ABBREVIATED OPERATIONS IN ARITHMETIC. There are certain abbreviated or contracted methods of calculation in particular cases, in which the pupil may be profitably exercised after he shall have become familiar with the general Rules of Arithmetic. The most useful of these will be given under their appropriate heads. We shall first explain the methods of proving Addition, Subtraction, &c., by the following Property of the Number 9. 311. Any number divided by 9 will leave the same remainder as the sum of its digits divided by 9. Take any number, as 345, which is 300+40+5. 300=3x100=3X(99+1)=3×99+3; and 40=4× 10=4x(9+1)=4×9+4; then 345 3×99+4×9+3+4+5. Now it is evident that 3 times 99+4 times 9 is divisible by 9, without a remainder; and therefore the remainder left by 345-9 must be that which is left by the sum of the digits (3+4+5)÷9. Similar illustration will apply to any other number consisting of two or more places of figures. This property of 9, it may be remarked, is a consequence of 9's being one less than 10, the basis of numeration. The same property belongs to 3, because 3 is a measure of 9. Addition Proved by Rejecting 9s. § 312. To prove Addition,―reject all the 9s in the sum of the digits of each of the numbers which are added together, and set down each excess or remainder. Then reject the 9s in the sum of these excesses: the remainder must be equal to that found by rejecting the 9s in the sum of the numbers. Beginning with the upper line, we say, 3 and 4 are 7, and 7 are 14, which is 5 above 9; 5 and 8 are 13, 4 above 9; 4 and 3 are 7. In like manner we find the excesses 2 and 3. The sum of these excesses is 12, which is 3 above 9; and the excess above the 9s in 121071 is also 3. The two excesses last found being the same, namely, 3, the addition is proved to be correct. This is evident from considering, that in finding each of these two excesses, we find the excess above all the 9s in the numbers which are added together, (§ 311.) Subtraction proved by Rejecting 9s. §313. To prove Subtraction,―reject all the 9s in the sum of the digits of the remainder, and also of the subtrahend. Then reject the 9s in the sum of the excesses thus found: the remainder last found must be equal to the excess above the 9s in the minuend. The reason of this is evident from the method of proving Addition, the minuend being the sum of the remainder and subtrahend. The pupil may be required to give an Example of this method of proving Subtraction. Multiplication Proved by Rejecting 9s. $314. To prove Multiplication,-reject the 9s in the sum of the digits of each factor. Then reject the 9s in the product of the excesses thus found: the remainder must be equal to the excess above the 9s in the product of the two numbers. The excess above the 9s in the multiplicand is 5, and the excess above the 9s in the multiplier is 4. The product of these two excesses is 20, in which the excess above the 9s is 2; and 2 is also the excess above the 9s in the product 182, &c. The multiplication is thus proved to be correct. The reason of this method will appear from considering, that if we reject all the 9s in the multiplicand and multiplier, and all the 9's in the product of the two remainders, the effect must be to reject all the 9s contained in the product of the two numbers. Division Proved by Rejecting 9s. 315. To prove Division,―reject the 9s in the divisor, and also in the quotient; and then reject the 9s in the product of the excesses thus found: the remainder must be equal to the excess above the 9s in the dividend. If there be any remainder in the division, its excess above the 9s must be added to the product of the excesses above 9s in the divisor and quotient. This method of proving Division results from the one just given for Multiplication,—the dividend being equal to the product of the divisor and quotient, plus the remainder, if any. The pupil may be required to give an example of this method of proving Division. Note. The preceding, as well as all other methods of arithmetical proof, afford but a high probability, not an absolute certainty, of the correctness of arithmetical calculations. For an error in the proof may balance an error in the calculation itself, so that the two erroneous results will agree. The only absolute proof consists in the intuitive truth of each particular result arising from the operation, performed according to the principles of the science. Arithmetical Complement-used in Abbreviating the Subtraction of two or more Numbers. § 316. The arithmetical complement of a number, is the difference between that number and a unit with as many Os annexed as there are figures in the given number. Thus the Complement of 6 is 10-6 =4; and the Complement of 64 is 100—64—36. The Complement of a number will be found by subtracting its right hand figure from 10, and each of its other figures from 9. Any number maye subtracted by adding its complement to the minuend, and sutracting 1 from that figure in the result which is next on the left of the subtrahend. 3756 Thus, instead of subtracting the 24, we may add its complement 76, and subtract 1 from the resulting 8 on the left. By adding the complement of 24, we have 3756+(100) -24), which exceeds the difference 3756-24 by 100; we must therefore subtract 100, to find the true difference, and this is done by merely subtracting 1 from the hundreds figure. Two or more numbers may be subtracted at once, by mentally adding their complements to the minund, and subtracting 1 for each complement from that figure the result which is next on the left of each subtrahend. EXAMPLES. (I.) 4 3 7 6 (II.) 7 43 4 (III.) 5 8 6 0 (IV.) 9 7 6 3 3703. 951 -86 1 -7 6 4 -4 9 1 0 2 27 3761 -16 8743 In these examples the sign is prefixed to those numbers which are to be subtracted from the sum of the other numbers. In the first example we find the complements of 2 and 4 by by subtracting these figures from 10, and add these complements with 1 and 6. Carrying 2, we subtract the 0 and 6 from 9, and add the complements; and in like manner we add the third column. Finally, from the fourth resulting figure 5 we subtract 2, for the two complements that were taken in the third column. Let the pupil be required to perform the operations in the other Examples. ABBREVIATIONS IN MULTIPLICATION. One means of abbreviating Multiplication, is, to commit to memory the products of numbers beyond the common limit of the Multiplication Table. These products should be recited thu13 times 2 is 26; 13 times 3 is 39, and so on. The most useful part of the following Table, is from 13×2 to 13X9, and so on to 19×9; though the whole is well worth the labor of learning it. Supplement to the Multiplication Table. སྐ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 This Table having been well committed to memory, we may multiply at once by 13, 14, &c., as readily as by a single figure. Thus 34789 452257 13 times 9 is 117; 13 times 8 104, and 11 make 115, &c. Without using the preceding Table, we may apply the following method: § 317. To Multiply by 13, 14, &c. to 19. 1. Multiply by the units figure, and add to each partial product (after the first) that figure of the multiplicand which stands next on the right of the one multiplied. 2. Set down only the right hand figure of the last product; add the left to the last figure of the multiplicand, and set the sum on the left of the other product figures. |