fraction may be placed under the form x (a + 2/2) in which the presence, in the terms, of the common factor x, which becomes under the particular hypothesis, is manifest. Dividing both terms by this factor, we obtain, 1 = (1+dx)x 1+dx' which becomes =0, when x∞. The given expression may be when x=∞, thus falling into the class discussed. Or, we in which the presence of the factor x2, becoming infinite under the particular hypothesis, is evident. Striking out this factor, and making the required supposition, the value of the expression becomes Other cases might be treated similarly. The device of transforming the fraction or given expression, to render clear the presence of a common factor, is very often necessary, and skill in its employment is very important. One observation it is well to make before leaving this subject: Whenever a supposition is to be made that leads to results of co, 0, etc., we must beware of failing cases, concealed ratios, etc., and never pass over such results without careful inquiry concerning the circumstances that produce them. SECTION III. DISCUSSION OF PROBLEMS. 424. The Discussion of a problem is the operation of making every possible supposition upon the arbitrary quantities entering it, and explaining the results thus obtained. 425. Arbitrary Quantities are those to which any value may be assigned at pleasure. Arbitrary quantities are not unknown, but enter, as the known quantities, the general statement of a problem, and by assigning to them particular values we obtain the special cases of the problem. 426. An Indeterminate Problem is a problem that admits of an infinite number of solutions. A problem is always indeterminate when the conditions. are less in number than the unknown quantities involved, since the statement will then involve a less number of equations than of unknown quantities. PROBLEMS. 427. 1. Find a number that, increased by a, shall be equal to b. Let x represent the number. Then x+a=b (1) .'. x—b—a. Here we may have two cases, b>a or a>b. 1o. b>a. In this case the value of x is positive, and the solution gives the number required, without presenting any difficulty. 2o. ab; the value of x is negative. In this case the subtraction of a from b, in the strict arithmetical sense, is impossible, and hence the solution of the problem, understood in an arithmetical sense, is impossible; we cannot find a number that, increased by a second, shall be less than the second number. The value of x will, however, satisfy (1), and therefore is an answer to the problem taken in its algebraic or general sense, and we see at once that b-a added, in the algebraic sense, to a, will give b as the sum. TO ILLUSTRATE, 1°. Let b=73 and a 50. Then x-b-a-23, and we see at once that 50+23=73. 2°. Let b=73, and a=97. Then x-b-a--24. In this case the problem is one admitting no solution, if understood arithmetically; for, 97 being greater than 73, no number, added to 97, can give 73 as a sum. But, understood algebraically, the solution of the problem is not impossible, for -24, added algebraically to 97, gives 73 as the sum, and is therefore a true answer to the problem, satisfying the equation, and hence the conditions, thereof. 2. Find a number, such that if it be multiplied by a and the product be added to b, the sum will equal d increased by c times the number. Let x represent the number; then, from the conditions, ax+b=cx+d (1) .. x= Now the different values d-b a-c that may be assigned to a, b, c and d, give rise to several different cases. 10. Suppose d>b and a>c; the value of x will then be positive, and will satisfy the conditions of the problem without presenting any difficulty. We naturally expect that if a>c, as ax>cx, d will be greater than b. 20. Suppose d>b and a<c. The value of x will then be negative, since its terms will have contrary signs. In this case the solution of the problem, arithmetically considered, is impossible. For, since a<c, ax<cx, and the less product would require the addition of the greater of the quantities d, b, to give a sum equal to the sum of cx and the other, whereas the less quantity, b, is added thereto, and ax and b being less than cx and d, respectively, their sum is less than the sum of cx and d. Taking the problem in the algebraic sense, however, the solution is not impossible, for this negative value of x satisfies (1), just as well as the positive value. We can see, too, that, x being negative, of ax and cx, that one is algebraically greater which is numerically the less, and we should expect b, less than d, to be added to the algebraically greater (ax), of the quantities ax and cx, if d is to be added to the algebraically less, to make the sums equal. We may avoid even the apparent or arithmetical absurdity involved in taking this value of x, thus:-As x is negative, substitute for x in (1); whence, —ax+b= d-b —cx+d (2), and x= a value numerically equal to -a' d-b a but positive. This value satisfies (2). Modifying the enunciation of the problem to suit the new equation, it becomes: "Find a number such that, if it be multiplied by a, and the product be SUBTRACTED from b, the difference will equal d, DIMINISHED by times the number." The value of x, from (2), satisfies the conditions of this problem, even in the arithmetical sense. 3°. Suppose d<b and a>c. The value of x is negative, and the discussion is similar to the last. 4°. Suppose d<b and a<c. Though each term of the fractional value of x is negative, their quotient being positive this value of x satisfies the conditions of the problem, even in the arithmetical sense. This we should expect, since, as ac, axcx, and b, greater than d, ought to be added to the less of these, i. e., ax, d being added to the greater, to make the two sums equal. Though the value of x thus satisfies the equation, d being less than b, and a less than c, the subtractions (10), of b from d, and (20), of c from a, are impossible, arithmetically considered, (though simply giving negative remainders, algebraically considered). To avoid this difficulty, we may solve (1) thus: ax+b=cx+d, whence d-b cx-ax-b-d ... x= a result equal to but in b-d c-a α which the terms, (b―d and c-a), are both positive, and therefore involve no absurdity when arithmetically considered. 5°. Suppose d=b; a 0 = +0. The equation (1) becomes, under this hypothesis, ax+b=cx+b or ax=cx, which, a being greater or less than c, cannot be true unless x=±0, when this equation reduces to 0=0. Only one of the values +0 or 0, can be taken for either hypothesis. Here x= d-b =+∞. Under 6o. Suppose db, a=c. this hypothesis (1) becomes ax+b=ax+d; which, d and b being unequal, cannot be satisfied, unless x is so large that the inequality of d and b would not affect the relative values of the members; so large, in other words that the addition of d or b to ax would make no perceptible difference. Hence, x must be infinite, to satisfy this equation, and will be positive or negative, according to the supposition taken. The values of x in (5) and (6) are somewhat peculiar, if the problem be understood in its strict arithmetical sense. Still, x=0 and x==∞, are true answers to the problem. Taking x=0, if we understand 0 in its arithmetical sense, we have x= no quantity, i. e., there is no quantity that will satisfy the conditions of the problem, understood arithmetically, and, referring to the enunciation, we see that this is the case. Taking zero in its algebraic sense, of a quantity less than any assignable quantity, the solution presents no difficulty; for, ax and cx being infinitely small, the addition of either to b would have no effect, i. e., b±0=b±0 or b=b. |