For, if a:b::c:d, then (No. 108.) ad = bc, and dividing both by cd, we have, c::bid. a C = and, therefore, (No. 107.) a: d' 113. If four quantities be proportionals, they are also proportionals when taken inversely, that is the second: first : : fourth : third. For, since a:b::c:d, bc ad, (No. 108.) and dividing both 6 d by ac, we have a C -, and, therefore, (No. 107.) b:a::d:c. 114. If four quantities be proportionals, they are also proportionals componendo or dividendo, that is, the sum or difference of the first and second, is to the second, as the sum or difference of the third and fourth to the fourth. 115. If four quantities be proportionals, they are also proportionals convertendo, that is, the first is to its excess above the second, as the third to its excess above the fourth. For, since a:b::c:d, a-b:b::c-d:d, (No. 114.) and, alternately, (No. 112.) a-b:cd::b:d; and also from alternation, a:c::b:d, and hence (No. 111.) a:c::a-b: c—d, and, therefore, by again alternating, a: a-b::c: c-d. 116. When four quantities are proportionals, the sum of the first and second is to their difference, as the sum of the third and fourth to their difference. This readily appears from Nos. 114 and 115. For, (No. 114.) a+bb::c+d: d, therefore, (No. 112.) a+b:cd::b: d. And, (No. 115.) a: a-b::c: cd, therefore, (No. 112 ::: - .:c. C- d. But, (No. 112.) a: c:: b:d, hence, (No. 111.) a+b:c+d::a-b:c-d; and, therefore, alternately, a+b: a-b::c+d: c-d. 117. When any number of quantities are proportionals, as one antecedent is to its consequent, so is the sum of all the antecedents to the sum of all the consequents. Let a:b::cd::e:f, &c. Then shall a: b:: a+c+e+&c.: b+d+f+&c. For, since a:b::c: d, ad bc, (No. 108.;) in like manner, because a: b::e:f, af- be, now, ab ba; and, therefore, by adding equals to equals, we have ab+ad+af+&c. = ba+be+ be+&c., or, resolving into factors, a(b+d+f+&c.) = b(a+c+ e+&c.) and hence, (No. 110.) a: ba+c+e+&c.: b+d+f+ &c. 118. When four quantities are proportionals, if the first and second be multiplied or divided by any quantity, as also the third and fourth, the resulting quantities will be proportionals. ma nc For, if a:b::c:d, then= and = mb nd Hence, ma mb :: nc: nd, when m and n may be any quantities whatever, either integral or fractional. From the preceding theorem, it follows, that if the first and third, as also the second and fourth, be multiplied or divided by any quantity, the resulting quantities will be proportionals. 119. In two ranks of proportionals, if the corresponding terms be multiplied together, the resulting products will be proportionals. fore, (No. 107.) ae: bf:: cg: dh. The same demonstration may be applied to any number of ranks of proportional quantities. 120. If four quantities be proportionals, any like powers or roots of them will also be proportionals. an ". For, since a:b::c:d,=, and, therefore, = bn hence, (No. 107.) a"; b" :: c": da, where n may be either integral or fractional. 121. If there be any number of quantities, a, b, c, d, &c. in continued proportion, the first is to the third in the duplicate ratio of the first to the second; or the first to the fourth in the triplicate ratio of the first to the second, &c. For, since a:b :: b:c, a2 : b2 :: b2: c2, (No. 120.) but ac= 62, (No. 108.) therefore, a2: 62:: ac: c2, and hence, dividing the two last terms by c, we have a : c :: a2: b2, (No. 118.) Again, because a: è:: a2: b2. And, c:d::a: b. We have, (No. 119.) ac: cd:: a3: b3, or, dividing the two first terms by c, a: d::a3 : b3. In like manner, it may be shown, that a: e :: a1 : ba, &c. &c. 122. In the preceding theorems, it is supposed that one quantity is some determinate multiple, part, or parts of another, or that the fraction arising from the division of 'the one by the other, is some determinate fraction. This will always be the case when the two quantities have any common measure whatever. Let x be a common measure of a and b, and let a = mx, b = when m and n are whole numbers. a nx; then / = mx b nx m n 123. It often happens, however, that the quantities are incommensurable, or have no common measure whatever, as when one represents the side of a square, and the other its cannot be express α diagonal. In such cases, the value of a fraction whose numerator and deno n minator are whole numbers; a fraction, however, of this kind, may be found, which will express its value to any deaccuracy less than perfect. gree of Suppose x to be a measure of b, and let b = nx; also, let a be a greater than mæ, but less than (m+1)x; then is greater than and as x is diminished, will be increased, since na = b, and hence diminished. By diminishing x, therefore, the a n difference of and may be rendered less than any that can be assigned. ... b m 124. If a and b, as well as c and d, be incommensurable, m tween n b and +1, however the magnitudes m and ʼn are a = n increased, If they are not equal, they must b d have some assignable difference; and because each of them lies between m 1 and +1, this difference is less than, but n since, by supposition, n may be increased without limit, n may be diminished without limit, that is, it may become a less than any assignable quantity, therefore, and have no assignable difference, that is, and all the pre ceding propositions, respecting proportionals, are true of the four magnitudes a, b, c, and d.* 125. Quantities are said to be in Arithmetical Progression, when they increase or decrease by a common difference. Thus, 1, 3, 5, 7, 9, 11, &c.; 5, 3, 1, -1, -3, -5, &c. &c., are quantities in arithmetical progression. * Wood's Algebra, Art. 191. 126. If a be the first term, and b the commor of a series of terms in arithmetical progression, t cond term will be a+b, the third a+2b, the fou &c.; and since the coefficient of b in any term unity than the number of terms, counting from ning, the last term will therefore be a +(n-1) denotes the number of terms. 127. The sum of a series of quantities, in Ar Progression, is equal to the sum of the first and l or of any two terms equally distant from them, by half the number of terms. sum of the series a, a+b, a+2b, a+3b, &c. +a+b +a+2b, &c. -- a Let s Then, s = a And sa+(n-1)b+a+(n−2)b+a+(n−3)b, &c. Or, 2s — 2a+(n—1)b+2a+(n−1)b+2a+(n—1.)6 + terms, by adding the two series together. Hence, 28(2a+:(n−1)b) × n, and, therefore, s From the equation s = (2a+(n−1)b), any three of quantities s, a, b and n, being given, the fourth may b It will be fully as well, however, to substitute the giv n bers in the general formula, s = (2a+(n−1)b), and i equation determine the value of the unknown quantity. 128. If three quantities be in arithmetical progres sum of the extremes is double of the mean. For, if a, a+b, a+2b, be the three quantities, then i vious, that a+(a+2b)=(a+b)2. Hence, an arithmetical mean between two quantities i to half their sum. Ex. 1. Required the sum of 20 terms of the series 1, { 9, &c. Here, a = 1, b = 2, and ʼn = 20; hence, s = (2a+(n? — . =(2+(20—1)2) = (2+38)10 = 40 x 10 = 400. × |