Art. 10. PROPOSITION IV.* (Euc. v. 6.) ENUNCIATION. If a >b, prove that (a - b) R = aR − bR. .. (a - b) is a positive integer which may be called c. .. a=b+c. .. aR= (b+c) R =bR+cR. .. cRaRbR .. (a - b) RaRbR. Art. 11. EXAMPLE 4. [Prop. 2. If A and B are multiples of G, then the difference of A and B is a multiple of G. Art. 12. PROPOSITION V.† ENUNCIATION. To prove that r (sA) = rs (A) = sr (A) = s (r¿). Let a rectangle be drawn and divided into compartments standing in r columns and s rows. Place the magnitude A in each compartment. Then the sum of the magnitudes in any row is rA, and the sum of those in any column is sA. Since there are s rows, the sum of all the magnitudes is s (rA). Since there are r columns, the sum of all the magnitudes is r (sA). If the number of the magnitudes be counted, it is rs, or it may also be expressed as sr. Hence the sum can be written in either of the forms rs (A) or sr (A). But the sum of the magnitudes is the same in whatever way it is determined. .. r(sA)=rs(A) = sr (A) = s (rA). Art. 13. EXAMPLE 5. If A and B are multiples of G, then the sum and difference of rA and sВ are multiples of G. Both these results are contradictory to the hypothesis that rA > rB. Hence A must be greater than B. The second and third cases can be proved in like manner. [Prop. 1. 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 C A B D E 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. |