Square roots with pencil and paper: method 2

A little while ago I wrote about the Babylonian method for approximating square roots with pencil and paper. In that post I noted that the Babylonian method is quite efficient, but annoying in some ways since it makes you deal with big fractions and/or long division. Today I’m going to describe another method for computing square roots with pencil and paper which is less efficient, but explicitly deals with the decimal representation and hence more convenient in some ways.

Here’s how it works. Let’s say we’re trying to find the square root of 7. First, we write 7 as 7.00000… and group the digits by twos, like so:

7.\;00\;00\;00\;00\dots

The first step is to write down the first digit of the solution. This is easy: 2^2 = 4 (too small), but 3^2 = 9 (too big), so we write down 2, then subtract 4 from 7 and write the remainder, and bring down the next two digits (in this example, 00).

Now, here’s the part that seems like voodoo, but don’t worry, I’ll explain why it works later! For now I’ll just stick to the how. Our answer so far is 2. We double it, getting 4, and now we look for the largest digit d for which 4d \times d < 300. I'm abusing notation a bit here; by 4d I don't mean 4 times d, I mean “the number you get by pasting the digit d after the digit 4″. Some experimenting shows that 46 \times 6 = 276  300. So 6 is the digit we are looking for, and it’s the next digit in the answer. We write 6 along the top, and subtract 276 (which is 46 * 6) from 300, then bring down the next two digits:

Now we repeat the process: we first double the answer so far (26) to get 52, then we look for the largest digit d for which 52d \times d < 2400. Again, some experimenting shows that 524 \times 4 = 2096 and 525 \times 5 = 2625, so the digit we are looking for is 4. We write 4 along the top, subtract 524 \times 4 = 2096, and bring down the next two digits:

I’ll carry it one more step: twice 264 is 528, and 5285 \times 5 = 26425 < 30400 (whereas 5286 \times 6 is too big), so we get

So far, we have computed that \sqrt{7} \approx 2.645\dots. And indeed, we can check that 2.645^2 = 6.996\dots, but 2.646^2 = 7.001\dots So, why does this work? Well, suppose we have worked out so far that

a^2 \leq 7,

and we want to find another digit in this approximation. First, we can multiply both sides of the equality by 100:

(10a)^2 \leq 700

Now we want to find the value of b so that

(10a + b)^2 \leq 700

which will give us one more digit of precision. Expanding out the binomial on the left:

100a^2 + 20ab + b^2 \leq 700

Factoring and rearranging a bit:

(10 \times (2a) + b) b \leq 100(7 - a^2)

7 - a^2 is the remainder; it’s what’s listed at the bottom under the most recent subtraction. The factor of 100 corresponds to bringing down the next two digits. And what do we want on the left? We want a value of b for which (10 \times (2a) + b) b is less than this remainder. This looks familiar! The 2a is where we multiply the current answer by two, and then 10 \times (2a) + b corresponds to appending the digit b to 2a.

So, which is better, this method or the Babylonian method? I don’t know, they are very different. I can think of situations where I would want one or the other. I guess I’ll leave the determination up to you!

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8 Responses to Square roots with pencil and paper: method 2

  1. JM says:

    Do you know where this method come from?

  2. Brent says:

    No, I don’t! It seems to be sort of folklore. If anyone can figure out where it comes from I’d love to know!

  3. JM says:

    I did come across a third method of finding a square root in a handbook on how to use a Chinese abacus, though it can probably be done on paper too. Unfortunately, I only know how to use this to find the square roots of perfect squares.

    As in your method, divide the number into groups of 2 digits, so 1089 would be 10 89. There is another number involved called the root number, which starts at 1. First subtract the root number from the left hand group. Then add 2 to the root number and repeat. Doing this with 1089, we get 9 89, then 6 89, then 1 89. Now the root number is too big, so we multiply the previous root number (5) by 10, and then add 11. We take the next group of 2 and continue. The root number is now 61, and we have one group of 189. If the new root number were still too big, we would bring down another group of two and multiply the new root number by 10 and then subtract 9. In this case that would yield 601.

    Anyway, 189 minus 61 is 128. 128 minus 63 is 65. The next root number would be 65 too, so we’ve finished. Add 1 to 65, and then divide by 2. This gets us 33, which is the square root of 1089. This method can be used for numbers of any size, though it becomes increasingly harder as the numbers get higher.

    I suppose you could use this for non-perfect squares: for every digit past the decimal point you want to know, add 2 zeros to the number you want to find the square root of. Then, stop when the root number is bigger than the other number, and the final result (with the decimal point moved over the appropriate number of places) will be a good approximation.

  4. TwoPi says:

    You might find this geometric approach interesting; it is essentially the same process as you’ve explored here, but using geometry as a foundation for the process. (There’s also an extension to cube roots….)

  5. Brent says:

    JM: fascinating! I’m pretty sure that your method can be seen as equivalent to the one I’ve described, perhaps in a bit I’ll write about the connection.

    TwoPi: Cool, thanks! I now remember reading that when you posted it, but had forgotten about it. =)

  6. Kevin says:

    Nice article. I’ve never that method. Like to give it a try now that I’ve got some time.

  7. Scott says:

    Very informative post. I saw a teacher blow through how “he used to do square roots” in class once, but didn’t explain how it worked. Very cool stuff. I’m an aspiring math teacher at the moment at college park, and I think this kind of stuff is much more interesting than what I’ve been learning in linear algebra this semester….

  8. asmeurer says:

    Well, the Babylonian method converges double exponentially (not quadratically, as you say in the other post!), which means that you get roughly twice as many correct digits after each iteration. I haven’t analyzed this method, but I am sure that you can’t do much better. Also, the Babylonian method is what computers use to compute the square root. There are also probably some good ways to optimize the Babylonian method for rough pencil and paper (or head) calculations.

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