Sq. Roots the "Ole Fashion Way"

Students are required to learn the perfect squares, frequently up to 15 and sometimes the requirement is up to 25. They are also required to determine the square root of a non-perfect square to the nearest tenth.

The method typically used is to first estimate the square root of the imperfect square to the nearest tenth. Then, through trial and error, find the number (when squared) which approximates the non-perfect square best.

To further explain, say you were required to estimate √39.

Well, from the perfect square information you know it is between 6 and 7 which have squares of 36 and 49.

Since 39 is closer to 36 than it is to 49, your initial estimate might be about 6.2. But then to insure you had the correct number you would need to square 6.2 which is 38.44.

So, 6.2 is a little low, since 38.44 is less than 39. So next you check the next higher number (by tenths) which is 6.3. And the square of 6.3 is 39.69, which is too high.

Now the question is whether 38.44 or 39.69 is closer to 39 and, of course, the answer is 38.44 and therefore, to the nearest tenth the square root of 39 is 6.2.

In this instance, we made a reasonably good initial guess (6.2) and it only required squaring 6.2 and 6.3.

With an initial guess not quite so good, one could end up squaring 3 or 4 numbers before finding the final answer. There is another way to determine a square root - the "ole fashion way".

Apparently, the ole fashion method of taking a square root went out of vogue sometimes in the late 60s or early 70s. And furthermore, you might need to be about 55 years old (or more) to know the method even exist.

BUT the ole fashion way, I have found, is faster. To be specific, here are my comparisons for various numbers. I varied the order of the method and the specific number assigned to each method with a "coin toss" between the two.

      sqRt(Number)      Method     Result       Time
          15           ole fash      3.9      1min 42sec
          22          trial/error    4.7      1min 47sec

          46          trial/error    6.8      1min 27sec
          53           ole fash      7.3      1min 04sec

         130           ole fash      11.4          57 sec
         140          trial/error    11.9     1min 35sec

         515          trial/error    22.7     1min 50sec
         715           ole fash      26.7     1min 10sec

Therefore, I think it is worth the effort to explain the "ole fashion method" in detail.

Taking a square root the "ole fashion way" consist of an initial step followed by similar iterations,
with both the initial step and each iteration producing one digit of the final square root.
In each iterative step, a "funky division" is executed to produce the next digit of the square root.

But the very first step of the process is to "mark off" (by underlining) in groups of two digits the
number you wish to take the square root of. This process is started at the decimal point and worked
to the right (decimal part) and to the left. Again, mark off in groups of two digits by underlining.
If there is not an even number of decimal digits, just add a trailing zero. If there is an odd number
of whole digits (left of decimal) then the most left grouping will only contain one digit.

Next, for the most left grouping, determine by inspection what the largest perfect square is which is smaller
(or the same) as this grouping. For example, if the grouping was 42, then the largest perfect square would be 36
and the square root of 36 is 6. Write this number (6) above that grouping as the first digit in the final answer.
Now square that number and write it below that grouping. In the 6 / 42 example, 6 squared is 36 and write this
below the 42 in a manner similar to "long division".

At this point, an iterative process begins with each loop of the process producing the next digit of the final
square root answer. The steps of this iteration are:

1- Subtract the two numbers and write down the answer. In our 6 / 42 example above, 36 is below the 42, draw a horizontal line below the 36 (as in long division) and subtract them giving 6.

2 - Bring down the next grouping of numbers (which will be two digits) and place next to this "subtraction difference" (again as similar to long division).

3 - Now draw a "long division symbol" around this group of numbers. In our case (6 / 42) it would be around the 600. And this becomes the dividend of an upcoming "funky division".

4 - Now multiply the "final square root digits" (at this moment it is 6) by 20, giving 120. Write this 120 next to the 600 as if it was the divisor of the upcoming "funky division". Now how many times will 120 go into 600? Well, the answer is 5. BUT, this is where the "funky division" comes into play and when determining this digit you change the last digit of the divisor (which will always be a zero before you change it) to this number. So, in our case we would change the 120 to 125 and if we were to multiply 125 by 5 we get 625 which is larger than 600, so our digit of 5 is too large. This is VERY SIMILAR to long division, EXCEPT the last digit of the divisor is changed to our "predicted digit". So, our answer is really 4, since 5 is too big. So, multiply the 4 by 124 and get 496 and place this under the 600. And now we have completed this first iteration of the continuing process, so go back to step 1 and start the loop again. Of course, this time in step 1 you will be subtracting these two digits you just calculated.

OK - so above is the process. You may find a slightly different worded description of the same exact process at Gottschalk's Gestalts. Sometimes it is good to hear things from two different sources.

Now, we will use this method to take the square root of 813.08.

√813.0800000000

From the decimal, mark off in both directions by two digits. Looking at the left side of the decimal, the 13 is grouped together and the 8 is by itself. On the right side the 08 is grouped and then several 00s are grouped together.

The next step which produces the first digit of the final square root is to start at the left hand side which is the single 8. What is the largest perfect square which is equal to 8 or smaller than 8? Well, 3 squared is 9, which is too big. So, the largest perfect square is 4 and the square root of 4 is 2. So, place the 2 above the 8 as the first digit of the "final square root" answer and square the 2 which is 4 and place the 4 under the 8. This is the last time we will mention "largest square" in this computation of the square root. The remainder of the operation is like "funky division".

I will now refer to the elements as if they were division. So, the elements are divisor, dividend and quotient, with the "quotient" being the "final square root answer".

Subtract the 4 from the 8, giving 4, and bring down the next two digits which are paired off ... these are 13. So, now we have digits 413 . Treat these like a dividend and draw a "long division marker" around them. For the divisor, multiply the quotient (one digit of 2 right now) by 20, which gives you 40. Write this 40 next to the 413 as if it were the divisor.

Now the "funky division" takes place. So, how many times will 40 go into 413? Well, the answer is 10 with some remainder, but then there is this "funky" part of the operation. Whatever digit you get, write it as the next digit in the quotient and you must also write it in place of the zero in 40. And, as a hint, the number will never be larger than 9. So, try a 9 as the next digit in the quotient, and replace the 40 with a 49 as the divisor. If you multiply 9 * 49 then you get 441, which is bigger than the 413 and therefore the 9 is too big. Now try an 8. So, the divisor becomes 48 and the next digit of the quotient becomes 8, making the quotient 28. (And keep in mind when I say "quotient" we are really building digit-by-digit our final square root answer.) And 8 * 48 is 384, which is less than 413 so this works.

So, now - procedurally speaking - another iteration starts, so we will start a new paragraph.

Now, subtract 384 from 413, giving 29. Bring down the next two digits which are 08. So, now the dividend is 2908. And there is a 28 in the quotient. Multiply the 28 by 20 giving 560 and write it as the divisor next to the 2908 as the dividend. How many times will 560 go into 2908? Well, that is 5 and write this down as the next digit of the quotient (final square root). This makes the quotient be 28.5, as we had to bring up the decimal also. Now change the 560 to 565. Now multiply the last quotient digit (5) by 565, giving 2825 and place this under the 2908. Since 2825 is less than 2908 then the 5 is correct.

And another iteration begins - subtract the 2825 from 2908 giving 83. Bring down the next two digits (00) and place next to the 83, giving 8300 as the dividend. Multiply the quotient digits (285) by 20 giving 5700 and write it down as the divisor next to the 8300. How many times will 5700 go into 8300. Well, 1 of course. Write down the next quotient digit as 1 (giving 28.51) and change the divisor from 5700 to 5701. Multiply the last quotient digit by 5701 and write it down under the 8300.

And guess what, another iteration begins - subtract 5701 from 8300 giving 2599. Now bring down the next two digits (00) and write them down as the last two digits next to 2599 (giving 259900) and this will be the next dividend. Multiply 2851 (quotient digits) by 20 giving 57020 and write this down next to the 259900 as the divisor. So, how many times will 57070 go into 259900. Well it appears to be 4, so write it as the next quotient number giving 28.514 and change the 57020 to 57024. Multiply that last quotient digit (4) by 57024, giving 228096. Write this down under the 259900.

And another loop begins - subtract, giving 31804. Bring down next two digits (00) giving 3180400 as the dividend. Multiply the quotient digits (28514) by 20 giving 570280. How many times will 570280 go into 3180400? I'm guessing 5. The quotient becomes 28.5145. Change the dividend to 570285 and multiply this by the last quotient digit (5) giving 2851425.

And we will let this be the last loop. Subtract giving 328975. Bring down the next two digits (00) giving 32897500 as dividend. Multiply 285145 by 20 giving 5702900. How many times will 5702900 go into 32897500. I say 5. Quotient becomes 28.51455. The divisor becomes 5702905. Multiply the last quotient digit (5) by 5702905 giving 28514525, which is less than 32897500, which means the 5 is good.

Lets quit now as that is a square root of 28.5146 (rounded to 4 numbers). By the way, 28.5146 squared is 813.0824 , which is very close to 813.08. So, the method has worked. If you need a better answer, just iterate more times and get more digits.