Let a be the given number, m the ratio of the squares of its two parts, We shall then have, according to the enunciation, This may be resolved in two ways; we may either reduce it is a square, the numerator and denominator being each a square, we thence conclude at once, By resolving separately the two equations of the first degree comprehended in this formula, namely, By the first solution, the second part of the number proposed is 1 + √ m 1 + √ m are both, as the enunciation requires, less than the number pre Their signs being opposite, the number a is strictly no longer their sum, but their difference. If we make m = 1, that is, if we suppose that the squares of the two parts sought are equal, we have vm = 1; and the first solution will give two equal parts, a conclusion, that is self-evident, while the second solution gives for the results two infinite quantities (68), namely, This is necessary, for it is only by considering two quantities infinitely great, with respect to their difference a, that we can suppose the ratio of their squares equal to unity. Now, let there be the two quantities x, and x-a, the ratio of their squares will be dividing the two terms of this fraction by x2, we obtain but it is evident, that the greater the number x, the less will be 120. Now in order to compare the general method with that, which we have just employed, we develope the equation These values of x appear very different from those, which were found above; yet they may be reduced to the same; and in this consists the utility of the example, on which I am employed. It will serve to show the importance of those transformations, which different algebraic operations produce in the expression of quantities. We must first reduce the two fractions comprehended under the radical sign to a common denominator. This may be done by multiplying the two terms of the second by 1-m, we have then a2 m2 + a2 m (1 a2 m2 (1 — m) (1 — m) a2 m 1 m = + - m) The denominator being a square, it is only necessary to extract the root of the numerator, we then have It is evident that the square of a product is composed of the product of the squares of each of its factors, for example, bcdxbcd= b2 c2 d2, and consequently the root of b2 c2 d2 is simply the product of the roots b, c, and d, of the factors b2, c2, and d3. Applying this principle to the product a3 m, we see that its root is the product of a, the root of a2, by m, which denotes the root of √ a3 m = a√m. It follows from these different transformations, that m, or that These expressions, however simple, are still not the same as those given in the preceding article; if, moreover, we seek to verify them for the case, in which m=1, they become We find, in the second, the symbol of infinity, as in the preceding article, but the first presents this indeterminate form, %, of which we have already seen examples in articles 69 and 70; and before we pronounce upon its value, it is proper to examine, whether it does not belong to the case stated in art. 70; whether there is not some factor common to the numerator and denominator, which the supposition of m = 1 renders equal to zero. m a (√m — m) = 1- m It is here evident, that the numerator does not become 0, except by means of the factor m m; we must therefore examine, whether this last has not some factor in common with the denominator 1 m. In order to avoid the inconvenience, arising from the use of the radical sign, let us make m=n, then taking the squares, we have m=n2; the quantities, therefore, become but n―nn (1 — n), and 1 — n2 = (1 — n) (1 + n) (34); restoring to the place of n its value ✔m, we have a result the same, as that found in art. 119. In the same manner we may reduce the second value of x, observing that * The example, which I have given at some length, corresponds with a problem resolved by Clairaut, in his Algebra, the enunciation It will be seen without difficulty, that we might have avoided radical expressions in the preceding calculations, by taking m2 to represent the ratio, which the squares of the two parts of the proposed number have to each other; m would then have been the square root, which may always be considered as known, when the square is known; but we could not have perceived from the beginning the object of such a change in a given term, of which algebraists often avail themselves, in order to render calculations more simple. It is recommended to the learner, therefore, to go over the solution again, putting m3 in the place of m. Of the Extraction of the Square Root of Algebraic Quantities. 121. We have sufficiently illustrated, by the preceding example, the manner of conducting the solution of literal questions. We have given also an instance of a transformation, namely, that of am into am, which is worthy of particular attention; since, by means of it, we have been able to reduce the factors, contained under a radical sign, to the smallest number possible, and thus to simplify very much the extraction of the remaining part of the root. This transformation consists in taking the roots of all the factors which are squares, and writing them without the radical sign, as multipliers of the radical quantity, and retaining under the radical sign all those factors, which are not squares. This rule supposes, that the student is already able to determine, whether an algebraic quantity is a square, and is acquainted with the method of extracting the root of such a quantity. In order to this, it is necessary to distinguish simple quantities from polynomials. 122. It is evident, from the rule given for the exponents in of which is as follows; To find on the line, which joins any two luminous bodies, the point where these two bodies shine with equal light. I bave divested this problem of the physical circumstances, which are foreign to the object of this work, and which only divert the attention from the character of the algebraic expressions. These expressions are very remarkable in themselves, and for this reason I have developed them more fully, than they were done in the work referred to. |