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Proposition 3. Any finite-dimensional power-associative algebra, which is not a nilalgebra, contains an idempotent \( e\left( { \neq 0}\right) \) . | Proof: \( \mathfrak{A} \) contains an element \( x \) which is not nilpotent. The subalgebra \( F\left\lbrack x\right\rbrack \) of \( \mathfrak{A} \) generated by \( x \) is a finite-dimensional associative algebra which is not a nilalgebra. Then \( F\left\lbrack x\right\rbrack \) contains an idempotent \( e \) \( \lef... | Yes |
Theorem 5. Two Cayley algebras \( \mathfrak{C} \) and \( {\mathfrak{C}}^{\prime } \) are isomorphic if and only if their corresponding norm forms \( n\left( x\right) \) and \( {n}^{\prime }\left( {x}^{\prime }\right) \) are equivalent (that is, there is a linear mapping \( x \rightarrow {xH} \) of \( \mathfrak{C} \) in... | Proof: Suppose \( \mathfrak{C} \) and \( {\mathfrak{C}}^{\prime } \) are isomorphic, the isomorphism being \( H \) . Then (25) implies \( {\left( xH\right) }^{2} - t\left( x\right) \left( {xH}\right) + n\left( x\right) {1}^{\prime } = 0 \) where \( {1}^{\prime } = {1H} \) is the unity element of \( {\mathfrak{C}}^{\pri... | Yes |
Theorem 6. The radical \( \mathfrak{N} \) of any finite-dimensional Jordan algebra \( \mathfrak{J} \) over \( F \) of characteristic 0 is the radical \( {\mathfrak{J}}^{ \bot } \) of the trace form\n\n\[ \left( {x, y}\right) = \operatorname{trace}{R}_{xy} \] for all \( x, y \) in \( \mathfrak{J} \) . | Proof: Without any assumption on the characteristic of \( F \) it follows from (4) that \( \left( {x, y}\right) \) in (16) is a trace form: \( \left( {{xy}, z}\right) - \left( {x,{yz}}\right) = \) trace \( {R}_{\left( x, y, z\right) } = 0 \) since the trace of any commutator is 0 . Hence \( {\mathfrak{J}}^{ \bot } \) i... | Yes |
Theorem 7. Let \( \mathfrak{A} \) be a finite-dimensional algebra over \( F \) (of arbitrary characteristic) satisfying\n\n(i) there is a nondegenerate (associative) trace form \( \left( {x, y}\right) \) defined on \( \mathfrak{A} \), and\n\n(ii) \( {\mathfrak{I}}^{2} \neq 0 \) for every ideal \( \mathfrak{I} \neq 0 \)... | Proof: Let \( \mathfrak{S}\left( { \neq 0}\right) \) be a minimal ideal of \( \mathfrak{A} \) . Since \( \left( {x, y}\right) \) is a trace form, \( {\mathfrak{S}}^{ \bot } \) is an ideal of \( \mathfrak{A} \) . Hence the intersection \( \mathfrak{S} \cap {\mathfrak{S}}^{ \bot } \) is either 0 or \( \mathfrak{S} \), si... | Yes |
Theorem 8. Any semisimple (hence any simple) Jordan algebra \( \mathfrak{J} \) of finite dimension over \( F \) of characteristic 0 has a unity element 1 . | Proof: \( \mathfrak{J} \) has a principal idempotent \( e \) . Then \( {\mathfrak{J}}_{0} \) is a nilalgebra, so that \( \left( {x, y}\right) = \operatorname{trace}{R}_{{x}_{1}{y}_{1}} \) by \( \left( {20}^{\prime }\right) \) since \( \operatorname{trace}{R}_{{z}_{0}} = 0 \) by (8). Hence \( x \) in \( {\mathfrak{J}}_{... | No |
Theorem 11. Let \( \mathfrak{A} \) be a finite-dimensional power-associative algebra over \( F \) satisfying the following conditions:\n\n(i)\ni) there is an (associative) trace form \( \left( {x, y}\right) \) defined on \( \mathfrak{A} \) ;\n\n(ii) \( \left( {e, e}\right) \neq 0\; \) for any idempotent \( e \) in \( \... | Proof: By (i) we know from IV that \( {\mathfrak{A}}^{ \bot } \) is an ideal of \( \mathfrak{A} \) . If there were an idempotent \( e \) in \( {\mathfrak{A}}^{ \bot } \), then (ii) would imply \( \left( {e, e}\right) \neq 0 \), a contradiction. Hence \( {\mathfrak{A}}^{ \bot } \) is a nilideal: \( {\mathfrak{A}}^{ \bot... | Yes |
Theorem 2. If \( \overline{AB} \equiv \overline{CD} + \overline{PQ} \) and \( \overline{{C}^{\prime }{D}^{\prime }} \equiv \overline{CD} \), then\n\n\[ \overline{AB} \equiv \overline{{C}^{\prime }{D}^{\prime }} + \overline{PQ} \] | The proof is immediate. | No |
Theorem 3. If \( \overline{AB} > \overline{CD} \) and \( \overline{CD} > \overline{EF} \), then \( \overline{AB} > \overline{EF} \) . | To begin with \( \overline{AB} \equiv \overline{EF} \) is impossible. If then \( \overline{EF} > \overline{AB} \), let us put \( \overline{EF} \equiv \overline{EG} + \overline{GF} \), where \( \overline{EG} \equiv \overline{AB} \) . Then \[ \overline{CD} \equiv \overline{CH} + \overline{HD};\;\overline{CH} \equiv \over... | No |
Theorem 4. If \( C \) be a point of \( \left( {AB}\right) \), then every point \( D \) of \( \left( {AB}\right) \) is either a point of \( \left( {AC}\right) \) or of \( \left( {CB}\right) \) . | If \( \overline{AC} \equiv \overline{AD} \) we have \( C \) and \( D \) identical. If \( \overline{AC} > \overline{AD} \) we may find a point of \( \left( {AC}\right) \left\lbrack {\text{and so of}\left( {AB}\right) }\right\rbrack \) whose distance from \( A \) is congruent to \( \overline{AD} \), and this will be iden... | No |
Theorem 6. If \( \overline{AB} \equiv \overline{PQ} + \overline{RS} \) and \( \overline{{A}^{\prime }{B}^{\prime }} \equiv \overline{PQ} + \overline{RS} \), then \( \overline{{A}^{\prime }{B}^{\prime }} \equiv \overline{AB} \) . | The proof is left to the reader. | No |
Theorem 7. If \( \overline{AB} \equiv \overline{PQ} + \overline{RS} \) and \( \overline{AB} \equiv \overline{PQ} + \overline{LM} \), then \( \overline{RS} \equiv \overline{LM} \) . | For if \( \overline{AB} \equiv \overline{AC} + \overline{CB} \), and \( \overline{AC} \equiv \overline{PQ} \), then \( \overline{CB} \equiv \overline{RS} \equiv \overline{LM} \) . | No |
Theorem 13. If \( A, B, C, D, E \) form the configuration of points described in Axiom XVI, the point \( E \) is a point of \( \left( {DF}\right) \) . | Suppose that this were not the case. We should either have \( F \) as a point of \( \left( {DE}\right) \) or \( D \) as a point of \( \left( {EF}\right) \) . But then, in the first case, \( C \) would be a point of \( \left( {DB}\right) \) and in the second \( D \) would be a point of \( \left( {BC}\right) \), both of ... | Yes |
Theorem 14. If \( A, B, C \) be three non-collinear points, and \( D \) a point within \( \left( {AB}\right) \) while \( E \) is a point of the extension of \( \left( {BC}\right) \) beyond \( C \), then the line \( {DE} \) will contain a point \( F \) of \( \left( {AC}\right) \) . | Take \( G \), a point of \( \left( {ED}\right) \), different from \( E \) and \( D \) . Then \( {AG} \) will contain a point \( L \) of \( \left( {BE}\right) \), while \( G \) belongs to \( \left( {AL}\right) \) . If \( L \) and \( C \) be identical, \( G \) will be the point required. If \( L \) be a point of \( \left... | Yes |
Theorem 15. If \( A, B, C \) be three non-collinear points, no three points, one within each of their three segments, are collinear. | The proof is left to the reader. | No |
Theorem 17. If a plane be determined by the vertices of a triangle, the following points lie therein:\n\n(a) All points of every line determined by a vertex, and a point of the line of the other two vertices.\n\n(b) All points of every line which contains a point of each of two sides of the triangle.\n\n(c) All points ... | The proof will come at once from 16, and from the consideration that if we know two points of a line, every other point thereof is either a point of their segment, or of one of its extensions. The plane determined by three points as \( A \) , \( B, C \) shall be written the plane \( {ABC} \) . | No |
Theorem 23. All points of each of the following figures will lie in the space defined by the vertices of a given tetrahedron.\n\n(a) A plane containing an edge, and a point of the opposite edge.\n\n(b) A line containing a vertex, and a point of the plane of the opposite face.\n\n(c) A line containing a point of one edg... | The proof will come directly if we take the steps in the order indicated, and hold fast to 16 , and the definitions of line, plane, and space. | No |
Theorem 29. A space contains wholly every plane whereof it contains three non-collinear points. | Suppose that we have a plane containing the point \( E \) of the segment \( \left( {AB}\right) \) but no point of the segment \( \left( {BC}\right) \) . Take \( F \) and \( G \) two other points of the plane, not collinear with \( E \), and construct the including space by means of the tetrahedron whose vertices are \(... | No |
Theorem 32. If two points be on the same side of a plane, a point opposite to one is on the same side as the other; and if two points be on the same side, a point opposite to one is opposite to both. | The proof comes at once from 30. | No |
Theorem 33. If two planes have a common point they have a common line. | Let \( P \) be the common point. In the first plane take a line through \( P \) . If this be also a line of the second plane, the theorem is proved. If not, we may take two points of this line on opposite sides of the second plane. Now any other point of the first plane, not collinear with the three already chosen, wil... | Yes |
Theorem 1. If \( \overline{AB} \) and \( \overline{PQ} \) be any two distances whereof the second is not null, there will exist in the segment \( \left( {AB}\right) \) a finite or null number \( n \) of points \( {P}_{k} \) possessing the following properties:\n\n\[ \overline{PQ} \equiv \overline{A{P}_{1}} \equiv \over... | Suppose, firstly, that \( \overline{AB} < \overline{PQ} \) then, clearly, \( n = 0 \) . If, however, \( \overline{AB} \equiv \overline{PQ} \) then \( n = 1 \) and \( {P}_{1} \) is identical with \( B \) . There remains the third case where \( \overline{AB} > \overline{PQ} \) . Imagine the theorem to be untrue. We shall... | Yes |
Theorem 2. In any segment there is a single point whose distances from the extremities are congruent. | The proof is left to the reader. | No |
Theorem 3. If a not null distance \( \overline{AB} \) be given and a positive integer \( m \), it is possible to find \( m \) distinct points of the segment \( \left( {AB}\right) \) possessing the properties\n\n\[ \overline{A{P}_{1}} \equiv \overline{{P}_{j}{P}_{j + 1}};\;\overline{A{P}_{j + 1}} \equiv \overline{A{P}_{... | It is merely necessary to take \( k \) so that \( {2}^{k} > m + 1 \) and find \( \overline{A{P}_{1}} \equiv \frac{1}{{2}^{k}}\overline{AB} \). | No |
Theorem 5. If \( \overline{AB} \) and \( \overline{PQ} \) be given, whereof the latter is not null, we may find \( n \) so great that \( \frac{1}{n}\overline{AB} < \overline{PQ} \) . | The proof is left to the reader. | No |
Theorem 8. If \( r > n \) and if distances \( r\overline{PQ} \) and \( n\overline{PQ} \) exist, then \( r\overline{PQ} > n\overline{PQ} \). | When \( m \) and \( n \) are both rational, this comes immediately by reducing to a common denominator. When one or both of these numbers is irrational, we may find a number in the lower class of the larger which is larger than one in the upper class of the smaller, and then apply I, 3. | No |
Theorem 9. If \( \overline{AB} > \overline{CD} \), the measure of \( \overline{AB} \) in terms of any chosen not null distance is greater than that of \( \overline{CD} \) in terms of the same distance. | This comes at once by reduction ad absurdum. | No |
Theorem 17. Two triangles are congruent if two sides and the included angle of one be respectively congruent to two sides and the included angle of the other. | The truth of this is at once evident when we recall the definition of congruent angles, and 12. | No |
Theorem 19. If three half-lines lie in the same half-plane and have their common bound on the bound of this half-plane; then one belongs to the interior angle of the other two. | Let the half-lines be \( \left| {{AB},}\right| {AC}, \mid {AD} \) . Connect \( B \) with \( H \) and \( K \), points of the opposite half-lines bounding this half-plane. If \( \left| {{AC},}\right| {AD} \) contain points of the same two sides of the triangle \( {BHK} \) the theorem is at once evident; if one contain a ... | No |
Theorem 21. If two points be at congruent distances from two points coplanar with them, all points of the line of the first two are at congruent distances from the latter two. | For we may find a congruent transformation keeping the former points invariant, while the latter are interchanged. | No |
Theorem 22. If \( \mid A{A}_{1}{}^{\prime } \) be a half-line of the interior \( \measuredangle {BA}{A}_{1} \), then we cannot have a congruent transformation keeping \( \mid {AB} \) invariant and carrying \( \mid A{A}_{1} \) into \( \mid A{A}_{1}{}^{\prime } \) . | We may suppose that \( {A}_{1} \) and \( {A}_{1}{}^{\prime } \) are at congruent distances from \( A \) . Let \( H \) be the point of the segment \( \left( {{A}_{1}{A}_{1}{}^{\prime }}\right) \) equidistant from \( {A}_{1} \) and \( {A}_{1}{}^{\prime } \) . We may find a congruent transformation carrying \( A{A}_{1}H{A... | Yes |
Theorem 24. An angle is congruent to its vertical. | We have merely to look at the congruent transformation interchanging a side of one with a side of the other. | No |
Theorem 25. If two angles of a triangle be congruent, the triangle is isosceles. | This is an immediate result of 18. | No |
Theorem 26. An exterior angle of a triangle is comparable with either of the opposite interior angles. | Let us take the triangle \( {ABC} \), while \( D \) lies on the extension of \( \left( {BC}\right) \) beyond \( C \) . Let \( E \) be the middle point of \( \left( {AC}\right) \) and let \( {DE} \) meet \( \left( {AB}\right) \) in \( F \) . If \( \overline{DE} > \overline{EF} \) find \( G \) of \( \left( {DE}\right) \)... | No |
Theorem 27. Two angles of a triangle are comparable. | For they are comparable to the same exterior angle. | No |
Theorem 28. If in any triangle one angle be greater than a second, the side opposite the first is greater than that opposite the second. | Evidently these sides cannot be congruent. Let us then have the triangle ABG where \( \measuredangle {BAG} > \measuredangle {BGA} \) . We may, by the definition of congruence, find such a point \( {C}_{1} \) of \( \left( {BG}\right) \) that \( \measuredangle {C}_{1}{AG} \) is congruent to \( \measuredangle {C}_{1}{GA} ... | Yes |
Theorem 32. Two distinct lines cannot be coplanar with a third, and perpendicular to it at the same point. | Suppose, in fact, that we have \( {AC} \) and \( {AD} \) perpendicular to \( B{B}^{\prime } \) at \( A \) . We may assume \( \overline{AB} \equiv \overline{A{B}^{\prime }} \) so that by I. \( {31AD} \) will contain a single point \( E \) either of \( \left( {CB}\right) \) or of \( \left( {C{B}^{\prime }}\right) \) . Fo... | Yes |
Theorem 35. Through any point of a given line will pass one line perpendicular to it lying in any given plane through that line. | Let \( A \) be the chosen point, and \( C \) a point in the plane, not on the chosen line. Let us take two such points \( B,{B}^{\prime } \) on the given line, that \( A \) is the middle point of \( \left( {B{B}^{\prime }}\right) \) and \( \overline{B{B}^{\prime }} < \overline{CB},\overline{B{B}^{\prime }} < \overline{... | Yes |
Theorem 36. If a line be perpendicular to two others at their point of intersection, it is perpendicular to every line in their plane through that point. | The proof given in the usual textbooks will hold. | No |
Theorem 39. If \( P \) be a point within the triangle \( {ABC} \) and there exist a distance congruent to \( \overline{AB} + \overline{AC} \), then\n\n\[ \overline{AB} + \overline{AC} > \overline{PB} + \overline{PC} \] | To prove this let \( {BP} \) pass through \( D \) of \( \left( {AC}\right) \) . Then as \( \overline{AC} > \overline{AD} \) a distance exists congruent to \( \overline{AB} + \overline{AD} \), and \( \overline{AB} + \overline{AD} > \overline{BP} + \overline{PD} \) . As \( \overline{AB} + \overline{AD} > \overline{PD} \)... | Yes |
Theorem 40. Any two right angles are congruent. | Let these right angles be \( \measuredangle {AOC} \) and \( \measuredangle {A}^{\prime }{O}^{\prime }{C}^{\prime } \) . We may assume \( O \) to be the middle point of \( \left( {AB}\right) \) and \( {O}^{\prime } \) the middle point of \( \left( {{A}^{\prime }{B}^{\prime }}\right) \), where \( \overline{OA} \equiv \ov... | Yes |
Theorem 41. There exists a congruent transformation carrying any segment \( \left( {AB}\right) \) into any congruent segment \( \left( {{A}^{\prime }{B}^{\prime }}\right) \) and any half-plane bounded by \( {AB} \) into any half-plane bounded by \( {A}^{\prime }{B}^{\prime } \) . | We have merely to find \( O \) and \( {O}^{\prime } \) the middle points of \( \left( {AB}\right) \) and \( \left( {{A}^{\prime }{B}^{\prime }}\right) \) respectively, and \( C \) and \( {C}^{\prime } \) on the perpendiculars to \( {AB} \) and \( {A}^{\prime }{B}^{\prime } \), at \( O \) and \( {O}^{\prime } \) so that... | Yes |
Theorem 42. If \( \mid {OA} \) be a given half-line, there will exist in any chosen half-plane bounded by \( {OA} \) a unique half-line \( \mid {OB} \) making the \( \measuredangle {AOB} \) congruent to any chosen angle. | The proof of this theorem depends immediately upon the preceding one. | No |
Theorem 43. Two plane angles of a dihedral angle are congruent. | We have merely to take the congruent transformation which keeps invariant all points of the plane whose points are equidistant from the vertices of the plane angles. Such a transformation may properly be called a reflection in that plane. | No |
Theorem 44. If two dihedral angles be congruent, any two of their plane angles will be congruent, and conversely. | The proof is immediate. | No |
Theorem 1. If a point \( P \) of a segment \( \left( {AB}\right) \) may be taken at as small a distance from \( A \) as desired, and \( C \) be any other point, the \( \measuredangle {ACP} \) may be made less than any given angle. | If \( C \) be a point of \( {AB} \) the theorem is trivial. If not, we may, by III. 4, find \( \mid {CD} \) in the half-plane bounded by \( {CA} \) which contains \( B \), so that \( \measuredangle {ACD} \) is congruent to the given angle. If then \( \mid {AB} \) belong to the internal \( \measuredangle {ACD} \), we ha... | Yes |
Theorem 4. If in any triangle one side and an adjacent angle remain fixed, while the other side including this angle varies, then the measures of the third side, and of the variable angles will be continuous functions of the measure of the variable side first mentioned. | Of course a constant is here included as a special case of a continuous function. | No |
Theorem 5. If two lines \( {AB},{AC} \) be perpendicular to \( {BC} \), then all lines which contain \( A \) and points of \( {BC} \) are perpendicular to \( {BC} \), and all points of \( {BC} \) are at congruent distances from \( A \) . | To prove this let us first notice that our \( \bigtriangleup {ABC} \) is isosceles, and \( \overline{AB} \) will be congruent to every other perpendicular distance from \( A \) to \( {BC} \) . Such a distance will be the distance from \( A \) to the middle point of \( \left( {BC}\right) \) and, in fact, to every point ... | Yes |
Theorem 6. If a set of lines perpendicular to a line \( l \), meet a line \( m \), the distances of these points from a fixed point of \( m \), and the angles so formed with \( m \), will vary continuously with the distances from a fixed point of \( l \) to the intersections with these perpendiculars. | The proof comes easily from 2 and 5. | No |
Theorem 7. Saccheri’s \( {}^{11} \) . In an isosceles birectangular quadrilateral a line through the middle point of the side adjacent to both right angles, which is perpendicular to the line of that side, will be perpendicular to the line of the opposite side and pass through its middle point. The other two angles of ... | Let the quadrilateral be \( {ABCD} \), the right angles having their vertices at \( A \) and \( B \) . Then the perpendicular to \( {AB} \) at \( E \) the middle point of \( \left( {AB}\right) \) will surely contain \( F \) point of \( \left( {CD}\right) \) . It will be easy to pass a plane through this line perpendicu... | No |
Theorem 9. If there exist a single rectangle, every isosceles birectangular quadrilateral is a rectangle. | Let \( {ABCD} \) be the rectangle. The line perpendicular to \( {AB} \) at the middle point of \( \left( {AB}\right) \) will divide it into two smaller rectangles. Continuing this process we see that we can construct a rectangle whose adjacent sides may have any measures that can be indicated in the form \( \frac{m}{{2... | No |
Theorem 10. If there exist a single right triangle the sum of whose angles is congruent to a straight angle, the same is true of every right triangle. | Let \( \bigtriangleup {ABC} \) be the given triangle, the right angle being \( \measuredangle {ACB} \) so that the sum of the other two angles is congruent to a right angle. Let \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be any other right triangle, the right angle being \( \measuredangle {A}^{\prime ... | Yes |
Theorem 11. If there exist any right triangle where the sum of the angles is less than a straight angle, the same is true of all right triangles. | We see the truth of this by continuity. For we may pass from any right triangle to any other by means of a continuous change of first the one, and then the other of the sides which include the right angle. In this change, by 2 , the sum of the angles will either remain constant, or change continuously, but may never be... | Yes |
Theorem 12. If there exist a right triangle where the sum of the angles is greater than two right angles, the same is true of every right triangle. | This comes immediately by reductio ad absurdum. | No |
Theorem 13. If there exist any triangle where the sum of the angles is less than (congruent to) a straight angle, then in every triangle the sum of the angles is less than (congruent to) a straight angle. | Let us notice, to begin with, that our given \( \bigtriangleup {ABC} \) must have at least two angles, say \( \measuredangle {ABC} \) and \( \measuredangle {BAC} \) which are less than right angles. At each point of \( \left( {AB}\right) \) there will be a perpendicular to \( {AB} \) (in the plane \( {BC} \) ). If two ... | Yes |
Theorem 14. If there exist any triangle where the sum of the angles is greater than a straight angle, the same will be true of every triangle. | This comes at once by reductio ad absurdum. | No |
Theorem 15. If in any triangle a line be drawn from one vertex to a point of the opposite side, the sum of the discrepancies of the resulting triangles is congruent to the discrepancy of the given triangle. | The proof is immediate. Notice, hence, that if in any triangle one angle remain constant, while one or both of the other vertices tend to approach the vertex of the fixed angle, along fixed lines, the discrepancy of the triangle, when not zero, will diminish towards zero as a limit. We shall make this more clear by say... | No |
Theorem 17. If in any triangle one side may be made less than any assigned segment, while neither of the other sides becomes indefinitely large, the discrepancy may be made less than any assigned angle. | If neither angle adjacent to the diminishing side tend to approach a straight angle as a limit, it will remain less than some non-re-entrant angle, and 16 will apply to all such angles simultaneously. If it do tend to approach a straight angle, let the diminishing side be \( \left( {AB}\right) \), while \( \measuredang... | No |
Theorem 1. If \( {AD} \) and \( {AX} \) be two mutually perpendicular lines we may find such a point \( B \) on either half of \( {AX} \) bounded by \( A \), that, a line being drawn perpendicular to \( {AB} \) at any point \( P \) of \( \left( {AB}\right) \) we may find on the half thereof bounded by \( P \), which li... | Let \( E \) be a point of the extension of \( \left( {AD}\right) \) beyond \( D \) . Draw a line there perpendicular to \( {AD} \) . If \( B \) be a point of \( {AX} \) very close to \( A \), and if a line perpendicular to \( {AB} \) at \( P \) of \( \left( {AB}\right) \), meet the perpendicular at \( E \) at a point \... | Yes |
Theorem 1. The product of two translations is a translation. The assemblage of all translations is a group. | We see, to begin with, that every congruent transformation has an inverse. This premised, suppose that we have a translation whereby \( A \) goes into \( {A}^{\prime } \), and a second whereby \( {A}^{\prime } \) goes into \( {A}^{\prime \prime } \) . We wish to show that the product of these two is not a reflection. S... | Yes |
Theorem 2. The lines through Theorem \( {2}^{\prime } \) . The points on each each of three given non-collinear of three coplanar but not concurrent points, perpendicular to the line of lines, orthogonal to the intersection the other two, are concurrent. of the other two, are collinear. | Returning to a centre of gravity of the two points \( {BC} \), we see that a line through it perpendicular to the line \( {BC} \) will have the equation\n\n\[ \left| {\frac{\left( xy\right) }{\sqrt{\left( yy\right) }} + \frac{\left( yz\right) }{\sqrt{\left( zz\right) }}\;\frac{\left( yz\right) }{\sqrt{\left( yy\right) ... | No |
Theorem 8. If four non-coplanar points be given, the eight points which are severally at congruent distances from them form, with the original four, a desmic configuration. | As there are eight points at congruent distances from the four given points, so there will be eight planes at congruent distances from them, we have but to take the polars of the eight points with regard to the Absolute. In like manner, if we consider not the points \( \left( x\right) ,\left( y\right) ,\left( z\right) ... | Yes |
Theorem 10. The necessary and sufficient condition that two lines should be paratactic is that their distances or angles should be congruent. | This condition may be expressed analytically by equating to zero the discriminant of either of our equations (19), (20).\n\n\[ \left\{ {{\left\lbrack \left( p \mid {p}^{\prime }\right) + \left( \sum {p}_{ij}{p}_{ij}{}^{\prime }\right) \right\rbrack }^{2} - \sum {p}_{ij}{}^{2}\sum {p}_{ij}{}^{\prime 2}}\right\} \left\{ ... | Yes |
Theorem 3. The necessary and sufficient condition that two crosses of different layers should intersect orthogonally is that the corresponding line and point of the complex plane should be in united position. | If a cross be improper, the assemblage of all crosses cutting it orthogonally will be made up of all lines through the point of contact, and all lines in the plane of contact. This assemblage, reducible in point space, is irreducible in cross space. The collineation group of cross space, is the general group depending ... | No |
Theorem 5. The common perpendiculars to a line, in the general position, of a non-synectic congruence, and each adjacent line will generate a chain. | Let us find, in point coordinates, the equation of the surface obtained by splitting off from a chain its improper crosses. We easily see that there will be two crosses of the chain which intersect orthogonally; taking these and the axes to determine the coordinate system, we may express our chain in the simple form\n\... | Yes |
Theorem 7. The common perpendiculars to pairs of crosses of a chain congruence will generate a second chain congruence in the other layer. Each congruence is the locus of the axes of the \( {\infty }^{2} \) chains of the other; the two are said to be reciprocal to one another. | The reciprocal to the chain congruence (9) will have equations\n\n\[ \n{U}_{i} = p\left| \begin{array}{ll} {Y}_{i} & {Y}_{k} \\ {Z}_{j} & {Z}_{k} \end{array}\right| + q\left| \begin{array}{ll} {Z}_{j} & {Z}_{k} \\ {T}_{j} & {T}_{k} \end{array}\right| + r\left| \begin{array}{ll} {T}_{j} & {T}_{k} \\ {Y}_{i} & {Y}_{k} \e... | Yes |
Theorem 2. A tangent to a circle Theorem \( {2}^{\prime } \) . A point on a circle is perpendicular to the diameter is orthogonal to the point where the through the point of contact. tangent thereat meets the axis. | These simple theorems may be proved in a variety of ways. For instance every circle will be transformed into itself by a reflection in any diameter, hence the tangent where the diameter meets the curve must be perpendicular to the diameter. Or, again, if \( \overline{AB} \equiv \overline{AC} \), a line from \( A \) to ... | Yes |
Theorem 4. If two tangents to a circle (horocycle) make a constant angle, the locus of their point of intersection is a concentric circle (horocycle). | Let the tangents at \( P \) and \( {P}^{\prime } \) meet at \( Q \), the centre of the circle being \( A \) . Let \( {\Delta \phi } \) be the angle between the tangents, and let \( {P}^{\prime \prime } \) be the point on the tangent at \( P \) whose distance from \( P \) equals \( P{P}^{\prime } \), or, in the infinite... | Yes |
Theorem 5. If two circles intersect in four points, two common tangents called radical axes are concurrent with the axes of the circles and harmonically separated by them, and are perpendicular to one another and to the line of centres. The bisectors of angles of the tangents at a centre of similitude are the intersect... | If the equations of the two circles be\n\n\[ \n{\cos }^{2}\frac{{r}_{1}}{k}\left( {aa}\right) \left( {xx}\right) - {\left( ax\right) }^{2} = 0,\;{\cos }^{2}\frac{{r}_{2}}{k}\left( {bb}\right) \left( {xx}\right) - {\left( bx\right) }^{2} = 0, \]\n\nthe equations of the radical axes will be\n\n\[ \n\left( {\cos \frac{{r}... | Yes |
Theorem 6. If a set of circles through two points have the line of tangent to two given lines have the these points as a radical axis, the intersection of the lines as a centre of points of contact of tangents to all similitude, the envelope of tangents of them from a point of the line to them at the points where they ... | Consider the assemblage of all circles through two given points. If the line connecting the two points be a radical axis for two of these circles it will be perpendicular to their line of centres at one centre of gravity of the two points, and in every case a perpendicular from the centre of a circle on a secant will m... | No |
Theorem 10. Four circles may be constructed so that the points of contact of tangents common to them and to each of three given circles form two pairs of orthogonal points. | It is here assumed that no two of the given circles are concentric. There is no reason to expect that because two circles intersect at right angles in two points they will in the other two. Let the circles be\n\n\[ \n{\cos }^{2}\frac{{r}_{1}}{k}\left( {aa}\right) \left( {xx}\right) - {\left( ax\right) }^{2} = 0,\;{\cos... | Yes |
Theorem 3. Two focal lines of Theorem \( {3}^{\prime } \) . Two foci of a a central conic pass through each central quadric lie on each axis, vertex, and are perpendicular to the and are orthogonal to the opposite opposite axis. centre. | The coordinates of the focal lines \( {f}_{h},{f}_{h}{}^{\prime } \), through the centre \( {u}_{h} = 0 \), will be\n\n\[ \n{u}_{h} : {u}_{k} : {u}_{l} = 0 : \sqrt{{c}_{h} - {c}_{k}} : \pm \sqrt{{c}_{l} - {c}_{h}}. \n\]\n\n(2)\n\nThe coordinates of the foci \( {F}_{h},{F}_{h}{}^{\prime } \) on the opposite axis will be... | Yes |
Theorem 4. The ratio of the sines of the \( k \) th parts of the distances from a point of a central conic to a focus and to the corresponding directrix is constant. | \[ \sin \frac{\overline{P{F}_{h}}}{k}\sin \frac{\overline{P{F}_{h}{}^{\prime }}}{k} : \sin \frac{\overline{P{F}_{k}}}{k}\sin \frac{\overline{P{F}_{k}{}^{\prime }}}{k} : \sin \frac{\overline{P{F}_{l}}}{k}\sin \frac{\overline{P{F}_{l}{}^{\prime }}}{k} = \frac{1}{{c}_{h}\left( {{c}_{k} - {c}_{l}}\right) } : \frac{1}{{c}_{... | Yes |
Theorem 5. The sum of the distances from real points of an angles which the real tangents to an ellipse and the difference of the ellipse or convex hyperbola, or the distances from real points of a difference of the angles which the hyperbola or semi-hyperbola to two real tangents to a concave hyperbola real foci on th... | Reverting to our point \( \left( x\right) \) we see\n\n\[ \sin \frac{\overline{P{f}_{h}}}{k} = \frac{\sqrt{{c}_{h} - {c}_{k}}{x}_{k} + \sqrt{{c}_{l} - {c}_{h}}{x}_{l}}{\sqrt{\left( {{c}_{l} - {c}_{h}}\right) {x}_{l}{}^{2} - \left( {{c}_{h} - {c}_{k}}\right) {x}_{k}{}^{2}}\sqrt{\frac{-\left( {{c}_{k} - {c}_{l}}\right) }... | No |
Theorem 12. The locus of the Theorem \( {12}^{\prime } \) . The envelope of reflection of a real focus of an ellipse the reflection in a variable point of in a variable tangent, is a circle an ellipse, of a real focal line, is a whose centre is the corresponding circle whose axis is the corresponding focus. focal line. | Let \( \left( y\right) \) be the coordinates of a point \( P \) of our conic. The equation of a line through the centre \( {O}_{h} \) conjugate to the line \( {O}_{h}P \) will be\n\n\[ \n{c}_{k}{y}_{k}{x}_{k} + {c}_{l}{y}_{l}{x}_{l} = 0.\n\]\n\nThis will meet the conic in two points \( {P}^{\prime } \) having the coord... | No |
Theorem 16. The product of the tangents of the \( k \) th parts of the distances from a centre of a central conic to a point of the curve and to the tangent where the curve meets a diameter conjugate to that from the centre to the point of the curve, is constant. | \[ {y}_{l}{x}_{k} - {y}_{k}{x}_{l} = 0,\;{c}_{k}{y}_{k}{x}_{k} + {c}_{l}{y}_{l}{x}_{l} = 0. \] | No |
Theorem 3. A line will meet a Theorem \( {3}^{\prime } \) . The tangent planes central quadric and its focal cones to a central quadric and its focal in five pairs of points with the same conics through a line form five sets centres of gravity. of dihedral angles with the same bisectors. | The proof of these two theorems is immediate. | No |
Theorem 5. The sum of the squares of the cotangents of the \( k \) th parts of the distances from a centre of a central quadric to three intersections with the surface of three mutually perpendicular lines through that centre is constant. | \[ {\operatorname{ctn}}^{2}\frac{\overline{{O}_{h}Q}}{k} + {\operatorname{ctn}}^{2}\frac{\overline{{O}_{h}{Q}^{\prime }}}{k} + {\operatorname{ctn}}^{2}\frac{\overline{{O}_{h}{Q}^{\prime \prime }}}{k} = - \frac{\left( {c}_{k} + {c}_{l} + {c}_{m}\right) }{{c}_{h}}. \] | Yes |
Theorem 8. If from a set of confocal central quadrics a one-parameter set of linear generators be so chosen that all intersect the same \( {\infty }^{1} \) lines of curvature of \( {\infty }^{1} \) developables circumscribed to pairs of confocal quadrics of the system, then the tangent planes to any two of these develo... | Theorem 8 may also be easily proved by showing that the generators of a set of confocal quadrics form an isotropic congruence, whereof much more later. \( {}^{61} \) The general theorem concerning isotropic congruences upon which this depends will be proved in Chapter XVI, where also will be found a bibliography of the... | No |
Theorem 3. The area of a region of a plane is the sum of the areas of any two regions into which it may be divided provided that these two have no common area. | This follows immediately from the definition given above. | No |
Theorem 4. The area of a triangle is the quotient of the excess divided by the measure of curvature of space. | Let us give a second demonstration of this fundamental theorem with the aid of integration. It will be sufficient to do so in the case of a right triangle, and we shall take a right triangle with one angle at \( C \) the intersection of \( {x}_{1} = 0 \) , \( {x}_{2} = 0 \), the right angle being at \( B \) a point of ... | Yes |
Theorem 1. The curvature of a curve at any point is equal to that of its osculating circle, and is equal to the absolute value of the product of the square root of the curvature of space and the cotangent of the \( {k}^{\text{th }} \) part of the distance of each point of the circle from its centre. | Let us now suppose that the equations of our curve are written in the form\n\n\[ \n{x}_{i} = {x}_{i}\left( {t}_{0}\right) + \left( {t - {t}_{0}}\right) {x}_{i}{}^{\prime }\left( {t}_{0}\right) + \frac{{\left( t - {t}_{0}\right) }^{2}}{2}{x}_{i}{}^{\prime \prime }\left( {t}_{0}\right) + \ldots \n\] \n\n\[ \n{x}_{i}{}^{\... | Yes |
Theorem 10. In any triply orthogonal system of surfaces, the curves of intersection are lines of curvature. | Let the three families of surfaces be given by the equations\n\n\[ \n{x}_{i} = {f}_{i}\left( {uv}\right) ,\;{x}_{i} = {\phi }_{i}\left( {vw}\right) ,\;{x}_{i} = {\psi }_{i}\left( {wu}\right) , \n\]\n\n\[ \n\left( {xx}\right) = {k}^{2},\;\left( {x\frac{\partial x}{\partial u}}\right) = \left( {x\frac{\partial x}{\partia... | Yes |
Theorem 12. Meunier's. The curvature of a curve on a surface at any point is equal to the curvature of the normal section with the same tangent divided by the cosine of the angle which the principal normal makes with the normal to the surface. | Reverting to our previous expressions \( {r}_{1},{r}_{2} \) and taking the lines of curvature as parameter lines, the curvature of the normal sections through the tangents to the lines of curvature are\n\n\[ \frac{1}{k\tan \frac{{r}_{1}}{k}},\;\frac{1}{k\tan \frac{{r}_{2}}{k}} \]\n\n\[ d{x}_{i} = \tan \frac{{r}_{1}}{k}... | Yes |
Theorem 16. In any surface of constant total relative curvature, the torsion of every asymptotic line is constant and equal to a square root of the total relative curvature, and the necessary and sufficient condition that a surface should have constant total relative curvature is that the asymptotic lines of one set sh... | In speaking of the total curvature of a surface we have used the word relative. It is now time to explain why that adjective is chosen. Let us try to express our total relative curvature in terms of \( E, F, G \) and their derivatives. We have\n\n\[ \frac{1}{{k}^{2}\tan \frac{{r}_{1}}{k}\tan \frac{{r}_{2}}{k}} = \frac{... | Yes |
Theorem 18. A surface of total relative curvature zero is a developable. | Clearly every developable has total relative curvature zero. | No |
Theorem 22. The necessary and sufficient condition that it should be possible to assemble the normals to a surface into one parameter families of left (right) paratactics, is that the given surface should have Gaussian curvature zero. It will, then, be possible to assemble the normals into families of right (left) para... | We shall, as in euclidean space, define as the geodesic curvature at any point of a curve on our surface, the curvature of its orthogonal projection on the tangent plane at that point. Let us denote this by \( \frac{1}{{\rho }_{g}} \), while \( \sigma \) is the angle which the osculating plane makes with the tangent pl... | No |
Theorem 26. The necessary and sufficient condition that a surface should be minimal is that the mean relative curvature should be zero. | We see from (23) that the numerator of the expression for the relative mean curvature is the simultaneous invariant of (13) and (20), and vanishes when, and only when, the tangents to the asymptotic lines are harmonically separated by those to the isotropic ones, hence | No |
Theorem 1. A line of a general Theorem \( {1}^{\prime } \) . Through a line of analytic congruence contains four a general analytic congruence will limiting points, mutually orthogonal pass four limiting planes, mutually in pairs, and these, when real, perpendicular in pairs, and these, determine two real regions of th... | We shall now look more closely into the question of the reality of limiting points and places. We may so choose our coordinate system that the equations of the line in question shall be \( {x}_{1} = {x}_{2} = 0 \) . Reverting to equation (8) of Chapter \( \mathrm{X} \) the equation of the ruled quartic surface will be,... | No |
Theorem 2. In hyperbolic space the limiting points of an actual line are real, and the limiting planes imaginary. In elliptic space this may occur, or the planes may be all real and the points all imaginary. | Giving to \( {x}_{0} : {x}_{3} \) one of the values from \( \left( {16}^{\prime }\right) \) we see that\n\n\[ \frac{{x}_{0}{}^{2} + {x}_{3}{}^{2}}{{x}_{0}{x}_{3}} = \pm \frac{2\left( {{a}_{1} + {a}_{2}}\right) }{{a}_{1} - {a}_{2}}.\]\n\nSubstituting in \( \left( {15}^{\prime }\right) \) we have\n\n\[ {x}_{1} + {x}_{2} ... | No |
Theorem 6. The necessary and sufficient condition that a general congruence should be composed of normals is that the focal points should coincide with a pair of limiting points. | In a normal congruence let us suppose that \( \left( x\right) \) traces a surface to which the given lines are normal so that\n\n\[ \left( {ydx}\right) = - \left( {xdy}\right) = 0. \]\n\nLet us then put\n\n\[ \overline{{x}_{i}} = {x}_{i}\cos \frac{r}{k} + {y}_{i}\sin \frac{r}{k},\;\overline{{y}_{i}} = {x}_{i}\cos \frac... | Yes |
Theorem 7. If a constant distance be laid off on each normal to a surface from the foot, in such a way that the points on adjacent normals are on the same side of the tangent plane corresponding to either, the locus of the points so found is a surface with the same normals as the original one. | Let us suppose that we have a normal congruence determined by mutually orthogonal points \( \left( x\right) \) and \( \left( y\right) \), where \( {x}_{i} = {x}_{i}\left( {uv}\right) \) traces a surface, not one of the orthogonal trajectories of the congruence. We shall choose as parameter lines in this surface the iso... | Yes |
Theorem 8. If a normal congruence be subjected to any finite number of reflections or refractions, the resulting congruence is normal. | We shall now abandon the general congruence and assume that, contrary to (12)\n\n\[ e : \frac{f + {f}^{\prime }}{2} : g \equiv \left( {E - {E}^{\prime }}\right) : \left( {F - {F}^{\prime }}\right) : \left( {G - {G}^{\prime }}\right) .\n\]\n\n(24)\n\nThere are two sharply distinct sub-cases which must not be confused:\n... | No |
Theorem 13. Each ray of a common perpendicular to the lines of two rays will be represented by a pair of poles of two great circles which connect the pairs of representing points. | It is clear that an analytic congruence may be represented in the form\n\n\[ \n{}_{l}{X}_{i} = {}_{l}{X}_{i}\left( {uv}\right) ,\;{}_{r}{X}_{i} = {}_{r}{X}_{i}\left( {uv}\right) , \n\]\n\nor else, in general,\n\n\[ \n{}_{l}{X}_{i} = {}_{l}{X}_{i}\left( {{}_{r}{X}_{1r}{X}_{2r}{X}_{3}}\right) . \n\]\n\nTwo adjacent rays ... | No |
Theorem 15. The necessary and sufficient condition that a congruence should be isotropic is that the corresponding relation between the representing spheres should be directly conformal. | Let us take up the isotropic case more fully. Any directly conformal relation between the real domains of two euclidean spheres of radius unity may be represented by an analytic function of the complex variable. Let us give the coordinates of points of our representing spheres in the following parametric form:\n\n\[ \n... | Yes |
Theorem 17. The necessary and sufficient condition that an isotropic congruence should contain the opposite of each of its rays is that the focal surface should be irreducible. | It is very easy to observe the distinction between the two cases in the case\nof the linear function\n\\[ \n{u}_{1} = \frac{\alpha {z}_{1} + \beta }{\gamma {z}_{1} + \delta }\n\\]\n\nIf \( \beta = - \bar{\gamma },\delta = \bar{\alpha } \) ,(29) is identically satisfied. But here it will be seen that if we write\n\n\\[ ... | Yes |
Theorem 1. The assemblage of all coordinate transformations which represent the identical transformation of a multiply connected space form a group. | If \( \left( x\right) \) and \( \left( {x}^{\prime }\right) \) be two sets of coordinates for the same point the expression\n\n\[ \left| {{\cos }^{-1}\frac{\left( x{x}^{\prime }\right) }{\sqrt{\left( xx\right) }\sqrt{\left( {x}^{\prime }{x}^{\prime }\right) }}}\right| \]\n\ncannot sink below a definite minimum value gr... | No |
Theorem 1. \( {AB}\int {CD} \) and \( {AE}\int {CD} \), then \( {EB}\int {CD} \) . | For \( C \) and \( D \) determine but two separation classes on the line, and both \( B \) and \( E \) belong to that class which does not include \( A \) . | No |
Theorem 2. If five collinear points be given, a chosen pair of them will either separate two of the pairs formed by the other three or none of them. | Let the five points be \( A, B, C, D, E \) . Let \( {AC}\int {DE} \) . Then, if \( {BC}\int {DE} \) , \( {AB}\int {DE} \), and if \( {AB}\int {DE},{BC}\int {DE} \) . But if we had \( {BC}\int {DE} \) and \( {AB}\int {DE} \) , \( {ABC} \) would belong to the same separation class with regard to \( {DE} \), and hence \( ... | No |
Theorem 3. If \( {AC}\int {BD} \) and \( {AE}\int {CD} \), then \( {AE}\int {BD} \) . | To begin with \( {BC}\int {AD},{EC}\int {AD} \) ; hence \( {BE}\int {AD} \) . Again, if we had \( {AB}\int {ED} \), we should have \( {AB}\int {EC} \), i.e. \( {AE}\int {BC} \) . But we have \( {AE}\int {CD} \) , hence \( {AE}\int {BD} \) a contradiction with \( {AB}\int {ED} \) . As a result, since \( {BE}\int {AD} \)... | No |
Theorem 5. A plane will contain completely every line whereof it contains two points. | Let the plane be determined by the point \( A \) and the line \( {BC} \) . If the two given points of the given line belong to \( {BC} \) or be \( A \) and a point of \( {BC} \), the theorem is immediate. If not, let the line contain the points \( {B}^{\prime } \) and \( {C}^{\prime } \) of \( {AB} \) and \( {AC} \) re... | Yes |
Theorem 6. If \( A, B, C \) be three non-collinear points, then the planes determined by \( A \) and \( {BC} \), by \( B \) and \( {CA} \), and by \( C \) and \( {AB} \) are identical. | We have but to notice that the lines generating each plane lie wholly in each of the others. | No |
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