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the 2nd divisor, or the 2nd quotient. Also there will remain over from the division of each heap (A) the same number or 2nd remainder; therefore from the division of the whole of the counters there will remain a number equal to the product of the 2nd remainder and ist divisor together with the lst remainder. Now assuming the two remainders to be the greatest possible, viz. less by one than the respective divisors, the product of the 2nd remainder and Ist divisor will be less by the 1st divisor than the product of the two divisors; therefore adding the 1st remainder, we find the total remainder to be less by one at least than the product of the divisors. Hence the number of counters in each heap (B) is the greatest possible number that can be obtained by dividing the whole number into equal heaps, in number equal to the product of the two divisors. In other words the number of counters in each heap (B) represents the quotient from the division by the product of the two divisors. Hence to divide by a number composed of two factors, we may divide by each factor in succession, and the total remainder will be found by adding the 1st remainder to the product of the 1st divisor and the 2nd remainder.

If the divisor be composed of more than two factors, they may be reduced to two, and the quotient obtained by successive division by each of these. But the division by each of these may be performed in like manner by dividing by each of two factors, of which it may be composed; and the same may be said of any divisor, which is composite. Hence to divide by any composite number, we may divide by each factor in succession.

Also for the formation of the remainder, we have to consider that the result of any number of divisions is (as has been proved) the same as the result from division by a number equal to the product of the divisors, hence the final result is the same, as that from division by the last divisor, and by the product of all the rest. Therefore (by the first part of the Prop.) the remainder after 3 divisions is equal to the product of the 3rd remainder and the first 2 divisiors together with the remainder after two divisions. Similarly the remainder after 4 divisions is equal to the product of the 4th remainder and the first 3 divisors together with the remainder after 3 divisions. Thus the remainder after any number of divisions is evidently to be obtained by multiplying each remainder into the product of all the previous divisors, adding the products and the first remainder, which is the rule.

The same exhibited algebraically :-Let a 1, (l2, as, &c. be the successive divisors; ru, ru, r3, &c. the successive remainders from each division; R1, R2, Rz, &c. the total remainders after one, two, three, &c. divisions: then R, = ; R, = r2 X 21, +ri

R2 =r3 X az X ai + R2

=rg X a, X a, tra X a, tri &c. = &c.

N

Prop. 16.To prove the Rule for Division by any number

having ciphers on the right. Since units, tens, &c. on multiplication by 10, become respectively tens, hundreds, &c. therefore conversely tens, hundreds, &c. when divided by 10, become respectively units, tens, &c. Now if we take away the figure on the right of any number, so as to make the tens' figure occupy the units' place, &c. we thus convert the tens, hundreds, &c. into units, tens, &c. i.e. we effect the same result as by the division by 10. Hence to divide by 10 we have only to take away the figure on the right; the figures that then remain will form the quotient, and that taken away will be the remainder. In a similar manner it may be shewn that the division by 100, 1000, &c. may be performed by taking away from the right hand as many figures as there are O's in the divisor. Hence, and by the previous Prop. the division by any number having ciphers on the right may be performed by taking away (or marking off) as many figures on the right as there are O's in the divisor, and dividing the remaining figures of the dividend by the remaining figures of the divisor, after throwing away the ciphers. Thus to divide by 25000, mark off 3 figures from the right of the dividend (which is in fact dividing by 1000), and divide the remaining figures by 25. The remainder will then be obtained by affixing the ciphers taken away to the remainder from the division, and adding the figures cut off from the dividend. Prop. 17.Every factor of a number is a measure of the

same,

and

every measure is a factor. Since factors are numbers, which by their product compose a given number, therefore every factor, or the product of any number of co-factors, will divide the given number without remainder, the quotient being the product of the other co-factors. Therefore every factor of a number is also a measure of the same. Conversely, every measure of a number is a factor of the same. For the measure multiplied by the number of times it is contained in the given number becomes equal to the given number. Therefore the measure is one of those numbers, which by their product compose the given number, and is consequently a factor.

The same Prop. exhibited algebraically ;-Let N be the given number, a, b, c. &c. its factors. Then N=aXbXcX &c. = a X (6 XCX &c.) = (a Xb) X (C X &c.) .:: N contains a a number of times equal to (b XCX &c.) and

N contains a x b a number of times equal to (c X &c.); &c. &c. .. a, a X b, &c. are measures of N. Again let n measure N, or be contained in N, a times,

then N=nxa .. n is a factor of N.

Prop. 18.--If one number measure another, the factors of

the first are factors also of the second, The second may be obtained by multiplying by the first the number of times it is contained in the second. But if the first be a composite number, this Multiplication may be effected by multiplying by each of its factors in succession. Therefore each of these factors being one of the numbers, which by their product form the second, is a factor of the second.

The same Prop. exhibited algebraically:-Let the number n measure N; or let N contain n, a times; let n=bXCX &c.: then

N= a Xn=a Xib XCX &c.) =a X6 XCX &c. .: b, c, &c. are numbers which, together with a, by their product form N. i. b, c,

&c. are factors of N.

Prop. 19.-If one number measure another, the first mea

sures also every multiple of the second. Since the second number is a factor of its multiple, it is therefore a measure of the same (Prop. 17); consequently every factor of the second is a factor of its multiple (Prop. 18); but every measure of the second is a factor of it (Prop. 17), therefore every measure of the second is a factor and measure of every multiple of the second.

The same Prop. exhibited algebraically :-Let n measure or be contained in N, a times; so that N =n Xa; then

N Xm=(n Xa) Xm=n X (a X m).
Hence N X m contains n, (a X m) times, or n measures N X m.

Prop. 20.--If one number measure each of two others, it

measures also their sum or difference. Since the quotient of a sum may be obtained by dividing each part and adding the quotients, therefore if we divide each of two numbers by another, and add the quotients, we shall obtain the quotient of their sum. Now the quotient of each of the two given numbers divided by their measure is an exact quotient, therefore the sum of these quotients, i.e. the quotient of the sum of the two numbers, is also an exact quotient. Hence the given measure of the two numbers measures also their sum.

Again: since the larger of two numbers is equal to the sum of the smaller and their difference, therefore the quotient of the larger divided by any number, is equal to the sum of the quotients of the smaller, and their difference, divided by the same number. But if the divisor be the given measure the first and second of these quotients are exact, therefore the last

must also be ; for no whole number can be equal to the sum of a whole number, and a fraction. Hence the difference of the two numbers is measured by any measure of the two.

The same Prop. exhibited algebraically :-Let Y and M be the two numbers, m a common measure of them : let M=m X a, N=mXb: then

M+N=m Xa+mXb=mX (a + b)

.. m is contained in M + N, (a + b) times. Again, if M be greater than N, then

M = (M — N) + N
.. M: m =(M — N) = m+N;m

a = (M — N); mtb

- b = (M-- N); m
hence m is contained in M -- N, (a - b) times.

or

Prop. 21.-- To prove that the G.C.M. of several numbers

is the product of all the common prime factors. Since all the factors of any measure of a number are factors of the number itself, therefore all the factors of every common measure, and of the G.C.M. are factors of all the numbers. Hence all the prime factors of the G.C.M. are prime factors of all the numbers. Again since every factor, or the product of any number of co-factors, of a number is a measure of it, therefore the product of all the common prime factors is a common measure of all the numbers. It is also the G.C.M. For all the factors of the G.C.M. are among these factors, therefore the G.C.M. cannot contain more than these factors, and evidently it cannot contain less than all of them, for if it could there would be a common measure greater than the G.C.M. viz. the product of all the prime factors, which is impossible. Hence the G.C.M. is the product of all the prime factors common to all the numbers.

Cor. 1. Hence every common measure of several numbers is a measure of the G.C.M. For all the prime factors of every common measure are prime factors of all the numbers; but the G.C.M. contains all these prime factors, therefore the G.C.M. is a multiple of the product of any number of them; or the product of any nnmber of thein, i.e. any common ineasure of the numbers, is a measure of the G.C.M.

Also every nieasure of the G.C.M. being the product of sume of the common primefactors of the numbers, is a common measure of them.

Cor. 2. If any of the numbers be measured by any of the others, they may be omitted in the formation of the G.C.M. For evidently no more common factors can be obtained from the multiple than from the measure; and any factor which is common to the measures, and the others, is common also to the multiples.

Prop. 22.- To prove and explain the Rule for finding the

G.C.M. of two numbers, when their prime factors are

not easily obtainable. Since the G.C.M. cannot be greater than the less of the two numbers, and may be equal to it, therefore it is first ascertained whether the less be a measure of the greater, which is done by dividing the greater by the less. If there be no remainder, the G.C.M. has been found to be equal to the less. But if there be a remainder, it is considered as follows:-Every common measure of divisor and dividend is a measure of the remainder, and therefore a common measure of divisor and remainder. For every common measure of divisor and dividend measures also the product of quotient and divisor (Prop. 19), and the dividend, and therefore measures the difference of these, or the remainder (Prop. 20); and hence is a common measure of divisor and remainder.

Again, every common measure of divisor and remainder is a measure of the dividend, and therefore a common measure of the divisor and dividend. For every common measure of divisor and remainder measures the prodnct of quotient and divisor (Prop. 19), and the remainder, and therefore measures the sum of these, or the dividend (Prop. 20), and hence is a common measure of divisor and dividend.

Hence, every common measure of divisor and dividend being a common measure of divisor and remainder, and vice versa, it follows that the G.C.M. of divisor and dividend is a common measure of the divisor and remainder, and cannot be greater than their G.C.M., else the divisor and remainder would be measured by a number greater than their G.C. M. which is absurd : nor can it be less, else there would be a common measure greater than the greatest, viz. the G.C.M. of divisor and remainder. If then we find the G.C.M. of the less and remainder, we shall have found the G.C.M. of the two given numbers. The first step in this investigation is the same as in the first, viz. to ascertain by division whether the less be the measure of the greater. If it be, the G.C.M. is found, but if not, the same considerations, as before, will shew that the G.C.M. is the same as that of the 1st and 2nd remainders, to find which has now to be tried in the same manner as the others.

From this it is concluded that if the greater of the two numbers be divided by the less, and the 1st divisor by the 1st remainder, the 1st remainder by the 2nd remainder, and so on, each remainder in its turn becoming a divisor of the previous divisor, and the division being con-. tinued till there is no remainder, then the G.C.M. of the two numbers will be that of the 1st divisor and Ist remainder, or that of 1st and 2nd remainders, or that of 2nd and 3rd, &c. or that of the last but one, and the

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