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Thus, 2503=(25 x 10)3=  253 x 103=15,625 X 1,000 =15,625,000 ; 4003=(4 x 100)3=43 x 100o=64 x 1,000,000 =64,000,000; 7,0003 (7 X 1,000)3 =73 x 1,0003 - 343 x 1,000,000,000=343,000,000,000 ; &c.
227. The square of the sum of any two numbers exceeds the sum of their squares by twice the product of the numbers. The square of the sum of 9 and
9=81 7, for instance, contains (a) the
2x9x7=126 square of 9, (b) twice the product
72=49 of 9 by 7, and (c) the square of 7. (9+7), or 16o=256 For, on performing the multiplication in the manner shown below, we see that the first partial product contains 72 and 9x7; that the second partial product, contains 92 and 9x7; and that, consequently, the sum of the two partial products contains 9%, 72, and the double of 9x7:
9-1-7 9x7+72=7 times the multiplicand 9+9x7 =9 92+2x9x7+72=16 ,
=16%, or 9+7) General formula : (x+y)=ro+2xy +yo. 228. The cube of the sum of any two numbers consists of four parts, namely: (a) the cube of the first number, (b) three times the product of the square of the first number by the second, (c) three times the product of the first number by the square of the second, and (d) the cube of the second number. Thus, the cube of the sum of 9
93=729 and 7 contains (a)93, (b) 3X 92 X7,
3x92 X7=1,701 (C) 3X9X72, and (d) 73. In
3X9X72=1,323 order to understand this, we have
73= 343 merely to take 92+2X9X7+72-- (9+7)), or 163=4,096 that is, (9+7)---for multiplicand, and 9+7 for multiplier : 92+2x9x7+7
9+7 92 X7+2x9x72+73= 7 times the multiplicand 93+2x92x7+9x7 93+3x92 X7+3X9X72+73=16
=163, or (9+7).
General formula : (x+y)=r3+3xoy+3ry"+y3.
SQUARES AND CUBES OF THE FIRST TEN (INTEGRAL) NUMBERS: Numbers, 2 3 4 5
9 Squares, 4 9 16 25 36 49 64 81 100 Cubes,
8 27 64 125 216 343 512 729 1,COO
229. The finding of roots, when powers and indices are given, is called EvOLUTION.
We work an exercise in Evolution when, for instance, we ascertain that the square root of 529 is 23, or that the cube root of 4,913 is 17, or that the fourth root of 20,736 is 12; in other words, when we extract (or “evolve") the square root of 529, or the cube root of 4,913, or the fourth root of 20,736.
Square” numbers, that is, numbers which (like 1, 4, 9, 16, 25, &c.) have exact square roots—are comparatively few; and • cubé” numbers—that is, numbers which (like 1, 8, 27, 64, 125, &c.) have exact cube roots—are fewer still. So that, in the great majority of instances, what is termed a square root or a cube root is merely an approximation. The number 6 for example, has neither an exact square root nor an exact cube root. As 6 lies between 4 and 9, the square root of 6, if determinable, would lie between the square root of 4 and the square root of 9-that is, between 2 and 3. The square root of 6, therefore, if a definite number, would exceed 2 by some proper fraction, and would be convertible into an improper fraction having, when in its simplest form, x (say) for numerator, and y for denominator. Now, if a fraction in its simplest form
z? were the square root of 6, -X would be equal to 6; so
y" that yż would be a measure of x2. But this is impossible: because x, being prime to y, must (p. 141) be prime to y?; and y’, being prime to x, must be prime to x2. Consequently, 6 has not an exact square root; and it could be shown, in the same way, that 6 has not an exact cube root.
For the square root of 6, we can write 2, or 2.4, or 2.44, or 2:449, or 2-4494, or 2-44948, or 2.449489,* &c. ; the number of approximations-each closer than the preceding one-being unlimited. Here, then, is a repetition of the paradox noticed at pp.
* This number is the square root of 5'999996361121, which differs from 6 by less than '000004.
143-4: we can go on “continually approaching, without ever reaching," the square root of a number which is not a square number, or the cube root of a number which is not a "cube” number.
In the decimal (-449489 &c.) resulting from the extraction of the square root of 6, we have an illustration of a class of decimals which neither terminate nor circulate. The decimal referred to would, if terminate, be convertible into a decimal fraction (§ 141); and would, if a circulating decimal, be convertible into a vulgar fraction (SS 153, 154). But we have just seen that there is no fraction which, when added to 2, would give the square root of 6; so that the decimal :449489 &c. is interminate and non-circulating.
EXTRACTION OF THE SQUARE ROOT.
230. When the integral part of a number occupies either1 or 2 places, ) the integral part of ( place
Sthe square root of the J 2 places
&c. Thus, as every number whose integral part is expressed by either one figure or two must be less than 100, but not less than 1, the square root of every such number must be less than V 100, but not less than vī_that is, must be less than 1o, but not less than 1 ; so that the integral part of the square root will occupy one place. Again : as every number whose integral part is expressed by either three or four figures must be less than 10,000, but not less than 100, the square root of every
such number must be less than ✓ 10,000, but not less than 100that is, must be less than 100, but not less than 10; so that the integral part of the square root will occupy two places. In like manner, as every number whose integral part is expressed by either five or six figures must be less than 1,000,000, but not less than 10,000, the square root of every such number must be less than ✓ 1,000,000, but not less than ✓ 10,000——that is, must be less than 1,000, but not less than 100; so that the integral part of the square root will occupy three places. We see, therefore, without proceeding further, that if the integral part of a
number occupied n places, the integral part of the square root
nti of the number would occupy either places-according as n happened to be even or odd.
EXAMPLE I.—Extract N 1,444.
This number being expressed by four figures, its square root will ($ 230) be expressed by two-a tens' and a units' figure. In determining how many tens there are in the root, we disregard the last two figures(44)of 1,444_knowing that ($225) the square of any number of tens can contain nothing lower than'hundreds. As 14 lies between 9 and 16, or between 32 and 42, 1,400—and therefore 1,444-must lie between 30% and 40€; so that the tens' figure of the root is 3.
Putting u for the number of units in the root, we have V 1,444 = 30+u; or 1,444= (30+u)=302+2 x 30 x 1+u?. When, therefore, 30%, or 900, is subtracted from 1,444, there remains 544=2 X 30 Xu+u=60 xu+u? As 60 X u is evidently a much larger number than u?, it follows that 60 x u constitutes - the principal portion of 544, and that u must be nearly equal to the both part of 544, or to the 6th part of 54 -- that is, to 9. Now, if 9 be the units' figure of the root, we shall have 544= 60x9+92=60+9) X9=69x9. But 69 x9, or 621, exceeds 544; so that the units' figure of the root must be a lower one than 9. We therefore try 8; and, finding 544=60X8+82= (60+8)x8=68 x 8, we conclude that the units' figure is 8, and that the required root is (30+8=)38.
EXAMPLE II.-Extract V208,849. This number being expressed by six figures, its square root will ($ 230) be expressed by three-a hundreds’, a tens', and a units' figure. Remembering that (§ 225) the square of any number of hundreds terminates with four ciphers, we disregard thelast four figures (8,849) of 208,849, and consider only the left-hand pair (20), whilst determining the hundreds' figure of the root. As 20 lies between 16 and 25, or between 42 and 52,200,000—and therefore 208,849—must lie between 4002 and 5002; so that the hundreds’ figure of the root is 4. Putting x for the undiscovered part of the root, we have V 208,849=400+x; or 208,849=(400
+x)=4004+2 X 400 Xx+x”. When, therefore, 4003, or 160,000, is subtracted from 208,849, there remains 48,849=2 X 400 Xx+ x=800 Xx+x2. As 800 X x-being a much larger number than x?—-constitutes the principal portion of 48,849, we divide 48,849 by 800, or 488 by 8, and find that, the undiscovered part of the root being less than 70, the tens' figure cannot be higher than 6. Adding 60% to 800 X 60,-or multiplying (800-+6)860 by 60,- we obtain 51,600, which exceeds 48,849; so that the
tens' figure of the root must be lower than 6. We therefore try 5. Adding 50 to 800 X 50,-or multiplying (800+50=) 850 by 50,—we find that the result, 42,500, is contained in 48,849; for which reason we conclude that the tens' figure of the root is 5. The “known” part of the root is now (400+ 50=)450. Putting y for the part still undiscovered, we have V208,849=450+y; or 208,849=(450+y)=4502+2 X 450 X yty? When, therefore, 450o is subtracted from 208,849, the remainder, 6,349, is equal to 2 x 450 Xy+y=900 xy+y?:
[This remainder is most easily obtained when we subtract 42,500 from 48,849 : 450o=(400+50)=4002+2 X 400 X 50+ 5O=160,000+800x50+50°=160,000+850 x 50=6o,OOO +42,500; 208,849-160,000=48,849; 48,849-42,500=6,349.]
As 900 Xy—being very much larger than y-—-constitutes the principal portion of 6,349, the division of 6,349 by 900, or of 63 by 9, will probably give the remaining part of the root. The quotient being 7, we add 7 to 900 X7, or multiply (900+7=) 907 by 7 ; and, finding the result to be exactly 6,349, we conclude that the units' figure of the root is 7, and that the root
EXAMPLE III.- Extract v 864,735,219.
of its square root will ($ 230) be expressed by five (2+1);
that the most left-hand figure of the root will represent one or more groups of 10,000 each. As the square of any number of such groups ($ 225) terminates with eight ciphers, we disregard the last eight figures (64735219) of 864,735,219, and consider only the most left-hand one (8), whilst determining the most lefthand figure of the root. Seeing that 8 lies between 4 and 9, or between 22 and 3*, we know that 800,000,000—and therefore 864,735,219--must lie between 20,0002 and 30,0002. The most left-hand figure of the root, therefore, is 2. Putting :c for the remainder of the root, we have V864,735,2195 20,000 + x; or 864,735,219= (20,000 + x)=20,000? +2x 20,000 Xx+x2. When, therefore, 20,000?, or 400,000,000, is subtracted from 864,735,219, there' remains 464,735,219= 2 x 20,000 X x + x = 40,000 X x + x. As 40,000 x r-being a much larger number than x–constitutes the principal portion of 464,735,219, the division of 464,735,219 by 40,000, or of 46,473 by 4, enables us to approximate to the undiscovered portion of the root. Although the quotient exceeds 9,000, (4 being contained more than 9 times in 46,) we know that the next figure of the root cannot be higher than 9, which, therefore, we try. Adding 9,000° to 40,000 X 9,000,-or multiplying