We have, (Def. 10. 3,) AD=BC. Adding CD to each, (Ax. 2,) AD+CD=BC+CD. BC+CD. Dividing each by D, (Ax. 7,) A+C= D Hence, dividing each by B+D, A+C C A B+D D ̄B Again, take any two couplets of numbers having the same ratio, as 8: 4, and 6: 3; the ratio of 8+6: 4+3 is equal to that of 8 : 4, or 6 : 3. That is, the ratio of the sum of the antecedents to the sum of the consequents is equal to the ratio of either of the given couplets. Cor. If any number of couplets having the same ratio, be added together, the ratio of the sum of all the antecedents, to the sum of all the consequents, will be equal to the ratio of any one antecedent to its consequent. For by adding two couplets having equal ratios, a new couplet is formed having the same ratio as either of the given couplets. Hence a third couplet may be added, a fourth, a fifth, &c. Thus, the ratio of each of the couplets 4: 2, of 10: 5, of 8: 4,of 12 6, is 2; and the ratio of 4+10+8+12: 2+5+ 4+6, is also 2. If from the terms of any couplet two magnitudes having the same ratio, be subtracted, the ratio of the difference of the antecedents to the difference of the consequents, will be equal to the ratio of either antecedent to its consequent. Let the ratio of A: B, be the same as the ratio of C: D, then will the ratio of A-C: B-D, be equal to the Multiplying each by BD, (Ax. 6,) AD=BC. BC-CD Dividing each by D, (Ax. 7,) A-C= D A-C C A Hence, dividing each by B-D, B-DD B Again, take the two couplets, 8: 4, and 6: 3, the ratio of 8-6: 4-3, is equal to the ratio of 8: 4, or 6: 3. That is, the ratio of the difference of the antecedents to the difference of the consequents, is equal to the ratio of either of the given couplets. Cor. If the terms of two couplets having the same ratio, be added to each other, and also subtracted from each other, the ratio of the sum of the antecedents to the sum of the consequents, will be equal to the ratio of the difference of the antecedents to the difference of the consequents. For, the ratio of Also, 24 6 16 4; 24+16 24-16 = 6+4 6-4 PROPOSITION III. THEOREM. If the antecedent and consequent of one couplet are respectively equal to the antecedent and consequent of another, the ratios of the two couplets must be equal. In the two couplets A: B and C D, if A=C and B=D; then will BD' For, by hypothesis, A=C and B=D, and if equals are divided by equals, the quotients will be equal, (Ax. 7,) therefore, A C That is, the ratio of A: B, equals that of C: D.. PROPOSITION IV. THEOREM. If the ratios of two couplets are equal and their antecedents are equal, their consequents must be equal; and if the ratios are equal and the consequents are equal, the antecedents must be equal. Let the ratio of A: B be equal to that of C: D, and the antecedent A=C, then will B=D. A C Since by hypothesis, BD' Multiplying each by B and D, (Def. 10. 3,) AD=BC. AD÷A=BC÷C : that is, D=B. Again, let the ratio A: B, be equal to that of C: D, and the consequent B=D; then will, A=C. Since, (Def. 11. 3,) Multiplying each by B and D, A C B D' AD=BC. Hence dividing each by the equals B and D, AD÷D=BC÷B; that is A=C. PROPOSITION V. THEOREM. If four magnitudes are proportional, the product of the extremes is equal to the product of the means. If A: B : : CD, then is For, since, (Def. 11. 3,) AD=BC. A C B D Multiply in each by BD, (Def. 10. 3,) AD=BC. Again, 86: 12: 9, then 8x9 6x12. Hence, If four magnitudes are proportional, &c. Cor. 1. If the first term be greater than the second, the third will be greater than the fourth; if equal, equal; if less, less. For since proportion is an equality of ratios, (Def. 11. 3,) if one is a ratio of equality, (Def. 2. 3. note,) the other must be also; if one is a ratio of greater inequality, the other must be; and if one is a ratio of less inequality, the other must be the same. Cor. 2. Any factor may be transferred from one extreme to the other, or from one mean to the other, without affecting the proportion. If mA : B :: C: D, then A : B :: C: mD; for in each case, mAD=BC. So, if A mB C D, then A: B:: mC: D; for in each case AD=mBC. Cor. 3. Either extreme is equal to the product of the means divided by the other extreme; and either mean is equal to the product of the extremes divided by the other mean. Thus, 8 6: 12: 9, then 8=(6×12)÷9. So, 6=(8×9) 12. Scholium. From this principle is derived the rule of simple proportion in Arithmetic which is commonly called The Rule of Three. Three numbers are given to find a fourth, which is obtained by multiplying the second and third terms together and dividing the product by the first. If four magnitudes are such, that the product of two of them is equal to the product of the other two, these magnitudes will form a proportion, by placing the factors of one product for the extremes, and the factors of the other, for the means. Let A, B, C, D, be four magnitudes such that, A.D=B.C, Then will, A: B:: C: D. For, since by hypothesis AD=BC, AD BC = Dividing each by BD, (Ax. 7,) BD BD : D. Therefore, (Def. 4. 3,) A : B :: C Again, since 8x5=4×10, then is 8 : 4 : : 10 : 5. For 84 10÷5. Hence, If four magnitudes, &c. Scholium. Any two equal quantities, each of which is composed of two factors, will form a proportion, by placing the two factors of the one for the extremes, and the two factors of the other for the means. PROPOSITION VII. THEOREM. If three magnitudes are proportional, the product of the extremes is equal to the square of the mean. |