Calculating Square Roots of Any Number By Hand

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Calculating square roots by hand is not even what a practicing mathematician would call a leisurely pastime. The procedure is hardly, if ever, taught in school in this age of computers. It is, however, quite possible to determine the square root of a number to any degree of accuracy for which a person is willing to stretch the work, and the mathematics involved in the procedure is not complex. A brief glimpse of a hand-calculated square root problem appears very much like a long division problem, and some aspects of the procedure are similar.

Step 1 - The Setup

Write the number for which the square root is to be calculated under a square root sign, elongated if desired to simulate the appearance of long division. Locate and mark the decimal point of this number. Above the radical sign, place another decimal point directly above the decimal point of the original number. This second point will become the decimal point in the square root of the original number. Note that since the handwritten work will look very similar to a long division problem, care should be given in allowing for space to write products, take differences, and carry down digits just as in long division.

On both sides of the decimal point in the original number, use some sort of marking system to pair off adjacent digits of the number. On the whole number side (the left side of the decimal point), cease pairing off digits when there are no digits or 1 digit left at the front of the number. On the fractional side (the right side of the decimal point), pair off digits using zeroes as placeholders if necessary. As a brief example, if one was wanting to find the square root of 13234.251, it would be marked thus:

1 ' 32 ' 34 . 25 ' 10 ' 00 ' 00

The period indicates the location of the decimal point, and the apostrophes pair off adjacent digits on each side of the decimal. Each pairing of digits on the fractional (right) side will constitute one digit of accuracy for the final square root calculation, so if exceptional accuracy is desired, leave plenty of room on the right side of the paper for additional work. In the above example as it is presented, one would be able to calculate the square root of 13234.251 to 4 decimal places beyond the decimal point, with the remainder of the root being truncated (that is, simply dropped off). As many additional pairs of zeroes could be added to the end of the number as one wished, depending on the desired accuracy of the result.

Step 2 - First Calculation

The calculation begins. Look at the very first pairing at the front of the number, similar in a sense to the way one looks at a number when beginning long division. This first pairing will either consist of 1 or 2 digits - for procedural purposes, it makes no difference. Determine the largest perfect square number (1, 4, 9, 16, 25, 36, 49, 64, and 81 will be your only choices for this first calculation) which is less than or equal to the value of the first pairing. Write this square number underneath the first pairing in preparation for subtraction, and write the square root of this perfect square above the radical sign, directly above the first pairing.

Subtract the perfect square from the value of the first pairing and write the difference below. Carry the next pairing of digits in the original number down, and write it immediately to the right of the difference that was just written. Let the number comprised of all these digits be called D, for the sake of later referral in Step 3.

So far, the look of the problem should seem very much like a long division problem as promised. It helps to conform to this style of presentation, since keeping track of differences and products will be as vital here as in division.

Step 3 - The Trial-By-Error Repeating Step

The mathematics involved thus far has been relatively simple. The first spot of good news is that it only gets slightly more difficult in this third step. The second spot of good news is that with this somewhat more lengthy step, the procedure will begin to repeat indefinitely, until the degree of accuracy desired is finally obtained.

Multiply the number above the square root sign by 2, treating the product as a product of two whole numbers. Place this product to the left of the number D (as indicated in Step 2). This is as good a place to put it as anywhere - just keep it very handy as it will play a vital role in calculation. Leave space at the end of this product for one additional digit (i.e.: if the product was 238, write "238__", for example).

A number between 0 and 9 must be chosen, at random if one wishes. This number will be written in 2 places: once above the digit pairing that was just carried down, and once at the end of the product that was just calculated (2 times the number above the square root sign). To avoid cumbersome wording, for the sake of this instruction, let the letter X denote the value between 0 and 9 that was selected. Let the letter Y denote the number created after attaching the digit X to the end of the product of 2 and the number above the square root sign.

The product of X and Y is calculated. What is being sought is the largest value of the number X, for which the product of X and Y will still be less than D (as indicated at the end of step 2). This is purely trial and error - keep trying values of X until the correct one is found. Another glimmer of hope: there are only 10 choices for X, so hang in there!

Once the correct value of X is found and the product of X and Y is calculated, this product is subtracted from D to obtain a new difference. Then next pair of digits from the original number are then carried down, and Step 3 repeats, until the degree of accuracy desired is obtained.

An Example: Finding the Square Root of 13,234.251

The example number in question has already been partitioned:

1 ' 32 ' 34 . 25 ' 10 ' 00 ' 00

The first pairing is the single digit 1. The largest perfect square less than or equal to 1 is 1 itself, so we write 1 below this pairing, and write the square root of 1 (which again is 1) above the square root sign in the problem. The difference between the first pairing and 1 is 0, of course, so the next pairing of 32 is carried down to this difference. The number comprised of the digits of the difference together with the carried-down pairing is simply 32, which is the value of D, as named in Step 2.

Multiplying the number above the square root sign by 2 gives the product 2 (2*1). This is written to the left of 32, with space for a digit at the end. With reference to the naming of X and Y per Step 3, it turns out (by trial and error) than the value of X to be chosen is 1, which is written above the square root sign above the '32' digit pairing, and written again after the 2 (1 is the largets value of X that can work, since 2*22=44 would exceed the value of D).

Write 21 (1*21) under D (which in this step D=32). The difference is 11. Carry down the next pairing of 34, making the new value of D to be 1134. Now Step 3 repeats once more. Multiply the number above the square root sign (11) by 2 to get 22, and write this to the left of 1134, leaving space for that extra digit at the end.

Through trial and error, one will find the largest value of X satisfying Step 3 on this pass is X=5, so write 5 above the square root sign above the '34' pairing, and next to the number 22, making the value of Y to be 225. The product of 5 and 225 is 1125, and 1134-1125=9. Carry down the next pairing of 25 to make the new value of D to be 925.

Continuing with the example with the next 3 pairings should give the approximation of the square root to be 115.040, which is the value of the square root truncated to 3 decimal places. Checking this answer, one will find that 115.040*115.040=13234.2016 - and this approximation can be made as accurate as one wishes by adding extra '00' pairings and continuing with the Step 3 calculation. The following is a table-formatted semblance of what the problem might look like when written in long-hand. Note that the values of X chosen in the Trial-By-Error Repeating Step are in boldface in both places where they were required to be written.

    1   1   5 . 0   4   0   √1' 32' 34.25' 10' 00     121    32        21225   1134        11252300       925                 023004      92510               92016230080        49400                     0                  4940000 etc...

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