In order to illustrate this, take the equation x3 + xx−34x+56=0, wherein the signs change twice and the same sign occurs only once. We see thus that the equation has two positive roots and one negative root; and as the roots must be divisors of the last term 56, they must be included in the numbers ± 1, 2, 4, 7, 8, 14, 28, 56. 858. Let us, then, make r= 2, and then 8+468+56=0; whence it would appear that r=2 is a positive root, and therefore that r— - 2 is a divisor of the equation whereby the other two roots may easily be found; for, dividing by x-2, we have Making this quotient xx+3x-28-0, we find the other two roots, which will be z=−} ± √ +28=-±, that is, x=4 or x=-7; and considering the root beforehand, x=2, we perceive that the equation has two positive roots and one negative. We subjoin some examples. (I.) What numbers are those whose difference is 12, and whose product multiplied by their sum makes 14560? Let the less of the two numbers -x. The greater will be x + 12. Their product =xx+2x, multiplied by the sum 2x+12, gives 2x3 + 36xx + 144x = 14560. Divide by 2, and we have r3 + 18xx+72x=7280. The last term 7280 is too great to make trial of all the divisors, and it is divisible by 8, wherefore we shall make x=2y; because the new equation 8y3 + 72yy + 144y=7280, after the substitution, being divided by 8, becomes y3 + 9yy + 18y=910. To solve this, we only need try the divisors 1, 2, 5, 7. 10, 13, &c. of the number 910. Now it is evident that the first three are too small. Begin, therefore, by supposing y=7, and we find it is one of the roots, for the substitution gives 343 +441 +126=910. It follows, then, that x=14, and the other roots are found by dividing y3+9yy + 18y-910 by y-7, as follows: y-7)y3 +9yy+18y-910(yy +16y+130 y3-7yy 16yy+18y-910 130y-910 Supposing this quotient yy + 16y+130=0, we have yy=-16y-130, and thence y = −8 ± -66, which proves that the other two roots are impossible. The numbers sought, therefore, are 14 and 26, the product of which, 364, multiplied by their sum, 40, gives 1456. (II.) What numbers are those whose difference may be 18, and their sum multiplied by the difference of their cubes may produce the number 275184? Let the lesser number = 1. Then the greater will be r+18. The cube of the first will be r3, and the cube of the second =x3 + 54xx + 972x+ 5832. The difference of the cubes = 54rr+972x + 5832=54 (xx + 18x+108) multiplied by the sum 2r+18, or 2(x+9), gives the product 108 (x3 + 27 xx + 270x +972) = 275184. Dividing by 108, we have r3 + 27 xx + 270 +972=2548, or x3 + 27xx + 270c=1576. The divisors of 1576 are 1, 2, 4, 8, &c. Let us try r=4, and we shall find it will satisfy the terms of the equation. It remains then to divide by r-4 to find the other two roots. The quotient will be found to be ar+ 31x+394; making, therefore, rx=-31x-394, we find x=— 1376 961 4 that is, two imaginary roots. The numbers sought, therefore, are 4 and 22. - 31 + (III.) What numbers are those whose difference is 12, and the product of this difference by the sum of their cubes is 102144? Let the lesser number =r; the greater will be r+ 12. The cube of the first is 3; the cube of the second is r3 +36xx + 432x + 1728. The product of the sum of these cubes by the difference 12, is 12(2x3 + 36xx + 432r+ 1728)=102144. Dividing by 12 and 2, we have x3 +18xx+216x+864=4256, or x3 + 18xx+216x= 3392=8 x 8 x 53. Suppose x= 424. =2y, and substituting and dividing by 8, we have y3 + 9yy + 54y=8 x 53= The divisors of 424 are 1, 2, 4, 8, 53, &c. 1 and 2 are too small; but making y=4 we find 64+ 144 +216=424; so that y=4 and x=8, whence we conclude that the two numbers sought are 8 and 20. 859. We shall here close our brief explanation of the principal rules of algebra: they are to be considered more in the light of a preparation for reading and understanding the analytical reasoning and formula that we shall hereafter have to use, than as intended to perfect the architectural student in that whereof they treat. He that desires a more intimate acquaintance with the analytical process will of course apply himself to works expressly on the subject. Nevertheless our work could not have been considered complete without that which we have supplied. 860. It remains for us to give, under this chapter, a few applications of the use of decimals, whose nature has already been explained, and to close it with an explanation of duodecimals and the mode of working; to which we now proceed. DECIMALS. 861. In subsect. 783. et seq. the mode of converting vulgar into decimal fractions has been explained. We shall here more particularly apply them to the general subject of our work. Great facilities arise from their application, though there be many fractions of common occurrence which cannot be expressed in decimals without a great number of figures. The following table will show the mode of expressing in decimals the fractional parts of a foot, and will further illustrate the mode of writing down numbers in decimals: — We may here repeat, that the value of a decimal fraction is not altered by any ciphers on its right hand; thus, 2500 is of the same value as 25, but every cipher added between the number and the decimal point decreases the value of the decimal ten times: thus 25=}; 025; and 0025=0 The mode of finding the value of recurring decimals has already been given (subsect. 793.); we shall, therefore, proceed to the reduction of a decimal to its corresponding value in inferior denominations. For effecting this, the decimal must be multiplied by the number of parts its integer contains of the denomination to which it is to be reduced, and point off to the left as many figures in the product as there are places in the decimal. 862. Thus, to find the inches and parts equivalent to 5417 of a foot. that a foot contains 12 inches, we have ⚫5417 6.5004 inches; and 1 inch consisting of 12 parts Remembering 6.0048 parts; hence 5417 is equal to 6 inches, and 6.0048 parts. Again, to find the value in shillings and pence of 525 of a pound sterling, we have — 525 20 shillings in a pound 10.500 12 6000; that is, 10 shillings and 6 pence. Reciprocally, to find what decimal of a foot are 6 inches 60048, we have, first, Parts in an inch, 12) 60048 Parts in a foot, 12) 6.5004 5417, the decimal required. Again, to find what decimal of a pound sterling are ten shillings and sixpence: here we have Pence in a shilling 12) 6.0 Shillings in a pound 20)10.5 -525, the decimal required. 863. The addition of decimal fractions is performed by placing the different numbers with the points directly under each other, and then the addition is made as in whole numbers, observing to place the point in the sum under its place in the numbers. Example. Add the numbers 3·5675, 21 ·375, and 760-00875 together. 3.5675 21.375 760-00875 784.95125 864. The subtraction of decimal fractions is performed by placing the fractions with the points directly under each other, and subtracting as in whole numbers. Example. From 98.735 take 12.96785. 98.735 85.76715 865. The multiplication of decimal fractions is performed as in integers, taking care to place the decimal point in the product to the left of as many decimals as are contained in both factors. But if there be not as many places in the product as are contained in both factors, ciphers must be placed to the left to make up the deficiency. In this case there are three decimal places in the multiplicand, and one in the multiplier. Four decimals must therefore be cut off from the right. Example. - Multiply 07325 by ·5235. *07325 •5235 36625 21975 14650 36625 038346375 Here, because there are five places of decimals in the multiplicand and four in the multiplier, making, in all, nine places, and only eight places come from the multiplication, we must prefix a cipher to make up the nine places. 866. The division of decimal fractions is performed as in whole numbers, pointing off from the right of the quotient as many figures for decimals as the dividend has more decimal places than the divisor. If the quotient have not so many figures as the decimals in the dividend exceed those in the divisor, ciphers must be prefixed to the left to make up the deficiency before the point be placed. Example. - Divide 7-375 by 5-25. 5.25)7.3750000 (1.40476 5 25 2 125 2100 2500 2100 4000 3675 3250 3150 100 Now the dividend with the ciphers annexed has seven places of decimals, and the divisor only two; we must therefore cut off five places from the right hand for the decimals of the quotient. Example. Divide 5675 by 72.5. 72.5) 5675000 (007827 5075 6000 5800 2000 1450 5500 5075 425 867. The dividend has, with the ciphers that have been annexed, seven places of decimals, and the divisor only one place; hence we cut off from the right six places for the decimal of the quotient. But on examination it is found that there are only four significant figures obtained; two ciphers must, therefore, be prefixed to the quotient. DUODECIMALS. 868. Duodecimals are a series of denominations beginning with feet, wherein every inch in the lo ver denomination makes twelve in that next above it, and they form a series of fractions, whereof the denominations are understood, but not expressed. This method is chiefly in use among measurers of artificers' works, for computing the contents of work. The dimensions are taken in feet, inches, and twelfths of an inch, but not nearer, except in works of the greatest nicety. Feet and inches are marked with their initial letters, but twelfths or seconds by a double accent, thus 2", and thirds by a triple accent, thus 5"". 869. To multiply duodecimals together, write down the two dimensions so to be multiplied in such way that the place of feet may stand under the last place of the multiplicand; begin with the right hand denomination of the multiplier, and multiply it by every denomination of the multiplicand, throwing the twelve out of every product, and carrying as many units as there are twelves to the next. Placing the remainders, if any, under the multiplier, so that the like parts in the product may be under like parts of the multiplicand, proceed with every successive figure of the multiplier towards the left, in the same manner, always placing the first figure of the product under the multiplier. Then the sum of these partial products will be the whole product. In duodecimals there will be as many denominations below feet as in both the factors taken together. Example 1. Multiply 7 ft. 5 in. by 3 ft. 4 in. 870. In the first example there is only one place of duodecimals in each factor; there are therefore two places in the product. In the second example there are two places of duodecimals in the multiplicand and one in the multiplier, which make, together, three; there are therefore three denominations in the product. This method of placing the denominations of the factors gives the correct places of the product at once; since like parts of the product stand under like parts of the multiplicand. It also shows the affinity between duodecimals, decimals, and every series or scale of denominations whereof any number divided by the radix of the scale makes one of the next towards the left hand. The consideration is, moreover, useful in discovering readily the kind of product arising from the multiplication of any two single denominations together. 871. When the number of feet runs very high in the factors, it will be much better to write down the product of each multiplication, without casting out the twelve, and add together those of each denomination beginning on the right, and divide by 12, to carry to the next higher place, then add these, and so on, as often as there are places in the whole product. Example. Multiply 262 ft. 5 in. by 54 ft. 8 in. Thus, under inches, the products being set down and added, they amount to 2369, which, divided by twelve, gives 197 to carry to the place of feet, and 5 remainder. Then adding the feet together with the quantity carried, it gives the whole number of feet; while the operation is extremely simple and free from the troubles of either side operations or useless stress on the memory. 872. The division of the foot into 12 parts renders the application of the rules of practice very valuable in the computation of duodecimals. The practical rule is to set down the two dimensions one under the other, that is, feet under feet and inches under inches, and multiply each term in the multiplicand by the feet in the multiplier, beginning at the lowest; and, if the numbers be large, put down the inches without carrying 1 for every 12 from inches to feet. Then, instead of multiplying by the inches, take such aliquot parts of the multiplicand as the inches are of a foot; after which add the lines together, carrying 1 for every 12 inches. Example 1.-Multiply 7 ft. 5 in. by 3 ft. 4 in. The same examples have been used to show the relative advantage of the two methods. 873. The abridgment of the labours of practical men is always a matter of importance being identical with the saving of time which is lost in calculation, and which with the architect is of the utmost importance, when it is recollected what multifarious duties he has to discharge. Hence we doubt not that the following table of squares, cubes, and roots of numbers, up to 1000, will be most acceptable to him. An inspection of the table will at once instruct him in the method of using it. The first column shows the number, the second the square of such number, the third exhibits its cube. In the fourth column is found the square root of the number, and in the fifth its cube root. Thus, looking to the number 61 in the first column, we find its square to be 3721, its cube 226981, its square root 7.8102497, and its cube root 3.936497. Again, taking the number 784, we find its square to be 613089, its cube 481890304, its square root 28, and its cube root 9-220872. We presume that we need not further enlarge |