If the same multiples be taken of each of two magnitudes, they are called equimultiples of the magnitudes. Thus 2A and 2B are equimultiples of A and B. Find the smallest equimultiples of 4 and 5, which differ by more than 6. Art. 19. GEOMETRICAL ILLUSTRATIONS OF EQUIMULTIPLES. FIRST ILLUSTRATION. To construct equimultiples of a parallelogram and its base. Let ABCD be a parallelogram standing on the base AB. On AB produced take any number of lengths BE, EF, FG, GH each equal to AB, and through E, F, G, H draw parallels to BC cutting DC produced in K, L, M, N respectively. Then the parallelograms ABCD, BEKC, EFLK, FGML, GHNM are all equal because they stand on equal bases and are situated between the same parallels. Therefore the parallelogram AHND is the same multiple of the parallelogram ABCD as AH is of AB. Therefore parallelogram AHND, base AH are equimultiples of parallelogram ABCD, base AB. If AH=r(AB), then AHND=r(ABCD). Art. 20. SECOND ILLUSTRATION. To construct equimultiples of a triangle and its base. Let ABC be a triangle standing on the base AB. On AB produced take any number of lengths BD, DE, EF each equal to AB, and join CD, CE, CF. Then the triangles CAB, CBD, CDE, CEF all stand on equal bases, and have the same altitude. Therefore they are equal in area. Therefore the triangle AFC is the same multiple of the triangle ABC as AF is of AB. Therefore D Fig. 4. triangle AFC, base AF are equimultiples of triangle ABC, base AB. If AF=r (AB), then AAFC= r (▲ ABC). Art. 21. THIRD ILLUSTRATION. To construct equimultiples of the three magnitudes, the angle at the centre of a circle, the arc on which it stands, and the sector bounded by the arc and the sides of the angle. Let O be the centre of a circle. Let AOB be an angle at the centre standing on the arc AB. Now make any number of angles BOC, COD, DOE, EOF, FOG, GOH each equal to AOB. Then the arcs BC, CD, DE, EF, FG, GH are each equal to the arc AB, because they subtend equal angles at the centre of the circle. Hence the arc AH is the same multiple of the arc AB as the angle AOH is of the angle AOB. arc AH, angle AOH, sector AOH are equimultiples of arc AB, angle AOB, sector AOB. If arc AH=r(arc AB), then AÔH=r(AÔB), and sector AOH=r (sector AOB). Art. 22. FOURTH ILLUSTRATION. To construct equimultiples of the intercepts made by two parallel straight lines on two other straight lines. Let the straight lines OX, O Y be cut by the parallel lines AB, CD. To construct equimultiples of the intercepts AC, BD. Draw any straight line PQ parallel to AB, cutting OX at P and OY at Q. From P along PX set off any number of lengths PR, RS, ST each equal to AC, and draw TW parallel to AB to cut Oy at W, then PT and QW are equimultiples of AC and BD. For draw RZ parallel to PQ to cut OY at Z. Draw BE, QN parallel to OX cutting CD at E, and RZ at N respectively. bool In like manner if SV be drawn parallel to AB, ZV will be equal to BD; and if TW be drawn parallel to AB, VW will be equal to BD. Hence PR, RS, ST being each equal to AC, QZ, ZV, VW are each equal to BD. .. QW is the same multiple of BD If A and B are two magnitudes of the same kind, it is always possible to find a multiple of either which will exceed the other. This is usually known as the Axiom of Archimedes. But Euclid uses it in the Fifth Book, see Euc. v. 8, and it is also implied in the fourth definition of the Fifth Book. Art. 24. PROPOSITION VII. (Contained in Euc. v. 8.) ENUNCIATION. If X, Y, Z be three magnitudes of the same kind, and if X and Y be unequal, then it is always possible to find equimultiples of X and Y, such that some multiple of Z lies between them. Since X and Y are unequal, one of them must be the greater. Let Y be greater than X. It is to be proved that integers n and s exist, such that Since Y > X, therefore Y-X is a magnitude of the same kind as Z. Hence an integer n exists, such that Now let tZ be the greatest multiple of Z which does not exceed nX. |