To answer this question it is necessary to put together the numbers 5 and 3. It is evident that the first time a child undertakes to do this, he must take one of the numbers, as 5, and join the other to it a single unit at a time, thus 5 and 1 are 6, 6 and 1 are 7, 7 and 1 are 8; 8 is the sum of 5 and 3. A child is obliged to go through the process of adding by units every time he has occasion to put two numbers together, until he can remember the results. This however he soon learns to do if he has frequent occasion to put numbers together. Then he will say directly that 5 and 3 are 8, 7 and 4 are 11, &c.

Before much progress can be made in arithmetic, it is necessary to remember the sums of all the numbers from one tó ten, taken two by two in every possible manner. These are all that are absolutely necessary to be remembered. For when the numbers exceed ten, they are divided into two or more parts and expressed by two or more figures, neither of which can exceed nine. This will be illustrated by the examples which follow.

A man bought a coat for twenty-four dollars, and a hat for eight dollars. How much did they both come to?

Operation.

Coat 24 dolls.
Hat S doils.

Both 32 dolls.

Eight fold

In this example we have 8 dolls. to add to 24 dolls. Here are twenty dolls. and four dolls. and eight dolls. Eight and four are twelve, which are to be join

1 corresp. to

2

3

4

In the same manner if we had twelve figures, the places would have been in a thirteen fold ratio.

The ten fold ratio was probably suggested by counting the fingers. This is the most convenient ratio. If the ratio were less, it would require a larger number of places to express large numbers. If the ratio were larger, it would not require so many places indeed, but it would not be so easy to perform the operations as at present on account of the numbers in each place being so large.

ed to twenty. But twelve is the same as ten and two, therefore we may say twenty and ten are thirty and two are thirty

two.

A man bought a cow for 27 dolls. and a horse for 68 dolls. How much did he give for both?

Operation. Cow 27 dolls. Horse 68 dolls.

In this example, it is proposed to add together 27 and 68.

Now 27 is 2 tens

and 7 units; and 68 is 6 tens and 8 Both 95 dolls. units. 6 tens and 2 tens are 8 tens; and 8 units and 7 units are 15, which is 1 ten and 5 units; this joined to 8 tens makes 9 tens and 5 units, or 95.

A man bought ten barrels of cider for 35 dolls., and 7 barrels of flour for 42 dolls., a hogshead of molasses for 36 dolls., a chest of tea for 87 dolls., and 3 hundred weight of sugar for 24 dolls. What did the whole amount to?

Operation.

Cider

35 dolls.

Flour 42 dolls. Molasses 36 dolls. 87 dolls.

Tea Sugar

24 dolls.

In this example there are five numbers to be added together. We observe that each of these numbers consists of two figures. It will be most convenient to add together either all the units, or all the tens first, and then Amount 224 dolls. the other. Let us begin with the tens. 3 tens and 4 tens are seven tens, and 3 are 10 tens, and 8 are 18 tens, and 2 are 20 tens, or 200. Then adding the units, 5 and 2 are 7, and 6 are 13, and 7 are 20, and 4 are 24, that is, 2 tens and 4 units; this joined to 200 makes 224.

It would be still more convenient to begin with the units, in the following manner; 5 and 2 are 7, and 6 are 13, and 7 are 20, and 4 are 24, that is 2 tens and 4 units; we may now set down the 4 units, and reserving the 2 tens add them with the other tens, thus: 2 tens (which we reserved), and 3 tens are 5 tens, and 4 are 9 tens, and 3 are 12 tens, and 8 are 20 tens, and 2 are 22 tens, which written with the 4 units make 224 as before.

A general has three regiments under his command; in the first there are 478 men; in the second 564; and in the third 593. How many men are there in the whole?

Operation.

First reg. Second reg. Third reg.

478 men

564 men 593 men

1,635 men

In all

units as follows;

In this example, each of the numbers is divided into three parts, hundreds, tens, and units. To add these together it is most convenient to begin with the 8 and 4 are 12, and 3 are 15, that is, 1 ten and 5 units. We write down the 5 units, and reserving the 1 ten, add it with the tens. 1 ten (which we reserved) and 7 tens are 8 tens, and 6 are 14 tens, and 9 are 23 tens, that is, 2 hundreds and 3 tens. We write down the 3 tens, and reserving the 2 hundreds add them with the hundreds. 2 hundreds (which we reserved) and 4 hundreds are 6 hundreds, and 5 are 11 hundreds, and 5 are 16 hundreds, that is, 1 thousand and 6 hundreds. We write down the 6 hundreds in the hundreds' place, and the 1 thousand in the thousands' place.

The reserving of the tens, hundreds, &c. and adding them with the other tens, hundreds, &c. is called carrying. The principle of carrying is more fully illustrated in the following example.

A merchant had all his money in bills of the following description, one-dollar bills, ten-dollar bills, hundred-dollar bills, thousand-dollar bills, &c. each kind he kept in a separate box. Another merchant presented three notes for payment, one 2,673 dollars, another 849 dollars, and another 756 dollars. How much was the amount of all the notes; and how many bills of each sort did he pay, supposing he paid it with the least possible number of bills?

Operation.

:

7

4

9

5 6

The first note would require 2 of the thousand-dollar bills; 6 of of the hundred-dollar bills; 7 tendollar bills; and 3 one-dollar bills. The second note would require 8 of the hundred-dollar bills; 4 ten-dollar bills; and 9 onedollar bills. The third note would require 7 of the hundreddollar bills; 5 ten-dollar bills; and 6 one-dollar bills. Count

4 2 7 8

ing the one-dollar bills, we find 18 of them. This may be paid with 1 ten-dollar bill and 8 one-dollar bills; putting this 1 ten-dollar bill with the other ten-dollar bills, we find 17 of them. This may be paid with I hundred-dollar bill, and 7 ten-dollar bills; putting this one-hundred dollar bill with the other hundred-dollar bills, we find 22 of thern ; this may be paid with 2 of the thousand-dollar bills, and 2 of the hundred-dollar bills; putting the 2 thousand-dollar bills with the other thousand-dollar bills, we find 4 of them. Hence the three notes may be paid with 4 of the thousand-dollar bills, 2 of the hundred-dollar bills, 7 ten-dollar bills, and 8 onedollar bills, and the amount of the whole is 4,278 dollars.

Besides the figures, there are other signs used in arithmetic, which stand for words or sentences that frequently occur. These signs will be explained when there is occasion to use them.

A cross one mark being perpendicular, the other horizontal, is used to express, that one number is to be added to another. Two parallel horizontal lines = are used to express equality between two numbers. This sign is generally read is or are equal to. Example 5+ 3 = 8, is read 5 and 3 are 8; or 3 added to 5 is equal to 8; or 5 more 3 is equal to 8; or more frequently 5 plus 3 is equal to 8; plus being the Latin word for more. These four expressions signify precisely the same thing.

Any number consisting of several figures may sometimes be conveniently expressed in parts by the above method. Example, 23582000 + 300 + 50 + 8 = 1000 + 1200 +140+ 18.

A man owns three farms, the first is worth 4,673 dollars; the second, 5,764 dollars; and the third, 9,287 dollars. How many dollars are they all worth?

Perhaps the principle of carrying may be illustrated more plainly by separating the different orders of units from each other.

In this example the sum of the units is 14, the sum of the tens is 21 tens or 210, the sum of the hundreds is 15 hundreds or 1,500, the sum of the thousands is 18 thousands or 18,000; these numbers being put together make 19,724.

If we take this example and perform it by carrying the tens, the same result will be obtained, and it will be perceived that the only difference in the two methods is, that in this, we add the tens in their proper places as we proceed, and in the other, we put it off until we have added each column, and then add them in precisely the same places. Operation.

4,673 Here as before the sum of the units is 14, +5,764 but instead of writing 14 we write only the 4, +9,287 and reserving the 1 ten, we say 1 (ten, which we reserved) and 7 are 8, and 6 are 14, and 19,724 8 are 22 (tens) or 2 hundreds and 2 tens; setting down the 2 tens and reserving the hundreds, we say, 2 (hundreds, which we reserved) and 6 are 8, and 7 are 15, and 2 are 17 (hundreds) or 1 thousand and 7 hundreds; writing down the 7 hundreds, and reserving the 1 thousand, we say, 1 (thousand, which we reserved) and 4 are 5, and 5 are 10, and 9 are 19 (thousands) or 1 ten-thousand and 9 thousands; we write the 9 in its proper place, and since there is nothing more to add to the 1 (ten thousand) we write that down also, in its proper place. The answer is 19,724 dollars.

*It will be well for the learner to separate, in this way, several of the examples in Addition, because this method is frequently used for illustration in other parts of the book.