Unusual long division - square roots before calculators
However did we manage before calculators? Is there an efficient way
to do a square root if you have to do the work yourself?
Age
14 to 16
Challenge level
Problem
However did we manage before calculators? Let's think about that for a moment.
Just taking square roots as an example, every time we use Pythagoras to find the third side in a right-angled triangle we need to perform a square root. If it's the square root of a number like 4 that's easy : 2 . . plus or minus, if you like . . . but what if we needed the square root of 40?
Obviously it isn't 20, but what is it?
Now we do have a technique, one that we use in school from time to time, usually called 'trial and improvement', and it works like this: get close to the right answer, and then try to get closer still, being as systematic as possible, and continuing until we are close enough for our purposes.
Well that's good, it's nice to have a solution method, but before there were electronic calculators 'trial and improvement' could involve quite a lot of work, so people were motivated to find methods that used as little effort as possible.
Here's one of those methods, and your task is to decide how and why it works.
Click here to hear some audio or read the same explanation below the image.
Image
Finding the square root of 40:
The answer is 6.324 . . . taking the method as far as three decimal places.
Notice the position of the decimal point and step away from there in both directions, two digits at a time. Each of those pairs will lead to one digit of the answer, and remember the zeros continue indefinitely.
Start on the left. The first pair is 40. Find the largest square smaller than 40, that's 36. Subtract the 36 from the 40, which leaves 4, and enter a 6 as the first digit of your answer.
Next take the 4 and 'bring down' the next pair of digits (00 as it happens) to make 400. We are trying to decide the second digit of our answer and we find it like this:
Use the 6 but double it (12) and make that ten times bigger (120). Now find 'something'- a single digit, so that one hundred and twenty 'something' times that same 'something' is as large as possible but less than the 400.
123 times 3 makes 369, so three is the digit we want (124 times 4 would have been too big).
Subtract the 369 from the 400 (31) and 'bring down' the next pair of digits (so we are now aiming for 3100).
The digits we have so far in the answer are 6 and 3. Double 63 and then make it ten times bigger (1260). Use the same technique as before: find one thousand two hundred and sixty 'something', times 'something', that gets as close as possible to 3100 without exceeding it.
Our third digit will be 2, 1262 times 2 is 2524, 1263 times 3 would be too large.
Subtract to leave 576, bring down the next pair of digits, double the digits you already have, that's 632, which doubles to make 1264, now look for twelve thousand, six hundred and forty 'something' times 'something' to come as close as possible to 57,600 without exceeding it. That 'something' is 4 (check it), and so we continue until we have as many digits in our answer as we think we need.
Your task is to find the next digit (the answer is on the Hint page), check you can repeat the method for other square roots (use your calculator to check your answers). Try to account for the method and explain why it works?
Getting Started
The next digit of the answer is 5.
There are a number of ways of exploring around this area, don't rush.
You might try 'trial and improvement' as one way to work towards a square root of say 30, 50 or any other number.
Try this new method to find square roots for 30, 50, etc., then check with a calculator to see that it's working out correctly. Listen again to the audio as many times as you need. There's a lot of calculation to follow.
When you are ready, think about the harder questions:
-
Why multiply by 2 and by 10?
-
And why look for an end digit that is also the digit you use to multiply?
Student Solutions
Well done Carol in Leeds and others. This really is right at the top end of Stage 4 material and takes some following. Most people would need to go over the argument below several times, probably taking breaks and coming back again. There are quite a lot of numbers to keep track of so stay patient with yourself and here goes :
The square roots of $40$, or $4000$, or $0.4$ all contain the same run of digits, just shifted left or right.
When a number is reduced or increased by a factor of ten the square of that number is reduced or increased by a factor of one hundred.
$4000$ reduces by a factor of one hundred to become $40$ so a factor of ten will connect their square roots.
So if I knew what the square root of $4000$ was, the square root of $40$ would be the same but with the digits all one position lower (to the right). Now to work.
This method is about deciding, one by one, what each digit is.
Image
The square root of $40$ starts with a six and then becomes decimal, so the square root of $4000$ starts with sixty-something, before becoming decimal.
And it's that 'something' digit which I need to find.
I want the whole number whose square is as close as possible to $4000$.I know it's a two digit number and I know the first of those digits is six. If $x$ stands for the second digit then I'm trying to make $(60 + x)^2$ as close as possible to $4000$.
Written without the bracket $(60 + x)^2$ is $60^2 + 2.60.x + x^2$ so the challenge is to pick $x$ so that $2.60.x + x^2$ is as close as possible to $4000 - 3600$
$2.60.x + x^2$ is the same as $x(120 + x)$ and that matches the 'one hundred and twenty something, times something, to be close to $400$' in the second stage of working in the method.
The answer was three. $123 \times3$ is $369$
If $63$ is the whole number part of the square root of $40 00$, what about the square root of $40 00 00$ ? (the gaps in the number help me think)
The whole number part will have three digits : six, then three, then something else.
Reasoning like before, I need to get $(630 + x)^2$ as close as possible to $400000$
Writing that without the brackets is $630^2 + 2.630.x + x^2$
From which I can see that I need $1260x + x^2$ to be as close as possible to $400000 -630^2$
Or rather, I need to find the digit $x$ so that $x(1260 + x)$ is as close as possible to $3100$.
This corresponds to the third stage of working in the method. And the answer is two. $1262 \times2$ is $2524$
Now for something important : I notice that although $3100$ is the number in the working, this method for finding a square root doesn't do $400000 - 630^2$ directly to get it.Image
The subtraction isn't $630^2$ straight off, that's $396900$, but rather an accumulation of subtractions : $36(0000)$ and $369(00)$, the zeros are needed to keep the correct column positions.
Let's look at all the subtractions shown in the working so far
: $36, 369, 2524,$ and $50576$
$123 \times3$ is $3(2 \times60 + 3)$ and came to $369 $
but $3(2 \times60 + 3)$ is the expanded bit of $(60 + 3)^2$ without the $60^2$
$1262 \times2$ is $2(2 \times630 + 2)$ and came to $2524 $
but $2(2 \times630 + 2)$ is the expanded bit of $(630 + 2)^2$ without the $630^2$
$12644 \times4$ is $4(2 \times6320 + 4)$ and came to $50576$
but $4(2 \times6320 + 4)$ is the expanded bit of $(6320 + 4)^2$ without the $6320^2 $
Teachers' Resources
Why do this problem?
This problem takes more able students into the realm of 'non-calculator' methods that lie beyond the arithmetic they became familiar with when they were much younger. It usefully draws attention to the need for validation in any algorithm whether carried out electronically or 'by hand'.Possible approach
- Spend a little time looking at the validity of the standard method for 'Long Division'. Discuss the historical need for efficient algorithms before electronic calculators, when computation was manual, and point out that calculators and computers aren't 'magic' and there still has to be a valid algorithm.
- Find some square roots of two-digit numbers to 2dp by trial and improvement.
- Spend time understanding what this new method involves (maybe use the audio link on the Problem page ,while keeping the working still on view), practise and then organise a time trial.
- Alternate between this method's algorithm and trial & improvement, for the square roots of 30, 50, 60, 70, 80, and 90, all to two decimal places. Record the calculation time for each one and compare methods.
Key questions
-
How do we find the square root of 40 on a calculator that only does simple '4 rules' arithmetic?
-
What exactly is the method here?