repeated 6 times. Now let it be required to multiply the fraction α C by с α d adopting the same definition as above, we may say that what we did with unity to obtain we must now do with to obtain Ъ times с a с To obtain from unity the unit is divided into d equal с parts, and c of such parts are taken; therefore to obtain times α α the fraction is divided into d parts, and c of such parts are taken. a be divided into d equal parts, each of them is d, and if a c " c of such parts be taken the result is b ď The proof of the rule for the division of fractions may, in part, be illustrated by the same reasoning, for after inverting the divisor we "proceed as in multiplication." 7. We may call attention to the fact that the proof of the rule for the division of decimals given in some works on Algebra is unsatisfactory and imperfect, because it includes only the particular case when, neglecting the decimal point, the dividend is an exact multiple of the divisor. A full proof is given in Lund's new edition of Wood's Algebra. 8. We would not imply by what we have said on the method of finding the G.C.M. and the L.C.M. by means of the study of factors that the ordinary rule is of no importance. On the contrary, it is sometimes the readiest way of obtaining the G.C.M. and the L.C.M. Moreover, the theory of these points must be thoroughly mastered and understood; for questions are often asked on the theory or proof of the rule in each case, and on the following principles involved in the theory: : 9. Every measure of two expressions, A and B, divides D, their G.C.M., and every divisor of their G.C.M. measures them, 10. A factor which does not contain any factor common to both A and B, may be rejected at any stage of the process. 11. A factor which has no factor that the divisor has may be introduced into the dividend at any stage of the process. Suppose that A and B contain a common factor F, which is obvious on inspection, F will be a factor of the G.C.M., but need not be involved in the remaining processes of the usual rule. Similarly, if at any stage of the operation we perceive that a certain factor is contained in both the dividend and the divisor, we may strike out the factor and continue the operation with the remaining factors. The factor omitted must then be multiplied by the last divisor in the operation, and the product will be the G.C.M. 12. Every common multiple of A and B is a multiple of their L.C.M. 13. Let the student also think out the principles involved in the following statement, the truth of which may be seen by considering the question of factors as before suggested: The G.C.M. of two or more quantities is the L.C.M. of all the common measures of those quantities. 14. It may be observed that it is sometimes easier to find the G.C.M. by the common rule after reversing the order of the terms in each of the given expressions. Thus the G.C.M. of 371y3+26y2-50y+3 and 469y3 +75y2 – 103y - 21 is found more easily by writing them 3-50y+26y2+371y3 and -21 − 103y+75y2+469y3, and then proceeding by the ordinary process. factor of 21 but that 371 is not a factor of 469. Observe that 3 is a III. RATIO, PROPORTION, AND VARIATION. We give a few of the more important propositions, the proof of which should be well understood. 1. A ratio of greater inequality is diminished, and a ratio of less inequality is increased, by adding any quantity to both terms of the ratio. 2. When the difference between the antecedent and consequent is small compared with either of them, the ratio of the higher powers of the terms is found by doubling, trebling, &c., their difference. The utility of the latter rule will be sufficiently manifest when it is observed by what a troublesome process the several proposed ratios would be found without the rule in the following examples :Ex. 1. (1·5241)a : (1·524)* = 1·5240 + 4 × 0·0001 : 1·524 nearly= 1-5244 1.524 nearly. Ex. 2. 3/729: 3/728=7283: 728 nearly. 3. The following theorem respecting ratios is of very great use, especially if its wide application to problems in Proportion be understood: If α C e d f' (pan+qcn+ren` then each of these ratios is equal to pbn+qdn+rfn where p, q, r, n, are any quantities whatever. 1 n 4. To show that if quantities be proportional according to the Algebraical test, they will also be proportional according to the Geometrical test, and the converse. The converse is much more difficult to prove than the first part; for it must be proved for incommensurable quantities, as well as commensurable. 5. If four quantities be proportionals, the greatest and least of them together are greater than the other two together. 6. If a b c d, to prove that ma±nb: pa±qb:: mc±nd: pc±qd. 7. When any number of quantities are proportionals, as one antecedent is to its consequent, so is the sum of all the antecedents to the sum of all the consequents. 8. We will now exemplify a process somewhat different to the one usually employed in solving problems in proportion. It is capable of almost indefinite application. Especially does it enable us, when certain fractions are known to be equal, to determine other relations between the quantities involved in them. a3 c3 a+c Ex. 1. If a : b : : c : d prove that 13+3 = (1+0). α Let Hence a:b::c: =k..a=bk, and c=dk. a3 + c3 (bk)3+(dk)3 b3+d3 d b+d 3 (a +c)3= (bk + dk)3=(b +d)3; then, by alternation, a3 + c3 Let each fraction = -c c-a show that p+q+r=0. k, then pak-bk, q=bk — ck, and r=ck-ak; :.p+q+r=ak - bk+bk - ck+ck-ako. Ex. 3. If a:b::c:d prove that a: c :: ^/(a* + b1) : */ (ca+2a). ¿=k, then=k; :.a=bk and c=dk, and c dk d. Hence = */(a*+64) 4/(b*k*+b2) ^/ba 4/(k+1) __ b c ̄1√(c1+d1); or, a: c :: 2/(a*+b1) : 2/(c1+d1). 9. VARIATION is only a shortened form of expressing proportion, for when we say that x ∞ y, we mean that x bears to y the same ratio that any given value of x bears to the corresponding value of y; or xy: a given value of x the corresponding value of y. This fact is evident from the definition: One quantity is said to vary directly as another when the two quantities depend wholly upon each other, and in such a manner that if one be changed the other is changed in the same proportion. For instance, if a man arranges to work for a constant sum per day the amount of his wages varies as the number of days during which he works, for as the number of days increases or diminishes his wages will increase or diminish, and in the same proportion. Again, if the altitude of a triangle be constant, the area varies as the base; for if the base be increased or diminished the area is increased or diminished in the same proportion. 10. We may remark that the form xxy is sometimes called a proportional equation, expressing merely the proportionality of the two members of the equation, viz., the two variable and dependent quantities x and y. But the proportional equation x x y always supposes the proportionality of four quantities, viz., of any two values of x and the corresponding values of y. So that, although in considering varying and dependent quantities two terms only are expressed, the beginner should bear in mind that four are always implied, and that the processes he goes through in obtaining his conclusions are really processes with proportionals. One quantity is said to vary as another, not because the two increase or diminish together, but because as one increases or diminishes the other increases or diminishes in the same proportion. An illustration or two will make this point still clearer. Ex. 1. If A x B, and a, b be any pair of values of A and B, then for any OTHER values of A and B we have A B = m = a: b, where m is a constant value. That is, when one quantity varies as another, if any two pairs of values be taken of them the four will be proportionals, and we may write them in this case as a proportion, A: B:: a: b, or A: a :: Bb. This simply means that if any one value A be changed to any other value a, the value B will be changed to the value b in the same proportion. Ex. 2. If A vary inversely as B, and when A=2, B=12, what will B become when A=9? Ex. 3. If 6 horses can plough 17 acres in 2 days how many acres will 93 horses plough in 4 days? Now we may find the answer to a problem like this either by the Unitary Method already explained or by Compound Proportion put in the form Or we may solve the problem by means of the very important proposition in variation : If A x B when C is constant, and AC when B is constant, then A x BC when both are VARIABLE. This proposition, when applied to the above case and treated to illustrate Proportion, will stand thus: The number of acres (A) ploughed a number of horses (B) when the days (C) are constant ; or 6 93: 17 number of acres required when the days are not considered. And the number of acres (A) ploughed a number of days (C) when the horses (B) are constant; or 2: 4 :: 17: number of acres required when the horses are not considered. Therefore the number of acres (A) a number of horses (B) x number of days (C) when both horses (B) and days (C) are variable; that is A & BC when both B and C are variable; and .. 6×2 93 × 4: 17 acres number of acres required. 93 × 4 × 17 = 5923. 11. Let the student observe how the above proposition in Variation is the foundation of the so-called Double Rule of Three in Arithmetic, and how the separate statements in the above solution correspond to the ordinary separate statements of a Double Rule of Three sum, and he will agree with us in saying that Variation is only an abridgment of Proportion, in which two terms are expressed for each proportion instead of four, though four are always implied. When we say that A & B, we express the two terms; but we imply that the quantities depend upon each other in such a way that if A be changed to any other value a, then B must be changed to another value b, so that, fully expressed, we may say, as before, that A: a:: B: 6, and then we get the four terms really implied. This last result we A B may put in the form α or A =B, or A= mB, where m is equal ; and when we say that A & B we always imply |