6. What vulgar fraction is equivalent to 0.0123456789 ? CASE III. RULE. Secondly, find the vulgar fraction, which is equivalent to the circulating part of the decimal, by rule under Case II, of this Article; to the denominator of this fraction annex as many ciphers as there are decimals which precede the circulating part of the repetend; then add these two fractions together. Examples. 1. What vulgar fraction is equivalent to the compound repetend 0.343 ? . Ans. So todo=8=!%. 2. What vulgar fraction is equivalent to the compound repetend 0.087837 ? Ans. 3. What vulgar, fraction is equivalent to 0.083? Ans. th 4. What vulgar fraction is equivalent to the compound repetend 0.0357142857 ? Ans. zł. 5. What vulgar fraction is equivalent to the compound repetend 0.0714285? 6. What vulgar fraction is equivalent to 0.123456 ? CHAPTER IV. CONTINUED FRACTIONS. 55. If we divide both numerator and denominator of the fraction it by the numerator, we obtain, 351 1 351. Again, performing the same operation upon the fraction 2013, we find 263 1 ' 3511 +88 263; this value of Isi substituted in I. we get, 35 13 II. 351 III. 351 1 965=2+1 1+1 87 1. 87; this value substituted in III. we finally obtain Such fractions as the above are called continued fractions. In the last example, the parts 1, 1, 2, &c. are called the first, second, third, &c. partial fractions. If we seek for the greatest common measure of the numerator and denominator of the first fraction 35}, by the rule un. der Art. 9, we shall obtain, OPERATION. 702 263 176 1)87(87 87 .. . 0 Here we discover that the successive quotients are the same as the successive denominators of the partial fractions, which compose the continued fraction already drawn from Hence, to convert a vulgar fraction into a continued frac. tion, we have this RULE. Seek, by rule under Art. 9, the greatest common measure of the numerator and denominator of the given fraction; the re. ciprocals of the successive quotients will form the partial fractions which constitute the continued fraction required. Examples. OPERATION. 753 The partial fractions are g, as, 1, 2, , therefore we shall have 251_1 764 3+ 1 22+1 - 1+1 4+1 |