How many values has ? What values are imaginary? Verify by substituting these values in the first equation. Scholium III. When a problem embraces several unknown quantities, it may be solved by representing these quantities in different ways; but the elegance and facility of the solution will frequently depend upon the notation which we employ. The following rule, taken from the Cambridge Mathematics, (application of Algebra to Geometry,) will be serviceable. 66 If among the quantities which would, when taken each (62) for the unknown quantity, serve to determine all the other quantities, there are two which would in the same way answer this purpose, and it could be foreseen that each would lead to the same equation (the signs + and excepted); then we ought to employ neither of these, but take for the unknown quantity one which depends equally upon both; that is, their half sum, or their half difference, or a mean proportional between them, or, &c." Thus, suppose it were required to find two numbers such that their sum should be 4 and the sum of their 4th powers 82. Instead of taking x and y for the numbers, we may make them both depend equally upon x by putting 2+x= greater No., since the sum would be 4 = greater No. + less No., and or (2+x) = 16+ 32 x + 24x2+8x3 + x1, (2x)=16-32x +24x28x3 + x1; = 2x+2.24x2+2.16 (2+x) + (2x)=82, x2+24x2 = 41-16=25; and greater No. = 2 + x = 3, 1, 2 + (-25)*, or 2 — (— 25)*, : − less No. 2 — = 1, 3, 2-(-25), or 2+ (-25), = where it must be observed that if 3 be taken for the value of the greater number, 1 must be taken for that of the less, and so on for the three remaining corresponding values. As a second example, suppose the equations and given to find x + y + (x2 + a3) * + (y2 + a2)* = b, any y. We shall make x and y depend equally upon the same unknown quantity by putting b dividing by a and putting- =n, to avoid fractions a '+1)2 = n − Z z2+2z+1+1+ +1+1+1+1 (2n+2)x+ ; + = n2 - 2nz 2 1 USE OF THE DOUBLE SIGN. z2-2mz+m2. = m2 → 1 = (m + 1) (m − 1), z = m ± [(m + 1) (m − 1)]3. 55 Scholium IV. From notions derived from the practice of arithmetic, the student, on first being introduced to problems capable of double solution, is frequently inclined to question the propriety of the second value of x; and, consequently, to reject or neglect the second or minus sign, in taking the square root of a quadratic equation. Such neglect must never be allowed; for the double sign (±) is necessary—and, so far from being any thing like an excrescence or deformity in algebraical language, is a symbol of the greatest utility, as will appear hereafter when we come to apply the foregoing principles to Geometry. The following problem, illustrative of the use of the double sign, will suffice for the present. Required the point in the line joining the centres of the earth and moon where their attractions are equal. Let E be the earth, M the moon and C the point Fig. 1. m of equal attraction, and let the quantity of matter inthe earth be represented by e and that in the moon by m, the radius of the moon's orbit, or distance from the earth, by r, the distance C M by u and consequently, C E by r-u. Now it is a principle of physics that the attraction of a sphere is proportional to the quantity of matter directly and to the square of the distance to its centre inversely, or But the mass of the moon, inferred from her action in raising the tides, is of that of the earth ;* or the distance of the point of equal attraction from the moon, is nearly of the moon's radius, or of moon's radius, where it remains to interpret the sign in the second answer. If we subtract from u (that is, from C lay off a line in the direction of M) a quantity less than u, the point of equal attraction will approach M, and the more, the greater the line subtracted, so that, when the line subtracted is CM, u will = 0, and when > CM, as CC', u will become, being = MC – CC' = MC − (CM+MC') = − MC'. Therefore if u = + • r, gives the point C on the left or this side, -⚫r corresponds to the point C' on the right or beyond the moon. Indeed, had we supposed the point of equal attraction at C' instead of at C, there would have resulted e = m u2 whence u = -1+8'66 + r, instead of u == • T. 7.66 7.66 The two values of u are therefore perfectly applicable, and, in fact, the problem would not be completely solved without them. The second value has revealed a truth, viz. that there is a second point of equal attraction beyond the moon-which, though not contemplated in the statement of the problem, is yet a direct consequence of the equation drawn from it; and which, when once discovered by the aid of algebraical symbols, recommends itself to the understanding without them, since the attraction of the earth is greater than that of the moon. Suppose, for a simple illustration, the masses of E and M were as 9 and 4, and their distances to C' as 3 9 and 2; then their attractions at C' would be as sequently equal. 32 4 and and con 22' As a last example in which a single equation is equivalent to two independent equations, we give the following important theorem: * Poisson, Traité de Mécanique, p. 258. CONSTANTS AND VARIABLES. 57 PROPOSITION III. Whenever a problem gives an equality between constant (63) and variable terms, the variables being capable of indefinite diminution so as to become less than any assignable quantity; two independent equations will be formed, one between the constants, the other between the variables. represent any equation in which A, B, are the sums of the constant, those of the variable terms in each member, X and Y, being capable of continually decreasing and becoming less than any assignable quantity, that is less than any assignable part of A or B. Then will A = B, and X = Y; for, if possible, let A be by some quantity D, or or greater or less than B; first suppose A > B A=B+D; X=Y-D, Y = D+ X, whence it appears that Y can never be less than D, even should X become actually =0; therefore Y is greater than, or at least as great as, an assignable quantity D, which is contrary to the hypothesis, and this contradiction holds as long as D has any value, or as long as A is > B. Again suppose or AB, or A+ D=B; X=D+Y; whence X is always greater than, or at least as great as D, and this is also contrary to the hypothesis, according to which X, as well as Y, is to be capable of indefinite diminution. Therefore it is absurd to suppose A either greater or less than B, or otherwise than equal to B; A B, and consequently X Y, = which was to be proved. = The method of proof here employed is called that of "reductio ad absurdum," leading to an absurdity; where, instead of showing directly that the proposition must be true, we show that it can not be otherwise than true. As an example, let it be required to find the sum of the repeating decimal fraction 333 &c. = '3+03+003+... [ad infinitum]. |