according to their numerical values as affected by the signs + and -; in other terms, the numerical values are not taken absolutely, but relatively, so that the sum is the difference between the sum of all the positive terms and all the negative terms, which difference is to be affected by the sign of the greater sum. Thus, if we add a, b, and c, we shall have a + b c for the algebraic sum of the quantities. If a 5, b = 7, c = 9, a+b-c becomes 5+7-93, whereas the arithmetical sum of 5, 7, and 9 = 21, (5+ 7 + 9 = 21). Again, if a and b have the same numerical values as before, and we suppose that c = 30, then a + b − c = 5 + 7 — 3018, a negative result, but the arithmetical sum of 5, 7, and 30,5+7+30=42. Hence, when quantities are added on the supposition that they are all positive, we say that the result is an arithmetical sum, or taken in an arithmetical sense; but if the quantities added are all negative, or some positive and others negative, the result is called an algebraic sum, or to be taken according to the signs of the terms; that is, in an algebraic sense. (2.) It is also evident that the word Subtraction has a more extensive meaning in Algebra than in Arithmetic. For in Arithmetic the minuend and subtrahend are both supposed to be positive, and the subtrahend to be less than the minuend, so that the difference or remainder will of course be less than the minuend. But in Algebra, the minuend being positive, and the subtrahend negative, the difference or remainder must of necessity be greater than the minuend. It is also clear that, if the minuend is positive, and the subtrahend positive, and greater than the minuend, the remainder will be negative, and so on. It is plain that the difference between Subtraction in Arithmetic and Algebra arises from the consideration that in Arithmetic the quantities are taken absolutely, whereas in Algebra their numerical values are taken relatively; that is to say, as affected by the signs + and − . Hence, when the minuend and subtrahend are both positive, and the subtrahend less than the minuend, we say that Subtraction is arithmetical; but if these conditions do not hold, the Subtraction is called algebraic. Thus, if a is the minuend and 6 the subtrahend, then a b is the arithmetical difference when b is not greater than a; but if b is greater than a, then the difference is algebraic. If a = 7, and b = 5, then a -b becomes 7-5=2= the arithmetical difference; but if a = 9, and b = 13, then a b becomes 913-4 the algebraic difference. (3.) Again, if a is the minuend, then a (b) = a + b = the algebraic remainder; which shows that to subtract a negative quantity is the same as to add an equal positive quantity; for if we add the positive quantity b to a, we get a + b, as above. (4.) Finally, to show that any compound quantity is to be subtracted from any other quantity, we often write the quantity to be subtracted after the quantity from which it is to be subtracted, putting the quantity to be subtracted in a parenthesis, and the sign - before the parenthesis and after the quantity from which the compound quantity is to be subtracted. Also, if the quantity from which the compound quantity is to be taken is itself a compound quantity, it will often be necessary to write it in a parenthesis. Thus, to show that bc is to be taken from a, we write a − (b −c); and to show that b + c d is to be taken from m-n+q, we may write (m-n + q) − (b + c − d), which may also be written in the form (b+c-d-m + n − q). SECTION IV. MULTIPLICATION. (1.) WHEN any quantity is added to itself any number of times, it is said to be multiplied. The quantity added is called the multiplicand, and the number of times it is taken in the sum is named the multiplier. The sum obtained by the addition of the quantity to itself is termed the product. Thus, 7+7+7+7+7 = 7 × 535 is called the product of 7 by 5, 7 being the multiplicand, and 5 the mul tiplier. In the same way, if we have a + a + a +, etc., to b terms, we get by addition a + a + a +, etc., to b terms = baab the product; a being the multiplicand, and b the multiplier. (2.) We have assumed that ba = ab; but since it is very easily proved to be true, when a and b are whole numbers, it may not be improper in this place to give the proof. Assuming a + a + a +, etc., to b terms; if we take a unit from the first a and add it to a unit of the second a, and add the sum to a unit of the third a, and so on to b terms, we shall have b units, or b; and in the same way, if we take a unit from the remainder of the first a, and add it to a unit of the remainder of the second a, and so on to b terms, we shall have units, or b; and proceeding in like manner successively, from any one set of remainders to the next following remainders, we shall have b units, or b, as often as a contains units, or, which is the same thing, we shall have b+b+b+, etc., to a terms; consequently, we have a + a + a +, etc., to b terms b + b + b + etc., to a terms. Thus, if we use 8 for a and 3 for b, we shall have 8+8+8=7+7+7+36 +6 +6 +3+35+5+ 5+3 +3 + 3, and so on to 8+8+8=3 + 3 + 3 + 3 + 3 +3 +3 + 3. = = By addition, a + a + a +, etc., to b terms ba, and b+b+b+, etc., to a terms ab, and of course baab, as was to be proved. Consequently, in finding the product of any two integral numbers, either may be taken for the multiplicand, and the other for the multiplier, and the same product will result. (3.) From what has been done, it is easy to show that the product of any three integers is independent of the order in which they are taken for multiplication. For let a, b, c represent the integers; then the product of a by b will be expressed (as above) by abba, and if we multiply these equals by c we get (by Ax. II.) abc = bac. = In like manner, since be cb, and acca, we get bca = cba, and acb= cab; observing, according to custom, to write the multiplier after the multiplicand. Again, since ab = a + a + a +, etc., to b terms, if we mul tiply these equals by c, we get (by Ax. II.) abc = ac + ac + ac +, etc., to b terms; or, since acca, we get abc = ca + ca + ca+, etc., to b terms = cab; and in the same way we have baccba, bea abc, and cba = acb. = Hence (by Ax. IV.) we deduce abc = bac = cab = acb = cbabca; consequently the product is independent of the order in which the numbers are taken in the multiplication. (4.) Again, if we multiply the equal products abc = bac= cab=acbcbabca by the integer d, (by Ax. II.), we get the equal products abcd = bacd = cabd = acbd = cbad = bead. Since abc ab + ab +, etc., to c terms, if we multiply these equals by d, we get (by Ax. II.) abcd = abd + abd +, etc., to c terms; or, since abd = dab, we have abcd dab + dab +, etc., to c terms = dabc; and, in the same way, each of the products, bacd, cabd, etc., may be changed to the products dbac, deab, etc.; and in the same way these products may be changed so that the last letter in each shall become the first, and so on. = It is hence manifest that the product abcd is independent of the order in which the numbers a, b, c, d are taken in the multiplication. (5.) In the same way the product abcde, of the integers a, b, c, d, e, may be shown to be independent of the order in which the numbers are taken in the multiplication; and so on, for the product of any number of integers. (6.) If we have to find the product of the integers, M and N, when either of them, as N, is the product of other integers, as a, b, c, etc., then it will be sufficient to multiply M by any one of the integers a, b, c, etc., and the product by any one of the remaining integers, a, b, c, etc., until they have all been used. For since N equals abc, etc., it is clear that the product MN must be bc, etc., times the product Ma, and that the product Max be, etc., must be c, etc., times the product Mab, and so on, until all the integers a, b, c, etc., have been used. It is evidently of no consequence in what order the integers a, b, c, etc., are taken in the successive multiplications. Consequently, we shall have the product MN equal to the pro duct Mabc, etc., in which the integers may be taken in any order in the multiplication. (7.) Reversely, if we have to multiply the integer M by the integers a, b, c, etc., successively, when taken in any order, then it will be sufficient to multiply M by the product of the integers a, b, c, etc. For if we put the product abc, etc., equal to N, then N will clearly be an integer; and the product, MN (by the last article), will be found by multiplying M by a, and the product by b, and the last product by c, and so on, until all the integers, a, b, c, etc., have been used. (8.) By Art. 9, Sec. I., if a stands for any number or quantity, and n for any positive integer, we express the product aaa, etc., to n factors, which is called the nth power of a, by a". If we have a" and am, n and m being positive integers, the product of a" and am will be expressed by a" × am = an+m= the power of a, whose exponent is n + m = the sum of the exponents of the powers that are multiplied together. = For, since aaa, etc., to n factors = a", and aaa, etc., to m factors a", if we multiply these equals by each other we get (by Ax. II.) aaa, etc., to n factors × aaa, etc., to m factors = aaa, etc., to (n + m) factors = a+m, and of course an xam = an+m Similarly, the product of a", a", a", a', etc., is evidently equal to am+n+p+, etc.), because a must enter the product as a factor as often as there are units in m +n+p+, etc. Thus, aa a2, a × a2 × a3 = ao, a3 × a2 = = a12. Hence, any integral powers of the same number or quantity are multiplied together, by adding the indices of the powers, for the index of the number or quantity in the product. (9.) We will now proceed to determine the sign of a product from the signs of its factors. 1. If a and b, regarded as positive, stand for the multiplicand and multiplier; then, since a x ba+a+a+, etc., to b terms, ba ab, it is clear that the product is positive. = = 2. Leta, supposed negative, stand for the multiplicand, while b, considered as positive, is the multiplier; then we have aba-a-a, etc., to b terms, - baab, |