the quotients above the divisors, in order to facilitate the process.

DEMONSTRATION OF THE PRECEDING RULE.

We have to prove two things; first, that the number 221, thus found, is necessarily the common divisor of both terms of the fraction, or that it must divide both exactly. Secondly; that it is their greatest common divisor.

I say first, that 221 is a common divisor, for we see that it divides 663 without remainder; and as it does also divide itself exactly, it must likewise divide 884, which, in the principles of division, is equal to 663 × 221; but 2431, being, according to the same principles, equal to 884 x2+663, it is plain, that as 221 divides exactly both 884 and 663, it will also divide 2431, which is a compound of both. A similar reasoning will shew, that 221, will divide, without remainder, 3315, which is equal to 2431 x884, which is divided exactly by 221.

I say now, that the two terms of the fraction, cannot have a greater common divisor than 221; for, it is evident, in the first place, that the greatest common divisor of the two terms of a fraction, cannot be greater than the numerator of that fraction, which is the smaller of those two terms. Now, the numerator, in this case, not being an exact divisor of the denomi nator, but having a remainder, I say, that the common divisor cannot be greater than this remainder; for we can write the fraction under this form: 2431

; now it is obvious, that the greatest

2431 × 884 common divisor of these two terms, cannot be greater than 884, for it must be such as to divide both 2431 and 884; and 884 cannot be divided by a number

greater than itself. Having then tried this number, it has been found to divide 2431 with a remainder 663, and 2 for a quotient; then, in the place of the two terms of the fraction, we may write : 884x2+663

884 x2+663 +884

; then we see, by a discourse

similar to the former, that 663 is the greatest common divisor that can be tried. It is therefore easy to convince one's self that the greatest common divisor that can be used, is the divisor by which a quotient is obtained without remainder; consequently 221, being in that case, is the greatest common divisor of the proposed fraction.

COMPLEX NUMBERS.

A number is called complex, which contains units of different kinds, but all reducible to the same kind. Thus 12£. 6s. 8d. is a complex number, because it can be all reduced to pence, its lowest denomination. 15 fathoms 4 feet 7 inches, is likewise a complex number, reducible to inches.

ADDITION OF COMPLEX NUMBERS.

The addition of complex numbers is effected on the same principles as that of incomplex ones, by placing the units of the same kind, the ones beneath the others, so as to form vertical columns, and the addition is begun by the lowest units. An example will make this obvious.

The sum of pence has been found to be 26; we have set down 2, and carried 2 to the column of shillings, (24 pence making two shillings.) Passing to the 2d CO

lumn, we have said; 2, brought forward, and 8, are 10, and 2 are 12; set down 2, and carry 1, which, joined to the tens of shillings, give 3 tens; and, as 2 tens of shillings make a pound, we have set down 1, worth one ten, to the column of shillings, and carried 1, worth one pound, to that of pounds, and then continued the operation as for incomplex numbers.

SUBTRACTION OF INCOMPLEX NUMBERS.

This operation needs only be exposed to be understood.

SUBTRACTION.
Fath. ft. Inc.

329 2 .9
235 5

7

Not being able to subtract 5 feet from 2 feet, we have had recourse to the superior units, and borrowed from the figure 9 one unit, worth 6 feet, which, Differ. 93 3 2 joined to 2, gives 8; and then, from 8 take 5, and remains; the rest as in common subtraction.

MULTIPLICATION OF COMPLEX NUMBERS.

This is one of the most complicated operations in arithmetic. It comprises many cases which can be resolved by various abridged methods, very convenient in practice but, as they all fall under the general method, we shall content ourselves with explaining it here, and making the application of it to the most difficult case.

Be it proposed to multiply 325£. 17s. 11d. by 659 fathoms 5 feet 10 inches.

It may be asked how it happens, that pounds are to be multiplied by fathoms, this proposition implying a seeming absurdity. It does indeed, but we must observe;

First; that in every multiplication, the multiplier is always to be considered as an abstract number, since it only serves to indicate the number of times the multiplicand is to be added up to itself; and, if it is used to designate particular units, it is only to know the ratio of the fractionary parts of that number with the principal unit; here the fractionary parts are 5 fest 10 inches.

E

Secondly, That the questions, which give occasion for complex multiplications, are almost all of the same nature, and can all be stated as in the annexed example, by saying; if a fathom, of a certain kind of work, has cost 325. 17s. 11d. how much will cost 659 fathoms 5 feet 10 inches? To answer this question, it is evident we must repeat the 1st number as many times as the 2nd contains fathoms and parts of fathoms, that is, multiply the first by the second. Let us proceed, now, to the explication of the operation.

I have multiplied first 325. by 659; then to multiply the 17s, by the same number, I have considered that 17s. was composed of 2, 14, and 1s, which I have written on the left of the partial products in one vertical column; the reason of this subdivision will soon be understood. It must always be made in this manner, that is to say; we must always begin by 2s. because this produce is obtained with great facility, nothing more being required for it, than doubling the figure of the units of the multiplier, wri ting them to the column of shillings, and carrying the other figures of the multiplier to the column of pounds.

Here, then, we have said; twice 9 are 18, which we have set down for shillings, and the 65 have been written in the column of pounds. This is the reason of that operation.

If I had pound to multiply by 659, the product would be 659£; but as I have only two shillings to multiply by 659, and as two shillings are the 10th part of 1 pound, I must then take the 10th part of 659, considered as a number of pounds; It is what we have done; for 659£. being the same as 650£.+9£. It is evident that writing only 65, we have taken the 10th of 650; and besides 2s. being the 10th of 1£. the 10th of 9£. will be 9 times 2s. or 18s. This, well understood, will greatly facilitate the rest..