Before proceeding farther, we write down the repeating part of a quotient, with the remainders which are left after the several figures are formed. Let the fraction be ¹/₁₇, we have 0₁₀5₁₅8₁₄8₄2₆3₉5₅2₁₆9₇4₂1₃1₁₃7₁₁6₈4₁₂7₁ This may be read thus: 10 by 17, quotient 0, remainder 10; 10² by 17, quotient 05, remainder 15; 10³ by 17, quotient 058, remainder 14; and so on. It thus appears that 10¹⁶ by 17 leaves a remainder 1, which is according to the theorem. If we multiply 0588, &c. by _any number under_ 17, the same cycle is obtained with a different beginning. Thus, if we multiply by 13, we have 7647058823529411 beginning with what comes after remainder 13 in the first number. If we multiply by 7, we have 4117, &c. The reason is obvious: ¹/₁₇ × 13, or ¹³/₁₇, when turned into a decimal fraction, starts with the divisor 130, and we proceed just as we do in forming ¹/₁₇, when within four figures of the close of the cycle. It will also be seen, that in the last half of the cycle the quotient figures are complements to 9 of those in the first half, and that the remainders are complements to 17. Thus, in 0₁₀5₁₅8₁₄8₄, &c. and 9₇4₂1₃1₁₃, &c. we see 0 + 9 = 9, 5 + 4 = 9, 8 + 1 = 9, &c., and 10 + 7 = 17, 15 + 2 = 17, 14 + 3 = 17, &c. We may shew the necessity of this as follows: If the remainder 1 never occur till we come to use _a_ᵇ⁻¹, then, _b_ being prime, _b_-1 is even; let it be 2_k_. Accordingly, _a_²ᵏ-1 is divisible by _b_; but this is the product of _a_ᵏ-1 and _a_ᵏ + 1, one of which must be divisible by _b_. It cannot be _a_ᵏ-1, for then a power of _a_ preceding the (_b_-1)th would leave remainder 1, which is not the case in our instance: it must then be _a_ᵏ + 1, so that _a_ᵏ divided by _b_ leaves a remainder _b_-1; and the _k_th step concludes the first half of the process. Accordingly, in our instance, we see, _b_ being 17 and _a_ being 10, that remainder 16 occurs at the 8th step of the process. At the next step, the remainder is that yielded by 10(_b_-1), or 9_b_ + _b_-10, which gives the remainder _b_-10. But the first remainder of all was 10, and 10 + (_b_-10) = _b_. If ever this complemental character occur in any step, it must continue, which we shew as follows: Let _r_ be a remainder, and _b_-_r_ a subsequent remainder, the sum being _b_. At the next step after the first remainder, we divide 10_r_ by _b_, and, at the next step after the second remainder, we divide 10_b_-10_r_ by _b_. Now, since the sum of 10_r_ and 10_b_-10_r_ is divisible by _b_, the two remainders from these new steps must be such as added together will give _b_, and so on; and the _quotients_ added together must give 9, for the sum of the remainders 10_r_ and 10_b_-10_r_ yields a quotient 10, of which the two remainders give 1. If ¹/₅₉ and ¹/₆₁ be taken, the repeating parts will be found to contain 58 and 60 figures. Of these we write down only the first halves, as the reader may supply the rest by the complemental property just given.