ON THE EXTRACTION OF THE SQUARE ROOT.
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.
15127 15127 228826129
----- × ----- or ---------,
6765 6765 45765225
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.
[22] Meaning, of course, a really fractional number, such as ⅞ or
¹⁵/₁₁, not one which, though fractional in form, is whole in reality,
such as ¹⁰/₅ or ²⁷/₃.
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_ + 2_a_(_b_ + _c_ + _d_)
+ _bb_ + 2_b_(_c_ + _d_)
+ _cc_ + 2_cd_
+ _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 2_a_
by _each_ of the succeeding parts, _b_, _c_, and _d_, and adding the
results, we multiplied 2_a_ 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.
[23] By square number I mean, a number which has a square root. Thus,
25 is a square number, but 26 is not.
[24] The term ‘root’ is frequently used as an abbreviation of square
root.
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:
7,61,76(200 7,61,76(276
4 00 00 70 4
------- 6 ---
400)3,61,76 47)361
70)3 29 00 329
------- -----
400) 32 76 546)3276
140) 32 76 3276
6) ----- ----
0 0
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.
34,86,78,44,01(50000 34,86,78,44,01(59049
25 00 00 00 00 9000 25
-------------- 40 ----
100000) 9 86 78 44 01 9 109) 986
9000) 9 81 00 00 00 981
------------- -------
100000) 5 78 44 01 11804) 57844
18000) 4 72 16 00 47216
40) ----------- --------
100000) 1 06 28 01 118089)1062801
18000) 1 06 28 01 1062801
80) ---------- -------
9) 0 0
163. The rule is as follows: To extract the square root of a number;--
I. Beginning from the right hand, cut off periods of two figures each,
until not more than two are left.
II. Find the root of the nearest square number next below the number
in the first period. This root is the first figure of the required
root; subtract its square from the first period, which gives the first
remainder.
III. Annex the second period to the right of the remainder, which gives
the first dividend.
IV. Double the first figure of the root; see how often this is
contained in the number made by cutting one figure from the right of
the first dividend, attending to IX., if necessary; use the quotient as
the second figure of the root; annex it to the right of the double of
the first figure, and call this the first divisor.
V. Multiply the first divisor by the second figure of the root; if the
product be greater than the first dividend, use a lower number for the
second figure of the root, and for the last figure of the divisor,
until the multiplication just mentioned gives the product less than the
first dividend; subtract this from the first dividend, which gives the
second remainder.
VI. Annex the third period to the second remainder, which gives the
second dividend.
VII. Double the first two figures of the root;[25] see how often the
result is contained in the number made by cutting one figure from the
right of the second dividend; use the quotient as the third figure of
the root; annex it to the right of the double of the first two figures,
and call this the second divisor.
[25] Or, more simply, add the second figure of the root to the first
divisor.
VIII. Get a new remainder, as in V., and repeat the process until all
the periods are exhausted; if there be then no remainder, the square
root is found; if there be a remainder, the proposed number has no
square root, and the number found as its square root is the square root
of the proposed number diminished by the remainder.
IX. When it happens that the double of the figures of the root is not
contained at all in all the dividend except the last figure, or when,
being contained once, 1 is found to give more than the dividend, put a
cipher in the square root and in the divisor, and bring down the next
period; should the same thing still happen, put another cipher in the
root and divisor, and bring down another period; and so on.
EXERCISES.
Numbers proposed. | Square roots.
73441 | 271
2992900 | 1730
6414247921 | 80089
903687890625 | 950625
42420747482776576 | 205962976
13422659310152401 | 115856201
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.
Numbers. | Square roots.
·001728 | ·0415692194
64·34 | 8·02122185
8074 | 89·8554394
10 | 3·16227766
1·57 | 1·2529964086141667788495