**Ancient Methods of
Calculating Square Roots**

Physical
evidence exists that the Babylonians had a method of calculating the square
root of some numbers as early as 2000 years before the birth of Christ. In the Yale collection there exist an
artifact that shows the calculation of the square root of two to five decimal
places accurately. Some images of the
item, called YBC7289 can be found here. What may surprise many students is that the
ancient Babylonian method seems to be the same as the method frequently taught
in school text books. The method is
also called Newton’s method, and the divide-and-average method. The method is an iterative method which
involves the following steps: (I will illustrate the method with an example of
the square root of 250).

1) Guess a number for the square root I’ll guess badly, and start with 10

2) Divide the number by the guess 250 / 10 = 25

3) Average the original guess and the new
guess (25+10)/2
= 17.5

4) make this
average value your new “guess” and new guess = 17.5

go back to step
2…..

As you continue through the steps you get
the following

Guess |
Quotient |
Average |
Next
guess |

10 |
250/10
= 25 |
(25+10)/2=17.5 |
17.5 |

17.5 |
250/17.5
= 14.285 |
(17.5+14.285)/2=15.89 |
15.89 |

15.89 |
250/15.89
= 15.73 |
(15.89+15.73)/2=15.81 |
15.81 |

The process may be continued to whatever
level of accuracy is desired. Notice
that after three divide and averages we have 15.81, which is accurate to the
hundredths place. place. _{}.

A Chinese method appeared sometime shortly
after 200 BC that became very common in American textbooks. The first occurrence of this method that I
know of was in the fourth chapter of The *Jiuzhang suanshu* or *Nine
Chapters on the Mathematical Art* . The book is a practical handbook of mathematics
consisting of 246 problems which provide methods of solving common problems of
engineering, and trade. The book holds
a position in Chinese mathematical development similar to Euclids “Elements” in
western mathematics. An explanation of the Chinese method with illustrations
can be found here.

The method somehow
became the common approach to solving problems in the The image at right shows
a poem to be memorized by students to remember the steps. The poem appeared in the 1772 textbook, __Arithmetick,
both in the theory and practice : made plain and easy in all the common and
useful rules __, by John Hill. The same algorithm
existed into textbooks of the 1960’s and really only disappeared with the
emerging ubiquity of the hand held calculator.
Here is another look at the same algorithm with directions that may seem
clearer, and an example taken from the Wikipedia online dictionary:

Write the number in decimal
and divide it into pairs of digits starting from the decimal point. The numbers
are laid out similar to the long division algorithm and the final square root
will appear above the original number.

For each iteration:

1. Bring down the most significant pair of
digits not yet used and append them to any remainder. This is the *current
value* referred to in steps 2 and 3.

2. If *r* denotes the part of the result
found so far, determine the greatest digit *x* that does not make *y*
= *x*(20*r* + *x*) exceed the current value. Place the new digit
*x* on the quotient line.

3. Subtract *y* from the current value
to form a new remainder.

4. If the remainder is zero and there are no
more digits to bring down the algorithm has terminated. Otherwise continue with
step 1.

Example: What
is the square root of 152.2756?

` ____1__2._3__4_`

` | 01 52.27 56 1`

` x 01 1*1=1 1`

` ____ __`

` 00 52 22`

` 2x 00 44 22*2=44 2`

` _______ ___`

` 08 27 243`

` 24x 07 29 243*3=729 3`

` _______ ____`

` 98 56 2464`

` 246x 98 56 2464*4=9856 4`

` _______`

` 00 00 Algorithm terminates: answer is 12.34`

` `

I think I probably was taught, and tested,
on the use of this algorithm each year from grade 5 or 6 onward, but do not
remember ever having seen an explanation for why it worked. Below is my attempt to explain why it works
geometrically, using the example of _{} so we can compare
with the example above.

Geometrically,
the square root of a number asks for the side of a square whose area is the
given number. In this problem we want
to find the side of a square whose area is 250 square
units. The directions say to mark off
the places in two from the decimal place.
So we will write it as 2 50 . 00
00 The two represents an
area of 200, so we are first looking for the tens digit of the root, that is,
we want the largest square with sides of a multiple of ten that will fit inside
the square of area 250. This is ten, so
our first approximation is ten. The
figure at the right shows the square with an area of 250 and part of it covered
by our first, and not very good, guess of 10 for the square root. The remaining area to be covered is 150
square units. The same idea appears in
the first line of the algorithm

1
__ . __ __ The one in the tens place
represents ten
2 50
. 00
00

__ 1 __ and this one under the 2 hundred represents 10
squared or 100

1 50 and the 150 is the remaining amount to be
covered. Up to here it seems quite straightforward, but at this point the double
and guess stuff seems weird at first.
Look again at the table and notice that we have broken the sides of the
square into 10 + some remaining amount labeled r. The remaining 150 is made up of three parts, the top-left
rectangle with an area of 10 r, the bottom-right rectangle with an area of 10
r, and the top-right square with area of r^{2}.

This makes the 150
= 10r + 10 r + r^{2 } or, collecting the like terms 20 r + r^{2}. If we factor this we get r(20+r). It is these two congruent rectangles that
account for the doubling that appears in the next line of the algorithm. Since we are looking for the units digit, 20
+ r is a two digit numeral with 2 in the tens place, and r in the units place,
like 2r; and we want to select r so that when we multiply the number 2r by r we get a number close to
150. We guess at 6 and try 26 * 6
which is 156… too big, so we try 25*5 and settle for 125

1 5 . ___ ____

2 50 . 00 00

__1________________

__ 1 50 __

__25
1 25__

__ 25 00__

__ __

At this point we have enlarged our square
to a 15 x 15 square taking up 225 of the 250 square units, and we now have
another remainder such that (15+r)^{2 }= 250. Our unfilled 15
square units is again composed of two rectangles and a square, with areas of 15
r, 15 r, and r^{2}. Since we
have the units digit, the next digit will be in the tenths place, and so (30+r
) r must be __<__ 25. Since r is
in the tenths place, we can think of 30+r as a decimal number like 30.r and r as .r; (although the algorithm lines
up the decimals to avoid having to worry about decimal places. We just think 30_ times _ < 2500 where
the same digit must be in both blanks.
Using an 8 gives 308 x 8 for 2464

1 5 . ___8___ ____

2 50 . 00 00

__1________________

__ 1 50 __

__25__ __ __ __ 1
25__

__ 25 00__

308
__24 64__

__ 36
00 __The remainder of 36 represents 36 one hundredths
error in our approximation. Notice that
15.8^{2} = 249.64. The method
can be continued as far as needed for the accuracy required..

A third approach that dates back to antiquity was discovered in the Bakhshali manuscript, an early mathematical manuscript which was only discovered in 1881, but is believed to be a copy of an original that dates back to about 400 AD. The square root algorithm in this birch bark manuscript is given in three steps, and once more I will illustrate the method by finding, yet again, the square root of 250.

The method uses
four values, the number (250), the nearest smaller perfect square (225) and its
root (15) and the remainder (25) when the nearest square is subtracted from the
number.

1) Find the root of the largest square smaller (15)

2) add the root (15) to the fraction formed by dividing the remainder by twice
the root (25/30)

3) From this value subtract one half the quotient found by dividing the square
of this fraction (25/30) by the sum of the root and the fraction (15 + 25/30).

The value given is 15 + 25/30 - (1/2)(25/30)^{2}/(15+25/30). In
decimal form we get 15 + .8333333 - .5(.69444)/(15.83333) which becomes 15 +
.83333 - .0219298 which is approximately 15.811403; a very close approximation
to the true 15.811388.