC, denoted ∆ABC. The point set of ∆ABC, denoted by the same notation, is ∆ABC := ∠ABC ∩ ∠ACB ∩ ∠BAC. The interior of ∆ABC is the point set ˚ ∆ABC := ˚ ∠ABC ∩ ˚ ∠ACB ∩ ˚ ∠BAC. Two triangles are said to be congruent if there is a one-to-one correspondence between their vertices such that the corresponding sides are congruent and the corresponding angles are congruent. More specifically, if a triangle with vertices A, B, C is congruent to a triangle with vertices A0 , B 0 , C 0 by the one-to-one correspondence A to A0 , B to B 0 , and C to C 0 , written ∆ABC ∼ = ∆A0 B 0 C 0 (the order of vertices is material), then AB ∼ = A0 B 0 , AC ∼ = A0 C 0 , BC ∼ = B 0 C 0 , and ∠A ∼ = ∠A0 , ∠B ∼= ∠B 0 , ∠C ∼ = ∠C 0. Congruence Axiom 6 (Side-angle-side) (SAS). If two sides and the included angle of a triangle are congruent respectively to two sides and the included angle of another triangle, then we say that the two triangles are congruent. More precisely, given two triangles with vertices A, B, C and vertices A0 , B 0 , C 0. If AB ∼ = A0 B 0 , AC ∼ = A0 C 0 , and ∠A ∼ = ∠A0 , then ∼ ∆ABC = ∆A B C.0 0 0 Corollary 4.1. Given a triangle ∆ABC and a segment A0 B 0 ∼ = AB. Then there exists a unique point C 0 on each side of the line A0 B 0 such that ∆ABC ∼ = ∆A0 B 0 C 0. Proof. Choose a side of line A0 B 0. There exists one and only one ray r(A0 , P ) such that ∠B 0 A0 P ∼ = ∠BAC by CA4. Then there exists a unique point C 0 on r(A0 , P ) such that A0 C 0 ∼ = AC by CA1. Thus ∆B 0 A0 C 0 ∼= ∆BAC by SAS. Proposition 4.2. Given a triangle ∆ABC. If AB ∼ = AC, then ∠B ∼ = ∠C. Proof. Consider the one-to-one correspondence A ↔ A, B → C, C → B. We have AB ∼ = AC, ∠BAC ∼ = ∠CAB, AC ∼ = AB. Then ∆ABC ∼ = ∆ACB by SAS. Thus ∠B ∼ = ∠C by definition of congruence of triangles. Definition 9 (Supplementary angle, opposite angle, right angle). Supplementary angles and opposite angles are defined as before. A right angle is an angle which is congruent to its supplement. A closed half-plane is not an angle by our definition of angles; it is convenient to call it a flat angle. Proposition 4.3 (Supplementary, opposite, right angle congruence rules). (a) Sup- plements of congruent angles are congruent. (b) Opposite angles are congruent each other. (c) Any angle congruent to a right angle is a right angle. 10  B B’ C O A C’ O’ A’ D D’ Figure 8: Supplements of congruence angles are congruent Proof. Given two congruent angles ∠AOB ∼ = ∠A0 O0 B 0. Pick a point C on the opposite ray of r(O, A) with C 6= O. Pick a point C 0 on the opposite ray of r(O0 , A0 ) with C 0 6= O. We may assume OA ∼ = O0 A0 , OB ∼ = O0 B 0 , OC ∼ = O0 C 0. See Figure 8. (a) We need to show ∠BOC ∼ = ∠B 0 O0 C 0. Since OA ∼ = O0 A0 , ∠AOB ∼ = ∠A0 O0 B 0 , OB ∼ = O0 B 0 , then ∆AOB ∼ = A0 O0 B 0 by SAS. Then AC ∼ = A0 C 0 by CA3; AB ∼ = A0 B 0 and ∼ 0 0 0 ∠BAC = ∠B A C by definition of congruence triangles. Thus ∆BAC = B A C by SAS. ∼ 0 0 0 Since OC ∼ = O0 C 0 , ∠OCB ∼ = O0 C 0 B 0 and CB ∼= C 0 B 0 , then ∆OCB ∼ = O0 C 0 B 0 by SAS. We see ∠BOC ∼ 0 0 = ∠B O C. 0 (b) Consider opposite angles ∠AOB and ∠COD in the left of Figure 8. Both are sup- plementary to ∠BOC. So ∠AOB ∼ = ∠COD by (a). (c) Let ∠AOB be a right angle. Need to show that ∠A0 O0 B 0 is a right angle. Notice that ∠AOB ∼ = ∠BOC by definition of right angles, ∠B 0 O0 C 0 ∼ = ∠BOC by (a), and ∠AOB ∼ = 0 0 0 0 0 0 ∼ 0 0 0 ∠A O B by given condition. Then ∠A O B = ∠B O C by transitivity. This means that ∠A0 O0 B 0 is a right angle. Proposition 4.4 (Existence of perpendicular line). For each line l and each point P not on l, there exists a unique line m through P perpendicular to l. P A Q B P’ Figure 9: Construction of perpendicular lines Proof. Pick two distinct points A, B on l. Draw segment AP. Then there exists a unique ray r(A, C) on the opposite side of line l such that ∠BAP ∼ = ∠BAC. Mark a point P 0 on the ray r(A, C) such that AP ∼ = AP 0. Draw line P P 0 = m. We claim that m ⊥ l. See Figure 9. If A, P, P 0 are collinear, then A is the intersection of lines AB and P P 0. Clearly, ∠BAP and BAP 0 are congruent supplementary angles. So they are right angles and m ⊥ l. Assume that A, P, P 0 are not collinear. Since P, P 0 on opposite sides of l, then r(P, P 0 ) intersects l at a unique point Q. We have triangles ∆AP Q and ∆AP 0 Q. Since AP ∼ = AP 0 , ∠P AQ ∼ = ∠P 0 AQ, AQ ∼ = AQ, then ∆P AQ ∼ = ∆P 0 AQ by SAS. Thus ∠AQP ∼ = ∠AQP 0 by definition of congruence triangles, i.e., m ⊥ l. 11 Proposition 4.5 (Angle-side-angle criterion) (ASA). If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. C C’ C’’ A B A’ B’ Figure 10: Angle-side-angle criterion Proof. Given triangles ∆ABC, ∆A0 B 0 C 0 , and ∠BAC ∼ = ∠B 0 A0 C 0 , AB ∼ = A0 B 0 , ∠ABC ∼ = 0 0 0 00 0 0 ∠A B C. Draw the unique point C on the ray r(A , C ) such that AC = A C. Then ∼ 0 00 ∆ABC ∼ = ∆A0 B 0 C 00 by SAS. Then ∠ABC ∼ = ∠A0 B 0 C 00 by definition of congruence of trian- gles. Since ∠ABC ∼ = ∠A0 B 0 C 0 by given condition, then ∠A0 B 0 C 0 ∼ = ∠A0 B 0 C 00 by transitivity. 0 0 00 0 00 This means that B , C , C are collinear, i.e., C , C are on both lines B 0 C 0 and A0 C 0. Since intersection point of two lines is unique, we have C 0 = C 00. Hence 4ABC ∼ = 4A0 B 0 C 0. See Figure 10. Hence ∆ABC ∼ = ∆A0 B 0 C 0. Proposition 4.6 (Angle addition). Given two angles ∠AOC and ∠A0 O0 C 0. Let r(O, B) be a ray between rays r(O, A) and r(O, C). Let r(O0 , B 0 ) be a ray between rays r(O0 , A0 ) and r(O0 , C 0 ). If ∠AOB ∼ = ∠A0 O0 B 0 and ∠BOC ∼ = ∠B 0 O0 C 0 , then ∠AOC ∼ = ∠A0 O0 C 0. Proof. We may assume that OA ∼ = O0 A0 , OB ∼= O0 B 0 , OC ∼= O0 C 0 , and that B is a point on AC between A, C. But we did not assume that B 0 is a point on A0 C 0. See Figure 11. Then ∆AOB ∼ = A0 O0 B 0 and ∆BOC ∼ = ∆B 0 O0 C 0 by SAS. We see that the supplementary angles ∠OBA, ∠OBC are congruent to the angles ∠O0 B 0 A0 , ∠O0 B 0 C 0 respectively. Then the supplementary angle ∠O0 B 0 C 00 of ∠O0 B 0 A0 is congruent to ∠OBC by the Supplementary Angle Congruence Rule. Thus ∠O0 B 0 C 00 ∼ = ∠O0 B 0 C 0 by transitivity. Since ∠O0 B 0 C 00 and ∠O0 B 0 C 0 are on the same side of line O0 B 0 , it follows that ∠O0 B 0 C 00 = ∠O0 B 0 C 0 by CA3. So A0 , B 0 , C 0 , C 00 are collinear. Since AB ∼ = A0 B 0 , BC ∼ = B 0 C 0 , then AC ∼ = A0 C 0. Since ∠OAC ∼ = ∠O0 A0 C 0 , ∠OCA ∼ = ∠O0 C 0 A0 , we have ∆AOC ∼ = ∆A0 O0 C 0 by ASA. Therefore ∠AOC ∼ = ∠A0 O0 C 0. C C’ B B’ O O’ A A’ Figure 11: Angle subtraction Proposition 4.7 (Angle subtraction). Let r(O, B) be a ray between rays r(O, A) and r(O, C). Let r(O0 , B 0 ) be a ray between rays r(O0 , A0 ) and r(O0 , C 0 ). If ∠AOB ∼ = ∠A0 O0 B 0 , ∠AOC ∼ = ∠A0 O0 C 0 , then ∠BOC ∼ = ∠B 0 O0 C 0. 12 Proof. We may assume that OA ∼ = O0 A0 , OC ∼= O0 C 0 , and that B is a point on AC and B 0 is a point on A C. See Figure 11. Since ∠AOC ∼ 0 0 = ∠A0 O0 C 0 , we have ∆AOC ∼ = ∆A0 O0 C 0 by SAS. Thus AC ∼ = A0 C 0 and ∠OAB ∼ = ∠O0 A0 B 0. Since ∠AOB ∼ = ∠A0 O0 B 0 , OA ∼ = O 0 A0 , ∠OAB ∼ = ∠O0 A0 B 0 , then ∆OAB ∼ = ∆O0 A0 B 0 by ASA. Thus AB ∼ = A0 B 0. Since AC ∼ = A0 C 0 , then BC ∼ = B C by Proposition 3.2 (segment subtraction). Now OC ∼ 0 0 = O C , ∠OCB ∼ 0 0 = ∠O C B , CB = C B , we have ∆OCB = O C B by SAS. Therefore ∠BOC ∼ 0 0 0 ∼ 0 0 ∼ 0 0 0 = ∠B 0 O0 C 0. Proposition 4.8. Given a triangle ∆ABC. If ∠B ∼ = ∠C, then AB ∼ = AC. Proof. Let A 7→ A, B 7→ C, C 7→ B. Since ∠ABC ∼ = ∠ACB, BC ∼ = CB, ∠ACB ∼ = ∠ABC, ∼ ∼ then ∆ABC = ACB by ASA. Thus AB = AC by definition of congruence of triangles. Definition 10. An angle ∠AOB is less than an angle ∠A0 O0 C 0 , written ∠AOB < ∠A0 O0 C 0 , if there exists a ray r(O0 , B 0 ) between the rays r(O0 , A0 ) and r(O0 , C 0 ), such that ∠AOB ∼ = ∠A0 O0 B 0. Proposition 4.9 (Strict total order of angles). For any two angles ∠A and ∠B, one and only one of the three holds: ∠A < ∠B, ∠A ∼= ∠B, ∠B < ∠A (trichotomy). Moreover, (a) If ∠A ∼ = ∠B, ∠B < ∠C, then ∠A < ∠C. (b) If ∠A < ∠B, ∠B ∼ = ∠C, then ∠A < ∠C. (c) If ∠A < ∠B, ∠B < ∠C, then ∠A < ∠C. Proof. Given two angles ∠AOB and ∠A0 O0 B 0. There exists a unique open ray r̊(O0 , C 0 ) in the open half-plane H̊(O0 A0 , B 0 ) such that ∠AOB ∼ = ∠A0 O0 C 0. If C 0 is on the ray r(O0 , B 0 ), then r(O0 , C 0 ) = r(O0 , B 0 ) and ∠AOB ∼ = ∠A0 O0 B 0. If C 0 is not on the ray r(O0 , B 0 ), there are two cases. Case 1. Points C 0 , A0 are on the same side of O0 B 0. Since C 0 , B 0 are on the same side of O0 A0 , then C 0 is contained in the interior of ∠A0 O0 B 0 ; so is the open ray r̊(O0 , C 0 ). Thus ∠AOB < ∠A0 O0 B 0. Case 2. Points C 0 , A0 are on opposite sides of O0 B 0. Then r(O0 , B 0 ) meets A0 C 0 at P 0 between A0 and C 0 by Crossbar Theorem. Thus r̊(O0 , B 0 ) is contained in ˚ ∠A0 O0 C 0. By 0 0 0 definition ∠AOB > ∠A O B. (a) Let ∠AOB ∼ = ∠A0 O0 B 0 < ∠A00 O00 C 00. There exists a ray r(O00 , B 00 ) between rays r(O , A ) and r(O , C 00 ) such that ∠A0 O0 B 0 ∼ 00 00 00 = ∠A00 O00 B 00 by definition. Then ∠AOB ∼ = ∠A00 O00 B 00 by transitivity. Thus ∠AOB < ∠A00 O00 C 00. (b) Let ∠AOB < ∠A0 O0 C 0 ∼ = ∠A00 O00 C 00. There exists a ray r(O0 , B 0 ) between the rays r(O , A ) and r(O , C ) such that ∠AOB ∼ 0 0 0 0 = ∠A0 O0 B 0. Let r(O00 , B 00 ) be a ray between the rays r(O00 , A00 ) and r(O00 , C 00 ) such that ∠A0 O0 B 0 ∼= ∠A00 O00 B 00. Then ∠AOB ∼ = ∠A00 O00 B 00 00 00 00 by transitivity. Thus ∠AOB < ∠A O C. (c) Let ∠AOB < ∠A0 O0 C 0 < ∠A00 O00 D00. There exists a ray r(O00 , C 00 ) between the rays r(O00 , A00 ) and r(O00 , D00 ) such that ∠A0 O0 C 0 ∼ = ∠A00 O00 C 00. Then ∠AOB < ∠A00 O00 C 00 by (b). Thus there exists a ray r(O , B ) between the rays r(O00 , A00 ) and r(O00 , C 00 ) such that 00 00 ∠AOB ∼ = ∠A00 O00 B 00. Therefore ∠AOB < ∠A00 O00 D00. Proposition 4.10 (Side-side-side criterion) (SSS). Given triangles ∆ABC and ∆A0 B 0 C 0. If AB ∼ = A0 B 0 , AC ∼ = A0 C 0 , BC ∼ = B 0 C 0 , then ∆ABC ∼ = ∆A0 B 0 C 0. Proof. Let C 00 be the unique point on the opposite side of H̊(A0 B 0 , C 0 ) bounded by A0 B 0 such that ∆ABC ∼ = ∆A0 B 0 C 00. Draw the segment C 0 C 00. The line A0 B 0 meets C 0 C 00 at D0 between C and C. See Figure 12. Then A0 C 0 ∼ 0 00 = AC ∼= A0 C 00 and B 0 C 0 ∼= BC ∼ = B 0 C 00 , i.e., ∆A0 C 0 C 00 = ∠A0 C 00 C 0 and ∠B 0 C 0 C 00 ∼ and ∆B 0 C 0 C 00 are isosceles triangles. Hence ∠A0 C 0 C 00 ∼ = ∠B 0 C 00 C 0. 13  C C’ γ α A B A’ D’ B’ β δ C’’ Figure 12: Side-side-side criterion If A0 ∗ D0 ∗ B 0 , then the open ray r̊(C 0 , C 00 ) is contained in ˚ ∠A0 C 0 B 0 , and the open ray r̊(C 00 , C 0 ) is contained in ˚ ∠A0 C 00 B 0 by Crossbar Theorem; thus ∠A0 C 0 B 0 ∼ = ∠A0 C 00 B 0 by angle addition. If A0 ∗ B 0 ∗ D0 , then r̊(C 0 , B 0 ) is contained in ˚ ∠A0 C 0 D0 , and r̊(C 00 , B 0 ) is contained in ˚ 0 00 0 0 0 0 ∼ ∠A C D by Crossbar Theorem; thus ∠A C B = ∠A0 C 00 B 0 by angle subtraction. Since A0 C 0 ∼ = AC ∼ = A0 C 00 , ∠A0 C 0 B 0 ∼ = ∠A0 C 00 B 0 , B 0 C 0 ∼= BC ∼ = B 0 C 00 , we see that ∆A0 B 0 C 0 ∼ = 0 0 00 ∼ ∆A B C by SAS. Hence ∆ABC = ∆A B C by transitivity. 0 0 0 Theorem 4.11 (Euclid’s Fourth Postulate). All right angles are congruent to each other. C C’ D’ E’ A O B A’ O’ B’ Figure 13: Euclid’s Fourth Postulate Proof. Given angles ∠AOC ∼ = BOC and ∠A0 O0 C 0 ∼ = B 0 O0 C 0 ; see Figure 13. We need to show ∠AOC = ∠A O C. Let r̊(O , P ) be the unique open ray in H̊(A0 B 0 , C 0 ) such that ∠AOC ∼ ∼ 0 0 0 0 0 = 0 0 0 0 0 0 0 0 0 0 0 ∠A O P. It suffices to show that r̊(O , P ) = r̊(O , C ). Suppose r̊(O , P ) 6= r̊(O , C ). Then either r̊(O0 , P 0 ) = r̊(O0 , D0 ), which is contained in ˚∠A0 O0 C 0 , or r̊(O0 , P 0 ) = r̊(O0 , E 0 ), which is contained in ˚ ∠B 0 O0 C 0. In the former case, we have ∠A0 O0 D0 < ∠A0 O0 C 0 and ∠B 0 O0 C 0 < ∠B 0 O0 D0 by definition of order of angles. Since ∠A0 O0 C 0 ∼ = ∠B 0 O0 C 0 by right angle property, then ∠A0 O0 D0 < ∠B 0 O0 D0 by Proposition 4.9. Note that ∠BOC ∼ = ∠B 0 O0 D0 by Proposition 4.3(a), and ∠AOC ∼ = ∠A O D. Then ∠AOC < ∠BOC. However, ∠AOC ∼ 0 0 0 = ∠BOC by right angle property. In summary we have ∠AOC ∼ = ∠A0 O0 D0 < ∠A0 O0 C 0 ∼ = ∠B 0 O0 C 0 < ∠B 0 O0 D0 ∼ = ∠BOC ∼ = ∠AOC. So ∠AOC < ∠AOC; this is a contradiction. In the latter case, we have ∠AOC ∼ = ∠A0 O0 E 0 > ∠A0 O0 C 0 ∼ = ∠B 0 O0 C 0 > ∠B 0 O0 E 0 ∼ = ∠BOC ∼ = ∠AOC. So ∠AOC > ∠AOC; this is a contradiction. 14 5 Axioms of Continuity Dedekind’s Axiom (Continuity Axiom). If a line l is partitioned into two nonempty subsets Σ1 , Σ2 , i.e., l = Σ1 ∪ Σ1 and Σ1 ∩ Σ2 = ∅, such that no point of either subset is between two points of the other (equivalently both are convex), then there exists a unique point O on l such that one of Σ1 , Σ2 is a ray with vertex O and the other is an open ray with vertex O opposite to the other ray. The pair {Σ, Σ2 } is called a Dedekind cut of l. Given two distinct points A, B ∈ l. If A, B ∈ Σi , we have AB ⊂ Σi , i.e., Σi has no “hole.” Suppose we do not require Σ1 ∩ Σ2 = ∅ in Dedekind’s axiom. If A, B ∈ Σ1 ∩ Σ2 , then we must have A = B. So the intersection Σ1 ∩ Σ2 contains exactly one point O. Thus Σ1 , Σ2 are two rays with the vertex O. So, when Σ1 ∩ Σ2 = ∅ is imposed, we say that the partition {Σ1 , Σ2 } determines one point on l. Definition 11. A subset Ω of points is said to be convex provided that whenever two points P, Q are contained in Ω then the segment P Q is contained in Ω. For a line l with a total order ¹, the following subsets of l are convex sets, known as intervals: line l; rays r(O, −) = {P ∈ l : P ¹ O}, r(O, +) = {P ∈ l : O ¹ P }; open rays r̊(O, −) = {P ∈ l : P ≺ O}, r̊(O, +) = {P ∈ l : O ≺ P }; closed interval (segment) [A, B] = AB = {P ∈ l : A ¹ P ¹ B}; open interval (segment) (A, B) = {P ∈ l : A ≺ P ≺ B}; half-closed and half-open intervals (segments) [A, B) = {P ∈ l : A ¹ P ≺ B}, (A, B] = {P ∈ l : A ≺ P ¹ B}. The points O, A, B are called endpoints of I. For a line not satisfying Dedekind’s axiom, a convex subset of the line is not necessarily an interval. Proposition 5.1 (Extended Dedekind’s Axiom). Dedekind’s axiom is valid for any nonempty interval I of any line l. More precisely, if a nonempty interval I is partitioned into two nonempty convex sets Σ1 , Σ2 , then both Σ1 , Σ2 are intervals with an endpoint O, one is closed and the other is open at O. Proof. Let l be totally ordered so that left and right sides of I are meaningful. If I is empty or contains exactly one point, nothing is to be proved for the statement is irrelevant. We assume that I contains at least two points. Let Σ1 be on the left side of Σ2. Since I is nonempty, the complement l r I has one of the forms: (a) empty set ∅, (b) a left (open) ray Γ1 , (c) a right (open) ray Γ2 , (d) disjoint union of a left (open) ray Γ1 and a right (open) ray Γ2. Set Σ01 := Σ1 ∪ Γ1 and Σ02 := Σ2 ∪ Γ2. Then Σ01 , Σ02 form a Dedekind cut of l. By Dedekind’s axiom, there exists a unique point O on l such that Σ01 is the right closed (open) ray r(O, −) (r̊(O, −)) with vertex O, and Σ02 is the left open (closed) ray r̊(O, +) (r(O, +)) with vertex O. Hence Σ1 is a right-closed (right-open) interval with right endpoint O ∈ Σ1 , and Σ2 is left-open (left-closed) interval with left endpoint O. 15 Example. Let Q be the field of rational numbers. Then Q2 forms an affine plane, called rational affine plane under its points and lines defined by one linear equation. Dedekind’s axiom is not satisfied by the rational plane. Consider the x-axis l = {(a, 0) : a ∈ Q}. Let Σ1 = {(a, 0) : a ∈ Q, a2 > 3, a > 0} and Σ2 = l r Σ1. Then Σ1 , Σ2 form a Dedekind cut, that is, they satisfy the conditions of Dedekind’s axiom. √ However, neither Σ1 nor Σ2 is an (open) ray of l. In fact, Σ1 = {(a, 0) : a ∈ Q, a > 3} is not an interval in Q2 (since it has no left endpoint). Dedekind’s axiom implies all of the following axioms. Euclid’s Proposition. For each segment there exists an equilateral triangle having one of its sides to be the given segment. Definition 12. A point P is said to be inside a circle of radius OR with center O if OP < OR. Circular Continuity Principle. If each of two circles has one point inside but outside the other, then the two circles intersect at two points. Elementary Continuity Principle. If one endpoint of a segment is inside a circle and the other endpoint is outside, then the segment intersects the circle. Archimedes’[0 a:ki0 mi:di:z] Axiom. Given a segment AB and a ray r with vertex O. For each point P 6= O on r, there exist an integer n and a point Q on r, where OQ ∼ = n · AB, such that either Q = P or O ∗ P ∗ Q. Aristotle’s[0 æristotl] Axiom. Given an acute angle ∠AOB and a segment CD. There exists a point Y on the ray r(O, B) such that XY > CD, where X is the foot of Y on the ray r(O, A). Proposition 5.2 (Dedekind’s implies Archimedes’). Dedekind’s axiom implies Archimedes’ axiom. Proof. Given a segment AB and a ray r with vertex O. A point P ∈ r is said to be reachable by AB if P = O or there exist an positive integer n and a point Q, such that OQ ∼ = n · AB and O ∗ P ∗ Q. Let Σ1 be the set of points on r reachable by AB, and points on the opposite ray of r; so O ∈ Σ1. Let Σ2 be the complement of Σ1 in the line l that contains r; Σ2 is also the complement of Σ1 in r. We claim that Σ2 = ∅. (If so, all points on the ray r are reachable by AB, which is Archimedes’ axiom.) Suppose Σ2 6= ∅. We claim that {Σ1 , Σ2 } is a Dedekind cut of l. One the one hand, let P, Q ∈ Σ1 be distinct points. If both P, Q are on r or on the opposite ray of r, it is clear that P Q ⊂ Σ1. If P is on the opposite ray of r and Q ∈ r, then P O ⊂ Σ1 and OQ ⊂ Σ1 ; so P Q = P O ∪ OQ ⊂ Σ1. On the other hand, let P, Q ∈ Σ2 be distinct points. Suppose P Q 6⊂ Σ2 ; there exists a point R ∈ Σ1 such that P ∗ R ∗ Q; since R can be reached, so is P ; thus P ∈ Σ1 , which is a contradiction. Therefore {Σ1 , Σ2 } is a Dedekind cut of l, and determines a unique point O0 on l. Case 1. O0 ∈ Σ1. Then Σ1 is a ray with vertex O0. Since the opposite ray of r is contained in Σ1 , then O0 ∈ r and Σ2 is an open ray on r with vertex O0. Let O0 be reached by laying off n copies of AB starting from O. Then by laying off one more copy of AB on r starting from O0 , we get points of Σ2 being reached by AB. This is impossible. Case 2. O0 ∈ Σ2. Then Σ2 is a ray on r with vertex O0 6= O, and Σ1 is the opposite open ray with vertex O0. Laying off one copy of AB starting from O0 on the open ray Σ1 , we obtain a point P 0 in Σ1. Then any point Q0 such that P 0 ∗ Q0 ∗ O0 is reachable by AB. Thus by laying one more copy of AB starting from Q0 , the point O0 is reachable. So O0 ∈ Σ1. This is a contradiction. 16 Proposition 5.3 (Dedekind’s implies Elementary Continuity). Dedekind’s axiom im- plies Elementary Continuity Principle. Proof. Let γ be a circle with center O and radius OR. Let AB be a segment with A inside and B outside γ, i.e., OA < OR and OB > OR. Let Σ1 denote the set of points on AB inside γ, and Σ2 the subset of points on AB outside or on γ. Then Σ1 , Σ2 form a Dedekind cut for the segment AB by trichotomy of segments. Dedekind’s axiom implies that there exists a unique point P on AB such that Σ1 , Σ2 are intervals with endpoint P , one contains P and the other does not contain P. We claim that P is on γ, i.e., OP ∼= OR. Case 1. OP < OR. Then P ∈ Σ1. Take a point Q ∈ Σ2 such that |P Q| = (|OR| − |OP |)/2. By triangle inequality we have |OR| < |OQ| < |OP | + |P Q| = |OP |/2 + |OR|, which is a contradiction. Case 2. OP > OR. Then P ∈ Σ2. Take a point Q ∈ Σ1 such that A ∗ Q ∗ P and |P Q| ≤ (|OP | − |OR|)/2. Since |OQ| < |OR| and |QP | = |P Q|, then by triangle inequality |OP | ≤ |OQ| + |QP | < |OR| + (|OP | − |OR|)/2 = |OR|/2 + |OP |, which is a contradiction. So we must have OP ∼ = OR. Relationship between the Axioms of Continuity. Dedekind’s axiom ⇒ Archimedes’ axiom, Circlar Continuity Principle Archimedes’ axiom ⇒ Aristotle’s axiom Circlar Continuity Principle ⇒ Elemenatry Continuity Principle 17