Elements of Algebra

434 ELEMENTS PART II. $b/c = 3/2$, the second $d/e$ must be nearly equal to it; if, therefore, we make $d/e = 3/2$, we shall have the difference $z = 1/2$. We may also determine this second fraction by means of the first, generally; for since $3/2 - d/e = (3e-2d)/(2e)$, we must have $3e - 2d = 1$, and, consequently, $2d = 3e - 1$, and $d = e + (e-1)/2$. So that making $(e-1)/2 = m$, or $e = 2m + 1$, we shall have $d = 3m + 1$, and our second fraction will be $d/e = (3m+1)/(2m+1)$. In the same manner, we may determine the second fraction for any first whatever, as in the following Table of examples: $b/c$ | $3/2$ | $5/3$ | $7/5$ | $8/5$ | $11/4$ | $13/8$ | $17/7$ ---|---|---|---|---|---|---|--- $d/e$ | $(3m+1)/(2m+1)$ | $(5m+1)/(3m+1)$ | $(7m+2)/(5m+1)$ | $(8m+3)/(5m+2)$ | $(11m+3)/(4m+1)$ | $(13m+5)/(8m+3)$ | $(17m+5)/(7m+2)$ 4. When we have determined, in the manner required, the two fractions, $b/c$, and $d/e$, it will be easy to find a third also analogous to these. We have only to suppose $f=b+d$, and $g = c + e$, so that $f/g = (b+d)/(c+e)$; for the two first giving $be - cd = \pm 1$, we have $f/g - b/c = \pm 1/(c^2+ce)$; and subtracting likewise the second from the third, we shall have $f/g - d/e = (be-cd)/(e^2+ce) = \pm 1/(ce+e^2)$. 5. After having determined in this manner the three fractions, $b/c$, $d/e$, and $f/g$, it will be easy to resolve our question for three numbers, $x$, $y$, and $z$, by making the three formulæ $xy + a$, $xz + a$, and $yz + a$, become squares: since we have only to make $x = b^2 - ac^2$, $y = d^2 - ae^2$, and $z = f^2 - ag^2$. For example, in the foregoing Table,