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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Part III. 1s. 6d.
7. Hydrostatics.
8. Hydraulics.
9. Pneumatics.
_Lardner’s Popular Geology. (From “The Museum of Science
and Art.”)_ 201 Illustrations. 2s. 6d.
_Lardner’s Common Things Explained. Containing_:
Air--Earth--Fire--Water--Time--The Almanack--Clocks and
Watches--Spectacles--Colour--Kaleidoscope--Pumps--Man--The
Eye--The Printing Press--The Potter’s Art--Locomotion
and Transport--The Surface of the Earth, or First
Notions of Geography. (From “The Museum of Science and
Art.”) With 233 Illustrations. Complete, 5s., cloth
lettered.
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_Lardner’s Popular Physics. Containing_:
Magnitude and Minuteness--Atmosphere--Thunder
and Lightning--Terrestrial Heat--Meteoric
Stones--Popular Fallacies--Weather
Prognostics--Thermometer--Barometer--Safety
Lamp--Whitworth’s Micrometric Apparatus--Electro-Motive
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(From “The Museum of Science and Art.”) With 85
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_Lardner’s Popular Astronomy. Containing_:
How to Observe the Heavens--Latitudes and Longitudes
--TheEarth--The Sun--The Moon--The Planets: are they
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Adams’s Planet--The Tides--Lunar Influences--and
the Stellar Universe--Light--Comets--Cometary
Influences--Eclipses--Terrestrial Rotation--Lunar
Rotation--Astronomical Instruments. (From “The Museum of
Science and Art.”) 182 Illustrations. Complete, 4s. 6d.
cloth lettered.
⁂ _Sold also in Two Series_, 2s. 6d. _and_ 2s._each_.
_Lardner on the Microscope_. (From “The Museum of Science
and Art.”) 1 vol. 147 Engravings. 2s.
_Lardner on the Bee and White Ants; their Manners
and Habits_; with Illustrations of Animal Instinct
and Intelligence. (From “The Museum of Science and Art.”)
1 vol. 135 Illustrations. 2s., cloth lettered.
_Lardner on Steam and its Uses_; including the Steam
Engine and Locomotive, and Steam Navigation.
(From “The Museum of Science and Art.”) 1 vol.,
with 89 Illustrations. 2s.
_Lardner on the Electric Telegraph, Popularised_. With
100 Illustrations. (From “The Museum of Science and Art.”)
12mo., 250 pages. 2s., cloth lettered.
⁂ _The following Works from “Lardner’s Museum of Science
and Art_,” may also be had arranged as described,
handsomely half bound morocco, cloth sides.
Common Things. Two series in one vol. 7s. 6d.
Popular Astronomy. Two series in one vol. 7s. 0d.
Electric Telegraph, with Steam and its Uses. In one vol. 7s. 0d.
Microscope and Popular Physics. In one vol. 7s. 0d.
Popular Geology, and Bee and White Ants. In one vol. 7s. 6d.
_Lardner on the Steam Engine, Steam Navigation, Roads, and
Railways_. Explained and Illustrated. Eighth Edition.
With numerous Illustrations. 1 vol. large 12mo. 8s. 6d.
_A Guide to the Stars for every Night in the Year. In
Eight Planispheres_. With an Introduction.
8vo. 5s., cloth.
_Minasi’s Mechanical Diagrams. For the Use of Lecturers
and Schools._ 15 Sheets of Diagrams, coloured,
15s., illustrating the following subjects: 1 and 2.
Composition of Forces.--3. Equilibrium.--4 and 5.
Levers.--6. Steelyard, Brady Balance, and Danish
Balance.--7. Wheel and Axle.--8. Inclined Plane.--9, 10,
11. Pulleys.--12. Hunter’s Screw.--13 and 14. Toothed
Wheels.--15. Combination of the Mechanical Powers.
=LOGIC=.
_De Morgan’s Formal Logic; or, The Calculus of Inference_,
Necessary and Probable. 8vo. 6s. 6d.
_Neil’s Art of Reasoning:_ a Popular Exposition of the
Principles of Logic, Inductive and Deductive; with an
Introductory Outline of the History of Logic, and an
Appendix on recent Logical Developments, with Notes.
Crown 8vo. 4s. 6d., cloth.
=ENGLISH COMPOSITION=.
_Neil’s Elements of Rhetoric_; a Manual of the Laws of
Taste, including the Theory and Practice of Composition.
Crown 8vo. 4s. 6d., cl.
=DRAWING=.
_Lineal Drawing Copies for the earliest Instruction_.
Comprising upwards of 200 subjects on 24 sheets, mounted
on 12 pieces of thick pasteboard, in a Portfolio.
By the Author of “Drawing for Young Children.” 5s. 6d.
_Easy Drawing Copies for Elementary Instruction._
Simple Outlines without Perspective. 67 subjects,
in a Portfolio. By the Author of “Drawing for Young Children.”
6s. 6d.
_Sold also in Two Sets._
SET I. Twenty-six Subjects mounted on thick
pasteboard, in a Portfolio. 3s. 6d.
SET II. Forty-one Subjects mounted on thick
pasteboard, in a Portfolio. 3s. 6d.
The copies are sufficiently large and bold to be drawn from
by forty or fifty children at the same time.
=SINGING=.
_A Musical Gift from an Old Friend_, containing
Twenty-four New Songs for the Young. By W. E.
HICKSON, author of the Moral Songs of
“The Singing Master.” 8vo. 2s. 6d.
_The Singing Master. Containing First Lessons in Singing,
and the Notation of Music_; Rudiments of the Science
of Harmony; The First Class Tune Book; The Second Class
Tune Book; and the Hymn Tune Book. Sixth Edition. 8vo.
6s., cloth lettered.
_Sold also in Five Parts, any of which may be had separately._
I.--_First Lessons in Singing and the Notation of
Music._ Containing Nineteen Lessons in the Notation
and Art of Reading Music, as adapted for the Instruction
of Children, and especially for Class Teaching, with
Sixteen Vocal Exercises, arranged as simple two-part
harmonies. 8vo. 1s., sewed.
II.--_Rudiments of the Science of Harmony or Thorough
Bass._ Containing a general view of the principles of
Musical Composition, the Nature of Chords and Discords,
mode of applying them, and an Explanation of Musical
Terms connected with this branch of Science.
8vo. 1s., sewed.
III.--_The First Class Tune Book_. A Selection of
Thirty Single and Pleasing Airs, arranged with suitable
words for young children. 8vo. 1s., sewed.
IV.--_The Second Class Tune Book_. A Selection of Vocal
Music adapted for youth of different ages, and arranged
(with suitable words) as two or three-part harmonies.
8vo, 1s. 6d.
V.--_The Hymn Tune Book_. A Selection of Seventy
popular Hymn and Psalm Tunes, arranged with a view of
facilitating the progress of Children learning to sing
in parts. 8vo. 1s. 6d.
⁂ The Vocal Exercises, Moral Songs, and Hymns, with the
Music, may also be had, printed on Cards, price Twopence
each Card, or Twenty-five for Three Shillings.
=CHEMISTRY=.
_Gregory’s Hand-Book of Chemistry_. For the use of
Students. By WILLIAM GREGORY, M.D., late
Professor of Chemistry in the University of Edinburgh.
Fourth Edition, revised and enlarged. Illustrated by
Engravings on Wood. Complete in One Volume. Large 12mo.
18s. cloth.
⁂ _The Work may also be had in two Volumes, as under._
INORGANIC CHEMISTRY. Fourth Edition,
revised and enlarged. 6s. 6d. cloth.
ORGANIC CHEMISTRY. Fourth Edition,
very carefully revised, and greatly enlarged.
12s., cloth.
(Sold separately.)
_Chemistry for Schools_. By DR. LARDNER.
190 Illustrations. Large 12mo. 3s. 6d. cloth.
_Liebig’s Familiar Letters on Chemistry, in its Relations
to Physiology, Dietetics, Agriculture, Commerce, and
Political Economy._ Fourth Edition, revised and
enlarged, with additional Letters. Edited by DR. BLYTH.
Small 8vo. 7s. 6d. cloth.
_Liebig’s Letters on Modern Agriculture._ Small 8vo. 6s.
_Liebig’s Principles of Agricultural Chemistry; with
Special Reference to the late Researches made in England._
Small 8vo. 3s. 6d., cloth.
_Liebig’s Chemistry in its Applications to Agriculture and
Physiology._ Fourth Edition, revised.
8vo. 6s. 6d., cloth.
_Liebig’s Animal Chemistry; or, Chemistry in its
Application to Physiology and Pathology._
Third Edition. Part I. (the first half of the work).
8vo. 6s. 6d., cloth.
_Liebig’s Hand-Book of Organic Analysis;_ containing
a detailed Account of the various Methods used in
determining the Elementary Composition of Organic
Substances. Illustrated by 85 Woodcuts.
12mo. 5s., cloth.
_Bunsen’s Gasometry_; comprising the Leading Physical
and Chemical Properties of Gases, together with the
Methods of Gas Analysis. Fifty-nine Illustrations.
8vo. 8s. 6d., cloth.
_Wöhler’s Hand-Book of Inorganic Analysis;_ One
Hundred and Twenty-two Examples, illustrating the most
important processes for determining the Elementary
composition of Mineral substances. Edited by
DR. A. W. HOFMANN, Professor in the Royal
College of Chemistry. Large 12mo.
_Parnell on Dyeing and Calico Printing_. (Reprinted
from Parnell’s “Applied Chemistry in Manufactures, Arts,
and Domestic Economy, 1844.”) With Illustrations.
8vo. 7s., cloth.
=GENERAL LITERATURE=.
_De Morgans Book of Almanacs_. With an Index of
Reference by which the Almanac may be found for every
Year, whether in Old Style or New, from any Epoch,
Ancient or Modern, up to A.D. 2000. With
means of finding the Day of New or Full Moon, from
B.C. 2000 to A.D. 2000._ 5s.,
cloth lettered.
_Guesses at Truth. By Two Brothers. New Edition. With an
Index._ Complete 1 vol. Small 8vo. Handsomely bound
in cloth with red edges. 10s. 6d.
_Rudall’s Memoir of the Rev. James Crabb; late of
Southampton. With Portrait._
Large 12mo., 6s., cloth.
_Herschell (R. H.) The Jews;_ a brief Sketch of their
Present State and Future Expectations.
Fcap. 8vo. 1s. 6d., cloth.