By Operation 6, a positive quantity is turned into a negative quantity, by simply subtracting the quantity from a column which has only zero upon it. (The sign at the top of [Pg 62] would become—during this process.) Operation 7 will be unintelligible, unless it be remembered that if we were calculating for instead of , Operation 6 would have completed the computation of itself; in which case the engine, instead of continuing its processes, would have to put on ; and then either to stop altogether, or to begin Operations 1, 2 ... 7 all over again for value of (= 2), in order to enter on the computation of ; (having however taken care, previous to this recommencement, to make the number on , equal to two, by the addition of unity to the former on that column). Now Operation 7 must either bring out a result equal to zero (if ); or a result greater than zero, as in the present case; and the engine follows the one or the other of the two courses just explained, contingently on the one or the other result of Operation 7. In order fully to perceive the necessity of this experimental operation, it is important to keep in mind what was pointed out, that we are not treating a perfectly isolated and independent computation, but one out of a series of antecedent and prospective computations. Cards 8, 9, 10 produce . In Operation 9 we see an example of an upper index which again becomes a value after having passed front preceding values to zero. has successively been , , , , ; and, from the nature of the office which , performs in the calculation, its index will continue to go through further changes of the same description, which, if examined, will be found to be regular and periodic. Card 12 has to perform the same office as Card 7 did in the preceding section; since, if had been = 2, the 11th operation would have completed the computation of . Cards 13 to 20 make . Since always consists of factors, has three factors; and it will be seen that Cards 13, 14, 15, 16 make the second of these factors, and then multiply it with the first; and that 17, 18, 19, 20 make the third factor, and then multiply this with the product of the two former factors. Card 23 has the office of Cards 11 and 7 to perform, since if were = 3, the 21st and 22nd operations would complete the computation of . As our case is , the computation will continue one more stage; and we must now direct attention to the fact, that in order to compute it is merely necessary precisely to repeat the group of Operations 13 to 20; and then, in order to complete the computation of , to repeat Operations 21, 22. [Pg 63]