Elements of arithmetic

156. We have already remarked (66), that a number multiplied by itself produces what is called the square of that number. Thus, 169, or 13 × 13, is the square of 13. Conversely, 13 is called the square root of 169, and 5 is the square root of 25; and any number is the square root of another, which when multiplied by itself will produce that other. The square root is signified by the sign

√ or √   ; thus, √25 means the square root of 25, or 5; √16 + 9

means the square root of 16 + 9, and is 5, and must not be confounded with √16 + √9, which is 4 + 3, or 7.

157. The following equations are evident from the definition:

√a × √a = a
√aa = a
√ab × √ab = ab
(√a × √b) × (√a × √b) = √a × √a × √b × √b = ab

whence

√a × √b = √ab

158. It does not follow that a number has a square root because it has a square; thus, though 5 can be multiplied by itself, there is no number which multiplied by itself will produce 5. It is proved in algebra, that no fraction[22] multiplied by itself can produce a whole number, which may be found true in any number of instances; therefore 5 has neither a whole nor a fractional square root; that is, it has no square root at all. Nevertheless, there are methods of finding fractions whose squares shall be as near to 5 as we please, though not exactly equal to it. One of these methods gives ¹⁵¹²⁷/₆₇₆₅, whose square, viz.

differs from 5 by only ⁴/₄₅₇₆₅₂₂₅, which is less than ·0000001: hence we are enabled to use √5 in arithmetical and algebraical reasoning: but when we come to the practice of any problem, we must substitute for √5 one of the fractions whose square is nearly 5, and on the degree of accuracy we want, depends what fraction is to be used. For some purposes, ¹²³/₅₅ may be sufficient, as its square only differs from 5 by ⁴/₃₀₂₅; for others, the fraction first given might be necessary, or one whose square is even nearer to 5. We proceed to shew how to find the square root of a number, when it has one, and from thence how to find fractions whose squares shall be as near as we please to the number, when it has not. We premise, what is sufficiently evident, that of two numbers, the greater has the greater square; and that if one number lie between two others, its square lies between the squares of those others.

159. Let x be a number consisting of any number of parts, for example, four, viz. a, b, c, and d; that is, let

x = a + b + c + d

The square of this number, found as in (68), will be

• aa + 2a(b + c + d)
• + bb + 2b(c + d)
• + cc + 2cd
• + dd

The rule there found for squaring a number consisting of parts was: Square each part, and multiply all that come after by twice that part, the sum of all the results so obtained will be the square of the whole number. In the expression above obtained, instead of multiplying 2a by each of the succeeding parts, b, c, and d, and adding the results, we multiplied 2a by the sum of all the succeeding parts, which (52) is the same thing; and as the parts, however disposed, make up the number, we may reverse their order, putting the last first, &c.; and the rule for squaring will be: Square each part, and multiply all that come before by twice that part. Hence a reverse rule for extracting the square root presents itself with more than usual simplicity. It is: To extract the square root of a number N, choose a number A, and see if N will bear the subtraction of the square of A; if so, take the remainder, choose a second number B, and see if the remainder will bear the subtraction of the square of B, and twice B multiplied by the preceding part A: if it will, there is a second remainder. Choose a third number C, and see if the second remainder will bear the subtraction of the square of C, and twice C multiplied by A + B: go on in this way either until there is no remainder, or else until the remainder will not bear the subtraction arising from any new part, even though that part were the least number, which is 1. In the first case, the square root is the sum of A, B, C, &c.; in the second, there is no square root.

160. For example, I wish to know if 2025 has a square root. I choose 20 as the first part, and find that 400, the square of 20, subtracted from 2025, gives 1625, the first remainder. I again choose 20, whose square, together with twice itself, multiplied by the preceding part, is 20 × 20 + 2 × 20 × 20, or 1200; which subtracted from 1625, the first remainder, gives 425, the second remainder. I choose 7 for the third part, which appears to be too great, since 7 × 7, increased by 2 × 7 multiplied by the sum of the preceding parts 20 + 20, gives 609, which is more than 425. I therefore choose 5, which closes the process, since 5 × 5, together with 2 × 5 multiplied by 20 + 20, gives exactly 425. The square root of 2025 is therefore 20 + 20 + 5, or 45, which will be found, by trial, to be correct; since 45 × 45 = 2025. Again, I ask if 13340 has, or has not, a square root. Let 100 be the first part, whose square is 10000, and the first remainder is 3340. Let 10 be the second part. Here 10 × 10 + 2 × 10 × 100 is 2100, and the second remainder, or 3340-2100, is 1240. Let 5 be the third part; then 5 × 5 + 2 × 5 × (100 + 10) is 1125, which, subtracted from 1240, leaves 115. There is, then, no square root; for a single additional unit will give a subtraction of 1 × 1 + 2 × 1 × (100 + 10 + 5), or 231, which is greater than 115. But if the number proposed had been less by 115, each of the remainders would have been 115 less, and the last remainder would have been nothing. Therefore 13340-115, or 13225, has the square root 100 + 10 + 5, or 115; and the answer is, that 13340 has no square root, and that 13225 is the next number below it which has one, namely, 115.

161. It only remains to put the rule in such a shape as will guide us to those parts which it is most convenient to choose. It is evident (57) that any number which terminates with ciphers, as 4000, has double the number of ciphers in its square. Thus, 4000 × 4000 = 16000000; therefore, any square number,[23] as 49, with an even number of ciphers annexed, as 490000, is a square number. The root[24] of 490000 is 700. This being premised, take any number, for example, 76176; setting out from the right hand towards the left, cut off two figures; then two more, and so on, until one or two figures only are left: thus, 7,61,76. This number is greater than 7,00,00, of which the first figure is not a square number, the nearest square below it being 4. Hence, 4,00,00 is the nearest square number below 7,00,00, which has four ciphers, and its square root is 200. Let this be the first part chosen: its square subtracted from 76176 leaves 36176, the first remainder; and it is evident that we have obtained the highest number of the highest denomination which is to be found in the square root of 76176; for 300 is too great, its square, 9,00,00, being greater than 76176: and any denomination higher than hundreds has a square still greater. It remains, then, to choose a second part, as in the examples of (160), with the remainder 36176. This part cannot be as great as 100, by what has just been said; its highest denomination is therefore a number of tens. Let N stand for a number of tens, which is one of the simple numbers 1, 2, 3, &c.; that is, let the new part be 10N, whose square is 10N × 10N, or 100NN, and whose double multiplied by the former part is 20N × 200, or 4000N; the two together are 4000N + 100NN. Now, N must be so taken that this may not be greater than 36176: still more 4000N must not be greater than 36176. We may therefore try, for N, the number of times which 36176 contains 4000, or that which 36 contains 4. The remark in (80) applies here. Let us try 9 tens or 90. Then, 2 × 90 × 200 + 90 × 90, or 44100, is to be subtracted, which is too great, since the whole remainder is 36176. We then try 8 tens or 80, which gives 2 × 80 × 200 + 80 × 80, or 38400, which is likewise too great. On trying 7 tens, or 70, we find 2 × 70 × 200 + 70 × 70, or 32900, which subtracted from 36176 gives 3276, the second remainder. The rest of the square root can only be units. As before, let N be this number of units. Then, the sum of the preceding parts being 200 + 70, or 270, the number to be subtracted is 270 × 2N + NN, or 540N + NN. Hence, as before, 540N must be less than 3276, or N must not be greater than the number of times which 3276 contains 540, or (80) which 327 contains 54. We therefore try if 6 will do, which gives 2 × 6 × 270 + 6 × 6, or 3276, to be subtracted. This being exactly the second remainder, the third remainder is nothing, and the process is finished. The square root required is therefore 200 + 70 + 6, or 276.

The process of forming the numbers to be subtracted may be shortened thus. Let A be the sum of the parts already found, and N a new part: there must then be subtracted 2AN + NN, or (54) 2A + N multiplied by N. The rule, therefore, for forming it is: Double the sum of all the preceding parts, add the new part, and multiply the result by the new part.

162. The process of the last article is as follows:

In the first of these, the numbers are written at length, as we found them; in the second, as in (79), unnecessary ciphers are struck off, and the periods 61, 76, are not brought down, until, by the continuance of the process, they cease to have ciphers under them. The following is another example, to which the reasoning of the last article may be applied.

163. The rule is as follows: To extract the square root of a number;—

EXERCISES.

164. Since the square of a fraction is obtained by squaring the numerator and the denominator, the square root of a fraction is found by taking the square root of both. Thus, the square root of ²⁵/₆₄ is ⅝, since 5 × 5 is 25, and 8 × 8 is 64. If the numerator or denominator, or both, be not square numbers, it does not therefore follow that the fraction has no square root; for it may happen that multiplication or division by the same number may convert both the numerator and denominator into square numbers (108). Thus, ²⁷/₄₈, which appears at first to have no square root, has one in reality, since it is the same as ⁹/₁₆, whose square root is ¾.

165. We now proceed from (158), where it was stated that any number or fraction being given, a second may be found, whose square is as near to the first as we please. Thus, though we cannot solve the problem, “Find a fraction whose square is 2,” we can solve the following, “Find a fraction whose square shall not differ from 2 by so much as ·00000001.” Instead of this last, a still smaller fraction may be substituted; in fact, any one however small: and in this process we are said to approximate to the square root of 2. This can be done to any extent, as follows: Suppose we wish to find the square root of 2 within ¹/₅₇ of the truth; by which I mean, to find a fraction a/b whose square is less than 2, but such that the square of a/b + ¹/₅₇ is greater than 2. Multiply the numerator and denominator of ²/₁ by the square of 57, or 3249, which gives ⁶⁴⁹⁸/₃₂₄₉. On attempting to extract the square root of the numerator, I find (163) that there is a remainder 98, and that the square number next below 6498 is 6400, whose root is 80. Hence, the square of 80 is less than 6498, while that of 81 is greater. The square root of the denominator is of course 57. Hence, the square of ⁸⁰/⁵⁷ is less than ⁶⁴⁹⁸/₃₂₄₉, or 2, while that of ⁸¹/₅₇ is greater, and these two fractions only differ by ¹/₅₇; which was required to be done.

166. In practice, it is usual to find the square root true to a certain number of places of decimals. Thus, 1·4142 is the square root of 2 true to four places of decimals, since the square of 1·4142, or 1·99996164, is less than 2, while an increase of only 1 in the fourth decimal place, giving 1·4143, gives the square 2·00024449, which is greater than 2. To take a more general case: Suppose it required to find the square root of 1·637 true to four places of decimals. The fraction is ¹⁶³⁷/₁₀₀₀, whose square root is to be found within ·0001, or ¹/₁₀₀₀₀. Annex ciphers to the numerator and denominator, until the denominator becomes the square of ¹/₁₀₀₀₀, which gives ¹⁶³⁷⁰⁰⁰⁰⁰/₁₀₀₀₀₀₀₀₀, extract the square root of the numerator, as in (163), which shews that the square number nearest to it is 163700000-13564, whose root is 12794. Hence, ¹²⁷⁹⁴/₁₀₀₀₀, or 1·2794, gives a square less than 1·637, while 1·2795 gives a square greater. In fact, these two squares are 1·63686436 and 1·63712025.

167. The rule, then, for extracting the square root of a number or decimal to any number of places is: Annex ciphers until there are twice as many places following the units’ place as there are to be decimal places in the root; extract the nearest square root of this number, and mark off the given number of decimals. Or, more simply: Divide the number into periods, so that the units’ figure shall be the last of a period; proceed in the usual way; and if, when decimals follow the units’ place, there is one figure on the right, in a period by itself, annex a cipher in bringing down that period, and afterwards let each new period consist of two ciphers. Place the decimal point after that figure in forming which the period containing the units was used.

168. For example, what is the square root of (1⅜) to five places of decimals? This is (145) 1·375, and the process is the first example over leaf. The second example is the extraction of the root of ·081 to seven places, the first period being 08, from which the cipher is omitted as useless.

• 1,37,5(1·17260
• 1  
• 21)  37
•  21    
• 227) 1650
•  1589  
• 2342) 6100
• 4684
• 23446) 141600
• 140676    
• 23452)    92400
• 8,1(·2846049
• 4   
• 48)410
• 384
• 564) 2600
• 2256
• 5686) 34400
• 34116
• 569204) 2840000
• 2276816
• 569208)  56318400
• ·000002413672221(·001553599
• 1   
• 25) 141
• 125  
• 305) 1636
• 1525  
• 3103) 11172
• 9309  
• 31065) 186322
• 155325  
• 310709) 3099710
• 2796381   
• 30332900

169. When more than half the decimals required have been found, the others may be simply found by dividing the dividend by the divisor, as in (155). The extraction of the square root of 12 to ten places, which will be found in the next page, is an example. It must, however, be observed in this process, as in all others where decimals are obtained by approximation, that the last place cannot always be depended upon: on which account it is advisable to carry the process so far, that one or even two more decimals shall be obtained than are absolutely required to be correct.

• A
• 12(3·46410161513
• 9  
• 64) 300
• 256  
• 686) 4400
• 4116  
• 6924) 28400
• 27696  
• 69281)  70400
• 69281  
• 6928201) 11190000
• 6928201  
• 69282026) 4261799|00
• 4156921|56
• 692820321) 104877|4400
• 69282|0321
• 6928203225) 35595|407900
• 34641|016125
• 69282032301)  954|39177500
• 692|82032301
• 692820323023) 261|5714519900
• 207|8460969069
• 53|7253550831
•  
• B
• 692820323026)537253550831(77545870549
• 484974226118
• 52279324713
• 48497422611;
• 3781902102
• 3464101615
• 317800487
• 277128129
• 40672358
• 34641016
• 6031342
• 5542562
• 488780
• 484974
• 3806
• 3464
• 342
• 277
• 65
• 62
• 3

If from any remainder we cut off the ciphers, and all figures which would come under or on the right of these ciphers, by a vertical line, we find on the left of that line a contracted division, such as those in (155). Thus, after having found the root as far as 3·464101, we have the remainder 4261799, and the divisor 6928202. The figures on the left of the line are nothing more than the contracted division of this remainder by the divisor, with this difference, however, that we have to begin by striking a figure off the divisor, instead of using the whole divisor once, and then striking off the first figure. By this alone we might have doubled our number of decimal places, and got the additional figures 615137, the last 7 being obtained by carrying the contracted division one step further with the remainder 53. We have, then, this rule: When half the number of decimal places have been obtained, instead of annexing two ciphers to the remainder, strike off a figure from what would be the divisor if the process were continued at length, and divide the remainder by this contracted divisor, as in (155).

As an example, let us double the number of decimal places already obtained, which are contained in 3·46410161513. The remainder is 537253550831, the divisor 692820323026, and the process is as in (B). Hence the square root of 12 is,

3·4641016151377545870549;

which is true to the last figure, and a little too great; but the substitution of 8 instead of 9 on the right hand would make it too small.

EXERCISES.