▪

Professor Kato's method▪

Takashi Kojima's method

▪Steve Treadwell's modified method

ABACUS: MYSTERY OF THE BEAD

The Bead Unbaffled - An Abacus Manual

## Abacus Arithmetic

by Welton J. Crook

## Square Root

The extraction of the square root of a number may be accomplished easily enough by means of the abacus but the process is somewhat slow and tedious. It consists of a long series of additions and subtractions. The operation can be performed, no doubt, much more expeditiously by making use of a table of logarithms, even when competing with a fast and skillful abacus operator.

To Extract the Square RootThe general steps in extracting the square root with the aid of the abacus may be stated as follows:

a.Set up the number, the square root of which is to be extracted, on therighthand side of the abacus leaving a number of columns vacant, depending on whether or not several decimal places are required in the answer. The number, the square root of which is to be extracted, is called the "square" number. Starting at the decimal point it should be "pointed off" to form groups containing two numerals, in a manner similar to that performed in pencil and paper arithmetic.On the

lefthand side of the abacus, the number 1 is placed on column one. This number, on the left, is called the "root" number.Beginning with the left hand group of the "square" number, subtract the number 1. Then add 2 to the 1 of the "root" number, making 3 and subtract 3 from the residue of the left hand group of the square number. Continue adding 2 to the root number and subtracting the sum from the residue of the left hand group until the root number becomes larger than the residue of the left hand group of the square number. When this occurs add a zero to the root number and add 11. Now include the second group of two numerals with the residue of the first group of the square number and subtract from them the augmented root number. Continue adding 2 to the root number and subtracting from the square number as before and until the original square number is exhausted. If the square is not a perfect square and when the decimal point has been reached, further groups of 00 may be added and the process continued.

To the root number, which finally has been obtained, add 1 and divide by 2. The quotient will be the square root sought. The process of extracting the square root is much easier demonstrated by means of examples.

Example I___ \/484 = 22

Place 484 at the

rightside of the abacus, leaving column one vacant. Separate 484, the "square" number, into two groups, the first group containing one figure, 4, on column four and the other group containing two figures, 8 and 4 on column three and column two respectively.Place 1 on the first column on the

lefthand side of the abacus. This is the "root" number.Subtract 1 from the figure of the first "square" group, namely 4, (column four) leaving 3.

Add 2 to the root number (1) making 3 on the (left hand) column one. Subtract 3 from (right hand) column four, leaving zero. The first square group is exhausted.

Add 0 to the 3 in the "root" number and then add 11. 30 + 11 equals 41, which is the new root number. Left hand column one contains 4 and left hand column two contains 1. We now have 84 in the "square" number in columns three and two from the right. Subtract 41 from 84 leaving 43 in the "square" number.

Add 2 to the 41 in the "root" number, making 43. Subtract "root" number 43 from "square" number 43 leaving zero.

Add 1 to the "root" number, making 44. Divide 44 by 2 and obtain 22. This is the square root required.

Example II______ \/974169 = 987

Place 974169 on the right of the abacus and leave one column vacant. Mark off this "square" number in groups of two numerals. The first group will be 97, the second 41, and the third 69. Place 1 on column one on the left of the abacus. this is the "root" number.

Proceed as in Example I. Subtract 1 from 97, leaving 96. Add 2 to 1 to make 3 in the "root" number. Subtract 3 from 96 to leave 93 in the "square" number and so on, adding 2 to the "root" number and subtracting from the "square" number. Finally the "root" number becomes 17 and the number in the first group of the square number is only 16. Now add 0 to the "root" number, making 170 and then add 11, making 181, which is the new "root" number. Attach the second group, 41, of the square number to the 16 remaining in the first group. The new "square" group is now 1641.

Subtract 181 from 1641 leaving 1460. Add 2 to 181 making 183 for the "root" number. Subtract 183 from 1460 leaving 1277 and so on until the "root" number is 195 and the "square" number is only 137.

Add 0 to the "root" number 195 to make 1950 and then add 11, making 1961. Attach the third group, 69, of the "square" number to the remaining 137, making 13769. Subtract 1961 from 13769 leaving 11808. Continue adding 2 to the "root" number and subtracting from the "square" number. Finally, when the "root" number has become 1973, the "square" number has become exhausted.

Add 1 to the "root" number 1973 to make 1974 and divide by 2. 1974 ÷ 2 equals 987 which is the root required.

In connection with the operation of adding a zero to the "root" number and then adding 11, it may happen that when the next group of the "square" number is attached, the root number still may be too large. In such a case, add another group but instead of adding only one zero to the root number, add two zeros and then add 101 instead of 11.

Exercises:

______ \/30.714 = 5.542 ______ \/327.12 = 18.087