but as this cannot be done, we cannot exactly determine the units, except by trial. In the example above, I take 16, the double of the tens for a divisor, and see how many times it is contained in 82, the tens of the remainder, since the product of tens by units must be tens. I find a quotient of 5, which I place on the root, and also in the units place in the divisor. Multiplying the divisor, so increased, the product will be the double product of the tens by the units, the square of the units; for, 5×165 is the same as 5 × 160+5×5, or 5 times 16 tens, 5 times 5 units. If 5 × 165 had been greater than the dividend, it would have been necessary to have put a figure less than 5 in the root.

If the Os had been retained, instead of giving each figure its value depending on its place in the decimal notation, the operation would have been thus expressed :-

7225 || 80+5, the root = 85.

6400, square of the tens.

160+5825

800, double product of the tens and units. 25, square of the units.

825

00

When the number contains more places of figures, the same principles will apply, for (100)21.00.00 and (1000) * =1.00.00.00; and hence the square of the hundreds can have no significant figures in the first two periods, nor the square of the thousands in the first three periods.

For another example, let the square root of 399424 be extracted.

39.94.24 || 63.2

36

123 394
369

1262 | 25.24

25.24

In this example, the root will contain three places of figures, since the number itself has more than four places of figures,

and therefore will contain the square of hundreds. But in order to extract the root, we may consider it as composed of tens and units only; as for example, 342 may either be considered as 3 hundreds, 4 tens, and 2 units, or 34 tens, and 2 units. Having, therefore, separated the two right hand figures, as forming no part of the square of the tens, I proceed to extract the root of 39.94. For this purpose, 3994 may be considered as so many units, and the root extracted as in the preceding example. I find the root of this number to be 63, with a remainder of 25. Bringing down to this, the right hand period, I find, for the whole remainder, 2524. This must contain the double product of the units by the tens, the square of the units: 63 is the tens of the root; and the double of this, 126, will be the divisor to be used in finding the units. number contained seven or eight places of figures, there would then be four periods, and the root of the first three being extracted, it might be considered as so many tens, and the remainder of the root units, as in the preceding examples.

For the rule for extracting the square root of numbers, see Arithmetic, art. (56), page 67.

EXTRACTION OF THE CUBE ROOT OF NUMBERS.

(51.) Let a represent the tens of any number, and b the units; then, (a+b) will be that number, and

(a + b)3 = a3 +3a2b+3ab2+b3.

Hence it will appear, that the cube root of any number.is composed of the cube of its tens, + three times the square of the tens multiplied by the units, three times the tens multiplied by the square of the units, the cube of the units. If we take any number, 24 for example, we can put it under

the form 20+ 4, and raising it to the third power, we shall have

(20 + 4)3 = (20)3 +3(20)2 ×4+3(20 × (4)2 + 43)

8000+ 4800 +960 + 64 = 13824.

13.824 || 24

8

1200 | 5824

240

16

1456 | 5824

Now, in order to extract the cube root of 13.824, it will be necessary to discover what part of this number contains the cube of the tens, and since (10)3 1000, it is evident that the three right hand figures make no part of the cube of the tens. We must, therefore, find the cube of the tens in 13. Now the greatest cube in 13 is 8, which is the cube of 2. Putting 2 in the root, and subtracting its cube, we have 5824 remaining for the remaining terms; namely, three times the square of the tens multiplied by the units, three times the tens multiplied by the square of the units, and the cube of the units; but since the first of these terms is much the largest, we shall be able to obtain the units nearly, by dividing 5824 by 1200, which is three times the square of the tens. Having thus found the unit figure, which is 4, we can verify the result by forming with it and the tens, the three products represented by 3a2b+3ab2+b3; but this is the same as (3a2+3ab+b2)b. If, therefore, we add to 1200, the divisor = (3a2), three times 20 × 4, or 240 (3ab), and the square of 4, or 16 = (b2), and multiply the sum by 4= (b), the product will be 5824. If the product had been greater than the dividend, it would have been necessary to put a less figure in the units place. The ciphers in 1200 and 240 might have been omitted, but in that case, the figures must be written in the places corresponding to their values; thus,

12

24

16

1456

The reason for these values is obvious; for the square of

the tens can have no figure less than hundreds, and the product of the tens by the units, none less than tens.

If the number contains more places of figures, the same reasoning will apply as in the square root. The number must be divided into periods of three figures each; for we have (10) 1000, (100)3 =1.000.000, (1000) 1.000.000.000, from which it will appear, that the cube of tens can have no significant figure in the first period, the cube of hundreds none in the first two, and the cube of thousands none in the first three. The number of periods will, therefore, denote the number of figures in the root.

Let it be required to extract the cube root of 199176704.

199.176.704(584

125

Div. (5) × 3 = 25 × 375.00 | 74.176 1st dividend.

Complete divisor.

2d div. (58)3×3=3364×3=10092.00 | 4064704 2d dividend.

58×4×3=
(4)2=

2d complete divisor.

696.0

16

1016176 4064704

In this example, I first seek the root of the two left hand periods, which I find to be 58. I then call this 58 tens, and proceed to find the last figure of the root by taking three times the square of 58 tens, for a divisor, and bringing down the last period to the remainder, for a dividend. The divisor is completed in the same manner as in the last example. If the given number contained four periods, the same method would be followed, always considering the figures of the root, already found, so many tens, and the figure sought, so many units.

For the rule for extracting the cube root of numbers, see Arith. art. (57), p. 70.

3

EXAMPLES.

1. ✓7645373 = 197.

3

2. V52734375 = 375.

3

3. V334255384 = 694.

GENERAL METHOD OF EXTRACTING THE ROOTS OF NUMBERS OF ANY POWERS WHATEVER.

(52.) The roots of any powers whatever may be extracted by a method analogous to that given (48) for extracting the roots of algebraic quantities.

The first two terms of the developement of any power of a binomial, as (a+b)" is always a"+na"-1b (43). Hence it will appear, that if we make a = the tens of any number, and b= the units, we can approximate to the units by raising the tens to a power one less, or a"-1, and multiplying this power by n, the index of the root to be extracted, for a divisor. To find in what part of the number we are to look for the power of the tens, we are to point off the number in periods of n figures each; for, in raising 10 to any power, we add a cipher for every multiplication; thus, (10) 100, (10)3 1000, (10) 10000; and since in raising 10 to the nth power, it is to be multiplied by itself n-1 times, we shall have for the nth power of 10, 1 and n ciphers. Hence, the nth power of the tens of any number, can have no significant figure in the first n figures, on the right.

Let it now be required to extract the 4th root of 331776.

In this example, n= 4; I therefore point off four figures from the right, as containing no part of the fourth power of the tens. The greatest fourth power in the left hand period is 16, the fourth power of 2. I therefore conclude that 2 is the tens of the root: to find the units, I raise 2 to the third power, and multiply it by 4, the index of the root, for a divi

sor.

It then corresponds to na" of the developement of