Do you know how to calculate a square root using paper and pencil?

[CraigD]

Since about the 1970s, low-cost electronic calculators and computers have changed how we perform practical numeric calculations. In ways, this is beneficial – freed from time-consuming manual calculations, we’re able to concentrate on the meaning of the numeric data we’re manipulating, physical measurements, statistics, or pure mathematical abstractions, and arithmetic error is almost unheard of. In other ways, it’s detrimental, depriving more than a generation of exposure to traditional techniques some of which are millennia old, and reducing average mathematical intuitive ability in some areas, such as the understanding of numeration, estimation, and precision.

 

In an effort to gain a sense of how this change in mathematical tradition has and is unfolding, I’m envisioning a series of hypography polls, of which this is the first.

 

Here’s the poll question:

Without consulting any reference material – human, print, or electronic – do you know how to find the square root of any whole or terminating decimal number (eg: [math]sqrt{2}[/math]), to any number of digits precision, using only paper and pencil?

 

It would be additionally informative if responders posted their answer, age, profession, where they learned (or failed to learn) a square root-calculating technique, and any other pertinent information you can think of.

[CraigD]

#1 - Yes, I know how, and could do the calculation right now.

 

I'm a 46 year old American computer programmer. I believe I was taught a technique by a parent, relative, or adult friend, and later, in school.

[Turtle]

May I petition the court before voting with a point of order? If using the precision of compass & straight-edge in conjunction with the paper counts, then I can vote the first option, where the degree of precision is related to the quality of instruments, sharpness of pencil point, & steadiness of hand.:eek_big: ;) ;) If only the "similar looking to long division" method counts, then I must vote the second option.:)

[eric l]

I sure do (and I'm affraid not manyu people who don't will be replying to your poll)

 

I'm aged 59, Belgian, and have been active in application research as a chemical engineer. Due to mergers, I found myself without a job and am now active as a volunteer in a local non profit organization that tries to bridge the digital gap.

 

I learned it in high school (or what would be high school in the US) in what must be the 9th or 10th grade. From what I see here, this skill is now only taught in classes taking math as a major (6 hours a week or more).

 

In my professional carreer, I noced that I rarely needed it, even in the early days (we had our slide rulers). Also, even people who pretented to know how, often made an essential mistake. In fact, you have to divide the number into groups of two digits, but in such a way that the decimal point stands between two such groups and not wihtin one of them.

 

I never forgot the technique (as apparantly many people do) because, although this was not class stuff, I found out that it is simply an application of the formula

(a + :eek_big:² = a² + 2ab + b² = a*(a + 2b) + b²

Once you know WHY it's done that way, you won't forget HOW it's done ! Tha's why I believe that teaching the technique nowadays is only useful of you let the pupils discover why it's done that way.

[CraigD]

May I petition the court before voting with a point of order? If using the precision of compass & straight-edge in conjunction with the paper counts, then I can vote the first option, where the degree of precision is related to the quality of instruments, sharpness of pencil point, & steadiness of hand.:eek_big: ;) ;) If only the "similar looking to long division" method counts, then I must vote the second option.:)

I’d call that a #2 vote.

 

A poll concerning common compass & straightedge constructions would fit well into the series of polls I’m envisioning.

 

I suspect that many people who once knew the long-division like method have forgotten it due to disuse. What we do not use (frequently), we tend to forget.

[ronthepon]

Though I ad learnt it a couple of years ago, I have never used it when I had other choices. But hey, who won't?

 

I'm in the last year of school, BTW.

 

An excellent poll, CraigD

[Boerseun]

Number three for me. I have no idea how. I would love to, though.

 

Somebody care to explain? :eek_big:

[Jay-qu]

I didnt vote yes because this hardly counts, but I have done it before through trial and error :eek_big: took me a while

[C1ay]

I'm a 45 year old mechanical engineer that learned it from a WWII era math book I picked up at a yard sale. It is full of mathematical methods obsoleted by technology.

[Boerseun]

So - anybody care to explain the method? :lol:

[infamous]

When I was in high school it was taught to every student. That's been quite a while ago so, with the advent of computer technology and the passage of time I've forgotten the method. Having this technique brought back to mind, I'll need to refresh my memory and regain this valued mathematical tool............................Infy

[TheBigDog]

I am familiar with two methods. One that finds the precise answer, and an older one that is used recursivly to get closer and closer to the precise answer.

 

Bill

[CraigD]

Spoiler alert – don’t read before answering poll!

So - anybody care to explain the method? :lol:

Without explaining why it works, here’s the method I, and I suspect most people, know for “long division-like” calculation of a square root:

1. Write the starting number as 2-digit segments, beginning at the decimal point, zero-padding as needed (Eg: 15129 to 01 41 29, 5.125 to 05 12 50)
   
2. Find the greatest integer that squared is less than or equal to the first (leftmost) segment (eg: for 05, find 2). Write it in the left margin of a new line
   
3. Write the square of the found integer on the new line
   
4. Write the difference of the previous 2 line on a new line
   
5. Cary down the next 2 digits of the starting number (carry down “00” if there are none)
   
6. Double the found integer and write it in the left margin of the new line
   Your paper will now look like this:

     05.12 50
   2 04
   4  1 12

   
   

7. Find the greatest integer which, multiplied by the left of the number in the left margin with the integer concatenated to its right is less than or equal to the number on the last line (eg: of 112 and 4, find 2). Concatenate to the previous left margin number, and write it in the left margin of a new line.
   
8. Write the new left margin number multiplied by the found integer (its rightmost digit) on the new line
   
9. Write the difference of the previous 2 line on a new line
   
10. Cary down the next 2 digits of the starting number
    
11. Double the found integer and write it in the left margin of the new line
    
12. Repeat step 8 through 11 until you have calculated as many digits of the square root as you want
    

So, calculating 6 digits of the square root of 5.125, your paper will look like this (note the list of found integers written on the top margin, an optional step further resembling long division favored by some people):

  2 .2  3  8  6  4  6
 05.12 50
2 04
4  1 12
42   84
44   28 50
446  26 76
452   1 74 00
4523  1 35 69
4526    38 31 00
45268   36 21 44
45276    2 09 56 00
452764   1 81 10 56
452768     28 45 44 00
4527686    27 16 61 16
           1 28 82 84

[ronthepon]

Hey... good explaining, CraigD... I tried it once myself and gave up.

[C1ay]

Hey... good explaining, CraigD... I tried it once myself and gave up.

Yes, his explanation looks cleaner than the one I was working on at the same time but I'll post it anyhow in case it helps someone understand to see it differently.

 

Here's an example to find the square root of 65536.

 

Begin by dividing the number into number pairs from the left, i.e. 6,55,36 and place them under the radical:

 

  ________
/6,55,36

 

Now find the largest number you can square and still not be larger than the leftmost number(s), a 2 in this case. Square and subtract. Then bring down the next pair of numbers.

 

  2______
/6,55,36
 4
 -------
 2 55

 

Now you want to find a number such that it can be appended to the first number doubled and be used to multiply that number and still be smaller than the new dividend, i.e.

   2___5___
  /6,55,36
    4
    -------
45 /2 55
    2 25

Here you see that I've chosen 5 since 6*46 would be 276. Now subtract, bring down the next pair of numbers.

   2___5___
  /6,55,36
    4
    -------
45 /2 55
    2 25
    -------
50_/   3036

For the new divisor you need to double the number you have thus far for a square root, 25*2=50. Now you need to find a number that you can append to this and multiply this new number by and still be equal to or smaller than the new dividend, 3036. 6 looks like a good pick.

   2___5__6
  /6,55,36
    4
    -------
45 /2 55
    2 25
    -------
506/   3036
      3036
    -------

Checking you will see 256*256=65536

 

HTH,

[TheBigDog]

The other method is to take the number you are trying to find the root for, and take the largest integer that when squared is smaller than the first number. For 10 you would have 3 as that integer. You then divide 10 by 3 which gives 3.3333333 Next you take the average of 3 and 3.33333333 to get 3.1666666666. Then you divide 10 by 3.166666666 to get 3.1578947368

You then average 3.166666666 and 3.1578947368. You keep repeating this process. Each iteration adds two digits of precision to the answer.

 

Bill

[ronthepon]

The other method is to take the number you are trying to find the root for, and take the largest integer that when squared is smaller than the first number. For 10 you would have 3 as that integer. You then divide 10 by 3 which gives 3.3333333 Next you take the average of 3 and 3.33333333 to get 3.1666666666. Then you divide 10 by 3.166666666 to get 3.1578947368

You then average 3.166666666 and 3.1578947368. You keep repeating this process. Each iteration adds two digits of precision to the answer.

 

Bill

But how does this process work?

 

How does the other long division like one work?

 

Anyone got an answer?