that to which it was added in the first process. Thus, in the example above, if we begin at the bottom, 8 is added to 5, making 13; 4 to 13, making 17, &c. If in the proof, the first row be cut off, 4, which was before added to 13, is added to 8, making 12; 6, which was before added to 17, is added to 12, making 18, and so on; each figure, in the proof, being added to a different number, from that to which it was added in the first process. Thus, the chance of error, on account of the figures recurring in the same order, is very small. For, if we made a mistake in adding 4 to 13, it is very improbable that we shall make the same mistake in adding it to 8; and so, in the other cases.

Another method is, to make a second Addition, arranging the numbers differently; placing, for example, the number which was uppermost, in the middle, and that which was in the middle, at the top or bottom; so as to avoid adding them twice in the same order of arrangement. This is a very good method, but often long and tedious.

Another method is, to add each column, and set down its whole amount sepa rately, and then to add the amounts as they stand, thus,

But, after all, it is better to avoid errors by care and attention, than to detect them after they are made. If, however. any mode of proof be employed, there is one, which, for ingenuity, certainty, and expedition, is superior to any of those which have been mentioned. But, as it is somewhat complex in principle, it was thought proper to defer its insertion to this place. It is called, the proof by casting out of nines, or, more simply, the proof by nines. Before proceeding to it, however, we will attend to a few simple properties of numbers.

ART. I. 1. Two men start together, and travel the same way, one at the rate of 10, the other of 9 miles, an hour. At the end of one hour, the former will be 10, the latter, 9 miles from the starting point. Of course, they will be 1 mile apart. If they were, then, both to proceed at the rate of 9 miles an hour, they would always continue 1 mile apart. But, during the second hour, the forward man goes 10 miles, instead of 9, and, consequently, makes the distance between them a mile greater. So that in 2 hours, they are 2 miles apart. For the same reason, in 3 hours, they are 3 miles apart; in 4 hours they are 4 miles apart; and so on. Here we see, that each hour makes one mile more of distance between them. Therefore, the number of miles, between them, will always be equal to the number of hours.

2. I wish to measure a stick of timber. I use a pole 10 feet long, and make a mark at the end of every ten feet I measure. Another man follows me, using a pole 9 feet long. He, likewise, makes a mark at the end of every nine feet. As his pole is 1 foot shorter than mine, his first mark will be one foot behind mine. If he

should then take a 10 foot pole, every one of his marks would be one foot behind my marks. But, as he keeps on measuring with his 9 foot pole, his second mark will be 2 feet behind mine; his third, 3 feet; his fourth, 4 feet; and so on. Thus, at every meas. urement, his mark falls a foot farther behind mine. Therefore, our marks will be always just as many feet apart, as the number of

measurements.

3. From this illustration it will be very clear, that as 9 is one less than 10, so two 9s are 2 less than two 10s; three 9s, 3 less than three 10s; and so on. For one 10 consists of a 9 and a unit. And two 10s are one 10 and one 10. Then, if each of these 10s contain a 9 and a unit, both of them must contain two nines and two units. So, likewise, three 10s contain three 9s and 3 units; four 10s, four 9s and 4 units; and any number of 10s contains just as many Is as there are 108, and just as many units besides.

4. Therefore, if from one 10 I take 1 unit, one 9 remains; if from two 10s I take away 2 units, two 9s remain; if from three 10s, 3 units, three 9s remain; and, if from any number of 10s, I take the same number of units, the same number of 9s remains. That part of this idea which we wish to have distinctly remembered, may be concisely expressed as follows:

If from ANY NUMBER OF TENS, the SAME number of UNITS be taken away, THE REMAINDER WILL CONSIST OF EVEN 9s.

ART. II. It appears (1.4) that if from ten 10s, that is, from 100, I should take away 10 units, ten 9s would remain. Now these 10 units taken away, make one 10, which is equal to one 9 and 1 unit. Therefore, in the ten 10s or 100, there are ten 9s, and one 9 and 1 unit; that is, eleven 9s and 1 unit. Therefore, if from 100, I take away one unit, a number of even 9s, that is, eleven 9s, will remain. If, from 200 I should take 2 units, a number of even 9s would, likewise, remain. For 200 is 100 and 100; and if one unit be taken from each hundred, it will leave a number of even 9s in each case. But taking one unit from each, is taking 2 units from the whole. Therefore, if2 units be taken from 200, a number of even 9s remains ; if 3 units be taken from 300, a number of even 9s remains; and, if any number of units be taken from the same number of hundreds, a number of even 9s remains. Hence,

If from ANY NUMBER OF HUNDREDS, the SAME number of UNITS be taken away, THE REMAINDER WILL CONSIST OF EVEN 9s.

ART. III. 1. It appears (II.) that ten hundreds, or 1,000 consist of even 9s and 10 units. But these 10 units make one 10, which is equal to one 9 and 1 unit. Therefore, 1,000 consists of even 9s, and 1 unit. Hence, if from 1,000, I should take away 1 unit, a number of even 9s would remain. Then, as in hundreds, it may be shown, that, if 2 units be taken from 2,000, a number of even 9s will remain; if 3 units from 3,000; 4 units from 4,000; or, if any number of units be taken from the same number of thousands, a number of even 9s will remain. And,

If from ANY NUMBER OF THOUSANDS, the SAME number of UNITS be taken away, the REMAINDER WILL CONSIST OF EVEN 9s.

2. And, by the same method of proof it may be shown, that,

If from ANY NUMBER OF UNITS OF ANY ORDER, the SAME number of SIMPLE UNITS be taken away, THE REMAINDER WILL CONSIST OF EVEN 9s. ART. IV. 1. Let there be given any number, as 324. This consists of three hundreds

And two tens

And four units

Now it has been shown, (III. 3.) that,

300 contains even 9s and

And 20 contains even 9s and

And 4 contains no 9s and

300

The whole number, 324, then, consists of even 9s, and 3 units, and 2 units, and 4 units.

It is plain, then, that if 3 units and 2 units and 4 units, when added together, make 9, or any number of even 9s, the number 324, will consist of even 9s, with no remainder. 3 and 2 and 4 are 9. Therefore, 324 consists of even 9s, without remainder. But 3 and 2 and 4 are the figures which make 324. Therefore,

IF THE SUM OF the figures WHICH COMPOSE A NUMBER, CONSISTS OF EVEN 9s, THE NUMBER ITSELF CONSISTS OF EVEN 9s. Also,

IF THE SUM OF THE FIGURES WHICH COMPOSE A NUMBER, DOES NOT CONSIST OF EVEN 9s, THE NUMBER ITSELF DOES NOT CONSIST OF EVEN 9s, BUT LEAVES THE SAME REMAINDER AS THAT LEFT BY THE SUM OF ITS FIGURES.

Let there now be given an example in addition, as the following. Add

178

7

543 3

821 2

1542 3

After adding, draw a line up and down, on the right. Add the figures 1, 7 and 8, which compose the upper number. Their sum is 16-9+7. Reject the 9, and set down the 7, opposite the upper number, and outside the line. Add the figures of the second number, reject and set down, as before. Do the same with the third number. We know, therefore, that the upper number consists of even 9s and 7; the second, of even 9s and 3; and the third of even 9s and 2. Of course, if 7 and 3 and 2 added together, are even 9s, the sum of the numbers ought to consist of even 9s. But 7+3+2 =12-9+3. Therefore, the sum of the numbers ought to be even 9s, and 3 over. Set down this 3 opposite the amount, 1542.

It is plain, that if the example is correctly performed, this amount ought to contain just as many 9s as the numbers added, with the same excess, or remainder. What this remainder is, we shall find by adding the figures 1, 5, 4 and 2, which compose the amount, and rejecting the 9s from their sum. The sum is 12,-9+3. Reject the 9 and 3 remains. This remainder is exactly what it was shown above that it ought to be. Hence, we conclude, that the amount is correct, and that the addition has been correctly performed.

From what has been said, we derive the following rule for the proof of addition.

I. ADD THE FIGURES COMPOSING EACH OF THE GIVEN NUMBERS.

II. REJECT OR CAST OUT THE 98 FROM EACH SUM, AND SET DOWN

THE EXCESSES OPPOSITE THE SEVERAL NUmbers.

· III. ADD THESE EXCESSES, REJECT THE 9S FROM THEIR SUM, AND

PLACE THE REMAINDER OPPOSITE. THE AMOUNT.

IV. ADD THE FIGURES COMPOSING THE AMOUNT, REJECT THE 9s, AND, IF THE REMAINDER, THUS FOUND, AGREES WITH THE LAST, THE OPERATION MAY BE CONSIDERED AS CORRECTLY PERFORMED.

NOTE. It will be found most convenient and expeditious, to reject the 98 WHILE ADDING, that is, as often as you obtain an amount of 9 or more, reject the 9, and proceed with the remainder, if any, as before.

This rule may, possibly, make a false result appear correct. For, if we make one figure of the amount a unit or two too large, and another, as much too small, the sum of the figures, and of course, its excess of 9s, will remain the same. Or if there be any number of errors made, such, that half their amount shall be on the side of excess, and the other half on the side of deficiency, the excess of 9s will remain unaltered, and, of course, the result will seem correct. But it is very improbable that errors, exactly counterbalancing each other, will be made, in the same operation; and hence, this mode of proof may, perhaps, be better depended upon than any other.

It is, perhaps, needless to remark that errors may sometimes be made in proving, as well as in performing the first operation: "so that no mode of proof is a certain test of the accuracy of a result.

The property of the number 9 on which the above mode of proof depends, is not an essential property of the number; that is, it is not a property which belongs to it from its nature, but it is one which results from its place in the scheme of notation. It would belong to 11, if we were to reckon by twelves; and to 29, if we were to reckon by thirties. In other words, it always belongs to the number, which is one less than the radix of the system. The same property, also, belongs to the number 3; because 3 is contained in 9 an exact number of times. Of course, any number of 9s will be an exact number of 3s. We may, therefore, prove addition by casting out of 3s, in the same manner as by casting out of 9s.

MULTIPLICATION.

MENTAL EXERCISES.

§ X. 1. A shoemaker can make one pair of shoes in a day. How many can he make in 2 days? 2 times 1, or twice 1 are how many

?

How

2. A cabinet maker makes two tables in a day. many can he make in two days? 2 times 2, or twice 2 are how

many y?

3. If one inkstand has 3 pens in it, how many pens have two inkstands? 2 times 3 are how many?

4. If one camel has two humps on his back, how many humps have 4 camels? 4 times 2 are how many?

5. One boy has 3 apples. How many apples have 5 boys? 5 times 3 are how many?

6. If one chair has 4 legs, how many legs have 6 chairs? 6 times 4 are how many?

7. A window has 5 rows of panes, and 7 panes in each row. How many panes of glass in the whole window ? 5 times 7 are how many ?

8. One boy has 6 marbles.

How many marbles, have 8 boys? 8 times 6 are how many?

9. Three boys have 7 cents apiece. How many cents have all together? 3 times 7 are how many?

10. One boy has 5 school books. How many books have 4 boys? 4 times 5 are how many?

11. In one peck are 8 quarts. How many quarts in 3 pecks? 3 times 8 are how many?

12. One tree produced 6 bushels of apples. How many bushels would 5 trees produce? 5 times 6 are how many?

13. Two oranges cost 6 cents apiece. What did both cost? 2 times 6, or twice 6 are how many?

14. At 12 cents a pound, how much will 3 pounds of raisins cost? 3 times 12 are how many?

15. A chess board has 8 rows of squares, and 8 squares in a row, How many squares on the board? 8 times 8 are how many?

16. Five men drank a glass of ale apiece, and paid 3 cents a glass. How much was paid in all? 3 times 5

are how many?

17. If one lemon cost 4 cents, what will 2 cost? What will 3 cost? What will 4? 5? 6?7? 8? 9? 10?

18. At 3 dollars a barrel, what will 2 barrels of cider cost? What will 3? 4? 5? 6? 7? 8? 9? 10? 11? 12? 13? 14? 15? 16?

19. At 5 cents apiece, what will 2 story books cost? 3? 4? 5? 6? 7? 8? 9? 10? 11? 12?

20. How many are 3 times 5? 4 times 5? 6? 7? 8? 9? 21. Repeat the