more exactly. Suppose that it is required to find the square root of 66 to 1 within the Tooth part of unity. It is found (Art. 266) that, 268. If the denominator of a fraction whose square root is required involves one perfect square factor it is not necessary to multiply the terms of the fraction by the whole denominator, but only by the factor which is not a perfect square. Let it be required to find the square root of . The terms of this fraction may be multiplied by 48, by which means the denominator is rendered a perfect square. But since 48=16x3=42x3, if the terms of the 23 x 3 69 23 42x3 fraction are multiplied by 3, the result is 4x32-122, and the denominator is rendered a perfect square. 1 1 The square root of 69 to within the 1000th part of unity is 8.306; thereis the square root required to within the 12000th part fore 8.306 12 of unity. or 8306 12000 Therefore, to find the square root of a vulgar fraction whose denominator involves a perfect square factor, Rule. Multiply the terms of the fraction by the factor which is not a perfect square, and extract the square roots of the numerator and denominator as in Art. 262. 269. Extraction of the square root of a decimal fraction. The shortest process for the extraction of the square root of a decimal fraction is deducible from the last article. Suppose it required to extract the square root of the decimal fraction Now the denominator of this expression is not a square number, but it is equal to 100×10 or to 10 x 10. To render the denominator a square number it is sufficient to multiply the terms of the fraction 3425 1000 by 10. 100=100 ; √3425 185 =1.85; a result true to within the roth part of Now 34250=185 plus a fraction less than 1, and ..√3·425= unity. If a more exact degree of approximation is required, it is necessary to annex to 34250 one period of two zeros for each additional decimal figure required in the result. 270. To extract the square root of a decimal fraction. Rule. First, render the number of decimal figures even and equal to twice the number of decimal figures required in the root, (which is accomplished by annexing the proper number of zeros to the given decimal fraction). Second, make abstraction of the decimal point in the new number, and extract its square root to the nearest unit. Third, point off from the right of the root, the required number of decimal figures. This rule might be inferred from the process for the multiplication of decimal fractions; for the square of a decimal fraction, or the product of the fraction by itself, must contain twice as many decimal figures as are contained in the square root. When it is required to express decimally the square root of a vulgar fraction; Rule. Reduce the vulgar fraction to a decimal expressed by twice as many figures as are required in the root; extract the square root of this decimal; the result is the square root of the vulgar fraction expressed decimally. 270. Exercises in extraction of the square root of fractional expressions, and of numbers which are not perfect squares. 1st. Square root of 3?........... .......Ans. 1·732+. ....Ans. 2·449489+. ...Ans. 3·464101+. .......... Ans. §. .....Ans. 7. .......Ans. 34. ...Ans. 4·16833. ...Ans. 91. ..... Ans, 5·03. .......Ans. . ..........Ans. 2·423+. ............................... Ans. 0·035. .......Ans. 2·529822. ...........Ans. 3. ...Ans. 0·948683+. ....Ans. 0.3. .....Ans. 0.8. ...Ans. 2·52962. .Ans. 2529822. 20th. Square root of .000003418801 ?....................................... Ans. ·001849. OF THE CUBE AND CUBE ROOT OF NUMBERS. 271. If one factor is made to enter three times into a product, the product is termed the third power or cube of that factor, and the factor is called the cube root of the product. The cube of a number is indicated thus 53. 53=125 expresses that the cube of 5, or 5×5 × 5, is equal to 125. The cube root of a number is indicated by the symbol. Thus, 125=5, expresses that the cube root of 125 is 5. The formation of the cube of a whole number, or a fractional expression, depends on the rules for the multiplication of numbers whole or fractional. The cubes of the numbers from 1 to 10 inclusive are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000; and 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, are the cube roots of the corresponding numbers in the upper line. 272. From a comparison of these cube numbers with their cube roots, it appears that of numbers expressed by one, two, and three figures, nine only are perfect cubes or cube numbers; each of the 990 others having for cube root a whole number plus a fraction, which cannot be expressed exactly by means of unity. For let it be supposed that a whole number, n, has for exact cube root a a a a3 α an irreducible fractional expression, ; then, xq x z = ws = n. But a, b, being prime to each other, a3, b3, are also prime to each other a3 a (Art. 113). Whence cannot be equal to a whole number, nor to Vn. The cube roots of whole numbers which are not the exact cubes of other whole numbers cannot, therefore, be obtained exactly, for which reason they are termed incommensurable, irrational, or surd numbers. If a denote the tens of a number, and b the units, the number is expressed by a+b. Now by Art. 76 (a+b) (a+b) or (a+b)2=a2+2ab+b2, a+b -{a3+ a2b+2ab2+b3 it is found that (a+b)(a+b)(a+b) or (a+b)3=a3+3a2b+3ab2+b3. Whence the cube of a number consisting of tens and units is composed of the cube of the tens of the number, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units, Whence, if a and a+1 represent two consecutive whole numbers, the difference between the cube of a+1 and a is equal to three times the square of the less number, plus three times the less number, plus one, The difference between 913 and 903 is 3 × 902+3x90+1=24300+270+1=24571. From this example it may be inferred that there is a great difference between two consecutive cube numbers when the roots of these cubes are expressed by large numbers. 273. Process for the extraction of the cube root of a whole number. If the number contains not more than three figures its cube root is obtained by reference to Art. 271. Thus, the cube root of 27 is 3; the cube root of 235 is 6 plus a fraction; for 235 falls between 216=63 and 343=73; and the cube root of 950 is 9 plus a fraction, for 950 falls between 729=93 and 1000 103. Let the number whose cube root is required contain more than three figures, as 103823. 103823 falls between 1000=103 and 1000000=1003; its root therefore contains two figures, tens and units. Since 103=1000, the three last figures of the number can contain no part of the significant figures of the cube of the tens; it is therefore in the part 103 that these significant figures are found. Employing a comma to separate the last three figures, and referring to Article 271 for the root of the cube number equal to or next less than 103, it is found that 103 falls between 6443 and 125=53. Since also 103823 falls between 64000=403 and 125000=503, it is evident that the required cube root is composed of four tens and of a certain number of units less than ten. Subtracting 64, the cube of the tens of the root, from 103, and to the remainder 39 annexing the second period of the given number, the result, which is 39823, contains 3× 40o × b+3 × 40 ×b2+b3. Next, to find the numerical value of b in this example. The square of the tens giving hundreds, the significant figures of the triple square of the tens by the units must be found in 398, the part on the left of the last figures 23. Now 3×42=3x16=48, and 398+48=8+ a remainder; or 398234800=8+ a remainder. The quotient 8 is the figure expressing the units (b) of the root, or a figure expressing a larger number. 39823, besides three times the square of the tens of the root into the units, contains also the product of three times the tens by the square of the units together with the cube of the units. These partial products may be formed and subtracted successively from 39823, and the remainders; or their sum may be subtracted at once. It is, however, a less complicated process to form the cube of 48, and subtract it from the number 103823. 483=110592, which is greater than 103823, ... 48 is greater than the cube root of 103823. Trying the lower number 47, 473=103823. ..47=103823. 2d. Example. Extract the cube root of 47954. The number 47954 is comprehended between 1000 and 1000000, therefore its cube root falls between 10 and 100; that is, it contains tens and units. The cube of the tens is contained in the 47 thousands; and since 47 falls between 27=33 and 64=43; and 47954 also falls between 27000 and 64000, or between 30 and 403, it follows that 333 or 3 expresses the tens of the root. Subtracting 27, the cube of 3, from 47, the remainder is 20. To this remainder annexing 9, the first figure of the second period, the number 209 hundreds is composed of the triple square of the tens of the root by the units plus the hundreds arising from the other partial products. The triple square of the tens=3x32=27; dividing 209 by 27, the quotient is 7; 7 is the figure which expresses the units of the root, or it is a greater figure. Assuming that 7 expresses the units of the root, 37 is the root, and 373=50653. But 50653 is greater than 47954; therefore 37 is greater than 47954. Assuming the next lower number, 36, and forming its cube, the result is 46656. Subtracting 46656 from 47954, the remainder is 1298. Whence 47954 is not a perfect cube; but its cube root to the nearest unit is 36. In effect the difference between the given number and 363 is 1298, and the difference between 373 and 363 is 3× 362+3×36+1=3888+108+1=3997, a difference much greater than that between 363 and the given number. Form of calculation: 3d Example. Extract the cube root of 43725658. How many soever the number of figures by which the required root is expressed, it may be regarded as composed of units and tens only, the tens being expressed by more than one figure. Now, the cube of the tens gives at least a thousand; therefore the significant figures of the cube of the tens of the root are of necessity found in that part of the number which is on the left of the last three figures, 658. Next, if the root of the greatest cube contained in 43725, considered as expressing simple units, is extracted, the result is the whole number of tens contained in the required root. For let a be the root of the greatest cube contained in 43725; it follows that the root required has at least a number, a, of tens, since a3 × 1000 can in this case be subtracted from 43725000, and à fortiori from 43725658. Besides, the root cannot contain a+1 tens, for (a+1)3 being greater than 43725, (a+1)3× 1000 must exceed 43725000 by at least 1000; consequently (a+1)3×1000 is greater than 43725658. Therefore, in conclusion, the cube root required is composed of a tens plus a certain number of units less than 10. The question is consequently reduced to the extraction of the cube root of 43725. This number consisting of more than three figures, its cube root contains more than one figure; that is to say, it contains tens and units. To obtain the tens it is necessary to point off the three figures 725, and to extract the root of the greatest cube number contained in 43. (Note. It is evident that if the period on the left, in this instance 43, contained more than 3 figures, it would be necessary to point off three figures again from its right.) The greatest cube number contained in 43 is 27, the cube root of which is 3. This figure, then, expresses the tens of the cube root of 43725, or the hundreds of the whole root. Subtracting the cube of 3 from 43, the remainder is 43-27=16. Annexing the first figure of the 2d period to this remainder, the result is 167. Dividing 167 by three times the square of 3, or 27, the quotient is 6. 6, therefore, is the figure which expresses the units of the cube root of 43725, or a greater figure than that of the root. Making trial (by forming 363) it is found that the figure 6 is too great. Assuming the next lower figure, 5, and forming the cube of 35, which is 42875, it is found that 35 expresses the total number of tens of the required root, and that the second remainder is 43725-42875=850. To obtain the units of the root, the first figure of the 3d period of the given number is annexed to the second remainder, and the result, 8506, is divided by three times the square of the tens of the root, that is, by 3 x 352=3675: 8506+3675=2 |