BOOK II.
COMMERCIAL ARITHMETIC.
SECTION I.
WEIGHTS, MEASURES, &C.
212. In making the calculations which are necessary in commercial affairs, no more processes are required than those which have been explained in the preceding book. But there is still one thing wanted--not to insure the accuracy of our calculations, but to enable us to compare and judge of their results. We have hitherto made use of a single unit (15), and have treated of other quantities which are made up of a number of units, in Sections II., III., and IV., and of those which contain parts of that unit in Sections V. and VI. Thus, if we are talking of distances, and take a mile as the unit, any other length may be represented,[31] either by a certain number of miles, or a certain number of parts of a mile, and (1 meaning one mile) may be expressed either by a whole number or a fraction. But we can easily see that in many cases inconveniences would arise. Suppose, for example, I say, that the length of one room is ¹/₁₈₀ of a mile, and of another ¹/₁₇₄ of a mile, what idea can we form as to how much the second is longer than the first? It is necessary to have some smaller measure; and if we divide a mile into 1760 equal parts, and call each of these parts a yard, we shall find that the length of the first room is 9 yards and ⁷/₉ of a yard, and that of the second 10 yards and ¹⁰/₈₇ of a yard. From this we form a much better notion of these different lengths, but still not a very perfect one, on account of the fractions ⁷/₉ and ¹⁰/₈₇. To get a clearer idea of these, suppose the yard to be divided into three equal parts, and each of these parts to be called a foot; then ⁷/₉ of a yard contains 2⅓ feet, and ¹⁰/₈₇ of a yard contains ³⁰/₈₇ of a foot, or a little more than ⅓ of a foot. Therefore the length of the first room is now 9 yards, 2 feet, and ⅓ of a foot; that of the second is 10 yards and a little more than ⅓ of a foot. We see, then, the convenience of having large measures for large quantities, and smaller measures for small ones; but this is done for convenience only, for it is _possible_ to perform calculations upon any sort of quantity, with one measure alone, as certainly as with more than one; and not only possible, but more convenient, as far as the mere calculation is concerned.
[31] It is not true, that if we choose any quantity as a unit, _any_ other quantity of the same kind can be exactly represented either by a certain number of units, or of parts of a unit. To understand how this is proved, the pupil would require more knowledge than he can be supposed to have; but we can shew him that, for any thing he knows to the contrary, there may be quantities which are neither units nor parts of the unit. Take a mathematical line of one foot in length, divide it into ten parts, each of those parts into ten parts, and so on continually. If a point A be taken at hazard in the line, it does not appear self-evident that if the decimal division be continued ever so far, one of the points of division must at last fall exactly on A: neither would the same appear necessarily true if the division were made into sevenths, or elevenths, or in any other way. There may then possibly be a part of a foot which is no exact numerical fraction whatever of the foot; and this, in a higher branch of mathematics, is found to be the case times without number. What is meant in the words on which this note is written, is, that any part of a foot can be represented as nearly as we please by a numerical fraction of it; and this is sufficient for practical purposes.
The measures which are used in this country are not those which would have been chosen had they been made all at one time, and by a people well acquainted with arithmetic and natural philosophy. We proceed to shew how the results of the latter science are made useful in our system of measures. Whether the circumstances introduced are sufficiently well known to render the following methods exact enough for the recovery of _astronomical_ standards, may be matter of opinion; but no doubt can be entertained of their being amply correct for commercial purposes.
It is evidently desirable that weights and measures should always continue the same, and that posterity should be able to replace any one of them when the original measure is lost. It is true that a yard, which is now exact, is kept by the public authorities; but if this were burnt by accident,[32] how are those who shall live 500 years hence to know what was the length which their ancestors called a yard? To ensure them this knowledge, the measure must be derived from something which cannot be altered by man, either from design or accident. We find such a quantity in the time of the daily revolution of the earth, and also in the length of the year, both of which, as is shewn in astronomy, will remain the same, at least for an enormous number of centuries, unless some great and totally unknown change take place in the solar system. So long as astronomy is cultivated, it is impossible to suppose that either of these will be lost, and it is known that the latter is 365·24224 mean solar days, or about 365¼ of the average interval which elapses between noon and noon, that is, between the times when the sun is highest in the heavens. Our year is made to consist of 365 days, and the odd quarter is allowed for by adding one day to every fourth year, which gives what we call leap-year. This is the same as adding ¼ of a day to each year, and is rather too much, since the excess of the year above 365 days is not ·25 but ·24224 of a day. The difference is ·00776 of a day, which is the quantity by which our average year is too long. This amounts to a day in about 128 years, or to about 3 days in 4 centuries. The error is corrected by allowing only one out of four of the years which close the centuries to be leap-years. Thus, A.D. 1800 and 1900 are not leap-years, but 2000 is so.
[32] Since this was first written, the accident has happened. The _standard yard_ was so injured as to be rendered useless by the fire at the Houses of Parliament.
213. The day is therefore the first measure obtained, and is divided into 24 parts or hours, each of which is divided into 60 parts or minutes, and each of these again into 60 parts or seconds. One second, marked thus, 1″,[33] is therefore the 86400ᵗʰ part of a day, and the following is the
MEASURE OF TIME.[34]
60 _seconds_ are 1 _minute_ 1 m.
60 _minutes_ ” 1 _hour_ 1 h.
24 _hours_ ” 1 _day_ 1 d.
7 _days_ ” 1 _week_ 1 wk.
365 _days_ ” 1 _year_ 1 yr.
214. The _second_ having been obtained, a pendulum can be constructed which shall, when put in motion, perform one vibration in exactly one second, in the latitude of Greenwich.[35] If we were inventing measures, it would be convenient to call the length of this pendulum a yard, and make it the standard of all our measures of length. But as there is a yard already established, it will do equally well to tell the length of the pendulum in yards. It was found by commissioners appointed for the purpose, that this pendulum in London was 39·1393 inches, or about one yard, three inches, and ⁵/₃₆ of an inch. The following is the division of the yard.
MEASURES OF LENGTH.
The lowest measure is a barleycorn.[36]
3 _barleycorns_ are 1 _inch_ 1 in.
12 _inches_ 1 _foot_ 1 ft.
3 _feet_ 1 _yard_ 1 yd.
5½ _yards_ 1 _pole_ 1 po.
40 _poles_ or 220 _yards_ 1 _furlong_ 1 fur.
8 _furlongs_ or 1760 _yards_ 1 _mile_ 1 mi.
Also 6 _feet_ 1 _fathom_ 1 fth.
69⅓ _miles_ 1 _degree_ 1 deg. or 1°.
[33] The minute and second are often marked thus, 1′, 1″: but this notation is now almost entirely appropriated to the minute and second of _angular_ measure.
[34] The measures in italics are those which it is most necessary that the student should learn by heart.
[35] The lengths of the pendulums which will vibrate in one second are slightly different in different latitudes. Greenwich is chosen as the station of the Royal Observatory. We may add, that much doubt is now entertained as to the system of standards derived from nature being capable of that extreme accuracy which was once attributed to it.
[36] The inch is said to have been originally obtained by putting together three grains of barley.
A geographical mile is ¹/₆₀th of a degree, and three such miles are one nautical league.
In the measurement of cloth or linen the following are also used:
2¼ inches are 1 nail 1 nl. 4 nails 1 quarter (of a yard) 1 qr. 3 quarters 1 Flemish ell 1 Fl. e. 5 quarters 1 English ell 1 E. e. 6 quarters 1 French ell 1 Fr. e.
215. MEASURES OF SURFACE, OR SUPERFICIES.
All surfaces are measured by square inches, square feet, &c.; the square inch being a square whose side is an inch in length, and so on. The following measures may be deduced from the last, as will afterwards appear.
144 square inches are 1 square foot 1 sq. ft. 9 square feet 1 square yard 1 sq. yd. 30¼ square yards 1 square pole 1 sq. p. 40 square poles 1 rood 1 rd. 4 roods 1 acre 1 ac.
Thus, the acre contains 4840 square yards, which is ten times a square of 22 yards in length and breadth. This 22 yards is the length which land-surveyors’ chains are made to have, and the chain is divided into 100 links, each ·22 of a yard or 7·92 inches. An acre is then 10 square chains. It may also be noticed that a square whose side is 69⁴/₇ yards is nearly an acre, not exceeding it by ⅕ of a square foot.
216. MEASURES OF SOLIDITY OR CAPACITY.[37]
Cubes are solids having the figure of dice. A cubic inch is a cube each of whose sides is an inch, and so on.
1728 cubic inches are 1 cubic foot 1 c. ft. 27 cubic feet 1 cubic yard 1 c. yd.
[37] ‘Capacity’ is a term which cannot be better explained than by its use. When one measure holds more than another, it is said to be more capacious, or to have a greater capacity.
This measure is not much used, except in purely mathematical questions. In the measurements of different commodities various measures were used, which are now reduced, by act of parliament, to one. This is commonly called the imperial measure, and is as follows:
MEASURE OF LIQUIDS AND OF ALL DRY GOODS.
4 _gills_ are 1 _pint_ 1 pt. 2 _pints_ 1 _quart_ 1 qt. 4 _quarts_ 1 _gallon_ 1 gall. 2 _gallons_ 1 _peck_[38] 1 pk. 4 _pecks_ 1 _bushel_ 1 bu. 8 _bushels_ 1 _quarter_ 1 qr. 5 _quarters_ 1 _load_ 1 ld.
The gallon in this measure is about 277·274 cubic inches; that is, very nearly 277¼ cubic inches.[39]
217. The smallest weight in use is the grain, which is thus determined. A vessel whose interior is a cubic inch, when filled with water,[40] has its weight increased by 252·458 grains. Of the grains so determined, 7000 are a pound _averdupois_, and 5760 a pound _troy_. The first pound is always used, except in weighing precious metals and stones, and also medicines. It is divided as follows:
[38] This measure, and those which follow, are used for dry goods only.
[39] Since the publication of the third edition, the _heaped_ measure, which was part of the new system, has been abolished. The following paragraph from the third edition will serve for reference to it:
“The other imperial measure is applied to goods which it is customary to sell by _heaped measure_, and is as follows:
2 gallons 1 peck 4 pecks 1 bushel 3 bushels 1 sack 12 sacks 1 chaldron.
The gallon and bushel in this measure hold the same when only just filled, as in the last. The bushel, however, heaped up as directed by the act of parliament, is a little more than one-fourth greater than before.”
[40] Pure water, cleared from foreign substances by distillation, at a temperature of 62° Fahr.
AVERDUPOIS WEIGHT.
27¹¹/₃₂ _grains_ are 1 _dram_ 1 dr. 6 _drams_, or _drachms_ 1 _ounce_[41] 1 oz. 16 _ounces_ 1 _pound_ 1 lb. 28 _pounds_ 1 _quarter_ 1 qr. 4 _quarters_ 1 _hundred-weight_ 1 cwt. 20 _hundred-weight_ 1 _ton_ 1 ton.
[41] It is more common to divide the ounce into four quarters than into sixteen drams.
The pound averdupois contains 7000 grains. A cubic foot of water weighs 62·3210606 pounds averdupois, or 997·1369691 ounces.
For the precious metals and for medicines, the pound troy, containing 5760 grains, is used, but is differently divided in the two cases. The measures are as follow:
TROY WEIGHT.
24 _grains_ are 1 _pennyweight_ 1 dwt. 20 _pennyweights_ 1 _ounce_ 1 oz. 12 _ounces_ 1 _pound_ 1 lb.
The pound troy contains 5760 grains. A cubic foot of water weighs 75·7374 pounds troy, or 908·8488 ounces.
APOTHECARIES’ WEIGHT.
20 _grains_ are 1 _scruple_ ℈ 3 _scruples_ 1 _dram_ ʒ 8 _drams_ 1 _ounce_ ℥ 12 _ounces_ 1 _pound_ lb
218. The standard coins of copper, silver, and gold, are,--the penny, which is 10⅔ drams of copper; the shilling, which weighs 3 pennyweights 15 grains, of which 3 parts out of 40 are alloy, and the rest pure silver; and the sovereign, weighing 5 pennyweights and 3¼ grains, of which 1 part out of 12 is copper, and the rest pure gold.
MEASURES OF MONEY.
The lowest coin is a farthing, which is marked thus, ¼, being one fourth of a penny.
2 _farthings_ are 1 _halfpenny_ ½_d_. 2 _halfpence_ 1 _penny_ 1_d_. 12 _pence_ 1 _shilling_ 1_s_. 20 _shillings_ 1 _pound_[42] or _sovereign_ £1 21 _shillings_ 1 _guinea_.[43]
219. When any quantity is made up of several others, expressed in different units, such as £1. 14. 6, or 2cwt. 1qr. 3lbs., it is called a _compound quantity_. From these tables it is evident that any compound quantity of any substance can be measured in several different ways. For example, the sum of money which we call five pounds four shillings is also 104 shillings, or 1248 pence, or 4992 farthings. It is easy to reduce any quantity from one of these measurements to another; and the following examples will be sufficient to shew how to apply the same process, usually called REDUCTION, to all sorts of quantities.
I. How many farthings are there in £18. 12. 6¾?[44]
[42] The English pound is generally called a _pound sterling_, which distinguishes it from the weight called a pound, and also from foreign coins.
[43] The coin called a guinea is now no longer in use, but the name is still given, from custom, to 21 shillings. The pound, which was not a coin, but a note promising to pay 20 shillings to the bearer, is also disused for the present, and the sovereign supplies its place; but the name pound is still given to 20 shillings.
[44] Farthings are never written but as parts of a penny. Thus, three farthings being 3/4 of a penny, is written 3/4, or ¾. One halfpenny may be written either as 2/4 or ½; the latter is most common.
Since there are 20 shillings in a pound, there are, in £18, 18 × 20, or 360 shillings; therefore, £18. 12 is 360 + 12, or 372 shillings. Since there are 12 pence in a shilling, in 372 shillings there are 372 × 12, or 4464 pence; and, therefore, in £18. 12. 6 there are 4464 + 6, or 4470 pence.
Since there are 4 farthings in a penny, in 4470 pence there are 4470 × 4, or 17880 farthings; and, therefore, in £18. 12. 6¾ there are 17880 + 3, or 17883 farthings. The whole of this process may be written as follows:
£18 . 12 . 6¾ 20 -------- 360 + 12 = 372 12 ----- 4464 + 6 = 4470 4 ----- 17880 + 3 = 17883
II. In 17883 farthings, how many pounds, shillings, pence, and farthings are there?
Since 17883, divided by 4, gives the quotient 4470, and the remainder 3, 17883 farthings are 4470 pence and 3 farthings (218).
Since 4470, divided by 12, gives the quotient 372, and the remainder 6, 4470 pence is 372 shillings and 6 pence.
Since 372, divided by 20, gives the quotient 18, and the remainder 12, 372 shillings is 18 pounds and 12 shillings.
Therefore, 17883 farthings is 4470¾_d_., which is 372s. 6¾_d_., which is £18. 12. 6¾.
The process may be written as follows:
4)17883 ----- 12)4470 ... 3 ---- 20)372 ... 6 £18 . 12 . 6¾
EXERCISES.
A has £100. 4. 11½, and B has 64392 farthings. If A receive 1492 farthings, and B £1. 2. 3½, which will then have the most, and by how much?--_Answer_, A will have £33. 12. 3 more than B.
In the following table the quantities written opposite to each other are the same: each line furnishes two exercises.
£15 . 18 . 9½ | 15302 farthings. 115ˡᵇˢ 1ᵒᶻ 8ᵈᵚᵗ | 663072 grains. 3ˡᵇˢ 14ᵒᶻ 9ᵈʳ | 1001 drams. 3ᵐ 149 yds 2ᶠᵗ 9 in | 195477 inches. 19ᵇᵘ 2ᵖᵏˢ 1 gall 2 qᵗˢ | 1260 pints. 16 ʰ 23ᵐ 47ˢ | 59027 seconds.
220. The same may be done where the number first expressed is fractional. For example, how many shillings and pence are there in ⁴/₁₅ of a pound? Now, ⁴/₁₅ of a pound is ⁴/₁₅ of 20 shillings; ⁴/₁₅ of 20 is
4 × 20 4 × 4 16 ------, or ----- (110), or ---, 15 3 3
or (105) 5⅓ of a shilling. Again, ⅓ of a shilling is ⅓ of 12 pence, or 4 pence. Therefore, £⁴/₁₅ = 5_s._ 4_d._
Also, ·23 of a day is ·23 × 24 in hours, or 5ʰ·52; and ·52 of an hour is ·52 × 60 in minutes, or 3ᵐ·2; and ·2 of a minute is ·2 × 60 in seconds, or 12ˢ; whence ·23 of a day is 5ʰ 31ᵐ 12ˢ.
Again, suppose it required to find what part of a pound 6_s_. 8_d_. is. Since 6_s._ 8_d._ is 80 pence, and since the whole pound contains 20 × 12 or 240 pence, 6_s._ 8_d._ is made by dividing the pound into 240 parts, and taking 80 of them. It is therefore £⁸⁰/₂₄₀ (107), but ⁸⁰/₂₄₀ = ⅓ (108); therefore, 6_s._ 8_d._ = £⅓.
EXERCISES.
⅖ of a day is 9ʰ 36ᵐ ·12841 of a day 3ʰ 4ᵐ 54ᔆ·624[45] ·257 of a cwt. 28ˡᵇˢ 12ᵒᶻ 8ᵈʳ·704 £·14936 2ˢ 11ᵈ 3ᶠ·3856
[45] When a decimal follows a whole number, the decimal is always of the same unit as the whole number. Thus, 5ᔆ·5 is five _seconds_ and five-tenths of a _second_. Thus, 0ᔆ·5 means five-tenths of a second; 0ʰ·3, three-tenths of an hour.
221, 222. I have thought it best to refer the mode of converting shillings, pence, and farthings into decimals of a pound to the Appendix (See Appendix _On Decimal Money_). I should strongly recommend the reader to make himself perfectly familiar with the modes given in that Appendix. To prevent the subsequent sections from being altered in their numbering, I have numbered this paragraph as above.
223. The rule of addition[46] of two compound quantities of the same sort will be evident from the following example. Suppose it required to add £192. 14. 2½ to £64. 13. 11¾. The sum of these two is the whole of that which arises from adding their several parts. Now
¾_d._ + ½_d._ = ⁵/₄_d._ = £0 . 0 . 1¼ (219)
11_d._ + 2_d._ = 13_d._ = 0 . 1 . 1 13_s._ + 14_s._ = 27_s._ = 1 . 7 . 0 £64 + £192 = 256 . 0 . 0 ----------- The sum of all of which is £257. 8 . 2¼
This may be done at once, and written as follows:
£192 . 14 . 2½ 64 . 13 . 11¾ ---------------- £257 . 8 . 2¼
[46] Before reading this article and the next, articles (29) and (42) should be read again carefully.
Begin by adding together the farthings, and reduce the result to pence and farthings. Set down the last only, carry the first to the line of pence, and add the pence in both lines to it. Reduce the sum to shillings and pence; set down the last only, and carry the first to the line of shillings, and so on. The same method must be followed when the quantities are of any other sort; and if the tables be kept in memory, the process will be easy.
224. SUBTRACTION is performed on the same principle as in (40), namely, that the difference of two quantities is not altered by adding the same quantity to both. Suppose it required to subtract £19 . 13. 10¾ from £24. 5. 7½. Write these quantities under one another thus:
£24. 5. 7½ 19. 13. 10¾
Since ¾ cannot be taken from ½ or ²/₄, add 1_d._ to both quantities, which will not alter their difference; or, which is the same thing, add 4 farthings to the first, and 1_d._ to the second. The pence and farthings in the two lines then stand thus: 7⁶/₄_d._ and 11¾_d._ Now subtract ¾ from ⁶/₄, and the difference is ¾ which must be written under the farthings. Again, since 11_d._ cannot be subtracted from 7_d._, add 1_s._ to both quantities by adding 12_d._ to the first, and 1_s._ to the second. The pence in the first line are then 19, and in the second 11, and the difference is 8, which write under the pence. Since the shillings in the lower line were increased by 1, there are now 14_s._ in the lower, and 5_s._ in the upper one. Add 20_s._ to the upper and £1 to the lower line, and the subtraction of the shillings in the second from those in the first leaves 11_s._ Again, there are now £20 in the lower, and £24 in the upper line, the difference of which is £4; therefore the whole difference of the two sums is £4. 11. 8¾. If we write down the two sums with all the additions which have been made, the process will stand thus:
£24 . 25 . 19⁶/₄ 20 . 14 . 11¾ ------------------ Difference £4 . 11 . 8¾
225. The same method may be applied to any of the quantities in the tables. The following is another example:
From 7 cwt. 2 qrs. 21 lbs. 14 oz. Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz.
After alterations have been made similar to those in the last article, the question becomes:
From 7 cwt. 6 qrs. 49 lbs. 14 oz. Subtract 3 cwt. 4 qrs. 27 lbs. 12 oz. ---------------------------- The difference is 4 cwt. 2 qrs. 22 lbs. 2 oz.
In this example, and almost every other, the process may be a little shortened in the following way. Here we do not subtract 27 lbs. from 21 lbs., which is impossible, but we increase 21 lbs. by 1 qr. or 28 lbs. and then subtract 27 lbs. from the sum. It would be shorter, and lead to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and add the difference to 21 lbs.
226. EXERCISES.
A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾, £6473. 0. 0, and £49. 14. 4½; and the following debts to pay: £200 . 19. 6¼, £305. 16. 11, £22, and £19. 6. 0½. How much will remain after paying the debts?
_Answer_, £6190. 7. 4¾.
There are four towns, in the order A, B, C, and D. If a man can go from A to B in 5ʰ 20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ 0ᵐ 17ˢ, how long will he be in going from B to D, and from C to D?
_Answer_, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ.
227. In order to perform the process of MULTIPLICATION, it must be recollected that, as in (52), if a quantity be divided into several parts, and each of these parts be multiplied by a number, and the products be added, the result is the same as would arise from multiplying the whole quantity by that number.
It is required to multiply £7. 13. 6¼ by 13. The first quantity is made up of 7 pounds, 13 shillings, 6 pence, and 1 farthing. And
1 farth. × 13 is 13 farth. or £0 . 0 . 3¼ (219) 6 pence × 13 is 78 pence, or 0 . 6 . 6 13 shill. × 13 is 169 shill. or 8 . 9 . 0 7 pounds × 13 is 91 pounds, or 91 . 0 . 0 -------------- The sum of all these is £99 . 15 . 9¼
which is therefore £7. 13. 6¼ × 13.
This process is usually written as follows:
£7 . 13 . 6¼ 13 ------------- £99 . 15 . 9¼
228. DIVISION is performed upon the same principle as in (74), viz. that if a quantity be divided into any number of parts, and each part be divided by any number, the different quotients added together will make up the quotient of the whole quantity divided by that number. Suppose it required to divide £99. 15. 9¼ by 13. Since 99 divided by 13 gives the quotient 7, and the remainder 8, the quantity is made up of £13 × 7, or £91, and £8. 15. 9¼. The quotient of the first, 13 being the divisor, is £7: it remains to find that of the second. Since £8 is 160_s._, £8. 15. 9¼ is 175_s._ 9¼_d._, and 175 divided by 13 gives the quotient 13, and the remainder 6; that is, 175_s._ 9¼_d._ is made up of 169_s._ and 6_s._ 9¼_d._, the quotient of the first of which is 13_s._, and it remains to find that of the second. Since 6_s._ is 72_d._, 6_s._ 9¼_d._ is 81¼_d._, and 81 divided by 13 gives the quotient 6 and remainder 3; that is, 81¼_d._ is 78_d._ and 3¼_d._, of the first of which the quotient is 6_d._ Again, since 3_d._ is ¹²/₄, or 12 farthings, 3¼_d._ is 13 farthings, the quotient of which is 1 farthing, or ¼, without remainder. We have then divided £99. 15. 9¼ into four parts, each of which is divisible by 13, viz. £91, 169_s._, 78_d._, and 13 farthings; so that the thirteenth part of this quantity is £7. 13. 6¼. The whole process may be written down as follows; and the same sort of process may be applied to the exercises which follow:
£ _s._ _d._ £ _s._ _d._ 13)99 15 9¼( 7 13 6¼ 91 -- 8 20 --- 160 + 15 = 175 13 --- 45 39 -- 6 12 -- 72 + 9 = 81 78 -- 3 4 -- 12 + 1 = 13 13 -- 0
Here, each of the numbers 99, 175, 81, and 13, is divided by 13 in the usual way, though the divisor is only written before the first of them.
EXERCISES.
2 cwt. 1 qr. 21 lbs. 7 oz. × 53 = 129 cwt. 1 qr. 16 lbs. 3 oz. 2ᵈ 4ʰ 3ᵐ 27ˢ × 109 = 236ᵈ 10ʰ 16ᵐ 3ˢ £27 . 10 . 8 × 569 = £15666 . 9 . 4 £7 . 4 . 8 × 123 = £889 . 14 £166 × ₈/₃₃ = £40 . 4 . 10⁶/₃₃ £187 . 6 . 7 × ³/₁₀₀ = £5 . 12 . 4¾ ²/₂₅ 4_s._ 6½_d._ × 1121 = £254 . 11 . 2½ 4_s._ 4_d._ × 4260 = 6_s._ 6_d._ × 2840
229. Suppose it required to find how many times 1s. 4¼_d._ is contained in £3. 19. 10¾. The way to do this is to find the number of farthings in each. By 219, in the first there are 65, and in the second 3835 farthings. Now, 3835 contains 65 59 times; and therefore the second quantity is 59 times as great as the first. In the case, however, of pounds, shillings, and pence, it would be best to use decimals of a pound, which will give a sufficiently exact answer. Thus 1s. 4¼_d._ is £·067, and £3. 19. 10¾ is £3·994, and 3·994 divided by ·067 is 3994 by 67, or 59⁴¹/₆₇. This is an extreme case, for the smaller the divisor, the greater the effect of an error in a given place of decimals.
EXERCISES.
How many times does 6 cwt. 2 qrs. contain 1 qr. 14 lbs. 1 oz.? and 1ᵈ 2ʰ 0ᵐ 47ˢ contain 3ᵐ 46ˢ?
_Answer_, 17·30758 and 414·367257.
If 2 cwt. 3 qrs. 1 lb. cost £150. 13. 10, how much does 1 lb. cost?
_Answer_, 9_s._ 9_d._ ¹³/₃₀₉.
A grocer mixes 2 cwt. 15 lbs. of sugar at 11_d._ per pound with 14 cwt. 3 lbs. at 5_d._ per pound. At how much per pound must he sell the mixture so as not to lose by mixing them?
_Answer_, 5_d._ ¾ ¹⁵³/₉₀₅.
230. There is a convenient method of multiplication called PRACTICE. Suppose I ask, How much do 153 tons cost if each ton cost £2. 15. 7½? It is plain that if this sum be multiplied by 153, the product is the price of the whole. But this is also evident, that, if I buy 153 tons at £2. 15. 7½ each ton, payment may be made by first putting down £2 for each ton, then 10s. for each, then 5_s._, then 6_d._, and then 1½_d._ These sums together make up £2. 15. 7½, and the reason for this separation of £2. 15 . 7½ into different parts will be soon apparent. The process may be carried on as follows:
1. 153 tons, at £2 each ton, will cost £306 0 0
2. Since 10s. is £½, 153 tons, at 10_s._ each, will cost £15³/₂, which is 76 10 0
3. Since 5_s._ is ½ of 10_s._, 153 tons, at 5_s._, will cost half as much as the same number at 10_s._ each, that is, ½ of £76 . 10, which is 38 5 0
4. Since 6_d._ is ⅒ of 5_s._, 153 tons, at 6_d._ each, will cost ⅒ of what the same number costs at 5_s._each, that is, ⅒ of £38 . 5, which is 3 16 6
5. Since 1½ or 3 halfpence is ¼ of 6_d._ or 12 halfpence, 153 tons, at 1½_d._ each, will cost ¼ of what the same number costs at 6_d._ each, that is, ¼ of£3 . 16 . 6, which is 0 19 1½ ----------- The sum of all these quantities is 425 10 7½ which is, therefore, £2 . 15 . 7½ × 153.
The whole process may be written down as follows:
or what 153 tons would | £153 0 0 cost at £1 per ton. | ----------- ----------- £2 is 2 × £1 | 306 0 0 2 0 0 10_s._ is ½ of £1 | 76 10 0 0 10 0 5_s._ is ½ of 10_s._ | 38 5 0 0 5 0 6_d._ is ⅒ of 5_s._ | 3 16 6 0 0 6 1½_d._ is ¼ of 6_d._ | 0 19 1½ 0 0 1½ | ------------ ---------- Sum | £425 10 7½ £2 15 7½
ANOTHER EXAMPLE.
What do 1735 lbs. cost at 9_s._ 10¾_d._ per lb.? The price 9_s._ 10¾_d_. is made up of 5_s._, 4_s._, 10_d._, ½_d._, and ¼_d._; of which 5_s._ is ¼ of £1, 4_s._ is ⅕ of £1, 10_d._ is ⅙ of 5_s._, ½_d._ is ¹/₂₀ of 10_d._, and ¼_d._ is ½ of ½_d._ Follow the same method as in the last example, which gives the following:
or what 1735 tons would | £1735 0 0 cost at £1 per lb. | ------------ ----------- 5_s._ is ¼ of £1 | 433 15 0 0 5 0 | 4_s._ is ⅕ of £1 | 347 0 0 0 4 0 | 10_d._ is ⅙ of 5_s._ | 72 5 10 0 0 10 | ½_d._ is ¹/₂₀ of 10_d._ | 3 12 3½ 0 0 0½ | ¼_d._ is ½ of ½_d._ | 1 16 1¾ 0 0 0¼ | ------------- ------------ by addition ... | £858 9 3¼ £0 9 10¾
In all cases, the price must first be divided into a number of parts, each of which is a simple fraction[47] of some one which goes before. No rule can be given for doing this, but practice will enable the student immediately to find out the best method for each case. When that is done, he must find how much the whole quantity would cost if each of these parts were the price, and then add the results together.
[47] Any fraction of a unit, whose numerator is unity, is generally called an _aliquot part_ of that unit. Thus, 2_s._ and 10_s._ are both aliquot parts of a pound, being £⅒ and £½.
EXERCISES.
What is the cost of
243 cwt. at £14 . 18 . 8¼ per cwt.?--_Answer_, £3629 . 1 . 0¾.
169 bushels at £2 . 1 . 3¼ per bushel?--_Answer_, £348 . 14 . 9¼.
273 qrs. at 19_s._ 2_d._ per quarter?--_Answer_, £261 . 12. 6.
2627 sacks at 7_s._ 8½_d._ per sack?--_Answer_, £1012 . 9 . 9½.
231. Throughout this section it must be observed, that the rules can be applied to cases where the quantities given are expressed in common or decimal fractions, instead of the measures in the tables. The following are examples:
What is the price of 272·3479 cwt. at £2. 1. 3½ per cwt.?
_Answer_, £562·2849, or £562. 5. 8¼.
66½lbs. at 1_s._ 4½_d._ per lb. cost £4. 11. 5¼.
How many pounds, shillings, and pence, will 279·301 acres let for if each acre lets for £3·1076?--_Answer_, £867·9558, or £867. 19. 1¼.
What does ¼ of ³/₁₃ of 17 bush. cost at ⅙ of ⅔ of £17. 14 per bushel?
_Answer_, £2·3146, or £2. 6. 3½.
What is the cost of 19lbs. 8oz. 12dwt. 8gr. at £4. 4. 6 per ounce?--_Answer_, £999. 14. 1¼ ⅙.
232. It is often required to find to how much a certain sum per day will amount in a year. This may be shortly done, since it happens that the number of days in a year is 240 + 120 + 5; so that a penny per day is a pound, half a pound, and 5 pence per year. Hence the following rule: To find how much any sum per day amounts to in a year, turn it into pence and fractions of a penny; to this add the half of itself, and let the pence be pounds, and each farthing five shillings; then add five times the daily sum, and the total is the yearly amount. For example, what does 12_s._ 3¾_d._ amount to in a year? This is 147¾_d._, and its half is 73⅞_d._, which added to 147¾_d._ gives 221⅝_d._, which turned into pounds is £221. 12. 6. Also, 12_s._ 3¾_d._ × 5 is £3. 1. 6¾, which added to the former sum gives £224. 14. 0¾ for the yearly amount. In the same way the yearly amount of 2_s._ 3½_d._ is £41. 16. 5½; that of 6¾_d._ is £10. 5. 3¾; and that of 11_d._ is £16. 14. 7.
233. An inverse rule may be formed, sufficiently correct for every purpose, in the following way: If the year consisted of 360 days, or ³/₂ of 240, the subtraction of one-third from any sum per year would give the proportion which belongs to 240 days; and every pound so obtained would be one penny per day. But as the year is not 360, but 365 days, if we divide each day’s share into 365 parts, and take 5 away, the whole of the subtracted sum, or 360 × 5 such parts, will give 360 parts for each of the 5 days which we neglected at first. But 360 such parts are left behind for each of the 360 first days; therefore, this additional process divides the whole annual amount equally among the 365 days. Now, 5 parts out of 365 is one out of 73, or the 73d part of the first result must be subtracted from it to produce the true result. Unless the daily sum be very large, the 72d part will do equally well, which, as 72 farthings are 18 pence, is equivalent to subtracting at the rate of one farthing for 18_d._, or ½_d._ for 3_s._, or 10_d._ for £3. The rule, then, is as follows: To find how much per day will produce a given sum per year, turn the shillings, &c. in the given sum into decimals of a pound (221); subtract one-third; consider the result as pence; and diminish it by one farthing for every eighteen pence, or ten pence for every £3. For example, how much per day will give £224. 14. 0¾ per year? This is 224·703, and its third is 74·901, which subtracted from 224·703, gives 149·802, which, if they be pence, amounts to 12_s._ 5·802_d._, in which 1_s._ 6_d._ is contained 8 times. Subtract 8 farthings, or 2_d._, and we have 12_s._ 3·802_d._, which differs from the truth only about ¹/₂₀ of a farthing. In the same way, £100 per year is 5_s._ 5¾_d._ per day.
234. The following connexion between the measures of length and the measures of surface is the foundation of the application of arithmetic to geometry.
Suppose an oblong figure, A, B, C, D, as here drawn (which is called a _rectangle_ in geometry), with the side A B 6 inches, and the side A C 4 inches. Divide A B and C D (which are equal) each into 6 inches by the points _a, b, c, l, m_, &c.; and A C and B D (which are also equal) into 4 inches by the points _f, g, h, x, y_, and _z_. Join _a_ and l, _b_ and _m_, &c., and _f_ and _x_, &c. Then, the figure A B C D is divided into a number of squares; for a square is a rectangle whose sides are equal, and therefore A _a f_ E is square, since A _a_ is of the same length as A _f_, both being 1 inch. There are also four rows of these squares, with six squares in each row; that is, there are 6 × 4, or 24 squares altogether. Each of these squares has its sides 1 inch in length, and is what was called in (215) _a square inch_. By the same reasoning, if one side had contained 6 _yards_, and the other 4 _yards_, the surface would have contained 6 × 4 _square yards_; and so on.
235. Let us now suppose that the sides of A B C D, instead of being a whole number of inches, contain some inches and a fraction. For example, let A B be 3½ inches, or (114) ⁷/₂ of an inch, and let A C contain 2½ inches, or ⁹/₄ of an inch. Draw A E twice as long as A B, and A F four times as long as A C, and complete the rectangle A E F G. The rest of the figure needs no description. Then, since A E is twice A B, or twice ⁷/₂ inches, it is 7 inches. And since A F is four times A C, or four times ⁹/₄ inches, it is 9 inches. Therefore, the whole rectangle A E F G contains, by (234), 7 × 9 or 63 square inches. But the rectangle A E F G contains 8 rectangles, all of the same figure as A B C D; and therefore A B C D is one-eighth part of A E F G, and contains ⁶³/₈ square inches. But ⁶³/₈ is made by multiplying ⁹/₄ and ⁷/₂ together (118). From this and the last article it appears, that, whether the sides of a rectangle be a whole or a fractional number of inches, the number of square inches in its surface is the product of the numbers of inches in its sides. The square itself is a rectangle whose sides are all equal, and therefore the number of square inches which a square contains is found by multiplying the number of inches in its side by itself. For example, a square whose side is 13 inches in length contains 13 × 13 or 169 square inches.
236. EXERCISES.
What is the content, in square feet and inches, of a room whose sides are 42 ft. 5 inch. and 31 ft. 9 inch.? and supposing the piece from which its carpet is taken to be three quarters of a yard in breadth, what length of it must be cut off?--_Answer_, The content is 1346 square feet 105 square inches, and the length of carpet required is 598 feet 6⁵/₉ inches.
The sides of a rectangular field are 253 yards and a quarter of a mile; how many acres does it contain?--_Answer_, 23.
What is the difference between 18 _square miles_, and a square of 18 miles long, or 18 _miles square_?--_Answer_, 306 square miles.
237. It is by this rule that the measure in (215) is deduced from that in (214); for it is evident that twelve inches being a foot, the square foot is 12 × 12 or 144 square inches, and so on. In a similar way it may be shewn that the content in cubic inches of a cube, or parallelepiped,[48] may be found by multiplying together the number of inches in those three sides which meet in a point. Thus, a cube of 6 inches contains 6 × 6 × 6, or 216 cubic inches; a chest whose sides are 6, 8, and 5 feet, contains 6 × 8 × 5, or 240 cubic feet. By this rule the measure in (216) was deduced from that in (214).
[48] A parallelepiped, or more properly, a _rectangular_ parallelepiped, is a figure of the form of a brick; its sides, however, may be of any length; thus, the figure of a plank has the same name. A cube is a parallelepiped with equal sides, such as is a die.
SECTION II.
RULE OF THREE.
238. Suppose it required to find what 156 yards will cost, if 22 yards cost 17_s._ 4_d._ This quantity, reduced to pence, is 208_d._; and if 22 yards cost 208_d._, each yard costs ²⁰⁸/₂₂_d_. But 156 yards cost 156 times the price of one yard, and therefore cost
208 208 × 156 ---- × 156 pence, or --------- pence (117). 22 22
Again, if 25½ French francs be 20 shillings sterling, how many francs are in £20. 15? Since 25½ francs are 20 shillings, twice the number of francs must be twice the number of shillings; that is, 51 francs are 40 shillings, and one shilling is the fortieth part of 51 francs, or ⁵¹/₄₀ francs. But £20 15_s._ contain 415 shillings (219); and since 1 shilling is ⁵¹/₄₀ francs, 415 shillings is
51 × 415 ⁵¹/₄₀ × 415 francs, or (117) -------- francs. 40
239. Such questions as the last two belong to the most extensive rule in Commercial Arithmetic, which is called the RULE OF THREE, because in it three quantities are given, and a fourth is required to be found. From both the preceding examples the following rule may be deduced, which the same reasoning will shew to apply to all similar cases.
It must be observed, that in these questions there are two quantities which are of the same sort, and a third of another sort, of which last the answer must be. Thus, in the first question there are 22 and 156 yards and 208 pence, and the thing required to be found is a number of pence. In the second question there are 20 and 415 shillings and 25½ francs, and what is to be found is a number of francs. Write the three quantities in a line, putting that one last which is the only one of its kind, and that one first which is connected with the last in the question.[49] Put the third quantity in the middle. In the first question the quantities will be placed thus:
22 yds. 156 yds. 17_s._ 4_d._
In the second question they will be placed thus:
20_s._ £20 15_s._ 25½ francs.
[49] This generally comes in the same member of the sentence. In some cases the ingenuity of the student must be employed in detecting it. The reasoning of (238) is the best guide. The following may be very often applied. If it be evident that the answer must be less than the given quantity of its kind, multiply that given quantity by the less of the other two; if greater, by the greater. Thus, in the first question, 156 yards must cost more than 22; multiply, therefore, by 156.
Reduce the first and second quantities, if necessary, to quantities of the same denomination. Thus, in the second question, £20 15_s._ must be reduced to shillings (219). The third quantity may also be reduced to any other denomination, if convenient; or the first and third may be multiplied by any quantity we please, as was done in the second question; and, on looking at the answer in (238), and at (108), it will be seen that no change is made by that multiplication. Multiply the second and third quantities together, and divide by the first. The result is a quantity of the same sort as the third in the line, and is the answer required. Thus, to the first question the answer is (238)
208 × 156 17_s._ 4_d_. × 156 ----------pence, or, which is the same thing, -------------------. 22 22
240. The whole process in the first question is as follows:[50]
yds. yds. _s._ _d._ 22 : 156 ∷ 17 . 4 12 --- 208 pence. 156 ---- 1248 1040 208 ----- 22)32448(1474¾_d._ and ¹⁴/₂₂, or ⁷/₁₁ of a farthing, 22 or (219) £6 . 2 . 10¾-⁷/₁₁. --- 104 88 ---- 164 154 ---- 108 88 -- 20 (228) 4 -- 80 66 -- 14
[50] It is usual to place points, in the manner here shewn, between the quantities. Those who have read Section VIII. will see that the Rule of Three is no more than the process for finding the fourth term of a proportion from the other three.
The question might have been solved without reducing 17_s._ 4_d._ to pence, thus:
yds. yds. _s._ _d._ 22 : 156 ∷ 17 . 4 156 (227) ---------- 22)£135 . 4 . 0(£6 . 2 . 10¾-⁷/₁₁ (228) 132 --- 3 × 20 + 4 = 64 44 -- 20 × 12 = 240 220 --- 20 × 4 = 80 66 -- 14
The student must learn by practice which is the most convenient method for any particular case, as no rule can be given.
241. It may happen that the three given quantities are all of one denomination; nevertheless it will be found that two of them are of one, and the third of another sort. For example: What must an income of £400 pay towards an income-tax of 4_s._ 6_d._ in the pound? Here the three given quantities are, £400, 4_s._ 6_d._, and £1, which are all of the same species, viz. money. Nevertheless, the first and third are income; the second is a tax, and the answer is also a tax; and therefore, by (152), the quantities must be placed thus:
£1 : £400 ∷ 4_s._ 6_d._
242. The following exercises either depend directly upon this rule, or can be shewn to do so by a little consideration. There are many questions of the sort, which will require some exercise of ingenuity before the method of applying the rule can be found.
EXERCISES.
If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?
_Answer_, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.
If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be in going 23 miles?
_Answer_, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).
Two persons, A and B, are bankrupts, and owe exactly the same sum; A can pay 15_s._ 4½_d._ in the pound, and B only 7_s._ (6¾)_d._ At the same time A has in his possession £1304. 17 more than B; what do the debts of each amount to?
_Answer_, £3340 . 8 . 3 ¾ ⁹/₂₅.
For every (12½) acres which one country contains, a second contains (56¼). The second country contains 17,300 square miles. How much does the first contain? Again, for every 3 people in the first, there are 5 in the second; and there are in the first 27 people on every 20 acres. How many are there in each country?--_Answer_, The number of square miles in the first is 3844⁴/₉, and its population 3,321,600; and the population of the second is 5,536,000.
If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will (118¼) yds. cost, if the width be 1 yd.?
_Answer_, £332. 5. (2⁴/₁₇).
If £9. 3. 6 last six weeks, how long will £100 last?
_Answer_, (65¹⁴⁵/₃₆₇) weeks.
How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea, worth 10_d._ an ounce?
_Answer_, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.
243. Suppose the following question asked: How long will it take 15 men to do that which 45 men can finish in 10 days? It is evident that one man would take 45 × 10, or 450 days, to do the same thing, and that 15 men would do it in one-fifteenth part of the time which it employs one man, that is, in (450 ÷ 15) or 30 days. By this and similar reasoning the following questions can be solved.
EXERCISES.
If 15 oxen eat an acre of grass in 12 days, how long will it take 26 oxen to eat 14 acres? _Answer_, (96¹²/₁₃) days.
If 22 masons build a wall 5 feet high in 6 days, how long will it take 43 masons to build 10 feet? _Answer_, (6⁶/₄₃) days.
244. The questions in the preceding article form part of a more general class of questions, whose solution is called the DOUBLE RULE OF THREE, but which might, with more correctness, be called the Rule of _Five_, since five quantities are given, and a sixth is to be found. The following is an example: If 5 men can make 30 yards of cloth in 3 days, how long will it take 4 men to make 68 yards? The first thing to be done is to find out, from the first part of the question, the time it will take one man to make one yard. Now, since one man, in 3 days, will do the fifth part of what 5 men can do, he will in 3 days make ³⁰/₅, or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of a day. From this we are to find how long it will take 4 men to make 68 yards. Since one man makes a yard in
3 × 5 3 × 5 ----- of a day, he will make 68 yards in ----- × 68 days, 30 30
3 × 5 × 68 or (116) in ---------- days; and 4 men will do this in one-fourth 30
3 × 5 × 68 of the time, that is (123), in ---------- days, or in 8½ days. 30 × 4
Again, suppose the question to be: If 5 men can make 30 yards in 3 days, how much can 6 men do in 12 days? Here we must first find the quantity one man can do in one day, which appears, on reasoning similar to that in the last example, to be 30/(3 × 5) yards. Hence, 6 men, in one day, will make
6 × 30 12 × 6 × 30 ------ yards, and in 12 days will make ----------- or 144 yards. 5 × 3 5 × 3
From these examples the following rule may be drawn. Write the given quantities in two lines, keeping quantities of the same sort under one another, and those which are connected with each other, in the same line. In the two examples above given, the quantities must be written thus:
SECOND EXAMPLE.
Draw a curve through the middle of each line, and the extremities of the other. There will be three quantities on one curve and two on the other. Divide the product of the three by the product of the two, and the quotient is the answer to the question.
If necessary, the quantities in each line must be reduced to more simple denominations (219), as was done in the common Rule of Three (238).
EXERCISES.
If 6 horses can, in 2 days, plough 17 acres, how many acres will 93 horses plough in 4½ days?
_Answer_, 592⅞.
If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each of which are 40 and 50 yards, how long will 37 men be in digging 53 fields, the sides of each of which are 90 and 125½ yards?
2451 _Answer_, 75----- days. 20720
If the carriage of 60 cwt. through 20 miles cost £14 10_s._, what weight ought to be carried 30 miles for £5. 8. 9?
_Answer_, 15 cwt.
If £100 gain £5 in a year, how much will £850 gain in 3 years and 8 months?
_Answer_, £155. 16. 8.
SECTION III.
INTEREST, ETC.
245. In the questions contained in this Section, almost the only process which will be employed is the taking a fractional part of a sum of money, which has been done before in several cases. Suppose it required to take 7 parts out of 40 from £16, that is, to divide £16 into 40 equal parts, and take 7 of them. Each of these parts is
16 16 16 × 7 £----, and 7 of them make ---- × 7, or ------ pounds (116). 40 40 40
The process may be written as below:
£16 7 ----- 40)112(£2 . 16_s._ 80 -- 32 20 --- 640 40 --- 240 240 --- 0
Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.
56 . 13 . 7½ 13 ---------------- 100) 736 . 17 . 1½ ( £7 . 7 . 4 ¼ ¹/₄₁ 700 --- 36 × 20 + 17 = 737 700 --- 37 × 12 + 1 = 445 400 --- 45 × 4 × 2 = 182 100 --- 82
Let it be required to take 2½ parts out of a hundred from £3 12_s._ The result, by the same rule is
£3 12_s._ × 2½ 5 -------------------, or 123 £3 12_s._ × ---; 100 200
so that taking 2½ out of a hundred is the same as taking 5 parts out of 200.
EXERCISES.
Take 7⅓ parts out of 53 from £1 10_s._
129 _Answer_, 4_s._ 1---_d._ 159
Take 5 parts out of 100 from £107 13_s._ 4¾_d._
_Answer_, £5. 7. 8 and ³/₂₀ of a farthing.
£56 3_s._ 2_d._ is equally divided among 32 persons. How much does the share of 23 of them exceed that of the rest?
_Answer_, £24. 11. 4½ ½.
246. It is usual, in mercantile business, to mention the fraction which one sum is of another, by saying how many parts out of a hundred must be taken from the second in order to make the first. Thus, instead of saying that £16 12_s._ is the half of £33 4_s._, it is said that the first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200; because, if £200 be divided into 100 parts, 2½ of those parts are £5. Also, £13 is 150 per cent of £8. 13. 4, since the first is the second and half the second. Suppose it asked, How much per cent is 23 parts out of 56 of any sum? The question amounts to this: If he who has £56 gets £100 for them, how much will he who has 23 receive? This, by 238, is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per cent.
Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2 parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.
From which the method of reducing other fractions to the rate per cent is evident.
Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the first contains 1586_d._, and the second 2916_d._, the first is 1586 out of 2916 parts of the second; that is, by the last rule, it is ¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The more expeditious way of doing this is to reduce the shillings, &c. to decimals of a pound. Three decimal places will give the rate per cent to the nearest shilling, which is near enough for all practical purposes. For instance, in the last example, which is to find how much £6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives £54·38, or £54. 7. Greater correctness may be had, if necessary, as in the Appendix.
EXERCISES.
How much per cent is 198¼ out of 233 parts?--_Ans._ £85. 1. 8¾.
Goods which are bought for £193. 12, are sold for £216. 13. 4; how much per cent has been gained by them?
_Answer_, A little less than £11. 18. 6.
A sells goods for B to the amount of £230. 12, and is allowed a commission[51] of 3 per cent; what does that amount to?
_Answer_, £6 . 18. 4¼ ⁷/₂₅.
[51] Commission is what is allowed by one merchant to another for buying or selling goods for him, and is usually a per-centage on the whole sum employed. Brokerage is an allowance similar to commission, under a different name, principally used in the buying and selling of stock in the funds.
Insurance is a per-centage paid to those who engage to make good to the payers any loss they may sustain by accidents from fire, or storms, according to the agreement, up to a certain amount which is named, and is a per-centage upon this amount. Tare, tret, and cloff, are allowances made in selling goods by wholesale, for the weight of the boxes or barrels which contain them, waste, &c.; and are usually either the price of a certain number of pounds of the goods for each box or barrel, or a certain allowance on each cwt.
A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what does he receive?--_Answer_, £2. 2. 6.
A ship whose value is £15,423 is insured at 19⅔ per cent; what does the insurance amount to?--_Answer_, £3033. 3. 9½ ²/₅.
247. In reckoning how much a bankrupt is able to pay his creditors, as also to how much a tax or rate amounts, it is usual to find how many shillings in the pound is paid. Thus, if a person who owes £100 can only pay £50, he is said to pay 10_s._ in the pound. The rule is easily derived from the same reasoning as in 246. For example, £50 out of £82 is
50 50×20 £---- out of £1, or ----- shillings, 82 82
or 12_s._ 2½ ¹⁵/₄₁ in the pound.
248. INTEREST is money paid for the use of other money, and is always a per-centage upon the sum lent. It may be paid either yearly, half-yearly, or quarterly; but when it is said that £100 is lent at 4 per cent, it must be understood to mean 4 per cent per annum; that is, that 4 pounds are paid every year for the use of £100.
The sum lent is called the _principal_, and the interest upon it is of two kinds. If the borrower pay the interest as soon as, from the agreement, it becomes due, it is evident that he has to pay the same sum every year; and that the whole of the interest which he has to pay in any number of years is one year’s interest multiplied by the number of years. But if he do not pay the interest at once, but keeps it in his hands until he returns the principal, he will then have more of his creditor’s money in his hands every year, and if it were so agreed will have to pay interest upon each year’s interest for the time during which he keeps it after it becomes due. In the first case, the interest is called _simple_, and in the second _compound_. The interest and principal together are called the _amount_.
249. What is the simple interest of £1049. 16. 6 for 6 years and one-third, at 4½ per cent? This interest must be 6⅓ times the interest of the same sum for one year, which (245) is found by multiplying the sum by 4½, and dividing by 100. The process is as follows:
(230) (_a_) |£1049 . 16 . 6 +-------------- _a_ × 4 | 4199 . 6 . 0 _a_ × ½ | 524 . 18 . 3 +--------------
(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀
20 ---- (228) 4,84[52] 12 ------ 10,11[53]
(_b_) £47 . 4 . 10¹¹/₁₀₀ Int. for one yr. +------------------ _b_ × 6 | 283 . 9 . 0⁶⁶/₁₀₀ _b_ × ⅓ | 15 . 14 . 11³⁷/₁₀₀ +--------------------- £299 . 4 . 0³/₁₀₀ Int. for 6⅓ yrs.
[52] Here the 4_s._ from the dividend is taken in.
[53] Here the 3_d._ from the dividend is taken in.
EXERCISES.
What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per cent?
_Answer_, £60. 9, very nearly.
What is the difference between the interest of £50. 19 for 7 years at 3 per cent, and for 8 years at 2½ per cent? _Answer_, 10_s._ (2½)_d._
What is the interest of £157. 17. 6 for one year at 5 per cent?
_Answer_, £7. 17. 10½.
Shew that the interest of any sum for 9 years at 4 per cent is the same as that of the same sum for 4 years at 9 per cent.
250. In order to find the interest of any sum at compound interest, it is necessary to find the amount of the principal and interest at the end of every year; because in this case (248) it is the amount of both principal and interest at the end of the first year, upon which interest accumulates during the second year. Suppose, for example, it is required to find the interest, for 3 years, on £100, at 5 per cent, compound interest. The following is the process:
£100 First principal. 5 First year’s interest. --- 105 Amount at the end of the first year. (249) 5 . 5 Interest for the second year on £105. -------- 110 . 5 Amount at the end of two years. 5 . 10 . 3 Interest due for the third year. ------------ 115 . 15 . 3 Amount at the end of three years. 100 . 0 . 0 First principal. ------------ 15 . 15 . 3 Interest gained in the three years.
When the number of years is great, and the sum considerable, this process is very troublesome; on which account tables[54] are constructed to shew the amount of one pound, for different numbers of years, at different rates of interest. To make use of these tables in the present example, look into the column headed “5 per cent;” and opposite to the number 3, in the column headed “Number of years,” is found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now, £100 must become 100 times as great; and 1·157625 × 100 is 115·7625 (141); but (221) £·7625 is 15_s._ 3_d._; therefore the whole amount of £100 is £115. 15. 3, as before.
[54] Sufficient tables for all common purposes are contained in the article on Interest in the Penny Cyclopædia; and ample ones in the Treatise on Annuities and Reversions, in the Library of Useful Knowledge.
251. Suppose that a sum of money has lain at simple interest 4 years, at 5 per cent, and has, with its interest, amounted to £350; it is required to find what the sum was at first. Whatever the sum was, if we suppose it divided into 100 parts, 5 of those parts were added every year for 4 years, as interest; that is, 20 of those parts have been added to the first sum to make £350. If, therefore, £350 be divided into 120 parts, 100 of those parts are the principal which we want to find, and 20 parts are interest upon it; that is, the principal is £(350 × 100)/150, or £291. 13. 4.
252. Suppose that A was engaged to pay B £350 at the end of four years from this time, and that it is agreed between them that the debt shall be paid immediately; suppose, also, that money can be employed at 5 per cent, simple interest; it is plain that A ought not to pay the whole sum, £350, because, if he did, he would lose 4 years’ interest of the money, and B would gain it. It is fair, therefore, that he should only pay to B as much as will, _with interest_, amount in four years to £350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck off the debt in consideration of its being paid before the time. This is called DISCOUNT;[55] and £291. 13. 4 is called the _present value_ of £350 due four years hence, discount being at 5 per cent. The rule for finding the present value of a sum of money (251) is: Multiply the sum by 100, and divide the product by 100 increased by the product of the rate per cent and number of years. If the time that the debt has yet to run be expressed in years and months, or months only, the months must be reduced to the equivalent fraction of a year.
[55] This rule is obsolete in business. When a bill, for instance, of £100 having a year to run, is _discounted_ (as people now say) at 5 per cent, this means that 5 per cent of £100, or £5, is struck off.
EXERCISES.
What is the discount on a bill of £138. 14. 4, due 2 years hence, discount being at 4½ per cent?
_Answer_, £11. 9. 1.
What is the present value of £1031. 17, due 6 months hence, interest being at 3 per cent?
_Answer_, £1016. 12.
253. If we multiply by _a_ + _b_, or by _a_-_b_, when we should multiply by _a_, the result is wrong by the fraction
_b_ _b_ --- + _b_, or ---------, _a_ _a_ - _b_
of itself: being too great in the first case, and too small in the second. Again, if we divide by _a_ + _b_, where we should have divided by _a_, the result is too small by the fraction _b_/_a_ of itself; while, if we divide by _a_-_b_ instead of _a_, the result is too great by the same fraction of itself. Thus, if we divide by 20 instead of 17, the result is ³/₁₇ of itself too small; and if we divide by 360 instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.
If, then, we wish to find the interest of a sum of money for a portion of a year, and have not the assistance of tables, it will be found convenient to suppose the year to contain only 360 days, in which case its 73d part (the 72d part will generally do) must be subtracted from the result, to make the alteration of 360 into 365. The number 360 has so large a number of divisors, that the rule of Practice (230) may always be readily applied. Thus, it is required to find the portion which belongs to 274 days, the yearly interest being £18. 9. 10, or 18·491.
274 18·491 ------ 180 is ½ of 360 9·246 --- 94 90 is ½ of 180 4·623 -- 4 is ¹/₉₀ of 360 ·205 ------ 9)14·074 ------ 8)1·564 ----- ·196 13·878 = £13 . 17 . 7 _Answer._
But if the nearest farthing be wanted, the best way is to take 2-tenths of the number of days as a multiplier, and 73 as a divisor; since _m_ ÷ 365 is 2_m_ ÷ 730, or (²/₁₀)_m_ ÷ 73. Thus, in the preceding instance, we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068, which divided by 73 gives 13·881, very nearly agreeing with the former, and giving £13. 17. 7½, which is certainly within a farthing of the truth.
254. Suppose it required to divide £100 among three persons in such a way that their shares may be as 6, 5, and 9; that is, so that for every £6 which the first has, the second may have £5, and the third £9. It is plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first must have 6 of those parts, the second 5, and the third 9. Therefore (245) their shares are respectively,
100 × 6 100 × 5 100 × 9 £-------, £------- and £-------, or £30, £25, and £45. 20 20 20
EXERCISES.
Divide £394. 12 among four persons, so that their shares may be as 1, 6, 7, and 18.--_Answer_, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19. 3.
Divide £20 among 6 persons, so that the share of each may be as much as those of all who come before put together.--_Answer_, The first two have 12_s._ 6_d._; the third £1. 5; the fourth £2. 10; the fifth £5; and the sixth £10.
255. When two or more persons employ their money together, and gain or lose a certain sum, it is evidently not fair that the gain or loss should be equally divided among them all, unless each contributed the same sum. Suppose, for example, A contributes twice as much as B, and they gain £15, A ought to gain twice as much as B; that is, if the whole gain be divided into 3 parts, A ought to have two of them and B one, or A should gain £10 and B £5. Suppose that A, B, and C engage in an adventure, in which A embarks £250, B £130, and C £45. They gain £1000. How much of it ought each to have? Each one ought to gain as much for £1 as the others. Now, since there are 250 + 130 + 45, or 425 pounds embarked, which gain £1000, for each pound there is a gain of £¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain 1000 × ¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles, by the process in (245), the following questions may be answered.
A ship is to be insured, in which A has ventured £1928, and B £4963. The expense of insurance is £474. 10. 2. How much ought each to pay of it?
_Answer_, A must pay £132. 15. (2½).
A loss of £149 is to be made good by three persons, A, B, and C. Had there been a gain, A would have gained 4 times as much as B, and C as much as A and B together. How much of the loss must each bear?
_Answer_, A pays £59. 12, B £14. 18, and C £74. 10.
256. It may happen that several individuals employ several sums of money together for different times. In such a case, unless there be a special agreement to the contrary, it is right that the more time a sum is employed, the more profit should be made upon it. If, for example, A and B employ the same sum for the same purpose, but A’s money is employed twice as long as B’s, A ought to gain twice as much as B. The principle is, that one pound employed for one month, or one year, ought to give the same return to each. Suppose, for example, that A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months, and the gain is £100; how much ought each to have of it? Now, since A employs £3 for six months, he must gain 6 times as much as if he employed it one month only; that is, as much as if he employed £6 × 3, or £18, for one month; also, B gains as much as if he had employed £4 × 7 for one month; and C as if he had employed £12 × 2 for one month. If, then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of those parts. The shares of the three are, therefore,
6 × 3 × 100 4 × 7 × 100 £----------------------, £----------------------, 6 × 3 + 4 × 7 + 12 × 2 6 × 3 + 4 × 7 + 12 × 2
12 × 2 × 100 and £----------------------. 6 × 3 + 4 × 7 + 12 × 2
EXERCISES.
A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B £100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much must each receive of the gain?
_Answer_, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.
A, B, and C rent a house together for 2 years, at £150 per annum. A remains in it the whole time, B 16 months, and C 4½ months, during the occupancy of B. How much must each pay of the rent?[56]
_Answer_, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.
[56] This question does not at first appear to fall under the rule. A little thought will serve to shew that what probably will be the first idea of the proper method of solution is erroneous.
257. These are the principal rules employed in the application of arithmetic to commerce. There are others, which, as no one who understands the principles here laid down can fail to see, are virtually contained in those which have been given. Such is what is commonly called the Rule of Exchange, for such questions as the following: If 20 shillings be worth 25½ francs, in France, what is £160 worth? This may evidently be done by the Rule of Three. The rules here given are those which are most useful in common life; and the student who understands them need not fear that any ordinary question will be above his reach. But no student must imagine that from this or any other book of arithmetic he will learn precisely the modes of operation which are best adapted to the wants of the particular kind of business in which his future life may be passed. There is no such thing as a set of rules which are at once most convenient for a butcher and a banker’s clerk, a grocer and an actuary, a farmer and a bill-broker; but a person with a good knowledge of the _principles_ laid down in this work, will be able to examine and meet his own future wants, or, at worst, to catch with readiness the manner in which those who have gone before him have done so for themselves.
APPENDIX TO THE FIFTH EDITION OF
DE MORGAN’S ELEMENTS OF ARITHMETIC.
I. ON THE MODE OF COMPUTING.
The rules in the preceding work are given in the usual form, and the examples are worked in the usual manner. But if the student really wish to become a ready computer, he should strictly follow the methods laid down in this Appendix; and he may depend upon it that he will thereby save himself trouble in the end, as well as acquire habits of quick and accurate calculation.
I. In numeration learn to connect each primary decimal number, 10, 100, 1000, &c. not with the place in which the unit falls, but with the number of ciphers following. Call ten a _one-cipher_ number, a hundred a _two-cipher_ number, a million a _six-cipher_ number, and so on. If _five_ figures be cut off from a number, those that are left are hundred-thousands; for 100,000 is a _five-cipher_ number. Learn to connect tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, &c. with 1, 2, 3, 4, 5, 6, &c. in the mind. What is a _seventeen-cipher_ number? For every 6 in seventeen say _million_, for the remaining 5 say _hundred-thousand_: the answer is a hundred thousand millions of millions. If twelve places be cut off from the right of a number, what does the remaining number stand for?--_Answer_, As many millions of millions as there are units in it when standing by itself.
II. After learning to count forwards and backwards with rapidity, as in 1, 2, 3, 4, &c. or 30, 29, 28, 27, &c., learn to count forwards or backwards by twos, threes, &c. up to nines at least, beginning from any number. Thus, beginning from four and proceeding by sevens, we have 4, 11, 18, 25, 32, &c., along which series you must learn to go as easily as along the series 1, 2, 3, 4, &c.; that is, as quick as you can pronounce the words. The act of addition must be made in the mind without assistance: you must not permit yourself to say, 4 and 7 are 11, 11 and 7 are 18, &c.; but only 4, 11, 18, &c. And it would be desirable, though not so necessary, that you should go back as readily as forward; by sevens for instance, from sixty, as in 60, 53, 46, 39, &c.
III. Seeing a number and another both of one figure, learn to catch instantly the number you must add to the smaller to get the greater. Seeing 3 and 8, learn by practice to think of 5 without the necessity of saying 3 _from_ 8 _and there remains_ 5. And if the second number be the less, as 8 and 3, learn also by practice how to pass _up_ from 8 to the next number which ends with 3 (or 13), and to catch the necessary augmentation, _five_, without the necessity of formally undertaking in words to subtract 8 from 13. Take rows of numbers, such as
4 2 6 0 5 0 1 8 6 4
and practise this rule upon every figure and the next, not permitting yourself in this simple case ever to name the higher one. Thus, say 4 and 8 (4 first, 2 second, 4 from the next number that ends with 2, or 12, leaves 8), 2 and 4, 6 and 4, 0 and 5, 5 and 5, 0 and 1, 1 and 7, 8 and 8, 6 and 8.
IV. Study the same exercise as the last one with two figures and one. Thus, seeing 27 and 6, pass from 27 up to the next number that ends with 6 (or 36), catch the 9 through which you have to pass, and allow yourself to repeat as much as “27 and 9 are 36.” Thus, the row of figures 17729638109 will give the following practice: 17 and 0 are 17; 77 and 5 are 82; 72 and 7 are 79; 29 and 7 are 36; 96 and 7 are 103; 63 and 5 are 68; 38 and 3 are 41; 81 and 9 are 90; 10 and 9 are 19.
V. In a number of two figures, practise writing down the units at the moment that you are keeping the attention fixed upon the tens. In the preceding exercise, for instance, write down the results, repeating the tens with emphasis at the instant of writing down the units.
VI. Learn the multiplication table so well as to name the product the instant the factors are seen; that is, until 8 and 7, or 7 and 8, suggest 56 at once, without the necessity of saying “7 times 8 are 56.” Thus looking along a row of numbers, as 39706548, learn to name the products of every successive pair of digits as fast as you can repeat them, namely, 27, 63, 0, 0, 30, 20, 32.
VII. Having thoroughly mastered the last exercise, learn further, on seeing three numbers, to augment the product of the first and second by the third without any repetition of words. Practise until 3, 8, 4, for instance, suggest 3 times 8 and 4, or 28, without the necessity of saying “3 times 8 are 24, and 4 is 28.” Thus, 179236408 will suggest the following practice, 16, 65, 21, 12, 22, 24, 8.
VIII. Now, carry the last still further, as follows: Seeing four figures, as 2, 7, 6, 9, catch up the product of the first and second, increased by the third, as in the last, without a helping word; name the result, and add the next figure, name the whole result, laying emphasis upon the tens. Thus, 2, 7, 6, 9, must immediately suggest “20 and 9 are 29.” The row of figures 773698974 will give the instances 52 and 6 are 58; 27 and 9 are 36; 27 and 8 are 35; 62 and 9 are 71; 81 and 7 are 88; 79 and 4 are 83.
IX. Having four numbers, as 2, 4, 7, 9, vary the last exercise as follows: Catch the product of the first and second, increased by the third; but instead of adding the fourth, go up to the next number that ends with the fourth, as in exercise IV. Thus, 2, 4, 7, 9, are to suggest “15 and 4 are 19.” And the row of figures 1723968929 will afford the instances 9 and 4 are 13; 17 and 2 are 19; 15 and 1 are 16; 33 and 5 are 38; 62 and 7 are 69; 57 and 5 are 62; 74 and 5 are 79.
X. Learn to find rapidly the number of times a digit is contained in given units and tens, with the remainder. Thus, seeing 8 and 53, arrive at and repeat “6 and 5 over.” Common short division is the best practice. Thus, in dividing 236410792 by 7,
7)236410792 --------- 33772970, remainder 2.
All that is repeated should be 3 and 2; 3 and 5; 7 and 5; 7 and 2; 2 and 6; 9 and 4; 7 and 0; 0 and 2.
In performing the several rules, proceed as follows:
ADDITION. Not one word more than repeating the numbers written in the following process: the accented figure is the one to be written down; the doubly accented figure is carried (and don’t _say_ “carry 3,” but do it).
47963 1598 26316 54792 819 6686 ------ 138174
6, 15, 17, 23, 31, 3″ 4′; 11, 12, 21, 22, 31, 3″7′; 9, 17, 24, 27, 32, 4″1′; 10, 14, 20, 21, 2″8′; 7, 9, 1′3′.
In verifying additions, instead of the usual way of omitting one line, adding without it, and then adding the line omitted, verify each column by adding it both upwards and downwards.
SUBTRACTION. The following process is enough. The carriages, being always of _one_, need not be mentioned.
From 79436258190 Take 58645962738 ----------- 20790295452
8 and 2′, 4 and 5′, 7 and 4′, 3 and 5′, 6 and 9′, 10 and 2′, 6 and 0′, 4 and 9′, 7 and 7′, 9 and 0′, 5 and 2′. It is useless to stop and say, 8 and 2 make 10; for as soon as the 2 is obtained, there is no occasion to remember what it came from.
MULTIPLICATION. The following, put into words, is all that need be repeated in the multiplying part; the addition is then done as usual. The unaccented figures are carried.
670383 9876 ------- 4022298 18′, 49′, 22′, 2′, 42′, 4′0′, 4692681 21′, 58′, 26′, 2′, 49′, 4′6′, 5363064 24′, 66′, 30′, 3′, 56′, 5′3′, 6033447 27′, 74′, 34′, 3′, 63′, 6′0′. ---------- 6620702508
Verify each line of the multiplication and the final result by casting out the nines. (_Appendix_ II. p. 166.)
It would be almost as easy, for a person who has well practised the 8th exercise, to add each line to the one before in the process, thus:
670383 9876 ------- 4022298 50949108 587255508 6620702508
8; 21 and 9 are 30′; 59 and 2 are 61′; 27 and 2 are 29; 2 and 2 are 4′; 49 and 0 are 49′; 46 and 4 are 5′0′.
On the right is all the process of forming the second line, which completes the multiplication by 76, as the third line completes that by 876, and the fourth line that by 9876.
DIVISION. Make each multiplication and the following subtraction in one step, by help of the process in the 9th exercise, as follows:
27693)441972809662(15959730 165042 265778 165410 269459 202226 83756 6772
The number of words by which 26577 is obtained from 165402 (the multiplier being 5) is as follows: 15 and 7′ are 2″2; 47 and 7′ are 5″4; 35 and 5′ are 4″0; 39 and 6′ are 4″5; 14 and 2′ are 16.
The processes for extracting the square root, and for the solution of equations (_Appendix_ XI.), should be abbreviated in the same manner as the division.[57]
[57] The teacher will find further remarks on this subject in the _Companion to the Almanac_ for 1844, and in the _Supplement to the Penny Cyclopædia_, article _Computation_.
APPENDIX II.
ON VERIFICATION BY CASTING OUT NINES AND ELEVENS.
The process of _casting out the nines_, as it is called, is one which the young computer should learn and practise, as a check upon his computations. It is not a complete check, since if one figure were made too small, and another as much too great, it would not detect this double error; but as it is very unlikely that such a double error should take place, the check furnishes a strong presumption of accuracy.
The proposition upon which this method depends is the following: If _a, b, c, d_ be four numbers, such that
_a_ = _bc_ + _d_,
and if _m_ be any other number whatsoever, and if _a, b, c, d_, severally divided by _m_, give the remainders _p, q, r, s_, then
_p_ and _qr_ + _s_
give the same remainder when divided by _m_ (and perhaps are themselves equal).
For instance, 334 = 17 × 19 + 11;
divide these four numbers by 7, the remainders are 5, 3, 5, and 4. And 5 and 5 × 3 + 4, or 5 and 19, both leave the remainder 5 when divided by 7.
Any number, therefore, being used as a divisor, may be made a check upon the correctness of an operation. To provide a check which may be most fit for use, we must take a divisor the remainder to which is most easily found. The most convenient divisors are 3, 9, and 11, of which 9 is far the most useful.
As to the numbers 3 and 9, the remainder is always the same as that of the sum of the digits. For instance, required the remainder of 246120377 divided by 9. The sum of the digits is 2 + 4 + 6 + 1 + 2 + 0 + 3 + 7 + 7, or 32, which gives the remainder 5. But the easiest way of proceeding is by throwing out nines as fast as they arise in the sum. Thus, repeat 2, 6 (2 + 4), 12 (6 + 6), say 3 (throwing out 9), 4, 6, 9 (throw this away), 7, 14, (or throwing out the 9) 5. This is the remainder required, as would appear by dividing 246120377 by 9. A proof may be given thus: It is obvious that each of the numbers, 1, 10, 100, 1000, &c. divided by 9, leaves a remainder 1, since they are 1, 9 + 1, 99 + 1, &c. Consequently, 2, 20, 200, &c. leave the remainder 2; 3, 30, 300, the remainder 3; and so on. If, then, we divide, say 1764 by 9 in parcels, 1000 will be one more than an exact number of nines, 700 will be seven more, and 60 will be six more. So, then, from 1, 7, 6, 4, put together, and the nines taken out, comes the only remainder which can come from 1764.
To apply this process to a multiplication: It is asserted, in page 32, that
10004569 × 3163 = 31644451747.
In casting out the nines from the first, all that is necessary to repeat is, one, five, ten, one, _seven_; in the second, three, four, ten, one, _four_; in the third, three, four, ten, one, five, nine, four, nine, eight, twelve, three, ten, _one_. The remainders then are, 7, 4, 1. Now, 7 × 4 is 28, which, casting out the nines, gives 1, the same as the product.
Again, in page 43, it is asserted that
23796484 = 130000 × 183 + 6484.
Cast out the nines from 13000, 183, 6484, and we have 4, 3, and 4. Now, 4 × 3 + 4, with the nines cast out, gives 7; and so does 23796484.
To avoid having to remember the result of one side of the equation, or to write it down, in order to confront it with the result of the other side, proceed as follows: Having got the remainder of the more complicated side, into which two or more numbers enter, subtract it from 9, and carry the remainder into the simple side, in which there is only one number. Then the remainder of that side ought to be 0. Thus, having got 7 from the left-hand of the preceding, take 2, the rest of 9, forget 7, and carry in 2 as a beginning to the left-hand side, giving 2, 4, 7, 14, 5, 11, 2, 6, 14, 5, 9, 0.
Practice will enable the student to cast out nines with great rapidity.
This process of casting out the nines does not detect any errors in which the remainder to 9 happens to be correct. If a process be tedious, and some additional check be desirable, the method of casting out _elevens_ may be followed after that of casting out the nines. Observe that 10 + 1, 100-1, 1000 + 1, 10000-1, &c. are all divisible by eleven. From this the following rule for the remainder of division by 11 may be deduced, and readily used by those who know the algebraical process of subtraction. For those who have not got so far, it may be doubted whether the rule can be made easier than the actual division by 11.
Subtract the first figure from the second, the result from the third, the result from the fourth, and so on. The final result, or the rest of 11 if the figure be negative, is the remainder required. Thus, to divide 1642915 by 11, and find the remainder, we have 1 from 6, 5; 5 from 4,-1;-1 from 2, 3; 3 from 9, 6; 6 from 1,-5;-5 from 5, 10; and 10 is the remainder. But 164 gives-1, and 10 is the remainder; 164291 gives-5, and 6 is the remainder. With very little practice these remainders may be read as rapidly as the number itself. Thus, for 127619833424 need only be repeated, 1, 6, 0, 1, 8, 0, 3, 0, 4,-2, 6, and 6 is the remainder.
When a question has been tried both by nines and elevens, there can be no error unless it be one which makes the result wrong by a number of times 99 exactly.
APPENDIX III.
ON SCALES OF NOTATION.
We are so well accustomed to 10, 100, &c., as standing for ten, ten tens, &c., that we are not apt to remember that there is no reason why 10 might not stand for five, 100 for five fives, &c., or for twelve, twelve twelves, &c. Because we invent different columns of numbers, and let units in the different columns stand for collections of the units in the preceding columns, we are not therefore bound to allow of no collections except in tens.
If 10 stood for 2, that is, if every column had its unit double of the unit in the column on the right, what we now represent by 1, 2, 3, 4, 5, 6, &c., would be represented by 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, &c. This is the _binary_ scale. If we take the _ternary_ scale, in which 10 stands for 3, we have 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, &c. In the _quinary_ scale, in which 10 is five, 234 stands for 2 twenty-fives, 3 fives, and 4, or sixty-nine. If we take the _duodenary_ scale, in which 10 is twelve, we must invent new symbols for ten and eleven, because 10 and 11 now stand for twelve and thirteen; use the letters _t_ and _e_. Then 176 means 1 twelve-twelves, 7 twelves, and 6, or two hundred and thirty-four; and 1_te_ means two hundred and seventy-five.
The number which 10 stands for is called the _radix_ of the _scale of notation_. To change a number from one scale into another, divide the number, written as in the first scale, by the number which is to be the radix of the new scale; repeat this division again and again, and the remainders are the digits required. For example, what, in the quinary scale, is that number which, in the decimal scale, is 17036?
5)17036 ----- 5)3407 Remʳ. 1 ---- 5)681 2 --- 5)136 1 --- 5)27 1 -- 5)5 2 - 5)1 0 - 0 1
_Answer_ 1021121
Quinary. Decimal. _Verification_, 1000000 means 15625 20000 1250 1000 125 100 25 20 10 1 1 ------ ----- 1021121 17036
The reason of this rule is easy. Our process of division is nothing but telling off 17036 into 3407 fives and 1 over; we then find 3407 fives to be 681 fives of fives and 2 _fives_ over. Next we form 681 fives of fives into 136 fives of fives of fives and 1 five of fives over; and so on.
It is a useful exercise to multiply and divide numbers represented in other scales of notation than the common or decimal one. The rules are in all respects the same for all systems, _the number carried being always the radix of the system_. Thus, in the quinary system we carry fives instead of tens. I now give an example of multiplication and division:
Quinary. Decimal. 42143 means 2798 1234 194 ------ ----- 324232 11192 232034 25182 134341 2798 42143 --------- ------ 114332222 542812
Duodecimal. Decimal. 4_t_9)76_t_4_e_08(16687 705)22610744(32071 4_t_9 1460 ----- 5074 2814 1394 2546 689 ----- 28_te_ 2546 ------ 3650 3320 ----- 3308 2_t_33 ------ 495
Another way of turning a number from one scale into another is as follows: Multiply the first digit by the _old_ radix _in the new scale_, and add the next digit; multiply the result again by the old radix in the new scale, and take in the next digit, and so on to the end, always using the radix of the scale you want to leave, and the notation of the scale you want to end in.
Thus, suppose it required to turn 16687 (duodecimal) into the decimal scale, and 16432 (septenary) into the quaternary scale:
16687 16432 Duodecimals into Decimals. Septenaries into Quaternaries. 1 × 12 + 6 = 18 1 × 7 + 6 = 31 × 12 + 6 × 7 + 4 --- ---- 222 1133 × 12 + 8 × 7 + 3 ---- ----- 2672 22130 × 12 + 7 × 7 + 2 ------ ------- _Answer_ 32071 1021012
Owing to our division of a foot into 12 equal parts, the duodecimal scale often becomes very convenient. Let the square foot be also divided into 12 parts, each part is 12 square inches, and the 12th of the 12th is one square inch. Suppose, now, that the two sides of an oblong piece of ground are 176 feet 9 inches 7-12ths of an inch, and 65 feet 11 inches 5-12ths of an inch. Using the duodecimal scale, and _duodecimal fractions_, these numbers are 128·97 and 55·_e_5. Their product, the number of square feet required, is thus found:
128·97 55·_e_5 --------- 617_ee_ 116095 617_ee_ 617_ee_ ------------ 68_e_8144_e_
_Answer_, 68_e_8·144_e_ (duod.) square feet, or 11660 square feet 16 square inches ⁴/₁₂ and ¹¹/₁₄₄ of a square inch.
It would, however, be exact enough to allow 2-hundredths of a foot for every quarter of an inch, an additional hundredth for every 3 inches,[58] and 1-hundredth more if there be a 12th or 2-12ths above the quarter of an inch. Thus, 9⁷/₁₂ inches should be ·76 + ·03 + ·01, or ·80, and 11⁵/₁₂ would be ·95; and the preceding might then be found decimally as 176·8 × 65·95 as 11659·96 square feet, near enough for every practical purpose.
[58] And at discretion one hundredth more for a large fraction of three inches.
APPENDIX IV.
ON THE DEFINITION OF FRACTIONS.
The definition of a fraction given in the text shews that ⁷/₉, for instance, is the _ninth_ part of _seven_, which is shewn to be the same thing as _seven-ninths_ of a unit. But there are various modes of speech under which a fraction may be signified, all of which are more or less in use.
1. In ⁷/₉ we have the 9th part of 7.
2. 7-9ths of a unit.
3. The fraction which 7 is of 9.
4. The times and parts of a time (in this case part of a time only) which 7 contains 9.
5. The multiplier which turns _nines_ into _sevens_.
6. The _ratio_ of 7 to 9, or the _proportion_ of 7 to 9.
7. The multiplier which alters a number in the ratio of 9 to 7.
8. The 4th proportional to 9, 1, and 7.
The first two views are in the text. The third is deduced thus: If we divide 9 into 9 equal parts, each is 1, and 7 of the parts are 7; consequently the fraction which 7 is of 9 is ⁷/₉. The fourth view follows immediately: For _a time_ is only a word used to express one of the repetitions which take place in multiplication, and we allow ourselves, by an easy extension of language, to speak of a portion of a number as being that number taken a _part of a time_. The fifth view is nothing more than a change of words: A number reduced to ⁷/₉ of its amount has every 9 converted into a 7, and any fraction of a 9 which may remain over into the corresponding fraction of 7. This is completely proved when we prove the equation ⁷/₉ of _a_ = 7 times _a_/9. The sixth, seventh, and eighth views are illustrated in the chapter on proportion.
When the student comes to algebra, he will find that, in all the applications of that science, fractions such as _a_/_b_ most frequently require that _a_ and _b_ should be themselves supposed to be fractions. It is, therefore, of importance that he should learn to accommodate his views of a fraction to this more complicated case.
2½ Suppose we take -----. 4³/₅
We shall find that we have, in this case, a better idea of the views from and after the third inclusive, than of the first and second, which are certainly the most simple ways of conceiving ⁷/₉. We have no notion of the (4³/₅)th part of 2½,
1 ( 3 ) nor of 2 ---(4---)ths 2 ( 5 )
of a unit; indeed, we coin a new species of adjective when we talk of the (4³/₅)th part of anything. But we can readily imagine that 2½ is some fraction of 4³/₅; that the first is _some_ part of a time the second; that there must be _some_ multiplier which turns every 4³/₅ in a number into 2½; and so on. Let us now see whether we can invent a distinct mode of applying the first and second views to such a compound fraction as the above.
We can easily imagine a fourth part of a length, and a fifth part, meaning the lines of which 4 and 5 make up the length in question; and there is also in existence a length of which four lengths and two-fifths of a length make up the original length in question. For instance, we might say that 6, 6, 2 is a division of 14 into 2⅓ equal parts--2 equal parts, 6, 6, and a third of a part, 2. So we might agree to say, that the (2⅓)th, or (2⅓)rd, or (2⅓)st (the reader may coin the adjective as he pleases) part of 14 is 6. If we divide the line A B into eleven equal parts in C, D, E, &c., we must then say that A C is the 11th part,
A D the (5½)th, A E the (3⅔)th, A F the (2¾)th, A G the (2⅕)th, A H the (1⅚)th, A I the (1⁴/₇)th, A K the (1⅜)th, A L the (1²/₉)th, A M the (1⅒)th, and A B itself the 1st part of A B. The reader may refuse the language if he likes (though it is not so much in defiance of etymology as talking of _multiplying_ by ½); but when A B is called 1, he must either call A F 1/(2¾), or make one definition of one class of fractions and another of another. Whatever abbreviations they may choose, all persons will agree that _a_/_b_ is a direction to find such a fraction as, repeated _b_ times, will give 1, and then to take that fraction _a_ times.
So, to get 2½/4⅗, the simplest way is to divide the whole unit into 46 parts; 10 of these parts, repeated 4⅗ times, give the whole. The
4⅗th is then ¹⁰/₄₆, and 2½ such parts is ²⁵/₄₆, or A C. The student should try several examples of this mode of interpreting complex fractions.
But what are we to say when the denominator itself is less than unity, as in 3¼/⅖? Are we to have a (⅖)th part of a unit? and what is it? Had there been a 5 in the denominator, we should have taken the part of which 5 will make a unit. As there is ⅖ in the denominator, we must take the part of which ⅖ will be a unit. That part is larger than a unit; it is 2½ units; 2½ is that of which ⅖ is 1. The above fraction then directs us to repeat 2½ units 3¼ times. By extending our word ‘multiplication’ to the taking of a part of a time, all multiplications are also divisions, and all divisions multiplications, and all the terms connected with either are subject to be applied to the results of the other.
If 2⅓ yards cost 3½ shillings, how much does one yard cost? In such a case as this, the student looks at a more simple question. If 5 yards cost 10 shillings, he sees that each yard costs ¹⁰/₅, or 2 shillings, and, concluding that the same process will give the true result when the data are fractional, he forms 3½/2⅓, reduces it by rules to ³/₂ or 1½, and concludes that 1 yard costs 18 pence. The answer happens to be correct; but he is not to suppose that this rule of copying for fractions whatever is seen to be true of integers is one which requires no demonstration. In the above question we want money which, repeated 2⅓ times, shall give 3½ shillings. If we divide the shilling into 14 equal parts, 6 of these parts repeated 2⅓ times give the shilling. To get 3½ times as much by the same repetition, we must take 3½ of these 6 parts at each step, or 21 parts. Hence, ²¹/₁₄, or 1½, is the number of shillings in the price.
APPENDIX V.
ON CHARACTERISTICS.
When the student comes to use logarithms, he will find what follows very useful. In the mean while, I give it merely as furnishing a rapid rule for finding the place of a decimal point in the quotient before the division is commenced.
When a bar is written over a number, thus, 7︤ let the number be called negative, and let it be thus used: Let it be augmented by additions of its own species, and diminished by subtractions; thus, 7︤ and 2︤ give 9︤, and let 7︤ with 2︤ subtracted give 5︤. But let the _addition_ of a number without the bar _diminish_ the negative number, and the _subtraction increase_ it. Thus, 7︤ and 4 are 3︤, 7︤ and 12 make 5, 7︤ with 8 subtracted is 1︦5. In fact, consider 1, 2, 3, &c., as if they were gains, and 1︤, 2︤, 3︤, as if they were losses: let the addition of a gain or the removal of a loss be equivalent things, and also the removal of a gain and the addition of a loss. Thus, when we say that 4︤ diminished by 1︦1 gives 7, we say that a loss of 4 incurred at the moment when a loss of 11 is removed, is, on the whole, equivalent to a gain of 7; and saying that 4︤ diminished by 2 is 6︤, we say that a loss of 4, accompanied by the removal of a gain of 2, is altogether a loss of 6.
By the _characteristic_ of a number understand as follows: When there are places before the decimal point, it is one less than the number of such places. Thus, 3·214, 1·0083, 8 (which is 8·00 ...) 9·999, all have 0 for their characteristics. But 17·32, 48, 93·116, all have 1; 126·03 and 126 have 2; 11937264·666 has 7. But when there are no places before the decimal point, look at the first decimal place which is significant, and make the characteristic negative accordingly. Thus, ·612, ·121, ·9004, in all of which significance begins in the first decimal place, have the characteristic 1︤; but ·018 and ·099 have 2︤; ·00017 has 4︤; ·000000001 has 9︤.
To find the characteristic of a quotient, subtract the characteristic of the divisor from that of the dividend, carrying one before subtraction if the first significant figures of the divisor are greater than those of the dividend. For instance, in dividing 146·08 by ·00279. The characteristics are 2 and 3︤; and 2 with 3︤ removed would be 5. But on looking, we see that the first significant figures of the divisor, 27, taken by themselves, and without reference to their local value, mean a larger number than 14, the first two figures of the dividend. Consequently, to 3︤ we carry 1 before subtracting, and it then becomes 2︤, which, taken from 2, gives 4. And this 4 is the characteristic of the quotient, so that the quotient has 5 places before the decimal point. Or, if _abcdef_ be the first figures of the quotient, the decimal point must be thus placed, _abcde·f_. But if it had been to divide ·00279 by 146·08, no carriage would have been required; and 3︤ diminished by 2 is 5︤; that is, the first significant figure of the quotient is in the 5th place. The quotient, then, has ·0000 before any significant figure. A few applications of this rule will make it easy to do it in the head, and thus to assign the meaning of the first figure of the quotient even before it is found.
APPENDIX VI.
ON DECIMAL MONEY.
Of all the simplifications of commercial arithmetic, none is comparable to that of expressing shillings, pence, and farthings as decimals of a pound. The rules are thereby put almost upon as good a footing as if the country possessed the advantage of a real decimal coinage.
Any fraction of a pound sterling may be decimalised by rules which can be made to give the result at once.
Two shillings is £·100 | One shilling is £·050 | Sixpence is £·025 | One farthing is £·001 | 04⅙
Thus, every pair of shillings is a unit in the first decimal place; an odd shilling is a 50 in the second and third places; a farthing is so nearly the thousandth part of a pound, that to say one farthing is ·001, two farthings is ·002, &c., is so near the truth that it makes no error in the first three decimals till we arrive at sixpence, and then 24 farthings is exactly ·025 or 25 thousandths. But 25 farthings is ·026, 26 farthings is ·027, &c. Hence the rule for the _first three places_ is
_One in the first for every pair of shillings; 50 in the second and third for the odd shilling, if any; and 1 for every farthing additional, with 1 extra for sixpence._
Thus, 0_s._ 3½_d._ = £·014 0_s._ 7¾_d._ = £·032 1_s._ 2½_d._ = £·060 1_s._ 11¼_d._ = £·096 2_s._ 6_d._ = £·125 2_s._ 9½_d._ = £·139 3_s._ 2¾_d._ = £·161 13_s._ 10¾_d._ = £·694
In the fourth and fifth places, and those which follow, it is obvious that we have no produce from any farthings except those above sixpence. For at every sixpence, ·00004⅙ is converted into ·001, and this has been already accounted for. Consequently, to fill up the _fourth and fifth_ places,
_Take 4 for every farthing[59] above the last sixpence, and an additional 1 for every six farthings, or three halfpence._
[59] The student should remember all the multiples of 4 up to 4 × 25, or 100.
The remaining places arise altogether from ·00000⅙ for every farthing above the last three halfpence; for at every three halfpence complete, ·00000⅙ is converted into ·00001, and has been already accounted for. Consequently, to fill up _all the places after the fifth_,
_Let the number of farthings above the last three halfpence be a numerator, 6 a denominator, and annex the figures of the corresponding decimal fraction._
It may be easily remembered that
The figures of ¹/₆ are 166666...
” ²/₆ ... 333333...
” ³/₆ ... 5
” ⁴/₆ ... 666666...
” ⁵/₆ ... 833333...
0_s._ 3½_d._ = ·014|58|3333...
0_s._ 7¾_d._ = ·032|29|1666...
1_s._ 2½_d._ = ·060|41|6666...
1_s._ 11¼_d._ = ·096|87|5
2_s._ 6_d._ = ·125|00|0000...
2_s._ 9½_d._ = ·139|58|3333...
3_s._ 2¾_d._ = ·161|45|83333...
13_s._ 10¾_d._ = ·694|79|1666...
The following examples will shew the use of this rule, if the student will also work them in the common way.
To turn pounds, &c., into farthings: Multiply the pounds by 960, or by 1000-40, or by 1000(1-⁴/₁₀₀); that is, from 1000 times the pounds subtract 4 per cent of itself. Thus, required the number of farthings in £1663. 11. 9¾.
1663·590625 × 1000 = 1663590·625 4 per cent of this, 66543·625 ----------- No. of farthings required, 1597047
What is 47½ per cent of £166. 13. 10 and ·6148 of £2971. 16. 9?
166·691 -------- 40 p. c. 66·6764 5 p. c. 8·3346 2½ p. c. 4·1673 -------- 79·1783 £79.3.6¾
2971·837 --------- ·6 1783·1022 ·01 29·7184 ·004 11·8873 ·0008 2·3775 --------- 1827·0854 £1827.1.8½
The inverse rule for turning the decimal of a pound into shillings, pence, and farthings, is obviously as follows:
_A pair of shillings for every unit in the first place; an odd shilling for 50 (if there be 50) in the second and third places; and a farthing for every thousandth left, after abating 1 if the number of thousandths so left exceed 24._
The direct rule (with three places) gives too little, the inverse rule too much, except at the end of a sixpence, when both are accurate. Thus, £·183 is rather less than 3_s._ 8_d._, and 6_s._ 4¾_d._ is rather greater than £319; or when the two do not exactly agree, the _common money is the greatest_. But £·125 and £·35 are exactly 2_s._ 6_d._ and 7_s._
Required the price of 17 cwt. 81 lb. 13½ oz. at £3.11.9¾ per cwt. true to the hundredth of a farthing.
3·590625 17 --------- 61·040625 lb. 56 ½ 1·795313 16 ⅐ ·512946 7 ⅛ ·224414 2 ⅛ ·064118 oz. 8 ¼ ·016029 4 ½ ·008015 1 ¼ ·002004 ½ ½ ·001002 --------- £63·664466 £63.13.3½
Three men, A, B, C, severally invest £191.12.7¾, £61.14.8, and £122.1.9½ in an adventure which yields £511.12.6½. How ought the proceeds to be divided among them?
A, 191·63229 B, 61·73333 C, 122·08958 Produce of £1. --------- 375·45520)511·62708(1·362686 136·17188 23·53532 1·00801 25710 3183 180
1·362686 1·362686 1·362686 92·236191 33·33716 85·980221 --------- --------- --------- 1·362686 8·17612 1·362686 1·226417 13627 272537 13627 9538 27254 8176 409 1090 409 41 122 27 4 7 3 -------- 1 1 8·41231 --------- --------- 1·663697 2·611346
261·1346 ... A’s share £261.2.8¼ 84·1231 ... B’s ... 84.2.5¾ 166·3697 ... C’s ... 166.7.4¾ -------- ------------- 511·6274 £511.2.6¾
If ever the fraction of a farthing be wanted, remember that the _coinage_-result is larger than the decimal of a pound, when we use only three places. From 1000 times the decimal take 4 per cent, and we get the exact number of farthings, and we need only look at the decimal then left to set the preceding right. Thus, in
134·6 123·1 369·7 5·38 4·92 14·79 ----- ------ ------ ·22 ·18 ·91
we see that (if we use four decimals only) the pence of the above results are nearly 8_d._ ·22 of a farthing, 5½_d._ ·18, and 4½_d._ ·91.
A man can pay £2376. 4. 4½, his debts being £3293. 11. 0¾. How much per cent can he pay, and how much in the pound?
3293·553)2376·2180(·7214756 70·7309 4·8598 1·5662 2488 183 18
_Answer_, £72. 2.11½ per cent. 0.14.5¼ per pound.
APPENDIX VII.
ON THE MAIN PRINCIPLE OF BOOK-KEEPING.
A brief notice of the principle on which accounts are kept (when they are _properly_ kept) may perhaps be useful to students who are learning book-keeping, as the treatises on that subject frequently give too little in the way of explanation.
Any person who is engaged in business must desire to know accurately, whenever an investigation of the state of his affairs is made.
1, What he had at the commencement of the account, or immediately after the last investigation was made; 2, What he has gained and lost in the interval in all the several branches of his business; 3, What he is now worth. From the first two of these things he obviously knows the third. In the interval between two investigations, he may at any one time desire to know how any one account stands.
An _account_ is a recital of all that has happened, in reference to any class of dealings, since the last investigation. It can only consist of receipts and expenditures, and so it is said to have two sides, a _debtor_ and a _creditor_ side.
All accounts are kept in _money_. If goods be bought, they are estimated by the money paid for them. If a debtor give a bill of exchange, being a promise to pay a certain sum at a certain time, it is put down as worth that sum of money. All the tools, furniture, horses, &c. used in the business are rated at their value in money. All the actual coin, bank-notes, &c., which are in or come in, being the only money in the books which really is money, is called _cash_.
The accounts are kept as if every different sort of account belonged to a separate person, and had an interest of its own, which every transaction either promotes or injures. If the student find that it helps him, he may imagine a clerk to every account: one to take charge of, and regulate, the actual cash; another for the bills which the house is to receive when due; another for those which it is to pay when due; another for the cloth (if the concern deal in cloth); another for the sugar (if it deal in sugar); one for every person who has an account with the house; one for the profits and losses; and so on.
All these clerks (or accounts) belonging to one merchant, must account to him in the end--must either produce all they have taken in charge, or relieve themselves by shewing to whom it went. For all that they have received, for every responsibility they have undertaken to _the concern itself_, they are bound, or are _debtors_; for everything which has passed out of charge, or about which they are relieved from answering _to the concern_, they are unbound, or are _creditors_. These words must be taken in a very wide sense by any one to whom book-keeping is not to be a mystery. Thus, whenever any account assumes responsibility to any parties _out of the concern_, it must be creditor in the books, and debtor whenever it discharges any other parties of their responsibility. But whenever an account removes responsibility from any other account in the same books it is debtor, and creditor whenever it imposes the same.
To whom are all these parties, or accounts, bound, and from whom are they released? Undoubtedly the merchant himself, or, more properly, the _balance-clerk_, presently mentioned. But it is customary to say that the accounts are debtors _to_ each other, and creditors _by_ each other. Thus, cash _debtor_ to bills receivable, means that the cash account (or the clerk who keeps it) is bound to answer for a sum which was paid on a bill of exchange due to the house. At full length it would be: “Mr. C (who keeps the cash-box) has received, and is answerable for, this sum which has been paid in by Mr. A, when he paid his bill of exchange.” On the other hand, the corresponding entry in the account of bills receivable runs--bills receivable, _creditor_ by cash. At full length: “Mr. B (who keeps the bills receivable) is freed from all responsibility for Mr. A’s bill, which he once held, by handing over to Mr. C, the cash-clerk, the money with which Mr. A took it up.” Bills receivable creditor _by_ cash is intelligible, but cash debtor _to_ bills receivable is a misnomer. The cash account is debtor _to the merchant by_ the sum received for the bill, and it should be cash debtor _by_ bill receivable. The fiction of debts, not one of which is ever paid to the party _to_ whom it is said to be owing, though of no consequence in practice, is a stumbling-block to the learner; but he must keep the phrase, and remember its true meaning.
The account which is made _debtor_, or bound, is said to be _debited_; that which is made _creditor_, or released, is said to be _credited_. All who receive must be _debited_; all who give must be _credited_.
No cancel is ever made. If cash received be afterwards repaid, the sum paid is not struck off the receipts (or debtor-side of the cash account), but a discharge, or credit, is written on the expenditure (or credit) side.
The book in which the accounts are kept is called a _ledger_. It has double columns, or else the debtor-side is on one page, and the creditor side on the opposite, of each account. The debtor-side is always the left. Other books are used, but they are only to help in keeping the ledger correct. Thus there may be a _waste-book_, in which all transactions are entered as they occur, in common language; a _journal_, in which the transactions described in the waste-book are entered at stated periods, in the language of the ledger. The items entered in the journal have references to the pages of the ledger to which they are carried, and the items in the ledger have also references to the pages of the journal from which they come; and by this mode of reference it is easy to make a great deal of abbreviation in the ledger. Thus, when it happens, in making up the journal to a certain date, that several different sums were paid or received at or near the same time, the totals may be entered in the ledger, and the cash account may be made debtor to, or creditor by, sundry accounts, or sundries; the sundry accounts being severally credited or debited for their shares of the whole. The only book that need be explained is the ledger. All the other books, and the manner in which they are kept, important as they may be, have nothing to do with the main principle of the method. Let us, then, suppose that all the items are entered at once in the ledger as they arise. It has appeared that every item is entered twice. If A pay on account of B, there is an entry, “A, creditor by B;” and another, “B, debtor to A.” This is what is called _double-entry_; and the consequence of it is, that the sum of all the debtor items in the whole book is equal to the sum of all the creditor items. For what is the first set but the second with the items in a different order? If it were convenient, one entry of each sum might be made a double-entry. The multiplication table is called a table of _double-entry_, because 42, for instance, though it occurs only once, appears in two different aspects, namely, as 6 times 7 and as 7 times 6. Suppose, for example, that there are five accounts, A, B, C, D, E, and that each account has one transaction of its own with every other account; and let the debits be in the _columns_, the credits in the _rows_, as follows:
-------------+------+------+------+------+------+ Debtor | A | B | C | D | E | -------------+------+------+------+------+------+ A, Creditor | | 23 | 19 | 32 | 4 | +------+------+------+------+------+ B, Creditor | 17 | | 6 | 11 | 25 | +------+------+------+------+------+ C, Creditor | 9 | 41 | | 10 | 2 | +------+------+------+------+------+ D, Creditor | 14 | 28 | 16 | | 3 | +------+------+------+------+------+ E, Creditor | 15 | 4 | 60 | 1 | | +------+------+------+------+------+
Here the 16 is supposed to appear in D’s account as D creditor by C, and in C’s account as C debtor to D. And to say that the sum of debtor items is the same as that of creditor items, is merely to say that the preceding numbers give the same sum, whether the rows or the columns be first added up.
If it be desired to close the ledger when it stands as above, the following is the way the accounts will stand: the lines in italics will presently be explained.
A, Debtor. | A, Creditor.| B, Debtor. | B, Creditor. To B 17 | By B 23 | To A 23 | By A 17 To C 9 | By C 19 | To C 41 | By C 6 To D 14 | By D 32 | To D 28 | By D 11 To E 15 | By E 4 | To E 4 | By E 25 To Balance 23 | | | By Balance 37 -- | -- | -- | -- 78 | 78 | 96 | 96 --------------+------------------+---------------+---------------+ C, Debtor. | C, Creditor. | D, Debtor. | D, Creditor. To A 19 | By A 9 | To A 32 | By A 14 To B 6 | By B 41 | To B 11 | By B 28 To D 16 | By D 10 | To C 10 | By C 16 To E 60 | By E 2 | To E 1 | By E 3 | By Balance 39 | To Balance 7 | --- | --- | -- | -- 101 | 101 | 61 | 61 --------------+------------------+-----------------+--------------- E, Debtor. | E, Creditor. | Balance, Debtor.| Balance, Cred. To A 4 | By A 15 | To B 37 | By A 23 To B 25 | By B 4 | To C 39 | By D 7 To C 2 | By C 60 | | By E 46 | | -- | -- To D 3 | By D 1 | 76 | 76 To Balance 46 | | | -- | -- | | 80 | 80 | |
In all the part of the above which is printed in Roman letters we see nothing but the preceding table repeated. But when all the accounts have been completed, and no more entries are left to be made, there remains the last process, which is termed _balancing the ledger_. To get an idea of this, suppose a new clerk, who goes round all the accounts, collecting debts and credits, and taking them all upon himself, that he alone may be entitled to claim the debts and to be responsible for the assets of the concern. To this new clerk, whom I will call the _balance-clerk_, every account gives up what it has, whether the same be debt or credit. The cash-clerk gives up all the cash; the clerks of the two kinds of bills give up all their documents, whether bills receivable or entries of bills payable (remember that any entry against which there is money set down in the books counts as money when given up, that is, as money due or money owing); the clerks of the several accounts of goods give up all their unsold remainders at cost prices; the clerks of the several personal accounts give up vouchers for the sums owing to or from the several parties; and so on. But where more has been paid out than received, the balance-clerk adjusts these accounts by giving instead of receiving; in fact, he so acts as to make the debtor and creditor sides of the accounts he visits equal in amount. For instance, the A account is indebted to the concern 55, while payments or discharges to the amount of 78 have been made by it. The balance-clerk accordingly hands over 23 to that account, for which it becomes debtor, while the balance enters itself as creditor to the same amount. But in the B account there is 96 of receipt, and only 59 of payment or discharge. The balance-clerk then receives 37 from this account, which is therefore credited by balance, while the balance acknowledges as much of debt. The balance account must, of course, exactly balance itself, if the accounts be all right; for of all the equal and opposite entries of which the ledger consists, so far as they do not balance one another, one goes into one side of the balance account, and the other into the other. Thus the balance account becomes a test of the accuracy of one part of the work: if its two sides do not give the same sums, either there have been entries which have not had their corresponding balancing entries correctly made, or else there has been error in the additions.
But since the balance account must thus always give the _same sum_ on both sides, and since _balance debtor_ implies what is favourable to the concern, and _balance creditor_ what is unfavourable, does it not appear as if this system could only be applied to cases in which there is neither loss nor gain? This brings us to the two accounts in which are entered all that the concern _began with_, and all that it _gains or loses_--the _stock account_, and the _profit-and-loss account_. In order to make all that there was to begin with a matter of double entry, the opening of the ledger supposes the merchant himself to put his several clerks in charge of their several departments. In the stock account, _stock_, which here stands for the owner of the books, is made creditor by all the property, and debtor by all the liabilities; while the several accounts are made debtors for all they take from the stock, and creditors by all the responsibilities they undertake. Suppose, for instance, there are £500 in cash at the commencement of the ledger. There will then appear that the merchant has handed over to the cash-box £500, and in the stock account will appear, “Stock creditor by cash, £500;” while in the cash account will appear, “Cash debtor to stock, £500.” Suppose that at the beginning there is a debt outstanding of £50 to Smith and Co., then there will appear in the stock account, “Stock debtor to Smith and Co. £50,” and in Smith and Co.’s account will appear “Smith and Co. creditors by stock, £50.” Thus there is double entry for all that the concern begins with by this contrivance of the stock account.
The account to which everything is placed for which an actual equivalent is not seen in the books is the _profit-and-loss_ account. This profit-and-loss account, or the clerk who keeps it, is made answerable for every loss, and the supposed cause of every gain. This account, then, becomes debtor for every loss, and creditor by every gain. If goods be damaged to the amount of £20 by accident, and a loss to that amount occur in their sale, say they cost £80 and sell for £60 cash, it is clear that there is an entry “Cash debtor to goods £60,” and “Goods creditor by cash £60.” Now, there is an entry of £80 somewhere to the debit of the goods for cash laid out, or bills given, for the whole of the goods. It would affect the accuracy of the accounts to take no notice of this; for when the balance-clerk comes to adjust this account, he would find he receives £20 less than he might have reckoned upon, without any explanation of the reason; and there would be a failure of the principle of double-entry. Since it is convenient that the balance account of the goods should merely represent the stock in hand at the close, the account of goods therefore lays the responsibility of £20 upon the profit-and-loss account, or there is the entry “Goods creditor by profit-and-loss, £20,” and also “Profit-and-loss debtor to goods, £20.” Again, in all payments which are not to bring in a specific return, such as house and trade expenses, wages, &c. these several accounts are supposed to adjust matters with the profit-and-loss account before the balance begins. Thus, suppose the outgoings from the mere premises occupied exceed anything those premises yield by £200, or the debits of the house account exceed its credits by £200, the account should be balanced by transferring the responsibility to the profit-and-loss account, under the entries “House expenses creditor by profit-and-loss, £200”, “Profit-and-loss debtor to house expenses, £200.” In this way the profit-and-loss account steps in from time to time before the balance account commences its operations, in order that that same balance account may consist of _nothing but the necessary matters of account for the next year’s ledger_.
This _transference of accounts_, or transfusion of one account into another, requires attentive consideration. The receiving account becomes creditor for the credits, and debtor for the debits, of the transmitting account. The rule, therefore, is: Make the transmitting account balance itself, and, on whichever side it is necessary to enter a balancing sum, make the account debtor or creditor, as the case may be, to the receiving account, and the latter creditor or debtor to the former. Thus, suppose account A is to be transferred to account B, and the latter is to arrange with the balance account. If the two stand as in Roman letters, the processes in Italic letters will occur before the final close.
A, Debtor. | A, Creditor. | B, Debtor. | B, Creditor. | | | To sundries £100|By sundries £500|To sundries £600|By sundries £400 To B . . . 400| |To Balance 200|By A . . . 400 ----| ----| ----| ---- £500| £500| £800| £800
And the entry in the balance account will be, “Creditor by B, £200,” shewing that, on these two accounts, the credits exceed the debits by £200.
Still, before the balance account is made up, it is desirable that the profit-and-loss account should be transferred to the stock account; for the profit and loss of this year is of no moment as a part of next year’s ledger, except in so far as it affects the stock at the commencement of the latter. Let this be done, and the balance account may then be made in the form required.
The stock account and the profit-and-loss account, the latter being the only direct channel of alteration for the former, differ in a peculiar manner[60] from the other preliminary accounts, and the balance account is a species of umpire. They represent the merchant: their interests are his interests; he is solvent upon the excess of their credits over their debits, insolvent upon the excess of their debits over their credits. It is exactly the reverse in all the other accounts. If a malicious person were to get at the ledger, and put on a cipher to the pounds in various items, with a view of making the concern appear worse than it really is, he would make his alterations on the _debtor_ sides of the stock and profit-and-loss accounts, and on the _creditor_ sides of all the others. Accordingly, in the balance account, the net stock, after the incorporation of the profit-and-loss account, appears on the _creditor_ side (if not, it should be called amount of _insolvency_, not _stock_), and the debts of the concern appear on the same side. But on the debit side of the balance account appear all the assets of the concern (for which the balance-clerk is debtor to the clerks from whom he has taken them).
[60] The treatises on book-keeping have described this difference in as peculiar a manner. They call these accounts the _fictitious accounts_. Now they represent the merchant himself; their credits are gain to the business, their debits losses or liabilities. If the terms real and fictitious are to be used at all, they are the _real_ accounts, end all the others are as _fictitious_ as the clerks whom we have supposed to keep them.
The young student must endeavour to get the enlarged view of the words debtor and creditor which is requisite, and must then learn by practice (for nothing else will give it) facility in allotting the actual entries in the waste-book to the proper sides of the proper accounts. I do not here pretend to give more than such a view of the subject as may assist him in studying a treatise on book-keeping, which he will probably find to contain little more than examples.
APPENDIX VIII.
ON THE REDUCTION OF FRACTIONS TO OTHERS OF NEARLY EQUAL VALUE.
There is a useful method of finding fractions which shall be nearly equal to a given fraction, and with which the computer ought to be acquainted. Proceed as in the rule for finding the greatest common measure of the numerator and denominator, and bring all the quotients into a line. Then write down,
1 2nd Quot. -------- ------------------------ 1st Quot. 1st Quot. × 2d Quot. + 1
Then take the third quotient, multiply the numerator and denominator of the second by it, and add to the products the preceding numerator and denominator. Form a third fraction with the results for a numerator and denominator. Then take the fourth quotient, and proceed with the third and second fractions in the same way; and so on till the quotients are exhausted. For example, let the fraction be ⁹¹³¹/₁₃₁₂₈.
9131)13128(1, 2 1137 3997(3, 1 551 586(1, 15 201 35(1, 2 26 9(1, 8 8 1
This is the process for finding the greatest common measure of 9131 and 13128 in its most compact form, and the quotients and fractions are:
1 2 3 1 1 15 1 2 1 8
1 2 7 9 16 249 265 779 1044 9131 --- --- --- --- ---- ----- ----- ----- ------ ------ 1 3 10 13 23 358 381 1120 1501 13128
It will be seen that we have thus a set of fractions ending with the original fraction itself, and formed by the above rule, as follows:
1 1 1st Fraction = -------- = --- 1st Quot. 1
2d Quot. 2 2d Fraction = ------------------------ = --- 1st Quot. × 2d Quot. + 1 3
2d Numʳ. × 3d Quot. + 1st Numʳ. 2 × 3 + 1 7 3d Fraction = ------------------------------- = --------- = --- 2d Denʳ. × 3d Quot. + 1st Denʳ. 3 × 3 + 1 10
3d Numʳ. × 4th Quot. + 2d Numʳ. 7 × 1 + 2 9 4th Fraction = ------------------------------- = --------- = ---; 3d Denʳ. × 4th Quot. + 2d Denʳ. 10 × 1 + 3 13 and so on. But we have done something more than merely reascend to the original fraction by means of the quotients. The set of fractions, ¹/₁, ²/₃, ⁷/₁₀, ⁹/₁₃, &c. are continually approaching in value to the original fraction, the first being too great, the second too small, the third too great, and so on alternately, but each one being nearer to the given fraction than any of those before it. Thus, ¹/₁ is too great, and ²/₃ is too small; but ²/₃ is not so much too small as ¹/₁ is too great. And again, ⁷/₁₀, though too great, is not so much too great as ²/₃ is too small.
Moreover, the difference of any of the fractions from the original fraction is never greater than a fraction having unity for its numerator and the product of the denominator and the next denominator for its denominator. Thus, ¹/₁ does not err by so much as ¹/₃, nor ²/₃ by so much as ¹/₃₀, nor ⁷/₁₀ by so much as ¹/₁₃₀, nor ⁹/₁₃ by so much as ¹/₂₉₉, &c.
Lastly, no fraction of a less numerator and denominator can come so near to the given fraction as any one of the fractions in the list. Thus, no fraction with a less numerator than 249, and a less denominator than 358, can come so near to
9131 249 ----- as ---. 13128 358
The reader may take any example for himself, and the test of the accuracy of the process is the ultimate return to the fraction begun with. Another test is as follows: The numerator of the difference of any two consecutive approximating fractions ought to be unity. Thus, in our instance, we have ¹⁶/₂₃ and ²⁴⁹/₃₅₈, which, with a common denominator, 23 × 358, have 5728 and 5727 for their numerators.
As another example, let us examine this question: The length of the year is 365·24224 days, which is called in common life 365¼ days. Take the fraction ²⁴²²⁴/₁₀₀₀₀₀, and proceed as in the rule.
24224)100000(4, 7, 1, 4, 9, 2 2496 3104 64 608 0 32
1 7 8 39 359 757 --- --- --- ---- ---- ---- 4 29 33 161 1482 3125
and ⁷⁵⁷/₃₁₂₅ is ·24224 in its lowest terms. Hence, it appears that the excess of the year over 365 days amounts to about 1 day in 4 years, which is not wrong by so much as 1 day in 116 years; more accurately, to 7 days in 29 years, which is not wrong by so much as 1 day in 957 years; more accurately still, to 8 days in 33 years, which is not wrong by so much as 1 day in 5313 years; and so on.
This method may be applied to finding fractions nearly equal to the square roots of integers, in the following manner:
__ √43 = 6 + ...
6 | 1 5 4 5 5 4 5 1 6 6 |1 5 4, &c. 1 | 7 6 3 9 2 9 3 6 7 1 |7 6 3, &c. --+----------------------+------ 6 | 1 1 3 1 5 1 3 1 1 1 2|1 1 3, &c.
Set down the number whose square root is wanted, say 43. This square root is 6 and a fraction. Set down the integer 6 in the first and third row, and 1 in the second row always. Form the successive rows each from the one before, in the following manner:
One row The next row has _b′_, _a′_, _c′_, formed in this order, being thus, _a_ _a′_ = excess of _b′c′_, already formed, over _a_. _b_ _b′_ = quotient of 43 - _a_² divided by _b_. _c_ _c′_ = integer in the quotient of 6 + _a_ divided by _b′_.
Thus the second row is formed from the first, as under:
6|1 = excess of 7 × 1 (both just found) over 6. 1|7 = 43 - 6 × 6 divided by 1. --+-- 6|1 = integer of 6 + 6 divided by 7 (just found).
The third row is formed from the second, thus:
1 5 = excess of 1 × 6 over 1. 7 6 = 43 - 1 × 1 divided by 7. 1 1 = integer of 6 + 1 divided by 6;
and so on. In process of time the second column, 1, 7, 1, occurs again, after which the several columns are repeated in the same order. As a final process, take the set in the lowest line (excluding the first, 6), namely, 1, 1, 3, 1, 5, 1, 3, &c. and use them by the rule given at the beginning of this article, as follows:
1 1 3 1 5 1 3 1 1, &c.
1 1 4 5 29 34 131 165 296 --- --- --- --- ---- ---- ---- ---- ---- 1 2 7 9 52 61 235 296 531 Hence, 6¹⁶⁵/₂₉₆ is very near the square root of 43, not erring by so much as
1 ---------. 296 × 531
If we try it, we shall find (⁶¹⁶⁵/₂₉₆) to be ¹⁹⁴¹/₂₉₆, the square of which is ³⁷⁶⁷⁴⁸¹/₈₇₆₁₆, or 43⁷/₈₇₆₁₆.
This rule is of use when it is frequently wanted to use one square root, and therefore desirable to ascertain whether any easy approximation exists by means of a common fraction. For example, √2 is often used.
_ √2 = 1 + ... 1|1 1 1|1 1 1|2 2 2 2 2 2 1 2 5 12 29 70 --- --- --- --- --- ---, &c. 2 5 12 29 70 169
Here it appears that
29 1 99 100 - 1 1---- does not err by --------; consequently, ---- or ------- is, 70 70 × 169 70 70
considering the ease of the operation, a fair approximation. In fact, ⁹⁹/₇₀ is 1·4142857 ... the truth being 1·4142135 ...
The following is an additional example:
__ √19 = 4 + ... 4 | 2 3 3 2 4 4 2 1 | 3 5 2 5 3 1 3 4 | 2 1 3 1 2 8 2 1 3 1 2, &c.
1 1 4 5 14 --- --- --- --- ---, &c. 2 3 11 14 39
APPENDIX IX.
ON SOME GENERAL PROPERTIES OF NUMBERS.
PROP. 1. If a fraction be reduced to its lowest terms, _so called_,[61] that is, if neither, numerator nor denominator be divisible by any integer greater than unity, then no fraction of a smaller numerator and denominator can have the same value.
[61] This theorem shews that what is _called_ reducing a fraction to its lowest terms (namely, dividing numerator and denominator by their greatest common measure), is correctly so called.
Let _a_/_b_ be a fraction in which _a_ and _b_ have no common measure greater than unity: and, if possible, let _c_/_d_ be a fraction of the same value, _c_ being less than _a_, and _d_ less than _b_. Now, since
_a_ _c_ _a_ _b_ --- = ---, we have --- = ---; _b_ _d_ _c_ _d_ let _m_ be the integer quotient of these last fractions (which must exist, since _a_ > _c_, _b_ > _d_), and let _e_ and _f_ be the remainders. Then
_a_ _mc_ + _e_ _c_ _mc_ --- or ---------- = --- = ---- _b_ _md_ + _f_ _d_ _md_
Hence,
_e_ _mc_ --- and ---- must be equal, for if not, _f_ _md_
_mc_ + _e_ _mc_ _e_ ---------- would lie between ---- and ---, _md_ + _f_ _md_ _f_
instead of being equal to the former. Hence,
_a_ _e_ --- = ---; _b_ _f_
so that if a fraction whose numerator and denominator have no common measure greater than unity, be equal to a fraction of lower numerator and denominator, it is equal to another in which the numerator and denominator are still lower. If we proceed with
_a_ _e_ --- = --- in a similar manner, we find _b_ _f_
_a_ _g_ --- = --- where _g_ < _e_, _h_ < _f_, _b_ _h_
and so on. Now, if there be any process which perpetually diminishes the terms of a fraction by one or more units at every step, it must at last bring either the numerator or denominator, or both, to 0. Let
_a_ _v_ --- = --- _b_ _w_
be one of the steps, and let _a_ = _kv_ + _x_, _b_ = _kw_ + _y_; so that
_kv_ + _x_ _v_ ---------- = ---. _kw_ + _y_ _w_
Now, if _x_ = 0 but not _y_, this is absurd, for it gives
_kv_ _kv_ ---------- = ----. _kw_ + _y_ _kw_
A similar absurdity follows if _y_ be 0, but not _x_; and if both _x_ and _y_ be = 0, then _a_ = _kv_, _b_ = _kw_, or _a_ and _b_ have a common measure, _k_. Now _k_ must be greater than 1, for _v_ and _w_ are less than _c_ and _d_, which by hypothesis are less than _a_ and _b_. Consequently _a_ and _b_ have a common measure _k_ greater than 1, which by hypothesis they have not. If, then, _a_ and _b_ be integers not divisible by any integer greater than 1, the fraction _a_/_b_ is really _in its lowest terms_. Also _a_ and _b_ are said to be _prime to one another_.
PROP. 2. If the product _ab_ be divisible by _c_, and if _c_ be prime to _b_, it must divide _a_. Let
_ab_ _b_ _d_ ---- = _d_, then --- = ---. _c_ _c_ _c_
Now _b_/_c_ is in its lowest terms; therefore, by the last proposition, _d_ and _a_ must have a common measure. Let the greatest common measure be _k_, and let _a_ = _kl_, _d_ = _km_. Then
_b_ _km_ _m_ _m_ --- = ---- = ---, and --- _c_ _kl_ _l_ _l_
is also in its lowest terms; but so is _b_/_c_; therefore we must have _m_ = _b_, _l_ = _c_, for otherwise a fraction in its lowest terms would be equal to another of lower terms. Therefore _a_ = _kc_, or _a_ is divisible by _c_. And from this it follows, that if a number be prime to two others, it is prime to their product. Let _a_ be prime to _b_ and _c_, then no measure of _a_ can measure either _b_ or _c_, and no such measure can measure the product _bc_; for any measure of _bc_ which is prime to one must measure the other.
PROP. 3. If _a_ be prime to _b_, it is prime to all the powers of _b_. Every measure[62] of _a_ is prime to _b_, and therefore does not divide _b_. Hence, by the last, no measure of _a_ divides _b_²; hence, _a_ is prime to _b_², and so is every measure of it; therefore, no measure of _a_ divides _bb_², consequently _a_ is prime to _b_³, and so on.
Hence, if _a_ be prime to _b_, _a_ cannot divide without remainder any power of _b_. This is the reason why no fraction can be made into a decimal unless its denominator be measured by no prime[63] numbers except 2 and 5. For if
_a_ _c_ --- = ---, _b_ 10ⁿ
which last is the general form of a decimal fraction, let
_a_ 10ⁿ_a_ --- be in its lowest terms; then ------ _b_ _b_
is an integer, whence (Prop. 2) _b_ must divide 10ⁿ, and so must all the divisors of _b_. If, then, among the divisors of _b_ there be any prime numbers except 2 and 5, we have a prime number (which is of course a number prime to 10) not dividing 10, but dividing one of its powers, which is absurd.
[62] For that which measures a measure is itself a measure; so that if a measure of _a_ could have a measure in common with _b_, _a_ itself would have a common measure with _b_.
[63] A prime number is one which is prime to all numbers except its own multiples, or has no divisors except 1 and itself.
PROP. 4. If _b_ be prime to _a_, all the multiples of _b_, as _b_, 2_b_, ... up to (_a_-1)_b_ must leave different remainders when divided by _a_. For if, _m_ being greater than _n_, and both less than _a_, we have _mb_ and _nb_ giving the same remainder, it follows that _mb_-_nb_, or (_m_-_n_)_b_, is divisible by _a_; whence (Prop. 2), a divides _m_-_n_, a number less than itself, which is absurd.
* * * * *
If a number be divided into its prime factors, or reduced to a product of prime numbers only (as in 360 = 2 × 2 × 2 × 3 × 3 × 5), and if _a_, _b_, _c_, &c. be the prime factors, and α, β, γ, &c. the number of times they severally enter, so that the number is _a_{^α} × _b_ᵝ × _c_ᵞ × &c., then this can be done in only one way: For any prime number _v_, not included in the above list, is prime to _a_, and therefore to _a_{^α}, to _b_ and therefore to _b_ᵝ and therefore to _a_{^α} × _b_ᵝ Proceeding in this way, we prove that _v_ is prime to the complete product above, or to the given number itself.
The number of divisors which the preceding number _a_{^α}_b_ᵝ_c_ᵞ ... can have, 0 and itself included, is (α + 1)(β+ 1)(γ + 1).... For _a_{^α} as the divisors 1, _a_, _a_² ... _a_{^α} and no others, α + 1 in all. Similarly, _b_ᵝ has β+ 1 divisors, and so on. Now as all the divisors are made by multiplying together one out of each set, their number (page 202) is (α + 1)(β + 1)(γ+ 1)....
If a number, _n_, be divisible by certain prime numbers, say 3, 5, 7, 11, then the third part of all the numbers up to _n_ is divisible by 3, the fifth part by 5, and so on. But more than this: when the multiples of 3 are omitted, exactly the fifth part of _those which remain_ are divisible by 5; for the fifth part of the whole are divisible by 5, and the fifth part of those which are removed are divisible by 5, therefore the fifth part of those which are left are divisible by 5. Again, because the seventh part of the whole are divisible by 7, and the seventh part of those which are divisible by 3, or by 5, or by 15, it follows that when all those which are multiples of 3 or 5, or both, are removed, the seventh part of those which remain are divisible by 7; and so on. Hence, the number of numbers not exceeding n, which are not divisible by 3, 5, 7, or 11, is ¹⁰/₁₁ of ⁶/₇ of ⁴/₅ of ²/₃ of n. Proceeding in this way, we find that the number of numbers which are prime to _n_, that is, which are not divisible by any one of its prime factors, _a_, _b_, _c_, ... is
_a_ - 1 _b_ - 1 _c_ - 1 _n_ ------- ------- ------- ... _a_ _b_ _c_
or _a_{^α-1} - 1}_b_ᵝ⁻¹_c_ᵞ⁻¹ ... (_a_ - 1)(_b_ - 1)(_c_ - 1)....
Thus, 360 being 2³3²5, its number of divisors is 4 × 3 × 2, or 24, and there are 2³3.1.2.4 or 96 numbers less than 360 which are prime to it.
PROP. 5. If _a_ be prime to _b_, then the terms of the series, _a_, _a_², _a_³, ... severally divided by _b_, must all leave different remainders, until 1 occurs as a remainder, after which the cycle of remainders will be again repeated.
Let _a_ + _b_ give the remainder _r_ (not unity); then _a_² ÷ _b_ gives the same remainder as _r__a_ + _b_, which (Prop. 4) cannot be _r_: let it be _s_. Then _a_ˢ ÷ _b_ gives the same remainder as _s__a_ ÷ _b_, which (Prop. 4) cannot be either _r_ or _s_, unless _s_ be 1: let it be _t_. Then _a_ᵗ ÷ _b_ gives the same remainder as _ta_ ÷ _b_; if _t_ be not 1, this cannot be either _r_, _s_, or _t_: let it be _u_. So we go on getting different remainders, until 1 occurs as a remainder; after which, at the next step, the remainder of _a_ ÷ _b_ is repeated. Now, 1 must come at last; for division by _b_ cannot give any remainders but 0, 1, 2, ... _b_- 1; and 0 never arrives (Prop. 3), so that as soon as _b_-2 _different_ remainders have occurred, no one of which is unity, the next, which must be different from all that precede, must be 1. If not before, then at _a_ᵇ⁻¹ we must have a remainder 1; after which the cycle will obviously be repeated.
Thus, 7, 7², 7³, 7⁴, &c. will, when divided by 5, be found to give the remainders 2, 4, 3, 1, &c.
PROP. 6. The difference of two _m_th powers is always divisible without remainder by the difference of the roots; or _a_ᵐ -_b_ᵐ is divisible by _a_-_b_; for
_a_ᵐ - _b_ᵐ = _a_ᵐ - _a_ᵐ⁻¹_b_ + _a_ᵐ⁻¹_b_ - _b_ᵐ
= _a_ᵐ⁻¹(_a_ - _b_) + _b_(_a_ᵐ⁻¹ - _b_ᵐ⁻¹)
From which, if _aᵐ⁻¹_-_bᵐ⁻¹_ is divisible by _a_ -_b_, so is _a_ᵐ-_b_ᵐ. But _a_-_b_ is divisible by _a_-_b_; so therefore is _a_²- _b_²; so therefore is _a_³-_b_³; and so on.
Therefore, if _a_ and _b_, divided by _c_, leave the same remainder, _a_² and _b_², _a_³ and _b_³, &c. severally divided by _c_, leave the same remainders; for this means that _a_-_b_ is divisible by _c_. But _a_ᵐ - _b_ᵐ is divisible by _a_-_b_, and therefore by every measure of _a_-_b_, or by _c_; but _a_ᵐ-_b_ᵐ cannot be divisible by _c_, unless _a_ᵐ and _b_ᵐ, severally divided by _c_, give the same remainder.
PROP. 7. If _b_ be a prime number, and _a_ be not divisible by _b_, then _a_ᵇ and (_a_-1)ᵇ + 1 leave the same remainder when divided by _b_. This proposition cannot be proved here, as it requires a little more of algebra than the reader of this work possesses.[64]
[64] Expand (_a_-1)ᵇ by the binomial theorem; shew that _when b is a prime number_ every coefficient which is not unity is divisible by _b_; and the proposition follows.
PROP. 8. In the last case, _a_ᵇ⁻¹ divided by _b_ leaves a remainder 1. From the last, _a_ᵇ-_a_ leaves the same remainder as (_a_-1)ᵇ + 1-_a_ or (_a_-1)ᵇ- (_a_-1); that is, the remainder of _a_ᵇ-_a_ is not altered if _a_ be reduced by a unit. By the same rule, it may be reduced another unit, and so on, still without any alteration of the remainder. At last it becomes 1ᵇ-1, or 0, the remainder of which is 0. Accordingly, _a_ᵇ-_a_, which is _a_(_a_ᵇ⁻¹- 1), is divisible by _b_; and since _b_ is prime to _a_, it must (Prop. 2) divide _a_ᵇ⁻¹-1; that is, _a_ᵇ⁻¹, divided by _b_, leaves a remainder 1, if _b_ be a prime number and _a_ be not divisible by _b_.
From the above it appears (Prop. 5 and 7), that if _a_ be prime to _b_, the set 1, _a_, _a_², _a_³, &c. successively divided by _b_, give a set of remainders beginning with 1, and in which 1 occurs again at _a_ᵇ⁻¹, if not before, and at _a_ᵇ⁻¹ certainly (whether before or not), if _b_ be a prime number. From the point at which 1 occurs, the cycle of remainders recommences, and 1 is always the beginning of a cycle. If, then, _a_ᵐ be the first power which gives 1 for remainder, _m_ must either be _b_-1, or a measure of it, _when b is a prime number_.
But if we divide the terms of the series _m_, _ma_, _ma_², _ma_³, &c. by _b_, _m_ being less than _b_, we have cycles of remainders beginning with _m_. If 1, _r_, _s_, _t_, &c. be the first set of remainders, then the second set is the set of remainders arising from _m_, _mr_, _ms_, _mt_, &c. If 1 never occur in the first set before _a_ᵇ⁻¹ (except at the beginning), then all the numbers under _b_-1 inclusive are found among the set 1, _r_, _s_, _t_, &c.; and if _m_ be prime to _b_ (Prop. 4), all the same numbers are found, in a different order, among the remainders of _m_, _mr_, &c. But should it happen that the set 1, _r_, _s_, _t_, &c. is not complete, then _m_, _mr_, _ms_, &c. may give a different set of remainders.
All these last theorems are constantly verified in the process for reducing a fraction to a decimal fraction. If _m_ be prime to _b_, or the fraction _m_/_b_ in its lowest terms, the process involves the successive division of _m_, _m_ × 10, _m_ × 10², &c. by _b_. This process can never come to an end unless some power of 10, say 10ⁿ, is divisible by _b_; which cannot be, if _b_ contain any prime factors except 2 and 5. In every other case the quotient repeats itself, the repeating part sometimes commencing from the first figure, sometimes from a later figure. Thus, ¹/₇ yields ·142857142857, &c., but ¹/₁₄ gives ·07(142857)(142857), &c., and ¹/₂₈ gives ·03(571428)(571428), &c.
In _m_/_b_, the quotient always repeats from the very beginning whenever _b_ is a prime number and _m_ is less than _b_; and the number of figures in the repeating part is then always _b_-1, or a measure of it. That it must be so, appears from the above propositions.
Before proceeding farther, we write down the repeating part of a quotient, with the remainders which are left after the several figures are formed. Let the fraction be ¹/₁₇, we have
0₁₀5₁₅8₁₄8₄2₆3₉5₅2₁₆9₇4₂1₃1₁₃7₁₁6₈4₁₂7₁
This may be read thus: 10 by 17, quotient 0, remainder 10; 10² by 17, quotient 05, remainder 15; 10³ by 17, quotient 058, remainder 14; and so on. It thus appears that 10¹⁶ by 17 leaves a remainder 1, which is according to the theorem.
If we multiply 0588, &c. by _any number under_ 17, the same cycle is obtained with a different beginning. Thus, if we multiply by 13, we have
7647058823529411
beginning with what comes after remainder 13 in the first number. If we multiply by 7, we have 4117, &c. The reason is obvious: ¹/₁₇ × 13, or ¹³/₁₇, when turned into a decimal fraction, starts with the divisor 130, and we proceed just as we do in forming ¹/₁₇, when within four figures of the close of the cycle.
It will also be seen, that in the last half of the cycle the quotient figures are complements to 9 of those in the first half, and that the remainders are complements to 17. Thus, in 0₁₀5₁₅8₁₄8₄, &c. and 9₇4₂1₃1₁₃, &c. we see 0 + 9 = 9, 5 + 4 = 9, 8 + 1 = 9, &c., and 10 + 7 = 17, 15 + 2 = 17, 14 + 3 = 17, &c. We may shew the necessity of this as follows: If the remainder 1 never occur till we come to use _a_ᵇ⁻¹, then, _b_ being prime, _b_-1 is even; let it be 2_k_. Accordingly, _a_²ᵏ-1 is divisible by _b_; but this is the product of _a_ᵏ-1 and _a_ᵏ + 1, one of which must be divisible by _b_. It cannot be _a_ᵏ-1, for then a power of _a_ preceding the (_b_-1)th would leave remainder 1, which is not the case in our instance: it must then be _a_ᵏ + 1, so that _a_ᵏ divided by _b_ leaves a remainder _b_-1; and the _k_th step concludes the first half of the process. Accordingly, in our instance, we see, _b_ being 17 and _a_ being 10, that remainder 16 occurs at the 8th step of the process. At the next step, the remainder is that yielded by 10(_b_-1), or 9_b_ + _b_-10, which gives the remainder _b_-10. But the first remainder of all was 10, and 10 + (_b_-10) = _b_. If ever this complemental character occur in any step, it must continue, which we shew as follows: Let _r_ be a remainder, and _b_-_r_ a subsequent remainder, the sum being _b_. At the next step after the first remainder, we divide 10_r_ by _b_, and, at the next step after the second remainder, we divide 10_b_-10_r_ by _b_. Now, since the sum of 10_r_ and 10_b_-10_r_ is divisible by _b_, the two remainders from these new steps must be such as added together will give _b_, and so on; and the _quotients_ added together must give 9, for the sum of the remainders 10_r_ and 10_b_-10_r_ yields a quotient 10, of which the two remainders give 1.
If ¹/₅₉ and ¹/₆₁ be taken, the repeating parts will be found to contain 58 and 60 figures. Of these we write down only the first halves, as the reader may supply the rest by the complemental property just given.
01694915254237288135593220338, &c.
016393442622950819672131147540, &c.
Here, then, are two numbers, the first of which multiplied by any number under 59, and the second by any number under 61, can have the products formed by carrying certain of the figures from one end to the other.
But, _b_ being still prime, it may happen that remainder 1 may occur before _b_-1 figures are obtained; in which case, as shewn, the number of figures must be a measure of _b_-1. For example, take ¹/₄₁. The repeating quotient, written as above, has only 5 figures, and 5 measures 41-1.
0₁₀2₁₈4₁₆3₃₇9₁
Now, this period, it will be found, has its figures merely transposed, if we multiply by 10, 18, 16, or 37. But if we multiply by any other number under 41, we convert this period into the period of another fraction whose denominator is 41. The following are 8 periods which may be found.
0₁₀2₁₈4₁₆3₃₇9₁ | 1₉2₈1₃₉9₂₁5₅ 0₂₀4₃₆8₃₂7₃₃8₂ | 1₁₉4₂₆6₁₄3₁₇4₆ 0₃₀7₁₃3₇1₂₉7₃ | 2₂₈6₃₄8₁₂2₃₈9₁₁ 0₄₀9₈₁7₂₃5₂₅6₄ | 3₂₇6₂₄5₃₅8₂₂5₁₅
To find _m_/41, look out for _m_ among the remainders, and take the period in which it is, beginning after the remainder. Thus, ³⁴/₄₁ is ·8292682926, &c., and ¹⁵/₄₁ is ·3658536585, &c. These periods are complemental, four and four, as 02439 and 97560, 07317 and 92682, &c. And if the first number, 02439, be multiplied by any number under 41, look for that number among the remainders, and the product is found in the period of that remainder by beginning after the remainder. Thus, 02439 multiplied by 23 gives 56097, and by 6 gives 14634.
The reader may try to decipher for himself how it is that, with no more figures than the following, we can extend the result of our division. The fraction of which the period is to be found is ¹/₈₇.
87)100(01149425 130 430 820 01149425 × 25 370 28735625 × 25 220 718390625 × 25 460 17959765625 × 25 25 448994140625 0114942528735625 718390625 1795976 5625 448994 ----------------------------+------ 0114942528735632183908045977|011494 |
APPENDIX X.
ON COMBINATIONS.
There are some things connected with combinations which I place in an appendix, because I intend to demonstrate them more briefly than the matters in the text.
Suppose a number of boxes, say 4, in each of which there are counters, say 5, 7, 3, and 11 severally. In how many ways can one counter be taken out of each box, the order of going to the boxes not being regarded. _Answer_, in 5 × 7 × 3 × 11 ways. For out of the first box we may draw a counter in 5 different ways, and to each such drawing we may annex a drawing from the second in 7 different ways--giving 5 × 7 ways of making a drawing from the first two. To each of these we may annex a drawing from the third box in 3 ways--giving 5 × 7 × 3 drawings from the first three; and so on. The following statements may now be easily demonstrated, and similar ones made as to other cases.
If the order of going to the boxes make a difference, and if _a_, _b_, _c_, _d_ be the numbers of counters in the several boxes, there are 4 × 2 × 3 × 1 × _a_ × _b_ × _c_ × _d_ distinct ways. If we want to draw, say 2 out of the first box, 3 out of the second, 1 out of the third, and 3 out of the fourth, and if the order of the boxes be not considered, the number of ways is
_a_ - 1 _b_ - 1 _b_ - 2 _d_ - 1 _d_ - 2 _a_------- × _b_------- -------- × _c_ × _d_------- ------- 2 2 3 2 3
If the order of going to the boxes be considered, we must multiply the preceding by 4 × 3 × 2 × 1. If the order of the drawings out of the boxes makes a difference, but not the order of the boxes, then the number of ways is
_a_(_a_-1)_b_(_b_-1)(_b_-2)_cd_(_d_-1)(_d_-2)
The nth power of _a_, or _a_ⁿ, represents the number of ways in which _a_ counters _differently marked_ can be distributed in _n_ boxes, order of placing them in each box not being considered. Suppose we want to distribute 4 differently-marked counters among 7 boxes. The first counter may go into either box, which gives 7 ways; the second counter may go into either; and any of the first 7 allotments may be combined with any one of the second 7, giving 7 × 7 distinct ways; the third counter varies each of these in 7 different ways, giving 7 × 7 × 7 in all; and so on. But if the counters be undistinguishable, the problem is a very different thing.
Required the number of ways in which a number can be compounded of other numbers, different orders counting as different ways. Thus, 1 + 3 + 1 and 1 + 1 + 3 are to be considered as distinct ways of making 5. It will be obvious, on a little examination, that each number can be composed in exactly twice as many ways as the preceding number. Take 8 for instance. If every possible way of making 7 be written down, 8 may be made either by increasing the last component by a unit, or by annexing a unit at the end. Thus, 1 + 3 + 2 + 1 may yield 1 + 3 + 2 + 2, or 1 + 3 + 2 + 1 + 1: and all the ways of making 8 will thus be obtained; for any way of making 8, say _a_ + _b_ + _c_ + _d_, must proceed from the following mode of making 7, _a_ + _b_ + _c_ + (_d_-1). Now, (_d_-1) is either 0--that is, _d_ is unity and is struck out--or (_d_-1) remains, a number 1 less than _d_. Hence it follows that the number of ways of making _n_ is 2ⁿ⁻¹. For there is obviously 1 way of making 1, 2 of making 2; then there must be, by our rule, 2² ways of making 3, 2³ ways of making 4; and so on.
{ 1 + 1 + 1 { 1 + 1 + 1 + 1 { 1 + 1 { { 1 + 1 + 2 { { 1 + 2 { 1 + 2 + 1 1 { { 1 + 3 { { 2 + 1 { 2 + 1 + 1 { 2 { { 2 + 2 { 3 { 3 + 1 { 4
This table exhibits the ways of making 1, 2, 3, and 4. Hence it follows (which I leave the reader to investigate) that there are twice as many ways of forming _a_ + _b_ as there are of forming _a_ and then annexing to it a formation of _b_; four times as many ways of forming _a_ + _b_ + _c_ as there are of annexing to a formation of _a_ formations of _b_ and of _c_; and so on. Also, in summing numbers which make up _a_ + _b_, there are ways in which _a_ is a rest, and ways in which it is not, and as many of one as of the other.
Required the number of ways in which a number can be compounded of odd numbers, different orders counting as different ways. If _a_ be the number of ways in which _n_ can be so made, and _b_ the number of ways in which _n_ + 1 can be made, then _a_ + _b_ must be the number of ways in which _n_ + 2 can be made; for every way of making 12 out of odd numbers is either a way of making 10 with the last number increased by 2, or a way of making 11 with a 1 annexed. Thus, 1 + 5 + 3 + 3 gives 12, formed from 1 + 5 + 3 + 1 giving 10. But 1 + 9 + 1 + 1 is formed from 1 + 9 + 1 giving 11. Consequently, the number of ways of forming 12 is the sum of the number of ways of forming 10 and of forming 11. Now, 1 can only be formed in 1 way, and 2 can only be formed in 1 way; hence 3 can only be formed in 1 + 1 or 2 ways, 4 in only 1 + 2 or 3 ways. If we take the series 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, &c. in which each number is the sum of the two preceding, then the _n_th number of this set is the number of ways (orders counting) in which _n_ can be formed of odd numbers. Thus, 10 can be formed in 55 ways, 11 in 89 ways, &c.
Shew that the number of ways in which _mk_ can be made of numbers divisible by _m_ (orders counting) is 2ᵏ⁻¹.
In the two series, 1 1 1 2 3 4 6 9 13 19 28, &c. 0 1 0 1 1 1 2 2 3 4 5, &c.,
the first has each new term after the third equal to the sum of the last and last but two; the second has each new term after the third equal to the sum of the last but one and last but two. Shew that the _n_th number in the first is the number of ways in which _n_ can be made up of numbers which, divided by 3, leave a remainder 1; and that the _n_th number in the second is the number of ways in which _n_ can be made up of numbers which, divided by 3, leave a remainder 2.
It is very easy to shew in how many ways a number can be made up of a given number of numbers, if different orders count as different ways. Suppose, for instance, we would know in how many ways 12 can be thus made of 7 numbers. If we write down 12 units, there are 11 intervals between unit and unit. There is no way of making 12 out of 7 numbers which does not answer to distributing 6 partition-marks in the intervals, 1 in each of 6, and collecting all the units which are not separated by partition-marks. Thus, 1 + 1 + 3 + 2 + 1 + 2 + 2, which is one way of making 12 out of 7 numbers, answers to
| | | | | | 1 | 1 | 111 | 11 | 1 | 11 | 11 | | | | | |
in which the partition-marks come in the 1st, 2d, 5th, 7th, 8th, and 10th of the 11 intervals. Consequently, to ask in how many ways 12 can be made of 7 numbers, is to ask in how many ways 6 partition-marks can be placed in 11 intervals; or, how many combinations or selections can be made of 6 out of 11. The answer is,
11 × 10 × 9 × 8 × 7 × 6 -----------------------, or 462. 1 × 2 × 3 × 4 × 5 × 6
Let us denote by _m_ₙ the number of ways in which _m_ things can be taken out of _n_ things, so that _m_ₙ is the abbreviation for
_n_ - 1 _n_ - 2 _n_ - _m_ + 1 _n_ × ------ × ------- ... as far as ------------- 2 3 _m_
Then _m_ₙ also represents the number of ways in which _m_ + 1 numbers can be put together to make _n_ + 1. What we proved above is, that 6₁₁ is the number of ways in which we can put together 7 numbers to make 12. There will now be no difficulty in proving the following:
2ⁿ = 1 + 1ₙ + 2ₙ + 3ₙ ... + _n_ₙ
In the preceding question, 0 did not enter into the list of numbers used. Thus, 3 + 1 + 0 + 0 was not considered as one of the ways of putting together four numbers to make 5. But let us now ask, what is the number of ways of putting together 7 numbers to make 12, allowing 0 to be in the list of numbers. There can be no more (nor fewer) ways of doing this than of putting 7 numbers together, among which 0 is _not_ included, to make 19. Take every way of making 12 (0 included), and put on 1 to each number, and we get a way of making 19 (0 not included). Take any way of making 19 (0 not included), and strike off 1 from each number, and we have one of the ways of making 12 (0 included). Accordingly, 6₁₈ is the number of ways of putting together 7 numbers (0 being allowed) to make 12. And (_m_- 1)ₙ₊ₘ₋₁ is the number of ways of putting together _m_ numbers to make _n_, 0 being included.
This last amounts to the solution of the following: In how many ways can _n_ counters (undistinguishable from each other) be distributed into _m_ boxes? And the following will now be easily proved: The number of ways of distributing _c_ undistinguishable counters into _b_ boxes is (_b_ - 1)_{_b_ + _c_ - 1}, if any box or boxes may be left empty. But if there must be 1 at least in each box, the number of ways is (_b_ - 1)_{_c_ - 1}; if there must be 2 at least in each box, it is (_b_ - 1)_{_c- b_-1}; if there must be 3 at least in each box, it is (_b_ - 1)_{_c_ - 2_b_ - 1}; and so on.
The number of ways in which _m odd_ numbers can be put together to make _n_, is the same as the number of ways in which _m_ even numbers (0 included) can be put together to make _n_-_m_; and this is the number of ways in which _m_ numbers (odd or even, 0 included) can be put together to make ½(_n_-_m_). Accordingly, the number of ways in which m odd numbers can be put together to make _n_ is the same as the number of combinations of _m_-1 things out of ½(_n_-_m_) + _m_-1, or ½(_n_ + _m_)-1. Unless _n_ and _m_ be both even or both odd, the problem is evidently impossible.
There are curious and useful relations existing between numbers of combinations, some of which may readily be exhibited, under the simple expression of _m_ₙ to stand for the number of ways in which _m_ things may be taken out of _n_. Suppose we have to take 5 out of 12: Let the 12 things be marked A, B, C, &c. and set apart one of them, A. Every collection of 5 out of the 12 either does or does not include A. The number of the latter sort must be 5₁₁; the number of the former sort must be 4₁₁, since it is the number of ways in which the _other four_ can be chosen out of all but A. Consequently, 5₁₂ must be 5₁₁ + 4₁₁, and thus we prove in every case,
_m_ₙ′ = _m_ₙ₋₁ + (_m_ - 1)ₙ₋₁
0ₙ and _n_ₙ both are 1; for there is but one way of taking _none_, and but one way of taking _all_. And again _m_ₙ and (_n_-_m_)ₙ are the same things. And if _m_ be greater than _n_, _m_ₙ is 0; for there are no ways of doing it. We make one of our preceding results more symmetrical if we write it thus,
2ⁿ = 0ₙ + 1ₙ + 2ₙ + ... + _n_ₙ
If we now write down the table of symbols in which the (_m_ + 1)th
0 1 2 3, &c. +-------------------------------------- 1 | 0₁ 1₁ 2₁ 3₁, &c. 2 | 0₂ 1₂ 2₂ 3₂, &c. 3 | 0₃ 1₃ 2₃ 3₃, &c. &c. | &c. &c. &c. &c.
number of the _n_th row represents _m_ₙ, the number of combinations of _m_ out of _n_, we see it proved above that the law of formation of this table is as follows: Each number is to be the sum of the number above it and the number preceding the number above it. Now, the first row must be 1, 1, 0, 0, 0, &c. and the first column must be 1, 1, 1, 1, &c. so that we have a table of the following kind, which may be carried as far as we please:
0 1 2 3 4 5 6 7 8 9 10 +---------------------------------------------------- 1 | 1 1 0 0 0 0 0 0 0 0 0 2 | 1 2 1 0 0 0 0 0 0 0 0 3 | 1 3 3 1 0 0 0 0 0 0 0 4 | 1 4 6 4 1 0 0 0 0 0 0 5 | 1 5 10 10 5 1 0 0 0 0 0 6 | 1 6 15 20 15 6 1 0 0 0 0 7 | 1 7 21 35 35 21 7 1 0 0 0 8 | 1 8 28 56 70 56 28 8 1 0 0 9 | 1 9 36 84 126 126 84 36 9 1 0 10 | 1 10 45 120 210 252 210 120 45 10 1
Thus, in the row 9, under the column headed 4, we see 126, which is 9 × 8 × 7 × 6 ÷ (1 × 2 × 3 × 4), the number of ways in which 4 can be chosen out of 9, which we represent by 4-{9}.
If we add the several rows, we have 1 + 1 or 2, 1 + 2 + 1 or 2², next 1 + 3 + 3 + 1 or 2³, &c. which verify a theorem already announced; and the law of formation shews us that the several columns are formed thus:
1 1 1 2 1 1 3 3 1 1 1 1 2 1 1 3 3 1 ----- ------- --------- 1 2 1 1 3 3 1 1 4 6 4 1, &c.
so that the sum in each row must be double of the sum in the preceding. But we can carry the consequences of this mode of formation further. If we make the powers of 1 + _x_ by actual algebraical multiplication, we see that the process makes the same oblique addition in the formation of the numerical multipliers of the powers of _x_.
1 + _x_ 1 + _x_ ------- 1 + _x_ _x_ + _x_² --------------- 1 + 2_x_ + _x_²
1 + 2_x_ + _x_² 1 + _x_ --------------- 1 + 2_x_ + _x_² _x_ + 2_x_² + _x_³ ----------------------- 1 + 3_x_ + 3_x_² + _x_³
Here are the second and third powers of 1 + _x_: the fourth, we can tell beforehand from the table, must be 1 + 4_x_ + 6_x_² + 4_x_³ + _x_⁴; and so on. Hence we have
(1 + _x_)ⁿ = 0ₙ + 1ₙ_x_ + 2ₙ_x_² + 3ₙ_x_³ + ... + _n_ₙ_x_ⁿ
which is usually written with the symbols 0ₙ, 1ₙ, &c. at length, thus,
_n_ - 1 _n_ - 1 _n_ - 2 (1 + _x_)ⁿ = 1 + _nx_ + _n_-------_x_² + _n_------- -------_x_³ + &c. 2 2 3
This is the simplest case of what in algebra is called the _binomial theorem_. If instead of 1 + _x_ we use _x_ + _a_, we get
(_x_+_a_)ⁿ = _x_ⁿ+1ₙ_ax_ⁿ⁻¹+2ₙ_a_²_x_ⁿ⁻²+3ₙ_a_³_x_ⁿ⁻³+... +_n_ₙ_a_ⁿ
We can make the same table in another form. If we take a row of ciphers beginning with unity, and setting down the first, add the next, and then the next, and so on, and then repeat the process with one step less, and then again with one step less, we have the following:
1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 3 4 5 6 1 3 6 10 15 1 4 10 20 1 5 15 1 6 1
In the oblique columns we see 1 1, 1 2 1, 1 3 3 1, &c. the same as in the original table, and formed by the same additions. If, before making the additions, we had always multiplied by _a_, we should have got the several components of the powers of 1 + _a_, thus,
1 0 0 0 0 1 _a_ _a_² _a_³ _a_⁴ 1 2_a_ 3_a_² 4_a_³ 1 3_a_ 6_a_² 1 4_a_ 1
where the oblique columns 1 + _a_, 1 + 2_a_ + _a_², 1 + 3_a_ + 3_a_² + _a_³, &c., give the several powers of 1 + _a_. If instead of beginning with 1, 0, 0, &c. we had begun with _p_, 0, 0, &c. we should have got _p_, _p_ × 4_a_, _p_ × 6_a_², &c. at the bottom of the several columns; and if we had written at the top _x_⁴, _x_³, _x_², _x_, 1, we should have had all the materials for forming _p_(_x_ + _a_)⁴ by multiplying the terms at the top and bottom of each column together, and adding the results.
Suppose we follow this mode of forming _p_(_x_ + _a_)³ + _q_(_x_ + _a_)² + _r_(_x_ + _a_) + _s_.
_x_³ _x_² _x_ 1 _x_² _x_ 1 _x_ 1 1 _p_ 0 0 0 _q_ 0 0 _r_ 0 3 _p_ _pa_ _pa_² _pa_³ _q_ _qa_ _qa_² _r_ _ra_ _p_ 2_pa_ 3_pa_² _q_ 2_qa_ _r_ _p_ 3_pa_ _q_ _p_
_px_³ + 3_pax_² + 3_pa_²_x_ + _pa_³ + _qx_² + 2_qax_ + _qa_² + _rx_ + _ra_ + _s_
= _px_³ + (3_pa_ + _q_)_x_² + (3_pa_² + 2_qa_ + _r_)_x_ + _pa_³ + _qa_² + _ra_ + _s_
Now, observe that all this might be done in one process, by entering _q_, _r_, and _s_ under their proper powers of _x_ in the first process, as follows
_x_³ _x_² _x_ 1 _p_ _q_ _r_ _s_ _p_ _pa_ + _q_ _pa_² + _qa_ + _r_ _pa_³ + _qa_² + _ra_ + _s_ _p_ 2_pa_ + _q_ 3_pa_² + 2_qa_ + _r_ _p_ 3_pa_ + _q_ _p_
This process[65] is the one used in Appendix XI., with the slight alteration of varying the sign of the last letter, and making subtractions instead of additions in the last column. As it stands, it is the most convenient mode of writing _x_ + _a_ instead of _x_ in a large class of algebraical expressions. For instance, what does 2_x_⁵ + _x_⁴ + 3_x_² + 7_x_ + 9 become when _x_ + 5 is written instead of _x_? The expression, made complete, is,
2_x_⁵ + 1_x_⁴ + 0_x_³ + 3_x_² + 7_x_ + 9
1 0 3 7 9 2 11 55 278 1397 6994 2 21 160 1078 6787 2 31 315 2653 2 41 520 2 51
_Answer_, 2_x_⁵ + 51_x_⁴ + 520_x_³ + 2653_x_² + 6787_x_ + 6994.
[65] The principle of this mode of demonstration of Horner’s method was stated in Young’s Algebra (1823), being the earliest elementary work in which that method was given.
APPENDIX XI.
ON HORNER’S METHOD OF SOLVING EQUATIONS.
The rule given in this chapter is inserted on account of its excellence as an exercise in computation. The examples chosen will require but little use of algebraical signs, that they may be understood by those who know no more of algebra than is contained in the present work.
To solve an equation such as
2_x_⁴ + _x_² - 3_x_ = 416793,
or, as it is usually written,
2_x_⁴ + _x_² - 3_x_ - 416793 = 0,
we must first ascertain by trial not only the first figure of the root, but also the denomination of it: if it be a 2, for instance, we must know whether it be 2, or 20, or 200, &c., or ·2, or ·02, or ·002, &c. This must be found by trial; and the shortest way of making the trial is as follows: Write the expression in its complete form. In the preceding case the form is not complete, and the complete form is
2_x_⁴ + 0_x_³ + 1_x_² - 3_x_ - 416793.
To find what this is when x is any number, for instance, 3000, the best way is to take the first multiplier (2), multiply it by 3000, and take in the next multiplier (0), multiply the result by 3000, and take in the next multiplier (1), and so on to the end, as follows:
2 × 3000 + 0 = 6000; 6000 × 3000 + 1 = 18000001
18000001 × 3000 - 3 = 54000002997
54000002997 × 3000 - 416793 = 162000008574207
Now try the value of the above when _x_ = 30. We have then, for the steps, 60 (2 × 30 + 0), 1801, 54027, and lastly,
1620810-416793,
or _x_ = 30 makes the first terms greater than 416793. Now try _x_ = 20 which gives 40, 801, 16017, and lastly,
320340-416793,
or _x_ = 20 makes the first terms less than 416793. Between 20 and 30, then, must be a value of _x_ which makes 2_x_⁴ + _x_²-3x equal to 416793. And this is the preliminary step of the process.
Having got thus far, write down the coefficients +2, 0, +1,-3, and -416793, each with its proper algebraical sign, except the last, in which let the sign be changed. This is the most convenient way when the last sign is-. But if the last sign be +, it may be more convenient to let it stand, and change all which come before. Thus, in solving _x_³-12_x_ + 1 = 0, we might write
-1 0 +12 1
whereas in the instance before us, we write
+2 0 +1 -3 416793
Having done this, take the highest figure of the root, properly named, which is 2 tens, or 20. Begin with the first column, multiply by 20, and join it to the number in the next column; multiply that by 20, and join it to the number in the next column; and so on. But when you come to the last column, subtract the product which comes out of the preceding column, or join it to the last column after changing its sign. When this has been done, repeat the process with the numbers which now stand in the columns, omitting the last, that is, the subtracting step; then repeat it again, going only as far as the last column but two, and so on, until the columns present a set of rows of the following appearance:
_a_ _b_ _c_ _d_ _e_ _f_ _g_ _h_ _i_ _k_ _l_ _m_ _n_ _o_ _p_
to the formation of which the following is the key:
_f_ = 20_a_ + _b_, _g_ = 20_f_ + _c_, _h_ = 20_g_ + _d_, _i_ = _e_ - 20_h_, _k_ = 20_a_ + _f_, _l_ = 20_k_ + _g_, _m_ = 20_l_ + _h_, _n_ = 20_a_ + _k_, _o_ = 20_n_ + _l_, _p_ = 20_a_ + _n_.
We call this _Horner’s Process_, from the name of its inventor. The result is as follows:
2 0 1 -3 416793 (20 40 801 16017 96453 80 2401 64037 120 4801 160
We have now before us the row
2 160 4801 64037 96453
which furnishes our means of guessing at the next, or units’ figure of the root.
Call the last column the _dividend_, the last but one the _divisor_, and all that come before _antecedents_. See how often the dividend contains the divisor; this gives the guess at the next figure. The guess is a true one,[66] if, on applying Horner’s process, the divisor result, augmented as it is by the antecedent processes, still go as many times in the dividend. For example, in the case before us, 96453 contains 64037 once; let 1 be put on its trial. Horner’s process is found to succeed, and we have for the second process,
2 160 4801 64037 96453 162 4963 69000 27453 164 5127 74127 166 5293 168
As soon as we come to the fractional portion of the root, the process assumes a more[67] methodical form.
The equation being of the _fourth_ degree, annex _four_ ciphers to the dividend, _three_ to the divisor, _two_ to the antecedent, and _one_ to the previous antecedent, leaving the first column as it is; then find the new figure by the dividend and divisor, as before,[68] and apply Horner’s process. Annex ciphers to the results, as before, and proceed in the same way. The annexing of the ciphers prevents our having any thing to do with decimal points, and enables us to use the quotient-figures without paying any attention to their _local_ values. The following exhibits the whole process from the beginning, carried as far as it is here intended to go before beginning the contraction, which will give more figures, as in the rule for the square root. The following, then, is the process as far as one decimal place:
[66] Various exceptions may arise when an equation has two nearly equal roots. But I do not here introduce algebraical difficulties; and a student might give himself a hundred examples, taken at hazard, without much chance of lighting upon one which gives any difficulty.
[67] This form might be also applied to the integer portions; but it is hardly needed in such instances as usually occur. See the article _Involution and Evolution_ in the _Supplement_ to the _Penny Cyclopædia_.
[68] After the second step, the trial will rarely fail to give the true figure.
2 0 1 -3 416793(213 40 801 16017 96453 80 2401 64037 ----- 120 4801 ----- 274530000 160 ---- 69000 47339778 --- 4963 74127000 --------- 162 5127 -------- 164 529300 75730074 166 ------ 77348376 1680 534358 ---- 539434 1686 544528 1692 1698 1704 ----
If we now begin the contraction, it is good to know beforehand on what number of additional root-figures we may reckon. We may be pretty certain of having nearly as many as there are figures in the divisor when we begin to contract--one less, or at least two less. Thus, there being now eight figures in the divisor, we may conclude that the contraction will give us at least six more figures. To begin the contraction, let the dividend stand, cut off one figure from the divisor, two from the column before that, three from the one before that, and so on. Thus, our contraction begins with
| | | | |0002 1|704 5445|28 7734837|6 47339778 | | | |
The first column is rendered quite useless here. Conduct the process as before, using only the figures which are not cut off. But it will be better to go as far as the first figure cut off, carrying from the second figure cut off. We shall then have as follows:
| | | 1|704 5445|28 7734837|6 47339778(6 | 5455|5 7767570|6 734354 5465|7 7800364|8 5475|9 | |
At the next contraction the column 1|704 becomes |001704, and is quite useless. The next step, separately written (which is not, however, necessary in working), is
| | 54|759 780036|48 734354(0 | |
Here the dividend 734354 does not contain the divisor 780036, and we, therefore, write 0 as a root figure and make another contraction, or begin with
| | |54759 78003|648 734354(9 | 78008|5 32277 78013|4 |
At the next contraction the first column becomes |0054759, and is quite useless, so that the remainder of the process is the contracted division.
| 7801|34)32277(4137 | 1072 292 58 3
and the root required is 21·36094137.
I now write down the complete process for another equation, one root of which lies between 3 and 4: it is
_x_³ - 10_x_ + 1 = 0
1 0 -10 -1(3·1110390520730990796 3 -1 2000 6 1700 209000 9 0 1791 19769000 9 1 188300 743369000000 9 2 189231 172311710273000 9 30 19016300 991247447681 9 31 19025631 39462875420 9 32 1903496300 0 0 1391491559 9 33 0 1903524299 0 9 58993123 9 33 1 1903552298 2 7 0 0 1886047 9 33 2 1903560698 0 5|9|1 172835 9 33 30 0 1903569097 8 5|6|3 1515 9 33 30 3 1903569144 5 2|2| 183 9 33 30 6 1903569191 1|8|8 12 9 33 30 90 1903569193 0|6| 1 9 33|30|99 1903569194|9|3| 9 33|31|08 | | | |09|33|31|17 | | | |
The student need not repeat the rows of figures so far as they come under one another: thus, it is not necessary to repeat 190356. But he must use his own discretion as to how much it would be safe for him to omit. I have set down the whole process here as a guide.
The following examples will serve for exercise:
1. 2_x_³ - 100_x_ - 7 = 0 _x_ = 7·10581133.
2. _x_⁴ + _x_³ + _x_² + _x_ = 6000 _x_ = 8·531437726.
3. _x_³ + 3_x_² - 4_x_ - 10 = 0 _x_ = 1·895694916504.
4. _x_³ + 100_x_² - 5_x_ - 2173 = 0 _x_ = 4·582246071058464. _ 5. ∛2 = 1·259921049894873164767210607278.[69]
6. _x_³ - 6_x_ = 100 _x_ = 5·071351748731.
7. _x_³ + 2_x_² + 3_x_ = 300 _x_ = 5·95525967122398.
8. _x_³ + _x_ = 1000 _x_ = 9·96666679.
9. 27000_x_³ + 27000_x_ = 26999999 _x_ = 9·9666666.....
10. _x_³ - 6_x_ = 100 _x_ = 5·0713517487.
11. _x_⁵ - 4_x_⁴ + 7_x_³ - 863 = 0 _x_ = 4·5195507.
12. _x_³ - 20_x_ + 8 = 0 _x_ = 4·66003769300087278.
13. _x_³ + _x_² + _x_ - 10 = 0 _x_ = 1·737370233.
14. _x_³ - 46_x_² - 36_x_ + 18 = 0 _x_ = 46·7616301847, or _x_ = ·3471623192.
15. _x_³ + 46_x_² - 36_x_ - 18 = 0 _x_ = 1·1087925037.
16. 8991_x_³ - 162838_x_² + 746271_x_ - 81000 = 0 _x_ = ·111222333444555....
17. 729_x_³ - 486_x_² + 99_x_ - 6 = 0 _x_ = ·1111..., or ·2222..., or ·3333....
18. 2_x_³ + 3_x_² - 4_x_ = 500 _x_ = 5·93481796231515279.
19. _x_³ + 2_x_² + _x_ - 150 = 0 _x_ = 4·6684090145541983253742991201705899.
20. _x_³ + _x_ = _x_² + 500 _x_ = 8·240963558144858526963.
21. _x_³ + 2_x_² + 3_x_ - 10000 = 0 _x_ = 20·852905526009.
22. _x_⁵ - 4_x_ - 2000 = 0 _x_ = 4·581400362.
23. 10_x_³ - 33_x_² - 11_x_ - 100 = 0 _x_ = 4·146797808584278785.
24. _x_⁴ + _x_³ + _x_² + _x_ = 127694 _x_ = 18·64482373095.
25. 10_x_³ + 11_x_² + 12_x_ = 100000 _x_ = 21·1655995554508805.
26. _x_³ + _x_ = 13 _x_ = 2·209753301208849.
27. _x_³ + _x_² - 4_x_ - 1600 = 0 _x_ = 11·482837157.
28. _x_³ - 2_x_ = 5 _x_ = 2·094551481542326591482386540579302963857306105628239.
29. _x_⁴ - 80_x_³ + 24_x_² - 6_x_ - 80379639 = 0 _x_ = 123.[70]
30. _x_³ - 242_x_² - 6315_x_ + 2577096 = 0 _x_ = 123.[71]
31. 2_x_⁴ - 3_x_³ + 6_x_ - 8 = 0 _x_ = 1·414213562373095048803.[72]
32. _x_⁴ - 19_x_³ + 132_x_² - 302_x_ + 200 = 0 _x_ = 1·02804, or 4, or 6·57653, or 7·39543[73].
33. 7_x_⁴ - 11_x_³ + 6_x_² + 5_x_ = 215 _x_ = 2·70648049385791.[74]
34. 7_x_⁵ + 6_x_⁴ + 5_x_³ + 4_x_² + 3_x_ = 11 _x_ = ·770768819622658522379296505.[75]
35. 4_x_⁶ + 7_x_⁵ + 9_x_⁴ + 6_x_³ + 5_x_² + 3_x_ = 792 _x_ = 2·0520421768796053652140434012812019734602755995 45541724214.[76]
36. 2187_x_⁴ - 2430_x_³ + 945_x_² - 150_x_ + 8 = 0 _x_ = ·1111...., or ·2222...., or ·3333...., or ·4444....
[69] The solution of _x_³ + 0_x_² + 0_x_-2 = 0.
[70] Taken from a paper on the subject, by Mr. Peter Gray, in the _Mechanics’ Magazine_.
[71] Taken from a paper on the subject, by Mr. Peter Gray, in the _Mechanics’ Magazine_.
[72] Taken from a paper on the subject, by Mr. Peter Gray, in the _Mechanics’ Magazine_.
[73] Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.
[74] Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.
[75] Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.
[76] Taken from the late Mr. Peter Nicholson’s Essay on Involution and Evolution.
APPENDIX XII.
RULES FOR THE APPLICATION OF ARITHMETIC TO GEOMETRY.
The student should make himself familiar with the most common terms of geometry, after which the following rules will present no difficulty. In them all, it must be understood, that when we talk of multiplying one line by another, we mean the repetition of one line as often as there are units of a given kind, as feet or inches, in another. In any other sense, it is absurd to talk of multiplying a quantity by another quantity. All quantities of the same kind should be represented in numbers of the same unit; thus, all the lines should be either feet and decimals of a foot, or inches and decimals of an inch, &c. And in whatever unit a length is represented, a surface is expressed in the corresponding square units, and a solid in the corresponding cubic units. This being understood, the rules apply to all sorts of units.
_To find the area of a rectangle._ Multiply together the units in two sides which meet, or multiply together two sides which meet; the product is the number of square units in the area. Thus, if 6 feet and 5 feet be the sides, the area is 6 × 5, or 30 square feet. Similarly, the area of a square of 6 feet long is 6 × 6, or 36 square feet (234).
_To find the area of a parallelogram._ Multiply one side by the perpendicular distance between it and the opposite side; the product is the area required in square units.
_To find the area of a trapezium._[77] Multiply either of the two sides which are not parallel by the perpendicular let fall upon it from the middle point of the other.
[77] A four-sided figure, which has two sides parallel, and two sides not parallel.
_To find the area of a triangle._ Multiply any side by the perpendicular let fall upon it from the opposite vertex, and take half the product. Or, halve the sum of the three sides, subtract the three sides severally from this half sum, multiply the four results together, and find the square root of the product. The result is the number of square units in the area; and twice this, divided by either side, is the perpendicular distance of that side from its opposite vertex.
_To find the radius of the internal circle which touches the three sides of a triangle._ Divide the area, found in the last paragraph, by half the sum of the sides.
_Given the two sides of a right-angled triangle, to find the hypothenuse._ Add the squares of the sides, and extract the square root of the sum.
_Given the hypothenuse and one of the sides, to find the other side._ Multiply the sum of the given lines by their difference, and extract the square root of the product.
_To find the circumference of a circle from its radius, very nearly._ Multiply twice the radius, or the diameter, by 3·1415927, taking as many decimal places as may be thought necessary. For a rough computation, multiply by 22 and divide by 7. For a very exact computation, in which decimals shall be avoided, multiply by 355 and divide by 113. See (131), last example.
_To find the arc of a circular sector, very nearly, knowing the radius and the angle._ Turn the angle into seconds,[78] multiply by the radius, and divide the product by 206265. The result will be the number of units in the arc.
[78] The right angle is divided into 90 equal parts called _degrees_, each degree into 60 equal parts called _minutes_, and each minute into 60 equal parts called _seconds_. Thus, 2° 15′ 40″ means 2 degrees, 15 minutes, and 40 seconds.
_To find the area of a circle from its radius, very nearly._ Multiply the square of the radius by 3·1415927.
_To find the area of a sector, very nearly, knowing the radius and the angle._ Turn the angle into seconds, multiply by the square of the radius, and divide by 206265 × 2, or 412530.
_To find the solid content of a rectangular parallelopiped._ Multiply together three sides which meet: the result is the number of cubic units required. If the figure be not rectangular, multiply the area of one of its planes by the perpendicular distance between it and its opposite plane.
_To find the solid content of a pyramid._ Multiply the area of the base by the perpendicular let fall from the vertex upon the base, and divide by 3.
_To find the solid content of a prism._ Multiply the area of the base by the perpendicular distance between the opposite bases.
_To find the surface of a sphere._ Multiply 4 times the square of the radius by 3·1415927.
_To find the solid content of a sphere._ Multiply the cube of the radius by 3·1415927 × ⁴/₃, or 4·18879.
_To find the surface of a right cone._ Take half the product of the circumference of the base and slanting side. _To find the solid content_, take one-third of the product of the base and the altitude.
_To find the surface of a right cylinder._ Multiply the circumference of the base by the altitude. _To find the solid content_, multiply the area of the base by the altitude.
The weight of a body may be found, when its solid content is known, if the weight of one cubic inch or foot of the body be known. But it is usual to form tables, not of the weights of a cubic unit of different bodies, but of the proportion which these weights bear to some one amongst them. The one chosen is usually distilled water, and the proportion just mentioned is called the _specific gravity_. Thus, the specific gravity of gold is 19·362, or a cubic foot of gold is 19·362 times as heavy as a cubic foot of distilled water. Suppose now the weight of a sphere of gold is required, whose radius is 4 inches. The content of this sphere is 4 × 4 × 4 × 4·1888, or 268·0832 cubic inches; and since, by (217), each cubic inch of water weighs 252·458 grains, each cubic inch of gold weighs 252·458 × 19·362, or 4888·091 grains; so that 268·0832 cubic inches of gold weigh 268·0832 × 4888·091 grains, or 227½ pounds troy nearly. Tables of specific gravities may be found in most works of chemistry and practical mechanics.
The cubic foot of water is 908·8488 troy ounces, 75·7374 troy pounds, 997·1369691 averdupois ounces, and 62·3210606 averdupois pounds. For all rough purposes it will do to consider the cubic foot of water as being 1000 common ounces, which reduces tables of specific gravities to common terms in an obvious way. Thus, when we read of a substance which has the specific gravity 4·1172, we may take it that a cubic foot of the substance weighs 4117 ounces. For greater correctness, diminish this result by 3 parts out of a thousand.
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CONTENTS:--The Planets; are they inhabited Worlds? Weather Prognostics. Popular Fallacies in Questions of Physical Science. Latitudes and Longitudes. Lunar Influences. Meteoric Stones and Shooting Stars. Railway Accidents. Light. Common Things.--Air. Locomotion in the United States. Cometary Influences. Common Things.--Water. The Potter’s Art. Common Things.--Fire. Locomotion and Transport, their Influence and Progress. The Moon. Common Things.--The Earth. The Electric Telegraph. Terrestrial Heat. The Sun. Earthquakes and Volcanoes. Barometer, Safety Lamp, and Whitworth’s Micrometric Apparatus. Steam. The Steam Engine. The Eye. The Atmosphere. Time. Common Things.--Pumps. Common Things.--Spectacles--The Kaleidoscope. Clocks and Watches. Microscopic Drawing and Engraving. The Locomotive. Thermometer. New Planets.--Leverrier and Adams’s Planet. Magnitude and Minuteness. Common Things.--The Almanack. Optical Images. How to Observe the Heavens. Common Things.--The Looking Glass. Stellar Universe. The Tides. Colour. Common Things.--Man. Magnifying Glasses. Instinct and Intelligence. The Solar Microscope. The Camera Lucida. The Magic Lantern. The Camera Obscura. The Microscope. The White Ants; their Manners and Habits. The Surface of the Earth, or First Notions of Geography. Science and Poetry. The Bee. Steam Navigation. Electro-Motive Power. Thunder, Lightning, and the Aurora Borealis. The Printing Press. The Crust of the Earth. Comets. The Stereoscope. The Pre-Adamite Earth. Eclipses. Sound.
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