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b=kd, then d measures both a and b; hence a±b=hd±k'd=d(h±k), and therefore d measures both a+b and a-b, the quotient being h+k in the former case, and h—k in the latter; and by lemma 1, d measures any multiples of a+b and a-b.

Now, let a and b be two polynomials, or the terms of a fraction, and let a divided by b leave a remainder c

b

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...d

d leave no remainder, as is shown

in the marginal scheme. Then we have, by the nature of division, these six equalities, viz.:

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b) a (m

m b
c) b (n

nc

d) c (p

p d

Where the equalities marked (4), (5), (6), are not deduced from those marked (1), (2), (3), but from the consideration that the dividend is always equal to the product of the divisor and quotient, increased by the remainder.

Now, by (6) it is obvious that d measures c, since c=p d; hence (Lemma 1) d measures n c, and it likewise measures itself; therefore (Lemma 2) d measures nc+d, which by (5) is equal to b; hence again d, measuring b and c, measures mb+c, by the Lemmas 1 and 2.

.. d measures a, which is equal to m b+c by (4).

Hence d measures both the polynomials a and b, and is consequently a common measure of these polynomials; but d is also the greatest common measure of a and b; for if d' is a greater common measure of a and b than d is, it is obvious that by (1) d' measures a-m b, or c; and d' measuring both b and c, it measures b—n c, or d by (2); hence d' measures d, which is absurd, since no quantity measures a quantity less than itself; therefore d is the greatest common measure.

Q. E. D.

Again, let a=h a' and b=h b'; then the greatest common measure of a and b will be h d', where d' is the greatest common measure of a' and b'. For let a' = d'm, and b' = d'en; then a ha'h d'm, and b = hb' = h d'n; then m and n contain no common factor, for d' is the greatest common measure of a' and b'; hence h d' is obviously the greatest common measure of hd' m and h d'n, or of a and b; and this proves the truth of Note 2.

b) a (m
m b

Moreover, if any divisor contains a factor, which is not a factor also of the corresponding dividend, it must be rejected before commencing the division. For as in the marginal scheme, let c contain a factor h, which is not a factor also of a and b; then rejecting h, the remaining factor c' is employed as a new divisor, instead of c, and For, by the nature of division, we have

so on.

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c=h c c) b (n

n c'

d=k d' d') c (p P d'

Now, by (6) we see that d' is a measure of c'; hence d', measuring n c' and kď, it also measures n c'+k d', or b; therefore d', measuring d' and b, measures m b+hc', or a; hence d' measures both a and b, and it is also the greatest common measure. For if d be a greater common measure than d', then d, measuring both a and b, measures a—m b, or hc' by (1); but d does not measure h, and it must therefore measure c'; hence d measures n c' and b; therefore it measures b―n d', or k ď' by (2); but again, d does not measure k, and hence it must measure d'; but d is greater than d', and cannot therefore measure it; hence d' is the greatest common measure. This is the proof of note 3, and in very nearly the same manner it is proved, that if any dividend be multiplied by a factor, which is not a factor of the greatest common measure, in order to make the leading term of the dividend divisible by that of the divisor, the final divisor, or resulting greatest common measure, will remain unchanged.

Thus it appears that, to avoid the difficulty of operating with fractional quotients, we can always remove from the divisor, or introduce into the dividend any factor which may obstruct the exact division of the leading coefficient of the dividend by that of the divisor. These remarks will be fully exemplified in the subsequent examples; and as the process for finding the greatest common measure of any two polynomials is now very important, being employed in the general solution of equations, we have endeavoured to explain the reasons of the several steps in the process, with perspicuity and clearness, as far as our limits will permit.

30. If the greatest common measure of three quantities be required, find the greatest common measure of two of them, and then that of this measure and the remaining quantity will be the greatest common measure of all three. For let a, b, c, be the quantities, and let d be the greatest common measure of a and b, and d' the greatest common measure of c and d; then any measure of d will evidently measure a and b, and whatever measures c and d will also measure a, b, c; hence the greatest common measure of c and d is also the greatest common measure of a, b, c, and therefore d' is the greatest common measure of a, b, c. This reasoning may be extended to any number of quantities.

31. If the two polynomials be the terms of a fraction, as, and d their greatest common measure, then we may put a=da', and b= db'; hence a' and consequently, by dividing both numerator and denominator

=

da'

b db

=

b

of a fraction by the greatest common measure of the terms of the fraction, the resulting fraction will be simplified to its utmost extent, and thus the proposed fraction will be reduced to its lowest terms.

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EXAMPLES.

(1.) What is the greatest common measure of 4x2 y3 z1 and 8x1 y3 z2? Here 4 is the greatest common measure of 4 and 8, and a2 y3 z2 is that of the literal parts; hence 4x2 y3 22 is the greatest common measure required.

H

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Hence x+y is the greatest common measure sought, and

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(x+y3)+(x+y) x2-xy + y2

(x2-y2)+(x+y)

=

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= reduced fraction.

(3.) Required the greatest common measure of the two polynomials
6a3— 6a2y+2ay2—2y3 . (a)
12a2-15ay +3y2

....

(b).

Here 6a36a2y+2ay2—2y3 = 2 (3a3-3a2y+ay2—y3)

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And therefore, by suppressing the factors 2 and 3, which have no common measure, we have to find the greatest common measure of

3a3—3a1y+ay2—y3 and 4a2—5ay+y2.

4a2-5ay+y2) 3a3— 3a2y+ ay3 — y3

4

12a3-12a2y+4ay3—4y3 (3a
12a3—15a2y+3ay2

3a2y+ ay2-4y3

4

12ay+4ay2-16y (3y

12a2y-15ay2+ 3y3

19ay2—19y3—19y2 ( a—y) Or, a-y) 4a2-5ay+y2 (4a-y

4a2-4ay

ay+y2
ay+y2.

Hence a-y

is the greatest common measure of the polynomials a and b.

(4.) Required the greatest common measure of the terms of the fraction

a6-a2x4

a+ax-ax-a3x3°

Here a2 is a simple factor of the numerator, and a3 is a factor of the denominator; hence a2 is the greatest common measure of these simple factors,

which must be reserved to be introduced into the greatest common measure of the other factors of the terms of the proposed fractions; viz.:

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Therefore a2 (a2—x2) is the greatest common measure; and hence

a6-a2x1 a+a3x—a1x2-a3x3

=

=

· a2+x2

(a-a2x1) a2 (a2-x2) (a®+a3x—a1x2—a3x3)÷a2 (a2—x3) a2+ax*

ADDITIONAL Examples.

(1.) Find the greatest common measure of 2a2x2, 4x2y2, and 6x3y.

(2.) Find the greatest common measure of the two polynomials a3—a2b+ 3ab2-3b3, and a2-5ab+4b2.

(3.) What is the greatest common measure of a3—xy2 and x2+2xy+y2? (4.) Find the greatest common measure of a+y3 and x13—y13.

(5.) Find the greatest common measure of the polynomials

(b−c) x2—b (2b−c) x +b3 . . (a)

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(b+c) x3−b (2b+c) x2+b3x. . . . . (b).

(6.) Find the greatest common measure of the polynomials

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32. We have already defined a multiple of a quantity to be any quantity that contains it exactly; and a common multiple of two or more quantities is a quantity that contains each of them exactly.

The least common multiple, of two or more quantities, is therefore the least quantity that contains each of them exactly.

33. To find the least common multiple of two quantities.

Divide the product of the two proposed quantities by their greatest common measure, and the quotient is the least common multiple of these

quantities; or divide one of the quantities by their greatest common measure, and multiply the quotient by the other.

Let a and b be two quantities, d their greatest common measure; and m their least common multiple; then let

α= hd, and b = kd;

and since d is the greatest common measure, h and k can have no common factor, and hence their least common multiple is hk; therefore hkd is the least common multiple of hd and k d; hence,

hk d2 hdxkd axb ab

m=hkd=

d

=

d

d

d

Q. E. D.

EXAMPLES.

(1.) Find the least common multiple of 2a2x and 8a3x3.

Here m =

ab 2a2x × 8a3x3

2a2x

= 8a3x3 = least common multiple.

(2.) Find the least common multiple of 4x2 (x2—y2) and 12x3 (x3—y3). Here d = 4x2 (x—y), and therefore we have

m=

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4x2 (x2-y2) × 12x3 (x23—y3) = 12x3(x+y) (x3—y1); 4x2 (x—y)

or m = 12x7+12xy-12x1y3-12x3y1.

(3.) Find the least common multiple of x2+2xy+y2 and x3—x y2. Here d = x+y, and therefore we get

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= x (x+y)(x2—y2) = least common multiple.

(4.) What is the least common multiple of x1-5x3 + 9x2 — 7x +2, and 24-6x2+8x-3?

By the process for finding the greatest common measure, we find

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x5—2x13—6x3+20x2-19x+6, the least common multiple.

(5.) Find the least common multiple of a2—2ab+b2, and aa—b1. (6.) Find the least common multiple of a2—b2, and a3+b3.

(7.) Find the least common multiple of x2-y2, and 23—y3.

(8.) Find the least common multiple of y2—8y+7, and y2+7y—8.

(1.) (a-b) (a1—b1).

(2.) (a-b) (a+b3).

ANSWERS.

(3.) (x+y) (x3—y3)
(4.) y-57y+56.

34. Every common multiple of two quantities, a and b, is a multiple of m, the least common multiple.

For let m' be a common multiple of a and b, then, because m' is greater than m; and if we suppose that m' is not a multiple of m, we have, as in the annexed scheme,

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