denomination near each other, or to write units under units, tens under tens, &c., in simple numbers, and pounds under pounds, shillings under shillings, yards under yards, feet under feet, &c., in compound. (6.) For the sake of uniformity, we usually place the minuend above the subtrahend, and the remainder beneath; beginning at the right hand to subtract. (c.) The following examples will illustrate the application of these principles : 1. What is the value of 4697 3265? Solution. — Beginning at the right hand, and considering the denomi nations separately, we have š units from 7 units leave 2 units ; 6 tens from 9 tens leave 3 tens; 2 hundreds from 6 hundreds leave 4 hundreds; 3 thousands from 4 thousands leave 1 thousand. Having thus subtracted the numbers in all the denominations, we know that the remainder must be 1 thousand, 4 hundred, 3 tens, and 2 units, or 1432. The work would be written thus: 4697 = Minuend. 1432 = Remainder. Second Example. - What is the value of £17 8 s. 9 d. £3 4 s. 6 d.? Solution. -- Beginning with the lowest denomination, we have 6 d. from 9 d.= 3 d.; 4 s. from 88.= 4 s.; £3 from £17 = £14. The answer is, therefore, £14 4 s. 3 d. Or, beginning at the left, £17 - £3 = £14; 8 8. — 48. = 4 8.; 9 d. -6d. = 3 d. Ans. £14 4 s. 3 dr The work would be written thus: £. d. 8. 14 4 3 Remainder. 63. Methods of Proof. (a.) From the nature of subtraction, it is evident, that if the minuend were divided into two parts, such that one should equal the subtrahend, the other would equal the remainder. (6.) We have, therefore, the following methods of testing the correctness of the work: First Method. — Add the remainder to the subtrahend, and if the sum thus obtained is equal to the minuend, the work is probably correct; but if it is not, there is an error in either the subtraction or the addition, and possibly in both. Second Method. - Subtract the remainder from the minuend, and if the result thus obtained equals the subtrahend, the work is probably correct. WRITTEN WORK AND PROOF OF THE FIRST EXAMPLE. 4697 Minuend. 1432 Remainder. 4697 Sum of Rem. and Sub. = Minuend. 3265 Diff. of Rem. and Min. = Subtrahend. 14 4 3 Remainder. 17 8 9 Sum of the Sub. and Rem. = Minuend. 3 4 6 Difference of Rem. and Min. = Subtrahende 64. Problems requiring no Reduction. What is the value of each of the following ? 1. 854736 721423 ? 2. 9863764 420423 ? 3. 2948769 1432526? 4. $5476.92 - $1261.40 ? 5. 736.87 yd. --- 415.24 yd.? 8. 6 T. 17 cwt. 2 qr. 26 lb. 13 oz. 11 dr. 2 T. 6 cwt. 15 lb. 8 oz. 3 dr.? 9. 16 lb 83 53 23 18 gr. – 5 th 43 2 3 1 6 gr. ? 10.1757 gal. 3 qt. 1 pt. 3 gi. — 1323 gal. 1 qt. 2 gi.? 11. 18 rd. 4 yd. 2 ft. 11 in. 6 rd. 2 yd. 1 ft. 5 in. ? 65. Simple Subtraction. Method when Reductions are necessary. When, as is often the case, a figure in the subtrahend represents a greater value than the corresponding figure of the minuend, we take one of a higher denomination in the minuend, reduce it to the required denomination, add its value to the value of the figure already expressed, and subtract the value of the subtrahend figure from the sum thus obtained. First Example. What is the difference between 62.7 and 35.86 ? WRITTEN WORK. 5 11.16,10 Minuend, changed in form. Explanation of Process. - As there are no hundredths expressed in the minuend, we reduce one of the 7 tenths to hundredths, leaving 6 tenths. 1 tenth 10 hundredths, from which subtracting 6 hundredths, leaves a remainder of 4 hundredths. As 8 tenths cannot be subtracted from 6 tenths, we reduce one of the 2 units to tenths, leaving 1 unit. 1 unit: 10 tenths, which added to the 6 tenths left in the tenths' place equal 16 tenths ; 8 tenths from 16 tenths = 8 tenths. As 5 units cannot be taken from the 1 unit left in the units' place, we reduce one of the 6 tens to units, leaving 5 tens. 1 ten =- 10 units, which added to 1 unit equal 11 units; 5 units from 11 units - = 6 uniis 3 teus from 5 tens leave 2 tens. The answer, then, is 2 tens, 6 units, 8 tenths, and 4 hundredths, or 26.84. This may be proved in the same way that the preceding examples were. 5 Questions on the above. Which expresses the larger number, 7 tenths, or 6 tenths and 10 hun. dredths, and why? 2 units and 6 tenths, or 1 unit and 16 tenths ? tens and 1 unit, or 4 tens and 11 units ? 6 tens, 2 units, and 7 tenths, or 5 tens, 11 units, 16 tenths, and 10 hundredths ? How then will the remainder, obtained by subtracting 35.86 froin 62.7, compare with the remainder obtained by subtracting it from 5 tens, Il units, 16 tenths, and 10 hundredths ? Second Example. How many are 83004 dollars minus 24765 dollars ? WRITTEN WORK. 4 = Minuend. 8 2 3 9 = Remainder. Explanation. — As we cannot take 5 dollars from 4 dollars, and as there are no tens or hundreds expressed in the minuend, we take 1 thousand from the 3 thousands, leaving 2 thousands; 1 thousand = 10 hundreds, and taking 1 of these hundreds to reduce to tens, we have 9 hundreds left. i hundred = 10 tens, and taking 1 of these tens to reduce to units we have 9 tens left. 1 ten = 10 units, which added to the 4 units in the units' place 14 units. Now, by subtracting, we have 4 thousands cannot be taken from 2 thousands, therefore we take 1 ten-thousand from the 8 ten-thousands in the minuend, leaving 7 tenthousands. 1 ten-thousand = 10 thousands, which added to the 2 thousands in the thousands' place : 12 thousands. 4 thousands from 12 thousands = 8 thousands. 2 ten-thousands from 7 ten-thousands = 5 ten-thousands. The remainder is, therefore, 58239 dollars. Questions upon the above. - How can it be shown that 3004 is equal tu 2 thousands, 9 hundreds, 9 tens, and 14 units ? That 83004 is equal to 7 ten-thousands, 12 thousands, 9 hundreds, 9 tens, and 14 units ? 66. Compound Subtraction. First Example. — Find the difference between 14 lb. 6 oz. 5 dwt. 7 gr. and 8 lb. 7 oz. 18 dwt. 23 gr. WRITTEN WORK. oz. 13 17 34 41 Minuend changed in form. Explanation. - As we cannot subtract 23 gr. from 17 gr., we take 1 dwt. from the 15 dwt., which reduced to grains and added to the 17 gr. equals 41 gr.; 23 gr. from 41 gr. = 18 gr. But as 18 dwt. cannot be taken from the 14 dwt. left in the minuend, we take 1 oz. from the 6 oz. and reduce it to pennyweights; 1 oz. = 20 dwt., which added to the 14 dwt. equal 34 dwt.; 18 dwt. from 34 dwt = 6 dwt. As 7 oz. cannot be taken from the 5 oz. left in the minuend, we take 1 lb. from the 14 lb. and reduce it to ounces ; 1 lb. = 12 oz., which added to the 5 oz. equal 17 oz.; 7 oz. from 17 oz. = 10 oz. 8 lb. from 13 lb. 5 lb. The answer is, therefore, 5 lb. 10 oz. 6 dwt. 8 gr., which may be proved as before. Questions. — Which expresses the greater quantity, 15 dwt. 17 gr., or 14 dwt. 41 gr., and why? 6 oz. 14 dwt., or 5 oz. 34 dwt. ? 14 lb. 5 oz., or 13 lb. 17 oz. ? 14 lb. 6 oz. 15 dwt. 17 gr., or 13 lb. 17 oz. 34 dwt. 41 gr. ? How would the remainder obtained by subtracting 8 lb. 7 oz. 18 dwt. 23 gr. from 14 lb. 6 oz. 15 dwt. 17 gr. compare with that obtained by subtracting it from 13 lb. 17 oz. 34 dwt. 41 gr. ? Second Example. — A farmer took 8 bu. 3 pk. 5 qt. of corn from a bin containing 17 bushels. How many bushels, pecks, and quarts remained ? Reasoning Process. — If the bin contained 18 bu., and he took out 8 bu. 3 pk. 5 qt., there would remain the difference between 17 bu. and 8 bu. 3 pk. 5 qt. This shows that 17 bu. is the minuend, and 8 bu. 3 pk 6 qt. the subtrahend. |