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Since we have dealt with different problems arising from astronomical calculation, let us now deal with the quadrature of numbers which are not perfect squares, to show clearly that it is possible to find the square root of any given non-square number. This is achieved in the following way:

Take the square root of the nearest^{41} perfect square and double it. Then take from the number whose square root you seek, the square you found nearest to it, and express the remainder as a fraction of the number obtained by doubling the square root^{42} of the square.^{43} For example, if eight were double the side of the square, take the fraction as eighths, if ten then as tenths, and so on.

Thus, suppose we wish to regard 18 as a square and find its root. Take a root of the perfect square nearest it, that is, of 16; its root is 4. Double this and obtain 8, and subtract 16 from 18 leaving 2; express this in terms of eighths and write that the square root of 18 is 4 and two eighths. Now two eighths equals a quarter and so the root is 4 and a quarter. To see that this in the correct answer, multiply 4 and a quarter by itself and so find the answer is 18.

Here are the steps: We say then that four-times 4 is 16 and four-times two eighths, that is, one quarter four times, equals four quarters equals one. Furthermore [four-times]^{44} two eighths is one, and one and one is two. Combine this with the 16 and it makes 18. But this method is too simplistic and is incomplete, and it has a lack of accuracy, for this answer is not the root multiplied by itself. If we multiply also a quarter times itself, we get not just 18, but 18 and one sixteenth. In what follows a more accurate method will be outlined, which we claim to be our own discovery, with the help of God. Let us turn our attention to this method for a while, showing how it can be used even for large numbers, so that it might become easier for us to understand.

We apply the aforementioned method to numbers from the unit to 99 and from a hundred to 9999, since the roots of all numbers between these two have two digits, and we should proceed by the following method:

Suppose we write the number 235 and seek to find its square root.

_{1} |
_{3} |
||||

2 | 3 | 5 | |||

1 | 5 | _{10} |
|||

2 | _{10} |

Take then the number which when multiplied by itself exactly produces the first digit two or is nearest to it. 2 is not correct since when multiplied by itself it gives 4. The number produced by the multiplication must be either equal to 2 or less than it. Hence the product should be one times itself, and we say one times one is again one. Subtract this from 2 and one is left over. Write this in small print above in the space between the 2 and the 3.... Now double the *one* which you found from the subtraction of one from 2. I refer here to the *one* which was the root not the square, for you should double that number which you found from the subtraction^{45}. One and one is two. Write this *two* below the 3, but not so that it lies on the same line as the unit previously written below the 2, but below that again in the third row. Now find the number which when multiplied by this cancels with 13, but when multiplied by itself cancels with the residual^{46}. I mean that the numbers must be less than or equal to 13 and the residual. This is what is always meant by 'cancel'.^{47} Six multiplied by two can be subtracted from 13, but when multiplied by itself produces a number which is greater than the residual. The remainder is 13^{48} and the square is 36. Hence we pass over the 6 and take 5 and say five-times 2 is 10.^{49} Write the 5 between the 3 and the 2 on the same row as the unit written below the 2. Taking the 10 from the 13 leaves 3. Write this in small print above in the space between the 3 and the 5. Again we multiply the 5 by itself giving 25 and subtract from 35 leaves 10.^{50} Write this outside of the row by itself. Now double the 5 and it becomes 10, just as at the beginning we doubled the unit below the 2. Write this in the third row in turn next to the two you wrote previously. Now combine the 20 and the 10 in the third row, since 2 lies in the tens column, this gives 30. Take half from it and we get 15, since 30 is double the root. The root then of 235 is 15 and ten thirtieths. The remainder is expressed in terms of that same thirtieth which is obtained by doubling the (integer part) of the square root. Now ten thirtieths is a third. Note that the (integer part) of the square root, that is 15, lies in the second row and its double is in the third. To check the result, proceed as follows: Multiply 15 by itself and by ten thirtieths and say, fifteen by 15 is 225, then fifteen times a third, that is ten thirtieths, gives 5 then 5 and 5 is 10. Combining this with 225 gives 235.

Peter G. Brown, "The Great Calculation According to the Indians, of Maximus Planudes - On Finding the Square Root of Any Number," *Loci* (March 2012)