Elements of Algebra

90 ELEMENTS SECT. II. then, $1-a)a^3 * (a^3 + \frac{a^4}{1-a})$ $\frac{a^3-a^4}{a^4}$ and again, $1-a)a^4 * (a^4 + \frac{a^5}{1-a})$ $\frac{a^4-a^5}{a^5}$, \ldots 291. This shews that the fraction $\frac{1}{1-a}$ may be exhibited under all the following forms: I. $1 + \frac{a}{1-a}$. II. $1 + a + \frac{a^2}{1-a}$; III. $1 + a + a^2 + \frac{a^3}{1-a}$. IV. $1 + a + a^2 + a^3 + \frac{a^4}{1-a}$; V. $1 + a + a^2 + a^3 + a^4 + \frac{a^5}{1-a}$, \ldots Now, by considering the first of these expressions, which is $1 + \frac{a}{1-a}$, and remembering that $1$ is the same as $\frac{1-a}{1-a}$, we have $1 + \frac{a}{1-a} = \frac{1-a}{1-a} + \frac{a}{1-a} = \frac{1-a+a}{1-a} = \frac{1}{1-a}$. If we follow the same process, with regard to the second expression, $1 + a + \frac{a^2}{1-a}$, that is to say, if we reduce the integral part $1+a$ to the same denominator, $1-a$, we shall have $\frac{1-a^2}{1-a}$, to which if we add $+\frac{a^2}{1-a}$, we shall have $\frac{1-a^2+a^2}{1-a}$, that is to say, $\frac{1}{1-a}$. In the third expression, $1 + a + a^2 + \frac{a^3}{1-a}$, the integers reduced to the denominator $1-a$ make $\frac{1-a^3}{1-a}$; and if we add to that the fraction $\frac{a^3}{1-a}$, we have $\frac{1}{1-a}$, as before; - therefore all these expressions are equal in value to $\frac{1}{1-a}$, the proposed fraction.