260. The powers of the nine digits, from the first to the sixth inclusive, are exhibited in the following Sursolids, or 5th powers, |1|32|243|1024| 3125| 7776| 16807| 3276859049 Square cubes, or 6th p. 164|729|4096|15625|46656|117649|262144|531441 343 512 729 2401 4096 2. Evolution. ANALYSIS. 261. The method of ascertaining, or extracting the roots of numbers, or powers, is called Evolution. The root of a number, or power, is a number, which, multiplied by itself continually, a certain number of times, will produce that power, and is named from the denomination of the power, as the square root, cube root, or 2d root, 3d root, &c. Thus 27 is the cube or 3d power of 3, and hence 3 is called the cube, or 3d, root of 27. 3 262. The square root of a quantity may be denoted by this character called the radical sign, placed before it, and the other roots by the same sign, with the index of the root placed over it, or by fractional indices placed on the right hand. Thus, 9, or 91, denotes the square root of 9, 27, or 273, denotes the cube root of 27, and 16, or 164, denotes the 4th root of 16. The latter method of denoting roots is preferable, inasmuch as by it we are able to denote roots and powers at the same time. Thus, 83 signifies that 8 is raised to the second power, and the cube root of that power extracted, or that the cube root of 8 is extracted, and this root raised to the second power; that is, the numerator of the index denotes the power, and the denominator the root of the number over which it stands. 263. Although every number must have a root, the roots of but very few numbers can be fully expressed by figures. We can, however, by the help of decimals approximate the roots of all sufficiently near for all practical purposes. Such roots as cannot be fully expressed by figures are denominated surds, or irrational numbers. 264. The least possible root, which is a whole number, is 1. The square of 1 is (1×) 1, which has one figure less than the number employed as factors; the cube of 1 is (1X1X1=) 1, two figures less than the number employed as factors, and so on. The least root consisting of two figures is 10, whose square is (10x10) 100, which has one figure less than the number of figures in the factors, and whose cube is (10×10× 10) 1000, two figures less than the number in the factors; and the same may be shown of the least roots consisting of 3, 4, &c. figures. Again, the greatest root consisting of only one figure, is 9, whose square is (9X×9) 81, which has just the number of figures in the factors, and whose cube is (9×9 X9) 729, just equal to the number of figures in the factors; and the greatest root consisting of two figures, is 99, whose square is (99×99=) 9801, &c., and the same may be shown of the greatest roots consisting of 3, 4, &c. figures. Hence it appears that the number of figures in the continued product of any number of factors cannot exceed the number of figures in those factors; nor fall short of the number of figures in the factors by the number of factors, wanting one. From this, it is clear that a square number, or the second power, can have but twice as many figures as its root, and only one less than twice as many; and that the third power can have only three times as many figures as its root, and only two less than three times as many, and so où for the higher powers. Therefore, 265. To discover the number of figures of which any root will consist. RULE.-Beginning at the right hand, distinguish the given number into portions, or periods, by dots, each portion consisting of as many figures as are denoted by the index of the root; by the number of dots will be shown the number of figures of which the root will consist. EXAMPLES. 1. How many figures in the square, cube, and biquadrate root of 348753421 ? 848753421 square root 5. 848753421 cube root 3. 848752421 biquadrate 3. 2. How many figures in the square and cube root of 68101 2.1416? 681012. 1416 square 5. 681012141600 cube 4. In distinguishing decimals, begin at the separatrix and proceed towards the right hand, and if the last period is incomplete, complete it by annexing the requisite number of ciphers. EXTRACTION OF THE SQUARE ROOT. ANALYSIS. 266. To extract the square root of a given number is to find a number, which, multiplied by itself, will produce the given number, or it is to find the length of the side of a square of which the given number expressen the area. 10 1. If 529 feet of boards be laid down in a square form, what will be the length of the sides of the square? or, in other words, what is the square root of 529 ? From what was shown (264), we know the root must consist of two figures, in as much as 529 consists of two periods. Now to understand the method of ascertaining these two figures, it may be well to consider how the square of a root consisting of two figures is formed. For this pur 23 23 9 square of units. 60 twice the product of 60 the tens by units. 400 square of the tens. 529 square of 23. 5 29 [20 4 00 1 29 pose we will take the number 23, and square it. By this operation, it appears that the square of a number consisting of tens and units is made up of the square of the units, plus twice the product of the tens, by the units, plus the square of the tens. See this exhibited in figure F. As 10x10=100, the square of the tens can never make a part of the two_right__hand figures of the whole square. Hence the square of the tens is always contained in the second period, or in the 5 of the present example. The greatest square in 5 is 4, and its root 2; hence, we conclude, that the tens in the root are 2-20, and 20×20= 400. But as the square of the tens can never contain significant figures below hundreds, we need only write the square of the figure denoting tens under the second period. From what precedes it appears that 400 of the 529 feet of boards are now disposed of 20 ft E 20 20 400 ft. in a square form, E measuring 20 feet on each side, and that 129 feet are to be added to this square in such manner as not to alter its form; and in order to do this, the additions must be made upon two sides of the square, E-20+ 20 40 feet. Now if 129, the number of feet to be added, be divided by 40, the length of the additions, or, dropping the cipher and 9, if 12 be divided by 4, the quotient will be the width of the additions; and as 4 in 12 is had 3 times, we conclude the addition will be 3 feet wide, and 40×3 120 feet, the quantity added upon the two sides. But since these additions are' no longer than the sides of the square, E, there must be a deficiency at 20 ft. the corner, as exhibited in F, whose sides are equal to the width of the additions, or 3 feet, and 3×3=9 feet, required to fill out the corner, so as to complete the square. The whole operation may be arranged as on the next page, where it will be seen, that we first find the root of the greatest square in the left hand period, place it in the form of a quotient, subtract the square from the period and to the remainder bring down the next period, which we divide, omitting the right hand figure, by double the root, and place the quotient for the second figure of the root; and the square of this 529 [ 23 43] 129 129 figure being necessary to preserve the form of the square, by filling the corner, we place it at the right of the divisor, in place of the cipher, which is always understood there, and then multiply the whole divisor by the last figure of the root. As we may conceive every root to be made up of tens and units, the above reasoning may be applied to any number whatever, and may be given in the following general 23X23 529 proof. RULE. 267. Distinguish the given numbers into periods; find the root of the greatest square number in the left hand period, and place the root in the manner of a quotient in division, and this will be the highest figure in the root required. Subtract the square of the root already found from the left hand period, and to the remainder bring down the next period for a dividend. Double the root already found for a divisor; seek how many times the divisor is contained in the dividend (excepting the right hand figure), and place the result for the next figure in the root, and also on the right of the divisor. Multiply the divisor by the figure in the root last found; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Double the root now found for a divisor, and proceed, as before, to find the next figure of the root, and so on, till all the periods are brought down. the side of a square, which | ameter of a circle 4 times as shall contain an aere, or 160 rods? Ans. 12.649 rods. large? Ans. 24. Circles are to one another as the squares of their diameter; therefore square the given diameters, multiply or divide it by the given proportion, as the required diameters is to be greater or less than the given diameter, and the square root of the product, or quotient, will be the diameter required? 14. The diameter of a circle is 121 feet; what is the diameter of a circle une half as | large? Ans. 85.5+ feet. 268. Having two sides of a right angled triangle given to find the other side. RULE.-Square the two given sides, and if they are the two sides which include the right angle, that is, the two shortest sides, add them together, and the square root of the sum will be the length of the longest side; if not, the two shortest; subtract the square of the less from that of the greater, and the square root of the remainder will be the length of the side required. (See demonstration, Part I. Art. 68.) |