should it happen that the set 1, r , s , t , &c. is not complete, then m , mr , ms , &c. may give a different set of remainders.
All these last theorems are constantly verified in the process for reducing a fraction to a decimal fraction. If m be prime to b, or the fraction m/b in its lowest terms, the process involves the successive division of m, m × 10, m × 10², &c. by b. This process can never come to an end unless some power of 10, say 10ⁿ, is divisible by b; which cannot be, if b contain any prime factors except 2 and 5. In every other case the quotient repeats itself, the repeating part sometimes commencing from the first figure, sometimes from a later
figure. Thus, ¹/₇ yields ·142857142857, &c., but ¹/₁₄ gives ·07(142857)(142857), &c., and ¹/₂₈ gives ·03(571428)(571428), &c.
In m/b, the quotient always repeats from the very beginning whenever b is a prime number and m is less than b; and the number of figures in the repeating part is then always b-1, or a measure of it. That it must be so, appears from the above propositions.
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