How To Calculate Any Square Root

An ancient formula is astoundingly accurate. Why didn't we learn this in school?

0:00 history
2:32 calculations
7:22 sqrt 2
10:19 intuition

Babylonian tablet YBC 7289
Babylonian numerals
Babylonian Method of Computing the Square Root, Olga Kosheleva
A 3D-print of ancient history
tecmath video square root
Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context, David Fowler, Eleanor Robson
Math! Science! History! It Didn't Belong to Pythagoras

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[ericfielding]

My parents (now in their eighties) were taught the square root by hand method in grade 4. I learned it in my 40s when reading a book on abacus calculations. I wasn't quite sure why the algorithm worked and meant to derive a proof eventually. What you presented here is essentially the same algorithm. So now I don't have to figure it out. THANKS!

[MegaMinerd]

For large numbers, divide by 100 before and multiply 10 after.
√999?
√9.99 ≈ 3 + .99/6
√999 ≈ 30 + 9.9/6 = 31 + 3.9/6 = 31.65
Actual value 31.6069

[adamrussell]

Here is another way of looking at the process. Think of the graph of y = x^2. It has a point on the curve (x, 17) and x is of course the square root of 17. We know that (4, 16) is pretty close. Mentally draw a tangent line at that point. The derivative of Y= x^2 is 2x, so the slope at our initial guess is 2*4=8. Delta x = delta Y / slope = 1/8. So the 2nd approximation is 4 + (17-16)/8.

[rubikguysocool]

This is the direct application of Newton's method to find the zero of a function:

Let f(x)=x²-a, so that we want to find x such that f(x)=0 <=> x²=a
we then have x_n+1 = x_n - f(x_n)/f'(x_n)

It becomes x_n+1 = x_n - (x_n²-a)/(2*x_n), which corresponds to the algorithm described in the video

[B_Ahmed]

You should mention that is exactly what you get if you apply Newtons' method.

[Bob]

Thanks for showing the table at 0:33. It shows clearly that the numbers from 1 to 59 were represented as two digits, where the leftmost digit shows the number of tens and the rightmost digit shows the number of ones. I.e., these numbers are represented very similar to the way we do, except that they didn't use a symbol for zero, they used different sets of digits for the tens and for the ones, and their maximum number of tens was 5.

[danmerget]

The method I learned in grade school is to take the average of a and s/a. This gives the same result as Presh's "a + (s - a^2)/(2a)", but I find "(a + s/a) / 2" to be more intuitive. If "a" is a bit low, then s/a will be a bit high (and vice versa), so halfway between them will be closer than either of them.

[MasterHigure]

Many have pointed out that this is what you get if you use Newton's method (a.k.a. Taylor approximation of order one). It is also the result of the following: Let x be a first approximation of √s. Then we would very much like to calculate the geometric mean of x and s/x, as that is exactly √s. But we can't. What we CAN do is to calculate the arithmetic mean. And that yields exactly this method.

As an added bonus, both Taylor approximation and the mean method have natural ways to bound their respective errors. Taylor has Taylor's theorem. And we know the geometric mean lies between the arithmetic and harmonic means.

[jeffw]

If you have all the two-digit squares memorized, you can add two zeros to the radical and make a better estimate. I would change 69 to 6900, and since I know that 83^2 is 6889, I know that my approximation will be 83. plus 11/83+84, or 11/167. So I have 83.0. I can continue calculating by trial and error, doubling the number of significant digits each time.
Once I calculated a square root to twenty-four significant digits, all in my head, which took me about three days.

[kappasphere]

I've implemented this algorithm a couple times before, but I never thought of it as something that can be calculated in your head. You have changed that today 🙂

[drpkmath]

Very nice! This reminds me of high school days! Enjoyed very much haha

[adaschma]

Isn't 111 closer to 11² = 121 than 10²=100? at ~5:54; So sqrt(111) ~~11 - 10/22 = 11 - 5 / 11 = 10.545454... which is obviously closer to 10.536... (sqrt(111)) than 10.55 (the approximation given in the video by starting with 10²)

[stuchly]

I have enjoyed this video thoroughly. Thank you very much for sharing.

[pjplaysdoom]

Another way to get a better value to approximate root(2) is to observe that root(98) = root(49)*root(2) = 7*root(2).
Using the method, root(98) = approx 10 - 2/20 = 99/10. Dividing by 7 gives root(2) = approx 99/70 = 1.4142857... which is already a lot better than 1.5.
Edit: If you use root(200) = 10*root(2) and 14*2 = 196 we get root(200) approx = 14 + 4/28 = 14.142857... so dividing by 10 for root(2) gives the same approximation as the root(98) example.

[okaro]

This is also essentially the same as the Newton's method. The 2 in the bottom can be explained by the fact that the derivative of x² is 2x.

[starpawsy]

I coded in Intel assembly a version of the Newton Raphson method. For the square root of a, take an estimate x. Then take a/x and average it with your original estimate. That is, a' = ( a + (x/a)) / 2 . Repeat until you are close enough. This method converges rapidly.

For the initial estimate, the Intel processor is very helpful. Floating point numbers are stored in quasi-scientific notation according to the IEEE standard. All you do is divide the exponent by two (right shift 1) and remove the 1 from the first bit of the mantissa. This is close enough for the method to converge rapidly. This method is about as fast as the supplied C library sqrt call, which uses a completely different algorithm that doesnt require a hardware floating point divide function.

[platypao]

as many pointed out, this can be seen as an average between the estimate (x) and S/x; an easiest way to visualize it is to consider these values as sides of a rectangle, having the same area S of the original square, and after every iteration the two sides converge to a common value, becoming a square.

[azrobbins]

I take exception that you are choosing to round some values and not round other values to make it seem like they were more accurate than they actually were.
For example, at 5:09, you show 8.312 and 8.307. You did not round the first number up, but you rounded the second number up. It should be 8.3125 and 8.3066, and if you rounded them both, would be 8.313 and 8.307...

[scotty]

Our past ancestors were so smart! The ways they went about solving problems show that thinking outside the box is key, not following some rigid process.

[lellab.]

I was taught how to calculate the square root by hand in middle school. I still remember it and use it to these days.