112. Fractions may be changed to others of equal value, with a common denominator, by multiplying each numerator by every denominator except its own, for the new numerator, and all the denominators together for the common denominator.

are fractions of the same value respectively with the former, having

the

common denominator bdf. For

=

adf a cbf C
bdf b' bdf d

= ;

edb e

=

bdf f

and

(Art. 101); the numerator and denominator of each fraction having been multiplied by the same quantity, viz. the product of the denominators of all the other fractions.

113. When the denominators of the proposed fractions are not prime to each other, find their Greatest Common Measure; multiply both the numerator and denominator of each fraction by the denominators of all the rest, divided respectively by their Greatest Common Measure; and the fractions will be reduced to a common denominator in lower terms than they would have been by proceeding according to the former rule.

114. To obtain them in the lowest terms each must be reduced to another of equal value, with the denominator which is the Least Common Multiple of all the denominators.

It becomes necessary therefore to investigate a rule for finding the Least Common Multiple of two or more quantities. And first,

115. To find the Least Common Multiple of any number of simple quantities.

To do this, observe the combinations of letters which form the several quantities, and resolve each quantity, by inspection, into its simple factors. The object then will be, to construct such a quantity as shall contain every factor found in all the proposed quantities, but no factor repeated which is not also similarly repeated in some one of them; for thus we shall obviously form a quantity, and the least quantity, which is divisible by each of the proposed quantities without remainder, that is, the Least

Common Multiple of them all. To this end, detach from each quantity all the factors, which are common to two or more of them, until the quantities are left prime to each other. The continued product of these common factors and prime results will be the Least Common Multiple required.

Thus, let the L. C. M. of 2a, 6ab, and 8ab be required. We see that the three quantities have a common factor 2a, which being detached leaves the quantities 1, 3b, and 46: of these again, the two latter have a common factor b, which being detached leaves the quantities 1, 3, and 4; and these are prime to each other. Therefore the L.C. M. required is

2a xbx 1 × 3 × 4 or 24ab.

OBS. Since the detaching of the Common Factors is the same thing as dividing the quantities by their Greatest Common Measures, it is clear that this method coincides with the arithmetical rule given in Art. 23.

It may also be observed that the preceding method is applicable to compound quantities, as well as simple, provided that each of the quantities can be readily resolved into its component factors Thus, if the L. C. M. of abad, and ab-ad be required, we see that the quantities have a common factor a, and when stripped of this become b+d, and b-d, which are prime to each other. Therefore the L.C. M. required is

a (b + d) (b − d) or a b2 – a ď2.

The following method is generally applicable to all quantities Simple or Compound.

116.

To find the Least Common Multiple of two quantities, or the least quantity which is divisible by each of them without remainder.

Let a and b be the two quantities, a their greatest common measure, m their least common multiple, and let m contain a, p times, and b, q times, that is, let m = pa = qb; then dividing the

two latter equal quantities by pb (Art. 82)

α

q

=

;

and since m

Ρ

is the least possible, p and q are the least possible; therefore

α

Р

in its lowest terms,* and consequently q ==; hence

a

*For, if not, let some other fraction be the fraction in its lowest terms; then since

a

multiplying these equal quantities by p'b (Art. 81) q'b=p'a, or there are common multiples of a and b less than pa and qb, which is impossible, since pa and qb are the least.

The rule here proved may be thus enunciated :-
:-

Find the G.C. M. of the two proposed quantities; divide one of them by this G. C.M.; and multiply the quotient thus obtained by the other quantity. The product is the Least Common Multiple required.

Ex. Required the Least Common Multiple of a* – x* and a3 - α x2 + x3.

The G. C. M. of these two quantities (See Art. 105, Ex. 2), is a2 – x2; and (a* − x1) ÷ (a3 − x2) = a2 + x2. ˆ Therefore the Least Common Multiple required

= (a2 + x2). (a3 — a3 x − ax2 + x3)
=a5 - a*x - α x2 + x3.

117. Every other common multiple of a and b is a multiple of m.

Let n be any other common multiple of the two quantities; and, if possible, let m be contained r times in n, with a remainder s, which is less than m; then n rm = s; and since a and b measure n and rm, they measure n they have a common multiple less the supposition.

118.

rm, or s (Art. 104); that is, than m, which is contrary to

To find the Least Common Multiple of three quantities a, b, c, find m the Least Common Multiple of a and b, and n the Least Common Multiple of m and c; then n is the Least Common Multiple sought.

For every common multiple of a and b is a multiple of m (Art. 117); therefore every common multiple of a, b, and c is a multiple of m and c; also every multiple of m and c is a multiple of a, b, and c; consequently the Least Common Multiple of m and c is the Least Common Multiple of a, b, and c.

And similarly if there be four or more quantities of which the Least Common Multiple is required.

Ex. Required the Least Com. Mult. of x3- a2x − a x2 + a3, x* — a*, and ax3 + a3 x − a2x2 — a1.

Here ax3+ a3 x − a2x2 — aa

= a (x3 + a2 x − a x2 — a3) ;

-

.. to find the G. C. M. of this quantity and the first, reject the factor a;

.. x-a is the G. C. M. of the first and last of the proposed quantities; and their least com. mult. is

(a x3 + a3x - a2x2 - a) (x2 - a3)...... (1).

The other quantity is

(x2 + a2) (x3 — a2)...............

·(2).

The G. c. M. of (1) and (2) is (x2 – a2) × the G. c. M. of ax3+ a3x — a2x2—aa and x2+a2. Rejecting the factor a in the former quantity,

.. the G. C. M. of (1) and (2) is (x2 + a2) (x2 − a2); .. least com. mult. required is

(a x2+a3x - a2x2 - a1) (x2 — ao),

or a x5 - a2x21 — a3 x + ao.

119. A more expeditious method of applying the preceding rule to find the Least Com. Mult., when it can readily be done, is that of resolving each quantity into its component factors, as follows:-taking the last Example,

(2) x1-a1= (x2+a2) (x2 — a2) = (x2 + a2) (x − a) (x +a).

(3)

a x3 + a3 x − a3 x3 — a1 − a x2 (x − a) + a3 (x − a)

= a (x2 + a2) (x − a).

Now the G. C. M. of (2) and (3) is (x2 + a2) (x − a) ;

.. least com. mult. of (2) and (3) is a (xa — a1)..... (4).

Again, the G. C. M. of (1) and (4) is x2 – a2 ;

..least com. mult. required is a (xa — a1) (x − a),

or a x5 — a2 x1 — a3 x + a®.