Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

much as possible, we should see if the dividend and divisor have not common factors, which may be cancelled (38). But when the terms of the fraction are polynomials, the common factors are not so easily found as when they are simple quantities. They are in general to be sought by a method analogous to that, which is given in arithmetic for finding the greatest common divisor of two numbers.

We cannot assign the relative magnitudes of algebraic expressions, as we do not give values to the letters which they contain; the denomination of greatest common divisor therefore, applied to these expressions, ought not to be taken altogether in the same sense as in arithmetic.

In algebra we are to understand by the greatest common divisor of two expressions, that which contains the most factors in all its terms, or which is of the highest degree (27). Its determination rests, as in arithmetic, upon this principle; Every common divisor to two quantities must divide the remainder after their division.

The demonstration given in arithmetic (art. 61) is rendered clearer by employing algebraic symbols. If we represent the common divisor by D, the two quantities proposed might be expressed by the products A D and B D, formed from the common divisor and the factor by which it is multiplied in each of the quantities. This being supposed, if Q stands for the entire quotient, and R for the remainder resulting from the division of AD by B D, we have A D = BD × Q + R (Arith. 61); dividing now the two members of the equation by D, we obtain


A =BQ + // ;
D

and since the first member, which in this case must be composed of the same terms, as the second, is entire, it must follow, that

R

D

is reduced to an expression without a divisor, that is to say, that R is divisible by D.

According to this principle, we begin, as in arithmetic, by inquiring whether one of the quantities is not itself the divisor of the other; if the division cannot be exactly performed, we divide the first divisor by the remainder, and so on; and that remainder, which will exactly divide the preceding, will be the greatest common divisor of the two quantites proposed. But it will be necessary in

the divisions indicated, to have regard to what belongs to the nature of algebraic quantities.

We are not, in the first place, to seek a common divisor of two algebraic quantities, except when they have common letters; and we must select from them a letter, with reference to which the proposed expressions are to be arranged, and that is to be taken for the dividend in which this letter has the highest exponent, the other being the divisor.

[merged small][ocr errors][merged small]

which are already arranged with reference to the letter a; we take the first for the dividend and the second for the divisor. A difficulty immediately presents itself, which we do not meet with in numbers, and this is, that the first term of the divisor will not exactly divide the first term of the dividend, on account of the factors 4 and b in the one, which are not in the other. But the letter b being common to all the terms of the divisor and not to those of the dividend, it follows (40) that b is a factor of the divisor, and that it is not of the dividend. Now every divisor common to two quantities can consist only of factors which are common to the one and to the other; if then there be such a divisor with respect to the two quantities proposed, it is to be looked for among the factors of the quantity 4 a2- 5 a b + b2, which remains of the quantity 4 a2 b-5ab2+b3, after suppressing b; so that the question reduces itself to finding the greatest common divisor of the two quantities

Sassa2b+ab2 — b3,
4a25ab+b2.

For the same reason that we may cancel in one of the proposed quantities the factor b which is not in the other, we may likewise introduce into this a new factor, provided it is not a factor of the first. By this step, the greatest common divisor, which can consist only of terms common to both, will not be affected. Availing myself of this principle, I multiply the quantity Sa2 b + a b2 -b3 by 4, which is not a factor of the quantity 4 a2-5 ab+b2, in order to render the first term of the one divisible by the first term of the other. I shall thus have for the dividend the quantity 12 a2 b+4 ab2 — 4 b3,

3 as

12 as

[ocr errors]

for the divisor the quantity

4a25ab+b2,

and the quotient will be sa.

Multiplying the divisor by this quotient, and subtracting the product from the dividend, I have for a remainder

[blocks in formation]

a quantity which, according to the principle stated at the commencement of this article, must have with 4 a2 — 5 ab+b2, the same greatest common divisor as the first.

Profiting by the remarks made above, I suppress the factor b, common to all the terms of this remainder, and multiply it by 4 in order to render the first term divisible by that of the divisor; I have then for a dividend the quantity

12 a2+4 ab- 16 b2,

and for a divisor the quantity

4a2-5 ab+b2;

and the quotient thence arising is 3.

Multiplying the divisor by the quotient, and subtracting the product from the dividend, we obtain the remainder

19 ab 1962,

and the question is reduced to finding the greatest common divisor to this quantity, and

[merged small][ocr errors]

But the letter a, with reference to which the division is made, not being in the remainder, except of the first degree, while it is of the second degree in the divisor, it is this which must be taken for the dividend, and the remainder must be made the divisor. Before beginning this new division I expunge from the divisor 19 ab 1962, the factor 19 b common to both the terms, and which is not a factor of the dividend; I have then for a dividend the quantity

and for a divisor

4a25ab+b2,

a-b.

The division leaves no remainder; so that ab is the greatest common divisor required.

By retracing these steps, we may prove à posteriori, that the quantity a—b must exactly divide the two quantities proposed, and that it is the most compounded of those which will do it. In dividing by ab the two quantities proposed

3 a3 3a2 b+ab2 —b3, 4 a2 b — 5 ab2+b3,

[ocr errors]

we resolve them as follows;

(Sa2+b2) (a—b), (4 ab-b2) (a—b).

49. When the quantity, which we take for a divisor, contains several terms having the letter, with reference to which the arrangement is made, of the same degree, there are precautions to be used, without which the operation would not terminate. See an example of this.

Let there be the quantities

a2 b + ac2 —d3, ab—ac+d2;

if we make the preparation as for common division

a2 b + ac2 — d3

—a2 ba2c-ad2

Rem. a2c+a c2-a d2-d3,

ab-ac+d2

a

by dividing first a2 b by a b, we have for the quotient a; multiplying the divisor by this quotient, and subtracting the products from the dividend, the remainder will contain a new term, in which a will be of the second degree, namely, a2 c, arising from the product of ac by a. Thus no progress has been made; for by taking the remainder

a2c+ ac2 a d2 — d3

-

for a dividend, and multipling by b to render the division possible by ab, we have

a2bc abc2abd2bd3

+

[blocks in formation]

rem. a2 c2+abc2acd2-abd2-bd3,

[blocks in formation]

a b―ac + d2

ac

a c produces still a term a2 c2, in which a is of

the second degree.

To avoid this inconvenience, it must be observed, that the divisor a b—a c + d2 = a (b —c) + d2, by uniting the terms ab-acin one; and for the sake of shortening the operation, making b―c=m, we have for a divisor a m+d2; but then the whole dividend must be multiplied by the factor m, to make a new dividend, the first term of which may be divided by am, the first term of the divisor; the operation then becomes

[blocks in formation]

a2 bm + ac2 m
a2 bm abd2

[blocks in formation]
[blocks in formation]

- d3 m

-abd2 c2 d2 - d3 m

The terms involving a2 now disappear from the dividend, and there remain only the terms which have the first power of a. To make these disappear we first divide the term ac2 m by am, and it gives for a quotient c2; multiplying the divisor by this quotient, and subtracting the products from the dividend, we obtain the second remainder. Taking this second remainder for a new dividend, and suppressing the factor d2, which is not a factor of the divisor, we have

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

The remainder b d-cam-dm3 of this last division, not involving a, it follows that if the proposed quantities have a common divisor, it is independent of the letter a.

c2 m

Having arrived at this point, we can continue the division no longer with reference to the letter a; but it will be observed, that if there be a common divisor, independent of a, to the quantities bl2 d m3 and a m + d2, it must divide separately the two parts a m and d of the divisor; for if a quantity is arranged with reference to the powers of the letter a, every divisor of this quantity, independent of a, must divide separately the quantities multiplied by the different powers of this letter.

To be convinced of this we need only observe, that in this case, each of the quantities proposed must be the product of a quantity depending on a, and of the common divisor which does not depend upon it. Now if we have, for example, the expression Aa+Ba3 + Ca2 + Da +E,

in which the letters A, B, C, D, E, designate any quantities whatever, independent of a, and it be multiplied by a quantity also independent of a, the product

MAa+MB a3 + MCa2 + MDa+ME,

arranged with reference to a, will contain still the same powers of a as before; but the coefficient of each of these powers will be a multiplier of M.

This being supposed, if we restore the quantity (b—c) in the place of m, we have the quantities

« ΠροηγούμενηΣυνέχεια »