APPENDIX.

Testing the Accuracy of Addition, Subtraction, Multiplication, and Division.

466. There is no better way of making sure of the correctness of an addition than adding "both ways"; and in subtraction the best test of accuracy is that the sum of the subtrahend and remainder is equal to the minuend.

In multiplication, the most thorough proof of accuracy is found if the product of the multiplier by the multiplicand is equal to the first product. In division, if the sum of the remainder added to the product of the quotient by the divisor is equal to the dividend, the work may be relied upon as correct. But these methods of proof require as much time as the original operation, hence the common use of the method called

Casting out Nines.

To cast out the nines of a number we may add all the terms of the number, and divide the sum by 9; the remainder will be the result sought. Thus, the sum of the terms of 4787763 is 42. Dividing this by 9, we obtain 6 as the excess of nines. But, since we wish to know only what the remainder is, we may drop the nines from the results as we proceed.

Thus, in the operation of casting out nines from 4787763, we may think the process indicated in light-faced italics, and speak the numbers printed in full-faced type, as follows:

9:

4 + 7 = 11, 11 − 9 = 2 ; 2 + 8 = 10, 10 − 9

= 6; 6+ 6 = 12, 12 - 9

written as it is pronounced.

= 1; 1 + 7 + 7 = 15, 15

3; 3 + 3 = 6. 6 being the excess of nines, is

In this process we skip the 9's, for there is no use of adding a nine and at once dropping it.

647)2514157(3885

8

Explanation.-Casting out the nines from the divisor and quotient separately, we find the remainder written over the last figure of each. We then multiply the one remainder by the other, cast out the nines of the product, and carry the excess to the remainder found by the division 562, and think 3 + 5+ 6 = 14, 1495, and 5 + 2 = 7. We finally cast out the nines from the dividend, and since we obtain 7 from this also we judge the work to be correct.

467. The principle on which this method is based is, that The remainder arising from dividiny any number by 9 is always the same as the remainder that arises from dividing the sum of all its terms by 9.

That this must be so is evident from the fact that, on being divided by 9, there is a remainder of 1 for every ten, hundred, or thousand that go to make up a number, thus:

Dividing 10, 100, or 1000 by 9, the remainder is always 1. Dividing 20, 200, or 2000 by 9, the remainder is always 2, etc. Hence, if we divide separately the parts of a number represented by its digits by 9, the remainders will be expressed by those digits: e. g., if we divide the 2000, 400, 70, and 8, in 2478, separately by 9, the remainders will be 2, 4, 7, and 8, the excess of 9's in the sum of which is evidently the same as in the sum of the digits or in the number itself.

468. Another excellent test of the correctness of an operation in division is that the remainder after division added to all the subtrahends produces a sum equal to the dividend.

Greatest Common Divisor. (See Art. 141.) Example. Let it be required to find the greatest common divisor of 91 and 224.

The process as given in Art. 141, page 142, is as follows: We divide the greater number by the less, and the divisor by the remainder, and so on till we find that will exactly divide the preceding divisor or remainder, as we choose to regard it. By trial, we find that 7 is a common divisor of 91 and 224. But, by reasoning, we might conclude that it must always be the case that the last remainder in such a succession of divisors will be a common divisor of the given numbers. The reasoning would be based on two principles:

91)224(2

182
42)91(2
84

7)42(6

42

1. That an exact divisor of any number must be an exact divisor of any multiple (number of times) that number.

If there is an exact whole number of times 7 apples in one heap, there would be an exact whole number of times 7 apples of the same size in any number of equal heaps.

2. That an exact divisor of two numbers will be an exact divisor of their sum or difference.

If there are 5 times 12 buttons on one string, and 2 times 12 buttons on another, there will be an exact whole number of times 12 buttons on one string more than the other, and an exact whole number of times 12 buttons on both.

With these principles in view, to show that the last divisor is a common divisor, we would reason thus:

Seven being a divisor of 42, it is, according to principle 1, a divisor of 84; and being a divisor of 84, it is (principle 2) a divisor of 91 (84 +7); and being a divisor of 91, it is a divisor (principle 1) of 182 (2 × 91); and being a divisor of 182 and 42, it is (principle 2) a divisor of their sum, 224. Hence, 7 must be an exact divisor of 91 and 224.

And to show that the last divisor is the greatest common divisor, we would reason as follows:

According to principle 1, the common divisor of 91 and 224 must be an exact divisor of 182 (2 × 91); and hence, being a common divisor of 182 and 224, it must be, according to principle 2, an exact divisor of their difference, 42; and, reasoning in a similar manner, being a divisor of 42, it must be a divisor of 9184, or 7. Hence, the greatest common divisor can not be greater than 7.

Thus, having found that 7 is a common divisor, and that the common divisor can not be greater than 7. we conclude that 7 is the greatest common divisor.

Repetends, or Circulating Decimals.

1. Decimals, equivalent to the common fractions, given in exercises 10 and 11, page 149, were easily found, but in 12, the division of the numerators by the denominators was interminable. The decimals produced might therefore be called interminate (not terminating).

2. But, if the division were carried far enough (never to a number of places in the quotient greater than the number represented by the divisor), a remainder would be obtained which had occurred before, and hence a figure or set of figures would be repeated in the same order in never-ending succession. Such a figure, or set of figures, was called a circulating decimal or repetend.

3. Let the pupil now reduce 1, or any number of 9ths less than 9, to a decimal; also, or any number of 99ths less than 99, to a decimal; also 1999, or any number of 999ths less than 999, to a decimal; also any number of 9999ths less than 9999, to a decimal, and let him note the several results. He will in every case obtain a repetend consisting of the same digits as the numerator.

4. From these experiments it may be safely concluded that any repetend is equivalent to a common fraction having the repetend for its numerator, and a number of 9's in the denominator equal to the number of places in the repetend. Hence the method of reducing repetends to common fractions becomes evident. Example.-1. Reduce 18 to a common fraction.

Progression.

469. An Arithmetical Progression is a series of numbers, increasing or decreasing by a constant difference.

1, 3, 5, 7, 9, 11, is an increasing, or ascending series.

12, 10, 8, 6, 4, 2, is a decreasing, or descending series. The constant difference in each is 2.

To find any term of an Arithmetical Series.

1. If the first term of an arithmetical series is 2, and the constant difference is 3, what is the fifth term?

Analysis. Since the constant difference is 3, the terms of the series are

Here we see that the fifth term is equal to the first term + the constant differ ence multiplied by a number one less than the number of the term required. 2 + (4 × 3) = 14, the fifth term.

2. If the first term of a descending arithmetical series is 30, and the constant difference is 5, what is the fourth term? Analysis. Since the constant difference is 5, the terms of the series are

Here we have the fourth term equal to the first term less the constant differ. ence multiplied by a number one less than the number of the term required. 30 (3 × 5) = 15, the fourth term.

Hence, having the first term and the constant difference, to find any required term we have the

470. Rule. To the first term add the product of the constant difference by a number one less than the term required, if the series is ascending; or, if the series is descending, subtract the product from the first term.

3. If the first term of an ascending series is 6, the constant difference 3, and the number of terms 300, what is the last term? The seventh term? The tenth term ?

4. The first term of a descending series is 110, constant dif ference 6. What is the seventh term ?