现代几何学的等同公理组

现代几何学的等同公理组
    几何图形的等同关系十分复杂、形式多样，需要6条几何公理来表达。
    例如，什么是角？角如何定义？多边形及其面积的测度，等等。
该公理组应用十分广范，尤其是在微积分中应用。
本文附件是M现代几何学的等同公理组英文陈述。
袁萌  陈启清   3月20日
附件：现代几何学的等同公理组（IV  1-6）   
  The axioms of this group define the idea of congruence or displacement. Segments stand in a certain relation to one another which is described by the word “congruent.” IV, I. If A, B are two points on a straight line a, and if A0 is a point upon the same or anotherstraightline a0,then,uponagivensideof A0 onthestraightline a0,wecanalways findoneandonlyonepoint B0 sothatthesegment AB(or BA)iscongruenttothesegment A0B0. We indicate this relation by writing AB≡ A0B0. Every segment is congruent to itself; that is, we always have AB≡ AB. We can state the above axiom briefly by saying that every segment can be laid off upon a given side of a given point of a given straight line in one and and only one way. IV, 2. If a segment AB is congruent to the segment A0B0 and also to the segment A00B00, then the segment A0B0 is congruent to the segment A00B00; that is, if AB ≡ A0B and AB≡ A00B00, then A0B0 ≡ A00B00. IV,3. Let ABand BCbetwosegmentsofastraightlineawhichhavenopointsincommon aside from the point B, and, furthermore, let A0B0 and B0C0 be two segments of the same or of another straight line a0 having, likewise, no point other than B0 in common. Then, if AB≡ A0B0 and BC ≡ B0C0, we have AC ≡ A0C0.
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Fig. 8.
Definitions. Let α be any arbitrary plane and h, k any two distinct half-rays lying in α and emanating from the point O so as to form a part of two different straight lines. We call the system formed by these two half-rays h, k an angle and represent it by the symbol ∠(h,k) or ∠(k,h). From axioms II, 1–5, it follows readily that the half-rays h and k, taken together with the point O, divide the remaining points of the plane a into two regions having the following property: If A is a point of one region and B a point of the other, then every broken line joining A and B either passes through O or has a point in common with one of the half-rays h, k. If, however, A, A0 both lie within the same region, then it is always possible to join these two points by a broken line which neither passes through O nor has a point in common with either of the half-rays h, k. One of these two regions is distinguished from the other in that the segment joining any two points of this region lies entirely within the region. The region so characterised is called the interior of the angle (h,k). To distinguish the other region from this, we call it the exterior of the angle (h,k). The half rays h and k are called the sides of the angle, and the point O is called the vertex of the angle. IV, 4. Let an angle (h,k) be given in the plane α and let a straight line a0 be given in a plane α0. Suppose also that, in the plane α, a definite side of the straight line a0 be assigned. Denote by h0 a half-ray of the straight line a0 emanating from a point O0 of this line. Then in the plane α0 there is one and only one half-ray k0 such that the angle (h,k), or (k,h), is congruent to the angle (h0,k0) and at the same time all interior points of the angle (h0,k0) lie upon the given side of a0. We express this relation by means of the notation ∠(h,k)≡∠(h0,k0) Every angle is congruent to itself; that is, ∠(h,k)≡∠(h,k) or ∠(h,k)≡∠(k,h) We say, briefly, that every angle in a given plane can be laid off upon a given side of a given half-ray in one and only one way.
IV, 5. If the angle (h,k) is congruent to the angle (h0,k0) and to the angle (h00,k00), then the angle (h0,k0) is congruent to the angle (h00,k00); that is to say, if ∠(h,k) ≡∠(h0,k0) and ∠(h,k)≡∠(h00,k00), then∠(h0,k0)≡∠(h00,k00).
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Suppose we have given a triangle ABC. Denote by h, k the two half-rays emanating from A and passing respectively through B and C. The angle (h,k) is then said to be the angle included by the sides AB and AC, or the one opposite to the side BC in the triangle ABC. Itcontainsalloftheinteriorpointsofthetriangle ABC andisrepresented by the symbol ∠BAC, or by ∠A. IV, 6. If, in the two triangles ABC and A0B0C0 the congruences AB≡ A0B0, AC ≡ A0C0, ∠BAC ≡∠B0A0C0 hold, then the congruences ∠ABC ≡∠A0B0C0and∠ACB≡∠A0C0B0 also hold.
Axioms IV, 1–3 contain statements concerning the congruence of segments of a straight line only. They may, therefore, be called the linear axioms of group IV. Axioms IV, 4, 5 contain statements relating to the congruence of angles. Axiom IV, 6 gives the connection between the congruence of segments and the congruence of angles. Axioms IV, 4–6 contain statements regarding the elements of plane geometry and may be called the plane axioms of group IV.
§7. CONSEQUENCES OF THE AXIOMS OF CONGRUENCE.
Suppose the segment AB is congruent to the segment A0B0. Since, according to axiom IV, 1, the segment AB is congruent to itself, it follows from axiom IV, 2 that A0B0 is congruent to AB; that is to say, if AB ≡ A0B0, then A0B0 ≡ AB. We say, then, that the two segments are congruent to one another. Let A, B, C, D,..., K, L and A0, B0, C0, D0,..., K0, L0 be two series of points on the straight lines a and a0, respectively, so that all the corresponding segments AB and A0B0, AC and A0C0, BC and B0C0,..., KL and K0L0 are respectively congruent, then the two series of points are said to be congruent to one another. A and A0, B and B0,..., L and L0 are called corresponding points of the two congruent series of points. From the linear axioms IV, 1–3, we can easily deduce the following theorems:
Theorem 9. If the first of two congruent series of points A, B, C, D,..., K, L and A0, B0, C0, D0,..., K0, L0 is so arranged that B lies between A and C, D,..., K, L, and C between A, B and D,..., K, L, etc., then the points A0, B0, C0, D0,..., K0, L0 of the second series are arranged in a similar way; that is to say, B0 lies between A0 and C0, D0,..., K0, L0, and C0 lies between A0, B0 and D0,..., K0, L0, etc.
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Let the angle (h,k) be congruent to the angle (h0,k0). Since, according to axiom IV, 4, the angle (h,k) is congruent to itself, it follows from axiom IV, 5 that the angle (h0,k0) is congruenttotheangle (h,k). Wesay,then,thattheangles (h,k)and (h0,k0)arecongruent to one another. Definitions. Two angles having the same vertex and one side in common, while the sides not common form a straight line, are called supplementary angles. Two angles having a common vertex and whose sides form straight lines are called vertical angles. An angle which is congruent to its supplementary angle is called a right angle. Two triangles ABC and A0B0C0 are said to be congruent to one another when all of the following congruences are fulfilled: AB≡ A0B0, AC ≡ A0C0, BC ≡ B0C0, ∠A≡∠A0, ∠B≡∠B0, ∠C ≡∠C0. Theorem 10. (First theorem of congruence for triangles). If, for the two triangles ABC and A0B0C0, the congruences AB≡ A0B0, AC ≡ A0C0, ∠A≡∠A0 hold, then the two triangles are congruent to each other.
Proof. From axiom IV, 6, it follows that the two congruences ∠B≡∠B0and∠C ≡∠C0 are fulfilled, and it is, therefore, sufficient to show that the two sides BC and B0C0 are congruent. We will assume the contrary to be true, namely, that BC and B0C0 are not congruent, and show that this leads to a contradiction. We take upon B0C0 a point D0 such that BC ≡ B0D0. The two triangles ABC and A0B0D0 have, then, two sides and the included angle of the one agreeing, respectively, to two sides and the included angle of the other. It follows from axiom IV, 6 that the two angles BAC and B0A0D0 are also congruent to each other. Consequently, by aid of axiom IV, 5, the two angles B0A0C0 and B0A0D0 must be congruent.
Fig. 9.
This, however, is impossible, since, by axiom IV, 4, an angle can be laid off in one and only one way on a given side of a given half-ray of a plane. From this contradiction the theorem follows. We can also easily demonstrate the following theorem:
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Theorem 11. (Second theorem of congruence for triangles). If in any two triangles one side and the two adjacent angles are respectively congruent, the triangles are congruent.
We are now in a position to demonstrate the following important proposition.
Theorem 12. If two angles ABC and A0B0C0 are congruent to each other, their supplementary angles CBD and C0B0D0 are also congruent.
Fig. 10.
Proof. Take the points A0, C0, D0 upon the sides passing through B0 in such a way that A0B0 ≡ AB, C0B0 ≡CB, D0B0 ≡ DB. Then, in the two triangles ABC and A0B0C0, the sides AB and BC are respectively congruent to A0B0 and C0B0. Moreover, since the angles included by these sides are congruent to each other by hypothesis, it follows from theorem 10 that these triangles are congruent; that is to say, we have the congruences AC ≡ A0C, ∠BAC ≡∠B0A0C0. On the other hand, since by axiom IV, 3 the segments AD and A0D0 are congruent to each other, it follows again from theorem 10 that the triangles CAD and C0A0D0 are congruent, and, consequently, we have the congruences: CD ≡C0D0, ∠ADC ≡∠A0D0C0. From these congruences and the consideration of the triangles BCD and B0C0D0, it follows by virtue of axiom IV, 6 that the angles CBD and C0B0D0 are congruent. As an immediate consequence of theorem 12, we have a similar theorem concerning the congruence of vertical angles.
Theorem 13. Let the angle (h, k) of the plane α be congruent to the angle (h0, k0) of theplane α0, and, furthermore, let l beahalf-rayintheplane α emanatingfromthe vertex of the angle (h, k) and lying within this angle. Then, there always exists in the plane α0 a half-ray l0 emanating from the vertex of the angle (h0, k0) and lying within this angle so that we have ∠(h,l)≡∠(h0,l0), ∠(k,l)≡∠(k0,l0).
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Fig. 11.
Proof. We will represent the vertices of the angles (h, k) and (h0, k0) by O and O0, respectively, and so select upon the sides h, k, h0, k0 the points A, B, A0, B0 that the congruences OA≡O0A0, OB≡O0B0 are fulfilled. Because of the congruence of the triangles OAB and O0A0B0, we have at once AB≡ A0B0, ∠OAB≡O0A0B0, ∠OBA≡∠O0B0A0. Let the straight line AB intersect l in C. Take the point C0 upon the segment A0B0 so that A0C0 ≡ AC. Then, O0C0 is the required half-ray. In fact, it follows directly from these congruences, by aid of axiom IV, 3, that BC ≡ B0C0. Furthermore, the triangles OAC and O0A0C0 are congruent to each other, and the same is true also of the triangles OCB and O0B0C0. With this our proposition is demonstrated. In a similar manner, we obtain the following proposition.
Theorem 14. Let h, k, l and h0, k0, l0 be two sets of three half-rays, where those of each set emanate from the same point and lie in the same plane. Then, if the congruences ∠(h,l)≡∠(h0,l0), ∠(k,l)≡∠(k0,l0) are fulfilled, the following congruence is also valid; viz.: ∠(h,k)≡∠(h0,k0). By aid of theorems 12 and 13, it is possible to deduce the following simple theorem, which Euclid held–although it seems to me wrongly–to be an axiom.
Theorem 15. All right angles are congruent to one another.
Proof. Let the angle BAD be congruent to its supplementary angle CAD, and, likewise, let the angle B0A0D0 be congruent to its supplementary angle C0A0D0. Hence the angles BAD, CAD, B0A0D0, and C0A0D0 are all right angles. We will assume that the contrary of our proposition is true, namely, that the right angle B0A0D0 is not congruent to the right angle BAD, and will show that this assumption leads to a contradiction. We layofftheangle B0A0D0 uponthehalf-ray ABinsuchamannerthattheside AD00 arising fromthisoperationfallseitherwithintheangle BADorwithintheangleCAD. Suppose, for example, the first of these possibilities to be true. Because of the congruence of the
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angles B0A0D0 and BAD00, it follows from theorem 12 that angle C0A0D0 is congruent to angle CAD00, and, as the angles B0A0D0 and C0A0D0 are congruent to each other, then, by IV, 5, the angle BAD00 must be congruent to CAD00.
Fig. 12.
Furthermore, since the angle BAD is congruent to the angle CAD, it is possible, by theorem 13, to find within the angle CAD a half-ray AD000 emanating from A, so that the angle BAD00 will be congruent to the angle CAD000, and also the angle DAD00 will be congruent to the angle DAD000. The angle BAD00 was shown to be congruent to the angle CAD00 and, hence, by axiom IV, 5, the angle CAD00, is congruent to the angle CAD000. This, however, is not possible; for, according to axiom IV, 4, an angle can be laid off in a plane upon a given side of a given half-ray in only one way. With this our proposition is demonstrated. We can now introduce, in accordance with common usage, the terms “acute angle” and “obtuse angle.” The theorem relating to the congruence of the base angles A and B of an equilateral triangle ABC followsimmediatelybytheapplicationofaxiomIV, 6 tothetriangles ABC and BAC. By aid of this theorem, in addition to theorem 14, we can easily demonstrate the following proposition.
Theorem 16. (Third theorem of congruence for triangles.) If two triangles have the three sides of one congruent respectively to the corresponding three sides of the other, the triangles are congruent.
Any finite number of points is called a figure. If all of the points lie in a plane, the figure is called a plane figure. Two figures are said to be congruent if their points can be arranged in a one-to-one correspondence so that the corresponding segments and the corresponding angles of the two figures are in every case congruent to each other. Congruent figures have, as may be seen from theorems 9 and 12, the following properties: Three points of a figure lying in a straight line are likewise in a straight line in every figure congruent to it. In congruent figures, the arrangement of the points in corresponding planes with respect to corresponding lines is always the same. The same is true of the sequence of corresponding points situated on corresponding lines. The most general theorems relating to congruences in a plane and in space may be expressed as follows:
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Theorem 17. If (A,B,C,...) and (A0,B0,C0,...) are congruent plane figures and P is a point in the plane of the first, then it is always possible to find a point P in the plane of the second figure so that (A,B,C,...,P) and (A0,B0,C0,...,P0) shall likewise be congruent figures. If the two figures have at least three points not lying in a straight line, then the selection of P0 can be made in only one way. Theorem 18. If (A,B,C,...) and (A0,B0,C0,... = are congruent figures and P represents any arbitrary point, then there can always be found a point P0 so that the two figures (A,B,C,...,P) and (A0,B0,C0,...,P0) shall likewise be congruent. If the figure (A,B,C,...,P) contains at least four points not lying in the same plane, then the determination of P0 can be made in but one way. Thistheoremcontainsanimportantresult;namely,thatallthefactsconcerningspace which have reference to congruence, that is to say, to displacements in space, are (by the addition of the axioms of groups I and II) exclusively the consequences of the six linear and plane axioms mentioned above. Hence, it is not necessary to assume the axiom of parallels in order to establish these facts. If we take, in, addition to the axioms of congruence, the axiom of parallels, we can then easily establish the following propositions:
Theorem 19. If two parallel lines are cut by a third straight line, the alternateinterior angles and also the exterior-interior angles are congruent Conversely, if the alternate-interior or the exterior-interior angles are congruent, the given lines are parallel. Theorem 20. The sum of the angles of a triangle is two right angles.
Definitions. If M is an arbitrary point in the plane α, the totality of all points A, for which the segments MA are congruent to one another, is called a circle. M is called the centre of the circle. From this definition can be easily deduced, with the help of the axioms of groups III and IV, the known properties of the circle; in particular, the possibility of constructing a circle through any three points not lying in a straight line, as also the congruence of all angles inscribed in the same segment of a circle, and the theorem relating to the angles of an inscribed quadrilateral.
§8. GROUP V. AXIOM OF CONTINUITY. (ARCHIMEDEAN AXIOM.)
This axiom makes possible the introduction into geometry of the idea of continuity. In order to state this axiom, we must first establish a convention concerning the equality of two segments. For this purpose, we can either base our idea of equality upon the axiomsrelat