Elements of Algebra

174 ELEMENTS SECT. III. $\frac{1}{8} = 0.125$; $\frac{1}{4} = 0.25$; $\frac{3}{8} = 0.375$; $\frac{4}{8} = 0.5$; $\frac{5}{8} = 0.625$; $\frac{6}{8} = 0.75$; $\frac{7}{8} = 0.875$, \ldots 535. If the denominator be 9, we have $\frac{1}{9} = 0.111, \ldots$ $\frac{2}{9} = 0.222, \ldots$ $\frac{3}{9} = 0.333, \ldots$ And if the denominator be 10, we have $\frac{1}{10} = 0.1$, $\frac{2}{10} = 0.2$, $\frac{3}{10} = 0.3$. This is evident from the nature of decimals, as also that $\frac{1}{100} = 0.01$; $\frac{37}{100} = 0.37$; $\frac{256}{1000} = 0.256$; $\frac{24}{10000} = 0.0024, \ldots$ 536. If 11 be the denominator of the given fraction, we shall have $\frac{1}{11} = 0.0909090, \ldots$ Now, suppose it were required to find the value of this decimal fraction: let us call it $s$, and we shall have $s = 0.090909,$ $10s = 00.909090,$ $100s = 9.09090.$ If, therefore, we subtract from the last the value of $s$, we shall have $99s = 9$, and consequently $s = \frac{9}{99} = \frac{1}{11}$: thus, also, $\frac{2}{11} = 0.181818, \ldots$ $\frac{3}{11} = 0.272727, \ldots$ $\frac{6}{11} = 0.545454, \ldots$ 537. There are a great number of decimal fractions, therefore, in which one, two, or more figures constantly recur, and which continue thus to infinity. Such fractions are curious, and we shall shew how their values may be easily found *. * These recurring decimals furnish many interesting researches; I had entered upon them, before I saw the present Algebra, and should perhaps have prosecuted my inquiry, had I not likewise found a Memoir in the *Philosophical Transactions* for 1769, entitled *The Theory of Circulating Fractions*. I shall content myself with stating here the reasoning with which I began. Let $\frac{n}{d}$ be any real fraction irreducible to lower terms. And suppose it were required to find how many decimal places we must reduce it to, before the same terms will return again. In order to determine this, I begin by supposing that $10n$ is greater than $d$; if that were not the case, and only $100n$ or $1000n>d$, it would be necessary to begin with trying to reduce $\frac{10n}{d}$ or $\frac{100n}{d}$, \ldots to less terms, or to a fraction $\frac{n^1}{d^1}$. This being established, I say that the same period can return only when the same remainder $n$ returns in the continual division.