# Computing Square Roots on Paper

To do a square root on an abacus, you use partitions to do a paper algorithm for square root using the abacus. The catch is that most people don’t even *remember* how to do square roots on paper, if they ever learned it at all. (In fact, in school, *I* didn’t learn the classical paper algorithm; we never really did roots on paper; the closest we did was using square root as an example of Newton’s method. Like so much of my basic math, I learned this from my father.)

So, for your entertainment and edification, today, I’ll describe the classical algorithm for computing square roots on paper. I’ll also show you just why it works.

Suppose you want to take the square root of *n*. Start by writing down *n*. The algorithm is going to work using *pairs* of digits, so you need to divide the number into *pairs* of digits. A pair *can’t* cross the decimal point, so start at the decimal, and break into pairs going both right and left. If you don’t have enough digits, you pad out the ends with zeros.

So let’s do that much with an example, Let’s say we wanted to take the square root of 513.5. The pairing would look like:

```    05 13. 20
```

Now, start with the first pair, and approximate: what’s the single digit that is *closest* to being the square root of that pair? Write that down *above* the first pair. Then take the square of that digit, and write it *beneath* the pair, and subtract.

So with our example, the first pair is “05”. The closest single digit approximation of the square root is “2”. So we’d write “2” above the “05”; and “4” beneath it, and subtract. The number on *top* of the radical is the current root estimate, which we’ll call *est*; the result of the subtraction is the current remainder.

```	      2
/--------
\/ 05 13. 20
4
----
1
```

That’s the setup. Now we get to the part where we iterate. We’ll take the next pair from the number we’re taking the square root of, pull it down, and write it next to the current remainder. The current remainder concatenated with those two digits is the *target number* for this iteration, which we’ll call *t*. What we want to do in the iteration is find a number *x* for which (20×*est*+n)×n ≤ t. That number *x* will be the next digit of the square root. So we’ll write *x* on top of the radical as the next digit; and subtract (20×*est*+x)×x from the target.

So we take the *est*, multiply it by 20; and write it down next to the current target:

```	      2
/--------
\/ 05 13. 20
4
----
(40)  1 13
```

Now, we need to approximate again. What’s the largest *x* for which (40+x)×x ≤ 113? 3×43=123, so 3 is too large. 2×42=84. So the next digit is two. So we write 2 on top of the radical, and subtract 84 from the target. Then we pull down the next two digits, and repeat.

```	      2 2
/--------------
\/ 05 13. 20
4
----
(40)  113
84
-----
(440)  2920

```

So we want to approximate: what number *x* is the largest *x* such that (440+x)×x ≤ 2920? That would be six. There’s one special thing about this step: this is were we cross the decimal point in the original number. So we copy the decimal point up above the radical; the next digit of our root is going to be after the decimal.

```	      2 2 . 6
/--------------
\/ 05 13. 20
4
----
(40)  113
84
-----
(440)  2920
2676
------
244

```

If we wanted to keep going, we could just add another trailing pair of zeros under the radical. Let’s do that once.

```	      2 2 . 6 5
/--------------
\/ 05 13. 20 00
4
----
(40)  113
84
-----
(440)  2920
2676
------
(4520)  24400
22625
-------
1775
```

We could keep going if we wanted. But that’s enough for now. To quickly check our result, what’s the square of 22.65? Whipping out my slide rule, I find that 22.65×22.65 is almost exactly 513.

——–

It’s pretty easy, and quite fast if you don’t want too many digits. As you compute more and more digits, figuring out the approximation starts to get a bit harder, because you’re using more and more digits in your target.

The obvious question is, why does this work? It’s actually pretty simple, and remarkably clever. It’s just another one of those places where you do a bit of algebra to see what’s going on.

At each step, we’re breaking the number *n* we’re taking the root of into two parts, *a* and *b*, such that ((a×10)+b)2 = n.

So, let’s expand that. (10a)2 + 20ab + b2 = n.

So, first step, we get a value for a. Then we compute an approximation of *b* using “20ab + b2≤n-(10a)2“.

That gives us a better approximation of the square root. We now repeat the process, using the *better* approximation as the new value of *a*, and approximate *b* using the inequality.

This is long enough that I’m not going to try to do it on the abacus as part of this post; I’ll show you how to do this algorithm using an abacus tomorrow.

1. #1 MiguelB
September 28, 2006

What we want to do in the iteration is find a number x for which (20×est+n)×n ≤ t.

Hm… you don’t actually use “x” in the equation? Don’t you mean “find a number n for which…”?

Very interesting post BTW. I learned this algorithm in the last year of elementary school, but never really had to use it afterwards.

2. #2 Barry Leiba
September 28, 2006

Corrections to some typos, and a clarification:

Typo: 513.5 should be 513.2

Typo: “find a number x for which (20×est+n)×n ≤ t” should be “find a number x for which (20×est+x)×x ≤ t”

Clarification: “what’s the single digit that is closest to being the square root of that pair?” … You mean “closest, without going over,” yes? If what we had was “80”, we’d pick 8, not 9, though 9 is closer.

3. #3 John Johnson
September 28, 2006

I actually did use it once. I forgot my calculator for a geometry test.

4. #4 Drekab
September 28, 2006

4th paragraph, 513.5? I was curious about how to do roots by hand a while back and found another algorithm, which was basically take a few guesses and average them out and see what you get. This way is a lot more fun and more rigorous, thanks.

5. #5 Ahcuah
September 28, 2006

Way back when, when I was in high school, I found out about this method from the Encyclopedia Brittanica, which not only said how to do it, but, as you did, also gives the reason for why it works.

After playing with it a while (and still remembering the algorithm, ahem, 35 years later), I immediately generalized it to extracting cube roots.

6. #6 Stephen
September 28, 2006

Eventually, i’ll get to square roots on your fingers. I’ve finished addition and subtraction, and plan multiplication soon.

What kind of abacus do you use? Soroban? Suan Pan?

Planning on describing Newtons anytime soon? I find Newton’s
quicker than the way you described (and which i was taught).

7. #7 dan
September 28, 2006

I remember learning how to do this in highschool in the middle 80’s… I also remember saying… isn’t this why they invented calculators… The teacher didn’t think it very funny however…

I also learned how to convert binary, decimal, octal, and hex to each base using paper and pen… (but that’s another story!!)

Cheers,
Dan

8. #8 Jokermage
September 28, 2006

This would have been so cool to learn in high school. I feel ignorant for not already knowing this trick.

I notice you can do this even with going over the target. You end up with a negative remainder, but the next iteration will correct the “overflow”. I tested this by estimating the square root of 2401 with an initial estimate of 5. The remainder ends up being -1. The -1 becomes -99 when you bring down the 01. You end up with (5*20)+(-1) * -1 = -99, which zeroes out the remainder. The column estimates are 5 and -1, which becomes 50-1 = 49.

It is a lot messier if you go over the target, but it still works.

9. #9 allan
September 28, 2006

I had a high school math teacher teach me this, simply because he disliked our “newfangled” calculators.

10. #10 Mark C. Chu-Carroll
September 28, 2006

Allan:

I agree with your math teacher 🙂

One of the things that my father taught me about math was that nothing makes you understand things like being able to figure it out yourself. I don’t really think that anyone should really be doing square roots this way by hand – there really *isn’t* a reason to do it by hand in practice when you’ve got a calculator. But *knowing how* to do it by hand, and understanding how and why it works – that’s an incredibly valuable thing. And if you only ever do your math by calculator, you’ll never get that deep understanding of how the numbers all fit together.

That’s why I love things like sliderules and abacus; because they help make math *tactile*. When you do math using either of them, you need to *understand* what you’re doing. With a calculator, it’s too easy to wind up just punching numbers in and getting an answer that means nothing to you. With a slipstick or an abacus, the numbers *mean something*.

11. #11 Mark C. Chu-Carroll
September 28, 2006

Stephen:

I like the chinese suan-pan. I find the extra beads helpful, both for some of the five’s complement tricks that they let you do, and for things like delaying carries. My biggest problem in a lot of manual calculation is that I have a tendency to lose my place; being able to stall a carry can really help me stay focused on where I am, and not drift off.

12. #12 Daniel Martin
September 28, 2006

See, now that’s much faster than the way I was taught, even if it is going to take a bit of practice to commit to memory properly.

What I was taught is basically Newton’s method as applied to square roots: to find the square root of t, make a guess x and repeatedly take the average of t/x and x as the new value of x. Stop when boredom sets in.

13. #13 Andres
September 28, 2006

Oh, thanks a lot, Mark! I shall print this and hang it on my wall. You know what? I’m a mathematician. Both my parents were math teachers. The only math they ever transmited to me (other than through their genes) was this, one time each. After both explanations, I forgot it. Know I’ll make sure I don’t, and who knows, perhaps some day I’ll explain it to a child… who will forget it, of course, so I hope your archive will still be around!

September 28, 2006

I concur with Mark on understanding, and I also think that the algorithmic approach that computers force us to explore is good here.

Until Newton’s method I’ve never saw any method of constructing squares. That type of education made me uninterested in constructive methods and interested in abstract methods. This uninterest made me (and still makes, when I forgot that shortcuts is the easiest way to get lost 🙂 miss the most obvious things.

For an easier approximative method, see Tommaso Dorigo’s tale of learning to construct squares. ( http://dorigo.wordpress.com/2006/09/12/me-a-genius-at-five/ )

15. #15 coturnix
September 28, 2006

Thank you for this. I learned this in 7th grade (yup, the glories of Eastern European education), but I forgot it. I always wanted to re-learn it. Now I have no excuse not to.

16. #16 Stu Savory
September 29, 2006

Just FYI, I have a page on how to get square roots geometrically, using compasses and straightedge only.

It is here http://www.savory.de/maths9.htm

Stu Savory

17. #17 AJS
September 29, 2006

This brings back a few memories! I did my maths O-level a year early: this was the last year that there were separate calculator and non-calculator papers, and we did the non-calculator one. I have seen the above method done once, but never managed to replicate it myself. Thanks for the explanation, it really makes sense now!

I also like the ruler and compass method. I’d forgotten that …..

Am I correct in thinking that Newton’s Method for finding square roots works on the basis that the nearer two numbers are, the closer will be the arithmetical mean and geometrical mean? i.e.
Lim [δx→0] sqrt(x(x + δx)) = x + .5 * δx
?

18. #18 Marlayna
September 30, 2006

Wow. I’m a student in electrical and computer engineering, going into my 4th year, and I this is the first time I ever hear of this method. And such a simple explanation, too! It only takes junior highschool algebra to understand why this works! No matrixes, no limits, no differentials, no integrals, no vectors, blah, blah… just simple highschool algebra. Why wasn’t I ever taught of this? Not even in my computational analysis class…!

Thank you Mark, that was way cool. 🙂

19. #19 Alon Levy
October 1, 2006

AJS, your limit isn’t precisely true, since the limit of SQRT(x(x+dx)) is just x. What is true, though, is that the limit of (SQRT(x(x+dx))-x)/dx is 0.5.

20. #20 Xanthir, FCD
October 1, 2006

Alon Levy: I’m not sure what you mean. The limit of (SQRT(x(x+dx)) is indeed x+.5*dx, as dx approaches 0. At the limit, yes, it’s x, but that’s clearly indicated in the formula that AJS provides.

21. #21 Jonathan
October 1, 2006

I too learned this algorithm in high school (late 70’s or early 80’s, thought it was neat, but never used it and forgot it).

Is the history of this algorithm known? Is it named?

I’m finding it very easy to use. I think when I teach my algebra students to find (a + b)^2, I will wait a week or so, then sneak this in as a reenforcing application.

Thanks

22. #22 GIBSON
December 13, 2006

Yeah even I came across this algorith in my school times.Initially I find it tough but later It became the easiest procedure for me to adopt.Please suggest me some online math game resources.