Neutral Geometry. April 18, 2013

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1 Neutral Geometry pril 18, Geometry without parallel axiom Let l, m be two distinct lines cut by a third line t at point on l and point Q on m. Let be a point on l and a point on m such that, are on the same side of t. Let be a point on the opposite open ray of r(, ), and a point on the opposite open ray of r(q, ). The four angles Q, Q, Q, Q are known as interior angles. The pair Q, Q and the other pair Q, Q are known as two pairs of alternate interior angles. See Figure 1. m R Q t l R Figure 1: ongruent alternate interior angles imply parallel Theorem 1.1 (lternate Interior ngle Theorem). If two distinct lines cut by a transversal have a pair of congruent alternate interior angles, then the two lines are parallel. roof. Let l, m be two lines cut by a transversal t such that Q = Q ; see Figure 1. We need to show that l, m do not meet. Suppose they meet at a point R, say, on the side of, bounded by t. raw a point R on the ray r(q, ) so that QR = R. Then Q R = QR by SS. Thus QR = Q R, i.e., Q = R Q. Since Q is supplementary to Q, and Q is supplementary to Q, we see that Q = Q by Supplementary ngle ongruence Rule. So Q = R Q. This means that R must be on the line l by etweenness xiom 1. Then it forces that l = m since there is only one line through two points R, R. This is a contradiction. Hence l, m do not meet, i.e., l m. orollary 1.2. Tow distinct lines perpendicular to a common line are parallel. In particular, there is exactly one line through a point perpendicular to a given line l. roof. Let l, m be two distinct lines perpendicular to a common line t. If l, m are cut by t at two distinct points, Q respectively, then all interior angles are right angles. Hence they are congruent by Right ngle ongruence Theorem. In particular, the alternate interior angles 1

2 are congruent. So l, m are parallel by lternate Interior ngle Theorem 1.1. If l, m are cut by t at the same point, we must have l = m, since all right angles are congruent and the two lines perpendicular to t must be the same. In the case that is not on the line l, suppose there are two lines m, n perpendicular to l and both pass through. Then m, n are parallel by lternate Interior ngle Theorem 1.1. This is contradictory to the fact that they meet at. In the case that is on l, the perpendicular line to l through is unique because all right angles are congruent and two perpendicular lines must be the same. orollary 1.3. Through a point not on a line l there exists at least one line m parallel to l. roof. Through the point not on line l there is a unique line m perpendicular to l. Through there is a unique line n perpendicular to m. The line m cut both lines l, n and all interior angles are right angles, hence they are congruent. Of course, the alternate interior angles are congruent. So l, n are parallel by lternate Interior ngle Theorem. Warning. Uniqueness of perpendicular line does not imply the uniqueness of parallel line. Recall the exterior angle of a triangle and its remote exterior angles. Theorem 1.4 (xterior ngle Theorem). n exterior angle of a triangle is greater than each of its remote interior angles. roof. Given a triangle and its exterior angle at vertex. See Figure 2. We show < first. Suppose =. Then lines, are parallel (since alternate interior angles are congruent). This is impossible. F G Figure 2: xterior angle is larger than its remote exterior angle Suppose >. Take an interior point F of such that F =. Then lines F, are parallel (since alternate interior angles are congruent). However, F meets at G between and by etween-ross Lemma. Since is part of line, this is contradictory to that F, are parallel. Hence we must have <. Likewise, the exterior angle >. Since = as opposite angles, we have > by linear order property of angles. roposition 1.5 (Side-angle-angle criterion) (S). Given triangles and. If =, =, and =, then =. roof. Suppose >. ick an interior point of such that =. The line meets at between and by etween-ross Lemma. Then = by S. Thus =. Since =, then =. Note that is an exterior angle of at. Then >, i.e., > by xterior ngle Theorem. This is a contradiction. Suppose <. similar argument leads to a contradiction. 2

3 Figure 3: Side-angle-angle criterion Warning. There is no angle-side-side criterion for congruence of triangles. Here is an example of two non-congruent triangles satisfying the angle-side-side conditions; see Figure 4. Figure 4: No angle-side-side criterion roposition 1.6. Given right triangles and with right angles and. If = and =, then =. roof. Suppose >. Let be a point between and such that =. See Figure 5. Then = by SS. Subsequently, = =. This means that is an isosceles triangle. Hence =. Now xterior ngle Theorem and right angle property imply Figure 5: Right triangle congruence criterion > = > = =, i.e., >, which is a contradiction. Likewise, < leads to a similar contradiction. We then have =. Hence = by SS. midpoint of a segment is a point between and such that =. It is easy to see that the midpoint of a segment is unique if it exists. roposition 1.7 (Midpoint Theorem). very segment has a unique midpoint. 3

4 F Figure 6: onstruction of midpoint roof. Given a triangle. We plan to find the midpoint of segment. xtend to a point such that. Then, are on opposite sides of line. Since > by xterior ngle Theorem, there exists a unique ray r(, ) between rays r(, ), r(, ) such that =. Note that, are on the same side of. Mark the unique point on r(, ) such that = by ongruence xiom 1. Note that,, are the same side. Since, are on opposite sides of, then, are on opposite sides of. So line meets the segment at a point F between and. To see that F is between and, it suffices to show that because of rossbar Theorem. In fact,,, are on the same side of. Since =, line is parallel to ; of course,,, are the same side of. Then. Now we have =, =, and F = F (opposite angles), then F = F by S. Hence F = F. The uniqueness is trivial. roposition 1.8 (isector Theorem). (a) very angle has a unique bisector. (b) very segment has a unique perpendicular bisector. roof. (a) Given an angle O with O = O. raw the segment, find the midpoint of, and draw segment O. Then O = O by SSS. Hence O = O. So ray r(o, ) is a bisector of O. The uniqueness is trivial. (b) Trivial. roposition 1.9 (ngle-opposite-side relation). For any triangle, we have > if and only if >. Figure 7: Side-angle proportion relation roof. (sufficiency): Find bisector r(, ) of, which meets at between and. Lay off on to have segment =. Then = by SS. So =. Note that > by xterior ngle Theorem for. Hence >. (necessity): If =, then is an isosceles triangle. So =, which is contradictory to >. If <, then >, which is contradictory to >. Hence > by trichotomy of segments. 4

5 orollary (a) If Q is a right angle for a triangle QR, then < Q, R < Q. (b) Given a point not on a line l, let Q be the foot of the perpendicular line through. Then Q < R for all points R on l other than Q. R Q l Figure 8: Shortest distance from a point to a line roof. Mark two points, on line l such that R Q. Then Q > QR by xterior ngle Theorem for RQ. Since Q = RQ, then QR < RQ. Subsequently, Q < R by the angle-opposite-side relation for triangle RQ. 2 Measures of segments and angles It is clear that segment congruence is an equivalence relation on the set of all segments. For each segment we may use [] to denote the congruence equivalence class of. Recall that line is totally ordered. We may view as a segment together with an order such that. For any segment, we can construct a new segment on the ray r(, ) by laying off on the ray r(, ), where, such that =. We write this new segment as +. Note that + is a segment on ray r(, ) with endpoint. Of course, + and + are certainly distinct segments. However, + = +. For the congruence equivalence classes [], [], we define If F is a segment, it is easy to see that We then have [] + [] := [ + ]. ( + ) + F = + ( + F ). [] + [] = [] + []. ([] + []) + [F ] = [] + ([] + [F ]). Let 1 denote the segment whose endpoints are and the midpoint of. For each 2 positive integer k, let k denote the segment obtained by laying off k copies of on the ray r(, ) starting from ; let us define the segment 1 2 := 1 ( ( ( ))) 1 1 k } {{ } k For integers p Z and positive integer q Z +, we write 2 p q := 2 p (q ) = q (2 p ), 2 p q[] := [2 p q ]. 5

6 Theorem 2.1. Fix a segment OI, called unit segment. There exists a unique mapping from the set of all segments to the set R + of positive real numbers,, satisfying the properties: (a) OI = 1. (b) = if and only if =. (c) if and only if = +. (d) < if and only if <. (e) For each positive real number a, there exists a segment such that = a. roof. (Sketch) Fix open ray r(o, I). ach segment is congruent to a segment O with unique point r(o, I). It suffices to assign for each point r(o, I) a positive real number. The right endpoints of the segments 2 p q OI are assigned to numbers 2 p q, known as dyatic rational numbers. If rchimedes axiom is satisfied, then every point in r(o, I) has a decimal expression with base 2, and the point is assigned to the real number with decimal expression. Given a real number a > 0. Let Σ 1 := { r(o, I) : OQ a} and Σ 2 := r(o, I) Σ 1. Then {Σ 1, Σ 2 } is a edekind cut of r(o, I). There exists a unique Q such that Σ 1 = [O, Q] and Σ 2 = (Q, ). We must have OQ = a. efinition 1. (a) Two angles O and O are said to be addable if there exists an angle O such that O and O = O. We define the partial addition O + O := O. (b) n half-plane is also known as a flat angle. congruent. We assume that all flat angles are Theorem 2.2 (egree measure of angles). There exists a unique mapping from the set of all angles to the interval (0, 180) of real numbers,, satisfying the properties: (a) = 90 if is a right angle. (b) = if and only if =. (c) If h(o, ) is contained in o O, then O = O + O. (d) < if and only if <. (e) If, are complementary, then + = 180. (f) For each positive real number a (0, 180), there exists an angle O such that O = 180. roof. (Sketch) Fix ray r(o, ) and consider angles O. Since each angle can be bisected, we denote by 1 O the angle O, where r(o, ) is the bisector of O. We then 2 have angles 1 2 O := 1 ( ( ( ))) 1 1 k O, k 1. } {{ } k If O is addable to (k 1) O, we define We thus have k O := (k 1) O + O 2 p q O = q(2 p O) as long as they are addable to each other. Let [ O] denote the equivalence class of angles congruent to O. Then 2 p q [ O] := [2 p q O]. 6

7 Now we fix a right angle ROT. ssign the angles 2 p q ROT to the real numbers 90 2 p q, that is, 2 p q ROT (2 p q ROT ) = 90 2 p q. edekind s axiom ensures that the assignment can be extended into a bijection from angle congruence classes onto the open interval (0, 180) of real numbers expressed in decimal form of base 2, satisfying (a)-(f). For instance, for part (e) about complementary angles,, we see that is a right angle so that 2 2 ( ) 2 = 90. orollary 2.3 (onsecutive Interior ngle Theorem). If two distinct lines cut by a transversal have a pair of consecutive interior angles whose angle sum is a flat angle, then the two lines are parallel. roof. We assume in Figure 1 that Q + Q is congruent to a flat angle. Q + Q is congruent is a flat angle, we see that Q = Q. So m l. Since roposition 2.4 (Triangle inequality). Let,, be three distinct non-collinear points. Then < +. roof. There exists a unique point such that and = by ongruence xiom 1. See Figure 9. Then = because of the isosceles triangle. Since > =, we have > =. Hence > by the angle-opposite-side relation. Since = +, we obtain + >. Subsequently, + > by measure of segments. Figure 9: Triangle inequality roposition 2.5 (Included angle-opposite-side relation of equal sides). Given triangles, and =, =. Then > if and only if >. F Figure 10: qual sides and its included angle-opposite-side relation 7

8 roof. : Let >. raw a ray r(, ) between r(, ) and r(, ) such that =. Let be a point on r(, ) such that =, and the intersection of r(, ) and such that. raw the angle bisector of and its intersection with at F between and. Then F = F by SS. Hence F = F. Note that F +F > by triangle inequality. Then F +F >. Since F + F = and =. We obtain >. : Let >. If = ; then = by SS; so =, which is a contradiction. If <, then > by what just proved previously, which is also a contradiction. Hence we have > by trichotomy of angles. 3 Saccheri-Legendre theorem Lemma 2. Given a triangle. Let be the midpoint of and a point on the ray r(, ) such that = and. raw the segment. Then (a) The angle sum of equals the angle sum of. (b) ither 1 2 or 1 2. Figure 11: Half-angle triangle roof. (a) Note that = by SS. Then =, =, i.e., =, =. The angle sum of is = + + +, which is the angle sum of. (b) Since = + and =, we have either 1 2 or 1 2. Lemma 3. The sum of any two angles of a triangle is smaller than a flat angle. roof. Given a triangle. The exterior angle of is larger than. Since the angle sum of and its exterior angle is a flat angle, then and are addable, and their addition is smaller than a flat angle. So their angle sum is less than 180. Theorem 3.1 (Saccheri-Legendre) (ngle-sum Theorem). The sum of degree measures of the three angles in any triangle is less than or equal to 180. roof. Suppose the angle sum of a triangle is greater than 180, say, ε with ε > 0. Then by Lemma 2 there exists another triangle 1 1 1, having the angle sum ε and Then there is a triangle 2 2 2, having the angle sum ε and , i.e., 2 1. ontinue this procedure, we obtain 2 2 triangles k k k having the angle sum ε and k 1, k 1. When k is 2 k large enough we have k < ε, then the sum of the other two angles of k k k will be greater than 180, which is contradictory to that the sum of any two angles of a triangle is less than

9 orollary 3.2. The sum of two angles of a triangle is less than or equal to its remote exterior angle. roof. Given a triangle. Let ext ( ) denote the exterior angle of at. Then = + ext ( ). So + ext ( ). efinition 4. quadrilateral is a collection of four points,,,, denoted, such that no three of them are collinear, and the interior of the four segments,,, are disjoint. quadrilateral is said to be convex if it has a pair of opposite sides, say, and, such that is contained in an open half-plane bounded by, and is contained in an open half-plane bounded by. Lemma 5. quadrilateral is convex if and only if the intersection :=, called the interior of, is nonempty. If is convex, then meets at. We define :=. roof. : Take a point. Then, are the on the same side of ;, are the on the same side of. So, are the same side of. Likewise,, are the same side of. Hence is convex by definition. : Let, be on the same side of, and, be on the same side of. raw segment to have triangles and. Then either, are on opposite sides of or, are on the same side of. ase 1. oints, are on opposite sides of. Then meets at a point such that. There are three subcases. Subcase This means that intersects at such that and. Since, we have and. Since, we have and. Hence. See the left of Figure 12. Subcase Then by rossbar Theorem. Thus. Hence ray r(, ) intersects at Q such that Q. This is contradictory to that, are on the same side of. See the middle of Figure 12. Subcase Then by rossbar Theorem. Thus. Hence ray r(, ) intersects at Q such that Q. This is contradictory to that, are on the same side of. See the right of Figure 12. Figure 12: onvex quadrilateral and possible cases impossible ase 2. oints, are on the same side of. Then either r(, ) is between r(, ) and r(, ), or r(, ) is between r(, ) and r(, ). 9

10 Subcase 2.1. Ray r(, ) is between r(, ) and r(, ). Then r(, ) meets at a point Q such that Q. If Q, then meets at Q, which is contradictory to definition of quadrilateral. See the left of Figure 13. If Q, then. Thus r(, ) meets, which is contradictory to that, are on the same side of. See the moddle of Figure 13. Q Q Q Figure 13: Other possible cases impossible Subcase 2.2. Ray r(, ) is between r(, ) and r(, ). Then r(, ) meets at a point Q such that Q. Then, are on opposite sides of, contradictory to that, are on the same side of. See the right of Figure 13. roposition 3.3. egree measure of angle sum of convex quadrilateral is at most 360. roof. Given a convex quadrilateral and consider triangles and by drawing segment. We have + =, + =. Then the angle sum of is congruent to the addition of the angle sum of and the angle sum of. Hence the angle sum of is less than or equal to ngle efect of Triangle efinition 6. The angle defect (or just defect) of a triangle is δ := 180. roposition 4.1 (dditivity of angle defect). Given a triangle and point between and on segment. raw segment to have triangles and. Then δ = δ + δ. onsequently, the angle sum of is equal to 180 if and only if the angle sums of both triangles and are equal to 180. Figure 14: dditivity of angle defect roof. Trivial. 10

11 Theorem 4.2 (Once-Then-ll Theorem). If there is one triangle whose angle sum is 180, then all triangles have angle sum equal to 180. This can be split into the following three statements. (a) If there exists a triangle whose angle sum is 180, then there exists a rectangle. (b) If there exists a rectangle, then every right triangle has angle sum equal to 180. (c) If all right triangles have angle sum equal to 180, then every triangle has angle sum equal to 180. roof. Let be a triangle. Then has at least two acute angles, say, and. There exists a unique line perpendicular to, meeting at. If, then > = right angle, which is a contradiction. Likewise, is impossible. Then we must have. Thus and are right triangles. (a) Let have angle sum equal to 180, i.e., its angle defect is zero. Then both right triangles and have angle sum equal 180. We construct a rectangle from the right triangle. Figure 15: triangle is cut to two right triangles raw ray r(, ) to be such that =. Mark a point on r(, ) such that =. Then = by SS. Thus =, =, so is a right angle. Since δ = 0, so is δ. Moreover, + = is a right angle. Hence the quadrilateral is a rectangle. (b) Given an arbitrary right triangle. It is easy to see that a rectangle can be doubled in either side. So we may assume that there exists a rectangle such that > and >. See Figure 16. learly, and are right triangles, having zero angle defect. Mark points on and Q on such that, Q, and =, Q =. Then Q =. Now the triangle has zero angle defect by additivity. Then Q has zero angle defect by additivity again. So has zero angle defect. Q Figure 16: ig rectangle (c) Now for an arbitrary triangle, we may assume that and are acute. Then can be divided into two right triangles and. Then and have zero angle defect. So is by additivity of angle defect. Hence has angle sum of degree measures equal to 180. orollary 4.3. If a triangle has positive angle defect, then all triangles have positive angle defect. roof. Trivial. 11

12 5 quivalence of arallel ostulates Unique arallel Line ostulate (UL). Through a point not on a line l there exists exactly one line m parallel to l. uclid s ostulate V (V). If two distinct lines l, m intersect a transversal t in such a way that the sum of the interior angles on one side of t is less than a flat angle, then l, m meet on the same side of t having the two interior angles. Hilbert s xiom of arallelism (H). Through a point not on a line l there exists at most one line m parallel to l. Theorem 5.1. uclid s ostulate V Hilbert s xiom of arallelism. l t R m Q Figure 17: quivalence of uclid s ostulate V and Hilbert s axiom roof. Given distinct lines l, m, t such that l, t meet at and m, t meet at Q. Mark point on l and point on m on the same side of t. Mark point on l and on m such that and Q. raw ray r(q, R ) on open half-pane H(t, ) such that R Q = Q. xtend r(q, R ) to line m. Then m is parallel to l. : Let m be an arbitrary line through Q and parallel to l. V implies Q + Q 180, Q + Q 180. Since Q + Q + Q + Q = 360, it follows that Q + Q = 180, Q + Q = 180. Since Q + Q = 180, then Q = Q. Note that Q = R Q. We thus have Q = R Q. So m = m, which is Hilbert axiom. : Let angle sum of consecutive interior angles be less that 180, i.e., Q + Q < 180. Since Q + Q = 180, then Q > Q. learly, m m. Since l H(m, ), r(q, ) H(m, ), then r(, ), r(q, ) do not meet. If r(, ), r(q, ) do not meet, then m is parallel to l. So m = m, contradiction to m m. Hence r(, ), r(q, ) meet. roposition 5.2. Under Hilbert s xiom of arallelism. Let two lines l, m be parallel and cut by a transversal t. Then (a) lternate interior angles are congruent. (b) orresponding angles are congruent. (c) The sum of degree measures of consecutive interior angles is 180. (d) The angle sum of degree measures of a triangle is

13 roof. (a) There exists a line m through and parallel to l, having congruent alternate interior angles. Then every line parallel to l must be this line m. So the alternate interior angles are congruent. (b) It follows from (a) and the fact that opposite angles are congruent. (c) It follows from the fact that the sum of supplementary angles is 180. (d) xtend the segment to and draw ray r(, ) such that =. Then r(, ) is parallel to. Thus =. Hence + + = + + = 180. Figure 18: ngle sum of a triangle is a flat angle Theorem 5.3. ngle sum of a triangle equal to 180 Hilbert s axiom of parallelism. roof. Given a line l and point not on l. Let t be the unique line through and perpendicular to l, meeting l at Q. Through there is a unique line m perpendicular to t. Then m is parallel to l. Let n be an arbitrary line through and parallel to l, but distinct from m. ick a point R on m such that n intersects the interior of Q R. Fix a point S on n such that S Q Y. Then angle R S is acute. See Figure 19. t R Y m S Z X n Q l Figure 19: xistence of rectangle implies unique parallel. Let X be a point on the open ray r(, S). rop perpendicular XY to m with foot Y on m, and perpendicular XZ to t with foot Z on t. Note that l, m, XZ are parallel to each other. So X, Z are on the same side of lines l, m respectively. We then have Z Q. nalogously, lines t, XY are parallel to each other. Then X, Y are on the same side of t and, Z are on the same side of XY. Hence the quadrilateral Y XZ is convex. Note that XZ, Z Y, Y X are right angles. Since every triangle has angle sum equal to 180, then Z X + XZ is a right angle. Since Z X + X Y is a right angle, we see that X Y = XZ. Thus XY = X Z by S. Subsequently, Z = XY. Now we can take X to be such that XY > Q by ristotle s axiom. Then Z > Q, which is impossible. (ristotle s axiom can be easily followed from rchimedes axiom when rectangle exists. However, it is no need to assume the existence of rectangle to obtain ristotle s axiom from rchimedes axiom.) 13

14 6 Saccheri quadrilaterals and Lambert quadrilaterals quadrilateral is said to be a Saccheri quadrilateral if its two base angles are right angles and the base-angle adjacent opposite sides are congruent. For instance, for quadrilateral with right angles, and = is a Saccheri quadrilateral. Figure 20: Saccheri quadrilateral quadrilateral is said to be a Lambert quadrilateral if it has at least three angles to be right angles. roposition 6.1. Let be a Saccheri quadrilateral with right angles,, and =. See Figure 20. Then =. roof. raw segments and. Note that = by SS. Then = and =. Subsequently, = by angle subtraction. Hence = by SS. Therefore =. roposition 6.2 (roperty of quadrilateral with two adjacent right angles). Let be a quadrilateral with two adjacent right angles and. See Figure 21. Then (a) < if and only if <. (b) = if and only if =. F Figure 21: is a Saccheri quadrilateral roof. (a) : ssume <. Find a point on such that = and ; see the left of Figure 21. It is clear that,, are on the same side of for is parallel to. We claim that, are on the same side of. Suppose it is not true, i.e., meets at point F such that F. Note that,, F are collinear. Then F (= F ) is larger than and F > by xterior ngle Theorem. Since, are right angles, so the sum of F, F are not a flat angle, which is a contradiction. Now we see that is in the interior of. Since, it follows that is also in the interior of. So the ray r(, ) is between rays r(, ) and r(, ). Hence =. learly, xterior ngle Theorem implies > = >. 14

15 : ssume <. If =, then = by roposition 6.1, which is contradictory to <. If <, then > by the previous argument, which is a contradictory to <. So we must have <. (b) Trivial. roposition 6.3 (roperty of quadrilateral with three right angles). Let be a Lambert quadrilateral with right angles,,. (a) Then is never obtuse. (b) If is a right angle, then the opposite sides of are congruent. (c) If is acute, then each side adjacent to is greater than its opposite side. roof. ssume that, and, are two pairs of opposite sides of. (a) Trivial because Lambert quadrilateral is convex and its angle sum is less than or equal to 360. (b) The quadrilateral is a rectangle. Then = and = by the property of quadrilateral with two adjacent right angles. (c) Then < and <. Hence < and < by the property of quadrilateral with two adjacent right angles. Lemma 7. Given a right triangle OXY such that Y is a right angle and O is acute. xtend OX to X such that O X X and OX = XX, extend Y X to Z such that Y X Z and X Z is perpendicular to XY, and extend OY to Y such that O Y Y and X Y is perpendicular to OY. See Figure 22. Then (a) X Y is at least a double of XY, i.e., X Y 2 XY. (b) OY is at most a double of OY, i.e., OY 2 OY. Z X X O Y Y Figure 22: ouble of hypotenuse roof. Since XO = XX, XY O = XZX = right angle, and OXY = X XZ, then OY X = X XZ by S. So XY = XZ and X Z = OY. Note that Y Y X Z is a Lambert quadrilateral with right angles Y Y Z, Y Y X, Y ZX. We have X Y ZY = 2 XY and Y Y ZX = OY by the property of quadrilateral with three right angles. Hence X Y 2 XY and OY 2 OY. rchimedes axiom implies ristotle s axiom. Given an acute angle XOY. For an arbitrary segment, there exists a point Y on ray r(o, Y ) such that X Y >, where X Y is perpendicular to OY with foot Y on ray r(o, Y ). roof. Note that there exists a positive integer n such that 2 n XY > by rchimedes s axiom. Let X 1 be a point on r(o, X) such that OX 1 = 2 OX, and let X1 Y 1 be a segment perpendicular to OY with foot Y 1 on OY. Then X 1 Y 1 2 XY. nalogously, let X 2 be a point on r(o, X) such that OX 2 = 2 OX1, and let X 2 Y 2 be the segment perpendicular to OY with foot of X 2 on OY. Then X 2 Y 2 2 X 1 Y XY. ontinue this procedure, 15

16 X 2 X X 1 O Y Y1 Y 2 Figure 23: rchimedes axiom implies ristotle s axiom we have points X k (k 1) on r(o, X) such that OX k = 2 OXk 1, and segments X k Y k perpendicular to OY with foot Y k on OY. Then X k Y k 2 X k 1 Y k 1 2 k XY, k 1. Hence there exits an integer n = 2 k such that X k Y k > n. 16

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