« ΠροηγούμενηΣυνέχεια »
braic quantities, " first divide each of them by their greatest simple common measure (if they have one); arrange their terms according to the dimensions of the same letter, and divide either, or both of them, by the greatest simple factor which it may contain ; then perform on them the same operation as that for finding the greatest common measure of two numbers, observing only, that the remainders which arise are to be divided by their greatest simple factors, and that the dividends may, if requisite, be multiplied by any simple quantity which will make the first term of the dividend
a multiple of the first term of the divisor. Lastly, multiply the e
compound common measure thus obtained by the simple one originally taken out, and the product will be the greatest common measure required."
CASE IV. To find the greatest common measure, or divisor, of the terms of a fraction, consisting of compound quantities, or to reduce a fraction to its lowest, or most simple terms.
Rule 1. Range the quantities according to the powers, or dimensions of some letter, as in Division, and by inspection expunge
the common factor or factors, if any, either in the divisor or divi- 1 dend; then divide that quantity which has the highest power by
the other, whether it be the numerator or denominator; and divide the last divisor by the last remainder, and so on till nothing remains; the last divisor will be the greatest common measure : but if such a divisor cannot be found, the fraction has no common measure. Having found the greatest common measure, divide the terms of the fraction by it, and the resulting fraction will be in its
Note. If any of the divisors become negative, they may have their signs changed without altering the truth of the result. Also,
if the first term of a divisor be not contained an exact number of 1 times in the first term of the dividend, the latter may be multiplied
by any quantity that will make the division complete. And lastly, any quantity which is common to all the terms of the dividend or divisor, may be expunged before the division is commenced.
6a?+7ax_3.x2 1. Find the greatest common measure of
6a2+1lax+322 then reduce the fraction to its lowest terms. 6a?+7ax—3x+)6a2+1lax+3x+(1
6a7ax-3x2 Now dividing this 4ax+6x? remainder by 2x, the quotient is 2a+3x ; and dividing the divisor 6a?+7ax—322 by this quotient, we have 2a+3x)6a77ax-3x+(30—2
therefore 2a+3x, the last divisor, is the greatest common measure 6a2+7ax-322
За—x required. 2a+3x).
Ans. 6a'+llax+3.2*3a+* In this example the dimension of the numerator is the same as that of the denominator; therefore, it makes no difference whether we divide the numerator by the denominator, or the denominator by the numerator. This will be evident from the following operation : 6a'+11ax+3x®)6a'+7ax—3x*(1
6a2+1lax+3x2 Now, dividing this -4ax—6x2 remainder by —2x, the quotient is 2a+3.x ; or, as the remainder above is now, negative. And, by changing all the signs, we obtained 4ax+6z", or 2a+31, for the new divisor. 2a+3.c)6a2+1lax+3x+(3a+2
2ax +3.c Hence 2a+3.x is found to be the greatest common measure, as before.
2. Required to find the greatest common measure of 2—24-2y 2_1 7a-23ab+682 casa
5a3—18a+b+11ab2-663 and then reduce each fraction to their lowest terms. -ky-2y)2-3xy+2y°(1
xxy2y Divide this —2xy +4yo remainder, each term of which can be divided by 2y, and the signs of the terms being changed, we obtain x-2y for a new divisor. 2-2y).x'—xy—24(x+y 2–
+xy-2y and since there is no remainder, therefore 2–2y, the last divisor, is the greatest common measure required.
2-2y—2y x+y -2y).
Answer. 2?—2xy+3y2 In this example, the dimensions of the numerator is the same as that of the denominator ; therefore it makes no difference whether we divide the numerator by the denominator, or the denominator by the numerator. The same remark will apply to the next example. This will be evident from the following operation. 743xy+2y)*—XY—24°(1
2xy—4y the remainder, which is now positive, and divide by +2y, and we obtain x-2y, for our new divisor, the same as before.
Here the greatest common measure is x8x43; therefore,
2x3_16x46 2 -81-3)
Ans. as required. 3x3_-24x-9 3
2x?_16x46 2(2-82-3) 2 Or thus,
as before. 3x3—243—9 3(23_8143) 3 axta, or 12—162—1, the least common measure. 2+1 sätta -2-)
axta 3d expression solved. Multiply the denominator by 7, we shall have 7-23ab+66)35a--126a’b+77ab2-4263(50–116
-lla'b+47 ab? -4263
--11a Multiply by 7
-77a*b-+253ab_-6608 Divide this remainder by 766 and 76ab?—22863 we have for the divisor 2-36)70-23ab-+-6(7a-26
--2ab +662 Therefore a 36 is the greatest -2ab+-662 common measure sought. 722-23ab+662
%11ab Now, for the last fraction in example second, the greatest common
x taxta2 measure of the fraction is x—a)
23+ax: +aʻxta at x2
x+a*x+a 3. Reduce
each to its a—axazi+
xtax-arma lowest terms. x+dos_a*x—a“)x*+a'r' ta'(1
This remainder, -ax+doxo+aor+2a+ which, being divided by a, and the signs of the terms being changed, becomes x-ax-aa-2a)x* tax-ax-a (x+2a
2ax: _2aRx_2a3c-4a Divide this remainder, 3a*r* +3aRx+3a* by 3a', we obtain a taxta for our new divisor. z+axta%)x-ar-a'z-—2a*(x-2a
x* ta'rta xaxta2
Ans. x'+avarma x2-a2 ama'x—ax:+29)a*—24(a
at-ar-axtax Divide this remaind. a'xta*ma-axe— &t by 2 a'+aʻx-ax?—28(a—a -axta (1
aa2x--axx Divide this remainder -2ax+2x which is now negative, by -4. -2a-x+2x
| 28 taz-ax-20 (a -2x
la2ax Divide this remainder aʻx—ax by x, and we obtain a'--* for our
, new divisor. a-r*)a+x*(-1
-a2 + x2 a2-x2
the answer a — a2x-ax2 + x3 4. Reduce the three following fractions, which are from Simpson's ' Algebra,' to their lowest terms, viz: 50%+10a*b+5al 28 tax? +bz?—2a*x+bax—2628
and a®6+ab2+2ab3 +64
abb Here, dividing first by the greatest simple divisors, 50% and b, we have a’+2ab+b, and a+2ab+2ab +63: and if the latter of these be divided by the former, the work will stand thus : a+2ab+b?)a®+2ab+2ab?+63(a
a +-2ab+ab where the remainder is tab? +68; which being divided by b?, its greatest simple divisor, gives a+b; by this divide a+2ab+, and the quotient will come out a+b, exactly; therefore, the last divisor, a+b, will exactly measure both quantities.
5a' +5a88 .. the fraction given will be reduced to az_a'x-ax+ao)at +0+0+0.-_*(ata
+2a+x? +0–22* at0_2*)—a*x-ax'+x"(a
-a-x+0+23 -a'z+0+x". From whence it appears that after
a?b+ab? +38: Again,
or d—' will measure both at—*, and amaʻxax+x*; and, by
a2+x2 dividing thereby, the fraction proposed is reduced to 23-ax:_8a+x+62°)x4–3ax8–8a+z+18a®z—8a(x—2a
2* — axl_8a*x+ 6aRx
-2ax® +2a+z? +16aor—-12a*
remainder —2a*va—4aox+4a*; which, divided by 2a’, gives me +2ax—2a’ for the next divisor. 2+2ax—2a*)x_ ax_80*x+6a(2—3a
2 + 2ax:_2a*r?
+4aRx2 +8a8—8a Now if, by proceeding in this manner, no compound divisor can be found, that is, if the last remainder be only a simple quantity, we may conclude the case proposed does not admit of any, but is already in its lowest terms. Thus, for instance, if the fraction
a +2a*x+3ax: +4208 proposed were to be
it is plain by inspection,
a2taxtra that it is not reducible by any simple divisor; but to know whether it may not, by a compound one, I proceed as above, and find the
I last remainder to be the simple quantity 7x2; whence I conclude that the fraction is already in its lowest terms.
Another observation may be here made, in relation to fractions that have in them more than two different letters. When one of the letters rises only to a single dimension, either in the numerator or in the denominator, it will be best to divide the said numerator or denominator (whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other ; this being done, let the greatest
; common divisor of these two parts be found; which will evidently be a divisor to the whole, and by which the division of the other quantity is to be tried; as in the following example, where the fraction given is 28 Fax: +ba?—2a*x+bax—2ba?
Here the deno2_bx+2ax—2ab minator being the least compounded, and b rising therein to a sin