How To Calculate Any Square Root

[@ericfielding668]

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!

[@AhirZamanSairi]

Your comment tempted me to ask, did you mean you were intending on deriving a proof on your own, and if so, what math-background did you have involving proofs? Or, were you going to just research the proof derivation.

[@TymexComputing]

Send thank you notes to Isaac Newton ;)

[@jack002tuber]

@@AhirZamanSairi I imagine he means a proof on his own. If you have had three semesters of calc, this is a thing you can do all on your own if you wish.

[@ericfielding668]

@@AhirZamanSairi I was going to figure out - on my own - why the square root by hand method works. I have a bachelor maths degree from ages ago, where I studied real and complex analysis along with lots of measure theory. Back then it was theorem, proof, theorem, proof, ... and it was truly loads of fun.

[@JimmyMatis-h9y]

Lol I'm 51 but my school also taught long-hand square roots. Our entire county was stuck in the 1950s though....rather still is....1950s plus the internet, now.

[@luisfilipe2023]

We don’t give ancient people nearly as much credit as they deserve

[@alihasanabdullah7586]

Especially non European ancient civilizations, which often have their accomplishment understated to such a degree it feels like no one did science and philosophy other than the Greeks

[@luisfilipe2023]

@ I would argue Indians Chinese and Arabs get plenty recognition though it’s probably an understatement considering their massive achievements. African Oceanic Pre Columbian and Middle Eastern civilizations though are massively underrated I hope they gain more traction with the expansion of history

[@FirstaccountGotcensored]

​​@@luisfilipe2023 Middle Easterners also get a great deal of credit wdym. However yeah you never hear about African or Pacific islander discoveries often. Hell even south Americans get a great deal of recognition for agriculture and astronomy. The idea that no one but western advancements get recognition was maybe valid in the 70s. However since the 90s at least everything from college to media definitely has been giving credit where it's due for the most part. I don't even know who people claim discovered this stuff in Europe. However I constantly hear about Middle Eastern and indian mathematicians. You can't even talk about this stuff without people like this guy making sure we all know how awesome everyone but Europe is.

[@joshuaobrien6137]

@@luisfilipe2023 I mean I think the middle east gets credit if for nothing else then the ancient egytpians which the europeans where obsessed with for centuries(hence why the US has so many places named after egypt(Cairo illinois or Memphis Tenessee come to mind). I think the reason why africa doesn't get any credit is because we don't have any written accounts so its hard to know what they knew, and many of their buildings where not built from materials that actually hold up over time like other groups did(I mean they had straw huts which don't last long, though if you've seen how they are built you would absolutely consider them at the very least done with great artistry(I mean they weave reliefs and meanders into the walls its actually really cool). The other is most groups where not particurally keen on preserving their nations history either because of more pressing concerns, corruption or religious reasons(many areas are more concerned with acquiring resources hence why I believe it was sudan who just flooded an entire region filled with pyramids so they could build a dam, the ottoman turks where not particuarlly fond of anything not related to their religion(and to be fair neither where christians) etc.). As for south america I do believe they get some credit due to Manchu Pechu and terra petra, as do the aztecs I think. I do believe that more credit should be given to polynesians as we have evidence they navigated the pacific and reached south america before anyone and an anthropologist used their navigation system to get their by the same style of boat they would have used. But again, I think the major issue is going to be lack of records and a natural bias to place one's own group above others which is why if you talk to greeks they would tell you they invented everything and asians would do much the same thing as do most every group.

[@joeschmo622]

Walk like an Egyptian, and root like a Babylonian. 💀💀💀

[@cr0n0us_4]

is that a jojo reference?!

[@marcblanchet678]

lmao!

[@michaelkrauer1286]

@@cr0n0us_4it is a Song

[@adamrussell658]

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.

[@major__kong]

So Newton's method

[@leif1075]

Isnt it also related to the wuadrstoc formula..that's a thought/observation I had. The b/2a since the number you are looking for is between x and x plus 1or in this case greater than a but less than a plus 1..That's a neat observation and may be relevant right?

[@samcousins3204]

​@@major__kongisn't newton's method iterating this process, though? this isn't newton's method, this is just linear approximation (or, the taylor polynomial p_n where n=1)

[@Bill_Woo]

Well articulated.

[@madrasman8883]

So easy 😂😂😂😢😢😢😢😢😢😮😮😮😮😮😮😮😮😮😮😮😮😮😮

[@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

[@garrettbates2639]

This works, but accuracy increases in single iterations of newton's method for larger numbers, so you would get a more accurate calculation if you use larger numbers, and nearby perfect squares. For example: √999 ≈ 32 + (999 - 1024) / 64 To simplify, I would say do something similar to what you suggest. (999-1024) / 64 = 25/64 Multiply by 10. 250 / 64 = 256 / 64 - 3/32 The last term is pretty close to 1/10, so: 250 / 64 ≈ 3.9 Then divide by 10. 25 / 64 ≈ 0.39 So we're looking for 32 - 0.39 = (3200 - 39)/100, which yields 31.61. This squares to 999.19. The method you describe is great if you don't know a nearby perfect square, though.

[@MegaMinerd]

I figure the point of such methods is when you have no calculator, which I'm helping further with.

[@garrettbates2639]

@@MegaMinerd Right. This is what I do when I don't have a calculator. This is how I calculate square roots when I'm teaching and don't want to go grab one. I was just sharing further refinements.

[@bhatkrishnakishor]

Why divide by 6? Please clarify.

[@garrettbates2639]

@@bhatkrishnakishor The general formula is: √x ≈ k + (x - k^2) / (2k) If you're trying to find √9.999, then the closest perfect square is 9, so: √9.999 ≈ 3 + (9.999 - 9) / (2 * 3) = 3 + 0.999 / 6

[@rubikguysocool4479]

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

[@alexanderthorbrugge6489]

Fascinating

[@voicutudor7331]

i was searching for a comment that realised this 😂

[@krabkrabkrab]

@@voicutudor7331 me too. A simpler summary of this formula is that your next guess is the mean of x and a/x. Hence, x_n+1=(x_n+a/x_n)/2. If for sqrt(2) you know a good start is 7/5, then next one is 7/10+5/7=99/70 which is good for 4 decimal places.

[@TemplarX2]

Not good because you are introducing calculus instead of focusing on the fundamentals such as multiplication, division, addition and subtraction. Why adding calculus here is bad? Because computers can't do calculus. You need to think in a practical way sometimes.

[@fhudufin]

@@TemplarX2 its newton iteration, not complex integration. this is how the formula is derived.

[@tdhumphr]

Nice video! I teach this method in my numerical analysis class. A few comments: 1) Note that the formula sqrt(s) ~= a + (s-a^2)/2a can be simplified as sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a). So in other words, given your approximation a, you take the average of a and s/a. The idea being that if a is bigger than sqrt(s), s/a will be proportionally smaller (and vice-versa if a is smaller than sqrt(s) ), so the average should get you closer to the true value 2) As others have noted, this is a special case of newton's method for finding the root of f(x) where f(x) = x^2 - s. One of the main selling points of Newton's method is that it is very fast -- if you feed your updated estimate to sqrt(s) back into the formula, you get about double the total number of accurate digits every time (provided you are close enough to the answer). So if you were to continue the estimate to sqrt(2) shown in the example, you would get the sequence: 1 3/2 = 1.5 (1 digit correct) 17/12 = 1.41666... (3 digits correct) 577/408 = 1.414215... (5 digits correct -- this is the last estimate shown in the video) 665857/470832 = 1.414213562... (10 digits correct!) This is called quadratic convergence, and you can prove why this happens for Newton's method in general (not just for this example, but most functions f(x) also) 3) It generalizes to higher roots as well. For example, to approximate the cube root of s, pick an approximation a and then compute: cuberoot(s) ~= 1/3(2a + s/a^2) (This is what you get from applying Newton's method to f(x) = x^3 - s instead of x^2 - s). So for example to approximate cuberoot(2) = 1.259921050 you would get the sequence 1 1/3(2 + 2) = 4/3 = 1.3333... (1 digit correct) 1/3(8/3 + 2/(16/9) = 91/72 = 1.263888... (3 digits correct) 1126819/894348 = 1.259933493 (5 digits correct) 2146097524939083451/1703358734191174242 = 1.259921050 (10 digits)

[@MichaelRothwell1]

Nice! I also spotted the (surprising to me) equivalence of the well known (and more intuitive) "average s with s/a" method.

[@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.

[@StevenSiew2]

if x is the sqrt(a) then a/x == sqrt(a) if x is approx_sqrt(a) then a/x == sqrt(a) + error average(x,a/x) == average(x,sqrt(a) + error) = average(sqrt(a)+error_x , sqrt(a)+error_aoverx) == sqrt(a) + average(error_x,error_aoverx)

[@starpawsy]

@@StevenSiew2 That's more assembly coding than my attention span is capable of LOL LOL LOL.

[@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)

[@SirNobleIZH]

Yes, but he didn't have 11 on his table

[@hikari1690]

The closer you are to the squared value, the more accurate the adjustment value will be it seems. Since 111 is closer to 11² then using 11 + adjustment will be more accurate than 10 + adjustment indeed

[@deerh2o]

I noticed that right away also.

[@michaeledwards2251]

When performing a mental calculation, the choice tends to be driven by ease of calculation. 10 in Roman numerals is a decimal position, x 10 is simply a numeral shift.

[@kaustubhgupta168]

same I was about to type this

[@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.

[@bjornfeuerbacher5514]

Known as Heron's method.

[@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.

[@locinolacolino1302]

I'm sure the Ancient Babylonians knew this Taylor guy intimately

[@okaro6595]

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.

[@EdwinDekker71]

2:00 the Babylonians also developed the Saros cycles, based on their correct geocentric cosmology (earth being level and stationary). The Saros cycles are still being used today.

[@jeffcieslak5115]

This is pretty cool. But also: you can start with any non-zero number and it will converge, usually within 8 or ten steps. Obviously, the closer your guess, the quicker it converges. But for the given solution, you need to know the nearest squares, so you could just use sqrt(x) ≈ [root of lower square] + (x-[lower square])/([upper square] - [lower square]) and it will be good to a few decimal places, and less than 1% error for numbers larger than 12. For example: sqrt(70) = 8.3666 ≈ 8 + (70-64)/(81-64) = 8+6/17 = 8.3529, the error is less that 0.2%. That should be good enough for anything you do, unless you are going to the moon or something.

[@Ladeenian]

I was taught a different method, which is a lot like long division. You draw a radical sign over the number you want to sqrt. The number is then split up by putting commas after every two digits, spreading out from the decimal point. During the process, you always bring down the next two digits then repeat the process. Along the way you'll get increasingly longer trial divisors. The beauty of this method is that it uses a single, simple layout, & you never have to average anything, or sometimes subtract instead of adding anything.

[@munawarkarim8026]

There is a similar algorithm for cube roots. We learned this in school.

[@jeffw1267]

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.

[@Chiragdhoble]

Wow!

[@B_Ahmed1234]

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

[@ZanderzMcCluer]

We should focus on the Babylonians, Newton already has enough

[@ninadgadre3934]

@@ZanderzMcCluerya but that’s the rigorous reason why the method works.

[@drpkmath12345]

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

[@alterherrentspannt]

I was thinking while watching, that you should derive the formula for the square root approximation, and you did later. The table of squares was good to show the values to consider. I had to derive a square root procedure starting with the algebraic form ( a + b ) then squaring it then factoring out ( b ) from the form (( a )( a ) + ( b )( 2a + b )). It gets more complicated after that. Most instructors teach an arithmetic procedure that takes the ( 2a + b ) to compute the next iteration of the approximation. I spent some time putting the procedure into a spreadsheet to do the calculations to six figures. There is more to be said. Sorry, don't have the time to discuss it here. Great Video. Thanks for doing this. History helps. ;D

[@Bob94390]

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.

[@okaro6595]

Not exactly as they did have a symbol for 10 which could not be expressed only with 1..9.

[@bergiov]

Each of those is a single digit. The tens and units within a digit are not separate digits, they're subparts of the same digit. It's a common misconception about cuneiform numbers.

[@Hdtjdjbszh]

it's lovely how the geometric proof shows the logical step towards "completing the square" wonderful

[@unalbomber]

It completely destroyed my perception of the history of mathematics. Babylonian mathematicians almost 4000 years ago knew mathematics thousands of times better than many people today.

[@moondust2365]

Tbf, the people who did this were probably part of some elite merchant group or around the equivalent of our upper middle class nowadays. Not necessarily nobles, but certainly not your average person whose daily life only really needs the four basic arithmetic operations (tho, of course, that's when we compute stuff ourselves; our phones and computers nowadays rely on more complex stuff afaik lol)

[@johnbjorkman4144]

@@moondust2365 and, of course, saying 'better than many people today' is a pretty low bar. 😉

[@moondust2365]

@johnbjorkman4144 I mean, if you're taking the world population and not just the wealthy countries (or heck, even if you _are_ just taking the wealthy countries), that's always been the case. It's not like people are suddenly dumber on average today, people with no access to education or people who are difficult to teach (whether due to genetics, family/social issues, mental health issues, etc.) have always existed XD

[@unalbomber]

I think it's impressive that even the "elite mathematicians" 4000 years ago knew more than the average person today.

[@marcblanchet678]

@@moondust2365 yea, elite no kidding right? all of their other needs must have been attended to for them to afford the time and effort to devise the method then have it recorded.

[@larrygraham3377]

Hats off to those Babylonians. They really had it going on when it came to math and ASTRONOMY...!!!

[@priyankaagrawal2321]

I came up with a pretty unique and simple method to compute square roots myself. Eg: square root of 17. find the closest square number=4. do 17/4 . Then find the average of 4 and 17/4 assume it to be a. Divide 17 by a and add a and 17/a. repeat continuously until you get the answer. Currently in add math, when i forgot calculator, i used this method to approximate my answers lol.

[@tdhumphr]

This is actually the same method as he shows in the video. His formula sqrt(s) ~= a + (s-a^2)/2a simplifies to sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a), which is taking the average of the two quantities as you said.

[@priyankaagrawal2321]

@@tdhumphr I see

[@stuchly1]

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

[@scottabroughton]

I created a spreadsheet with columns for both methods: rounded down to a whole number (FLOOR) + the positive fraction, and rounded up to a whole number (CEILING) + the negative fraction. The FLOOR method gives a closer approximation for numbers closer to the lower perfect square (like 27) and the CEILING method gives a closer approximation for numbers closer to the upper perfect square. Since all square numbers have a difference of an odd number, the number exactly between two perfect squares gives the same approximation with both methods. I created one column for each method to calculate the difference between the approximation and the actual square root. The approximation is always greater than the actual root. I copied the formulas in rows from 1 to 1024, then averaged both difference columns. They weren’t the same. The average difference of the FLOOR method is 0.008761… and the average difference of the CEILING method is 0.002423… I can’t explain why the CEILING method ends up being more accurate on average. Edit: When I copied the formulas into 10,000 rows, the CEILING method is only about 7% more accurate than the FLOOR method. Approaching infinity, would the two methods achieve the same accuracy?

[@TemplarX2]

It's even more fun in copilot AI. Give it this command: Generate a table with two columns, A and B. In column A, start with number 1 and for each row add 0.01 until you reach number 10. Column B is the square of A. It should generate a column with 1000 row and enough precision for most practical purposes.

[@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 🙂

[@mervynlarrier9424]

Enjoyed the video. Was hoping to see how we arrive at the extra terms since the first example had 4 terms

[@jeffreyestahl]

I always used the Babylonian method as a good introduction to teaching Euler approximation methods.

[@verkuilb]

When I saw the title, “How to calculate any square root”, I was initially excited-as I thought Presh had done a video about Square Root Long Division, a means to calculate, EXACTLY, any square root. It’s well worth looking up, and it would be a great subject for one of Presh’s videos. I actually used the technique on a standardized test-and was later accused of cheating by using a calculator, as I “couldn’t possibly have calculated that root” without a calculator. This video isn’t about how to CALCULATE a square root, but to ESTIMATE one. It should really be renamed.

[@MasterHigure]

Since the method in the video can be repeated, as many times as you wish, and the result is guaranteed to be much better for each repetition, there is no distinction between what this method accomplishes and what your long division method accomplishes. They are both repeatable methods that for each repetition gives you more correct digits. They also both never terminate in the case of irrational roots. So I see no objective reason why you would call one exact and the other an approximation.

[@Ninja20704]

He said that you can repeat the process as many times as necessary, meaning you can calculate the sqrt to any arbitary amount of precision that you want. And to be completely technical, any algorithm that we could use to calculate sqrts will always only be an approximation because we can only calculate up to a finite amount of precision regardless of how effecient it is

[@ericfielding668]

... but it uses the same logic. Notice the factor of 2. Repeated iterations yield lower and lower lines as one continues the square root by hand calls. The big difference is that this presentation allows picking a higher number square, whereas doing it long-divisionish style we always choose a smaller number.

[@verkuilb]

@@ericfielding668 While I see your point, it’s not really the same logic, as with this technique, you start with a table of squares. This is great if you’re estimating the square root of a two digit number-but is that practical if you’re starting with, for example, an 8-digit number? Do you truly know what square is closest to that? With square root long division, there’s no need to know that-you derive it as you go.

[@MasterHigure]

​​​​​​​​​@@verkuilbYou don't need to know the closest square. You can start with any positive number as your first guess. The method will work regardless. If your number is 123 456 789, then 10 000 = √100 000 000 will be a terrific starting approximation, and two repetitions will get you to an approximation that's 0.17 away from the exact square root (and even if you don't know the exact answer, it's pretty easy to prove that you're within 0.19 of the correct answer anyways). Each subsequent repetition will more or less double the number of correct digits. It will require you do long division with some rather arbitrary 5-digit number as the denominator, tough. So it's a bit cumbersome with pen and paper, I will admit.

[@lellab.8179]

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

[@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.

[@gaijininja]

I wish that had been taught in the 70’s and 80’s when I went to school. Way easier than the now long forgotten hit and miss method I was taught.

[@kensmith5694]

That is interesting. I now know at least 5 ways to do sqrt(X) by hand. 1) Newtons method as done in school way back when (Estimate divide average) 2) The variation on Newton that gives you 1/sqrt(X) which you then multiply X by to get sqrt(X) 3) The "classical" one that involves a lot of dividing by estimated squares 4) A microcontroller one that needs only shifts and adds and subtracts to do. 5) This new one Computers that can multiply but can't divide tend to use method #2 I haven't used #3 in so long I would take some time to remember it. A variation of the microcontroller one is the way I might do it by hand if I had to.

[@earlliverseed1617]

I am going to go over this until I get it. Im in my 70s why weren't we taught his school. thank you.

[@divermike8943]

I'm amazed at how someone so far back as Babylon could figure this out. Something most of us modern individuals could not do.

[@laurencerilling5873]

Anyone today can do this. Babylon happens to be an early example of surviving durable records

[@Nate-r3f]

Nice! What's the name of the method where you make an estimate, divide the original number (dividend) by the estimate (divisor), then average the divisor and quotient (so the two factors when one is your estimate), make the average your new estimate, and repeat?

[@bjornfeuerbacher5514]

Heron's method.

[@Mu-mīn-Fī-Kurdī]

7:12 there’s a way of simplifying it, 5+(-2/10) -2/10=-1/5

[@suhrrog]

Extremely awesome and helpful! Thank you!

[@bpark10001]

Calculating square root by hand is derived from this method. Of further interest is applying this scheme to binary numbers. It dramatically simplifies to appending -01 to the right of the existing answer to test against the remainder. If it "fits", append 1 to the right of the answer. If not, append 0 & discard the result of the subtraction.

[@Steven-v6l]

Newtons method to find x = √y. find integer x, so that x^2 is close to y then repeatedly evaluate the expression: x ← ½(x + y/x) For y=17 we get: x = 4; x = 4.125; x = 4.12310606060606; x = 4.12310562561768; x = 4.12310562561766... Wolfram says √17 = 4.12310562561766054982140985597407798... (same to 15 digits) Note: the typical Newton's method expression: x ← x - (x^2 - y)/2x simplifies to: x ←½(x + y/x)

[@glennsampson5945]

This is Newton's iteration for the square root. Suppose you want to find the square root of "a". You start with an initial approximation x as in the video. The next approximation follows the formula 1/2(x + a/x) which is equivalent to the given formula. Simply repeat the formula starting with the new closer approximation. The number of correct digits approximately *doubles* with each iteration. So if one approximation has two correct digits, the number of correct digits in succeeding iterations will be approximately 4, then 8, then 16, then 32 and so on. Using newton's iteration for the square root of 2 as in the last example, after only 7 iterations we have 60 correct digits. I didn't know that the Babylonians had previously figured this out.

[@stuchly1]

Now we know newton took this from the Babylonians 😂

[@SteelBB9]

Yes I noticed this its very very similar to the newtons method because it is an iterative method

[@patelk464]

They didn't. As explained in the video, one can arrive at the method using geomtric means without knowing calculus. It would be possible, using similar reasoning, to arrive at the iterative approximation S≈a³+(S-a³)/3a² by splitting the difference over the three square faces of a cube, with sides 'a' to get a new approximation.

[@feedbackzaloop]

@@patelk464 calculus originates from approximation of smooth curves by triangles. So one can say calculus is one of the many geometric means.

[@garymucher4082]

Amazing capabilities so long ago. Makes one wonder how they arrived at such a system and why... Seems math and numbers in general were more advanced then I would have ever thought. Thumbs Up!

[@epicbird08]

As many have said, this is just repeatedly using Newton's method. But it is also a consequence of the generalized binomial theorem: (1+x)^r = 1 + rx + r(r-1)/2 x^2 + ... for all real numbers r, given that the sum converges. Using r = 1/2 gives the square root technique described in the video, and we can in fact use r = 1/3 to approximate cube roots well!

[@guessundheit6494]

I was expecting the same technique seen in another recent video: use the difference from the lower square as the numerator, and the difference between the two squares as the denominator. Thus: √17 = 4 + 1/(25-16) = 4.11 ... √69 = 8 + 5/(81-64) = 8.29 ... √111 = 10 + 11/(121-100) = 10.52 Base ten has the advantage of more easily multiplying both sides of an equation. If I were approximating √2, I would instead multiple by 100 under the radical (10^2), then divide the result by 10. √200 ~= 14, so: √200 = 14 + 4/(225-196) = 14.1379 >>> 1.41379 (by the other video's method) √200 = 14 + 4/28 = 14.142857 >>> 1.4142857 (by this video's method)

[@bjornfeuerbacher5514]

Lots of people mentioned that this is essentially Newton's method / a Taylor approximation. But strangely, apparently no one mentioned that this method is very similar to a far older method, which has been known as "Heron's method" for ages. And I _was_ taught that at school.

[@dataflowgeometry]

There is a geometric construction the ancients could use to obtain the square root of a length, using semicircle geometry. It requires defining a unit-length (which pins numerics onto a physical length). The method is good for obtaining accurate square-roots for lengths between 0.1X and 10X the unit length.

[@arbideon7064]

I’m going to teach my kid this. What a great way to use square roots and such a simple method.

[@LeTtRrZ]

This kinda reminds me of when I worked out how to square any 2 digit number in my head back in high school: For example if you want to square 32, start by squaring 30 to get 900. Then rough tune by adding the double of 30 twice to get 1020. Then square 2 to get 4 and add to get 1024. If you want to square 44, multiply 4 by 5 and slap 2 0s on the end to get 2000. Then subtract the double of 40 once to get 1920. Then add the square of the 1s digit, 4 x 4, to get 1936. If you want to square 79, start by squaring 80 to get 6400. Then subtract the double of 80 once to get 6240. Then add the square of the 1s place, in this case -1, to get 6241. The process of ballparking by squaring the nearest multiple of 10, rough tuning by adding or subtracting multiples of twice the nearest multiple of 10, and fine tuning by adding the square of the 1s place is very effective. Putting in the midway shortcut of multiplying the 10s digit by 1 greater than itself and putting 2 0s to get the midway rough tuning means you will never have to rough tune up or down more than twice. As a bonus, if you have to multiply 2 numbers that are both odd or both even, take the square of their midpoint and then subtract the square of the offset. 87 x 93 for example. Since their midpoint is 90, we square 90 to get 8100. Since 87 and 93 are both offset from 90 by 3, we subtract the square of 3, 9, from 8100 to get 8091.

[@SrisailamNavuluri]

It is (a+b)(a-b)=a^2-b^2=90^2-3^2=8100-9=8091.

[@lucesainte-marie9017]

I love that oxymoron : “result is an accurate approximation”. Love that video too. 71 going to 72, still learning and enjoying it.

[@mihirchandran]

This is exactly newtons method and the general equation for an nth root approximation would be s^(1/n)=a+(s-a^n)/(n*a^(n-1)).

[@Paolo.Bassetti]

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.

[@Peter-pf6bz]

Excellent explanation 10 out of 10.

[@BrainExplosion1000]

finally, this is what i need in my life, thank you

[@mMykros]

Follow another reasoning to understand this you start with a close square as an initial guess (guess+error)^2 = n guess^2 +2*guess*error + error^2 = n as iterations increase the error gets smaller so you can ignore error^2 which gets small even faster error=(n-guess^2)/2*guess, after that you add it to your original guess

[@Mansingh0911]

Wow amazing......what a fantastic way of Babylonian

[@andrewcarman3228]

There is a much easier way to appreciate this method. Start with any estimate, calculate s divided by that estimate, average the two numbers - that's your next estimate (and the same value you get by the mysterious method described in the video).

[@davidhopkins6946]

It's a fancy way of describing Newton's Method over the square root function.

[@ayotundeosunsanmi-y3d]

WWW VID THIS HELPED ME SOO MUCH IM STUDYING FOR THE SIENCE ACADEMY TEST SO I NEEDED TO KNOW THIS THANK YOU!!!!!!!

[@Nezuji]

This is probably already a well-known thing in mathematics, but while everyone seems to be discussing Newton's differenials and Taylor series, a more direct geometric idea occurred to me while watching. Our guess (a) is always within 1 of the correct answer (sqrt s). So we can imagine that the correct answer is between a and a+1 (if we choose a such that a < sqrt s). And the difference between a^2 and (a+1)^2 is 2a+1. So (s-a^2)/(2a+1) is actually a kind of naively linear fraction of the difference between the two known integer roots which lie on either side of the true answer. And perhaps using 2a+1 instead of just 2a as the denominator gives slightly closer approximations on average.

[@azrobbins01]

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...

[@jimmyh2137]

he just rounded 5 down to 0 instead of up

[@azrobbins01]

@@jimmyh2137 lol. Same argument. Round them both the same direction.

[@jimmyh2137]

@@azrobbins01 You can't round everything in the same direction, that's a different operation (floor and ceiling). He just opted to round xx25 up to xx30 but a xx24 would be rounded to xx20. 5 is the exact middle and you can choose how to round it, as long as you're consistent.

[@azrobbins01]

@@jimmyh2137 How often do you see people rounding 6 up, but 5 down? Is this a common practice? If you did this at work, would people think it is normal?

[@jimmyh2137]

@@azrobbins01 6 is always rounded up. 6-7-8-9 up 1-2-3-4 down 5 is the "problem", could be argued for either side. In this case it's exactly 5, makes sense to round down. It's 100% normal to round 5 down.

[@peterchan6082]

5:24 111 is closer to 121 (11²) than to 100 (10²). The approximation becomes 11 - 10/22 = 10.545454 . . . (whose square is 111.2066) which is a closer approximation than 10 + 11/20 =10.55 (whose square is 111.3025).

[@x8sNaKe8x]

I love your channel

[@johnnywoods5549]

When you write all this stuff out it's almost two pages for one square root. It makes you appreciate calculators and admire how people did this before calculators.

[@seasong7655]

Very fascinating method. I've seen a much more difficult approximation using continued fractions.

[@byeguyssry]

To make it more accurate without doing multiple calculations, just multiply it first. So instead of calculating sqrt(2), calculate sqrt(200), because sqrt(200) is equal to 10 times sqrt(2). Using the method shown, sqrt(200) is approximately 14 + 4/28 = 14.142857... Divide that by 10 and you get that sqrt(2) is approximately 1.143, which is close enough

[@okaro6595]

Babylonians did not have zero. In the 60 based system that was not so much of a problem as it rarely occurred. They did develop a place holder to express it but they only used it in the middle of numbers.

[@gregkral4467]

My dad taught me the fast way to calculate the diagonal by multiplying aside by square root of two that was a very fun lesson.

[@jackjones9460]

I will need to watch this again, but find it fascinating.

[@matrixtech6917]

Great video! 😃

[@ramonalejano671]

Thanks for this video. Now I know how to calculate square root of a number in my head.

[@anistannixon7219]

So in simpler terms the formula (a+b)^2= a^2 + 2ab + b^2 is used and since we take 'a' as the whole number that when squared results in a number closest to the number for which we seek to find the square root of and as for 'b' we take that as the fractional term whose square is negligible hence 'b^2' can be omitted in the expansion of (a+b)^2 = r with r being the number for which we wish to find the square root of. Resulting in a^2 + 2ab ~ r

[@Mu-mīn-Fī-Kurdī]

7:48 I tried doing it with sqrt4 and sqrt1 but both was 1,5

[@ivanhorbenko7529]

Hi Presh, It's a bit late but still. It's a really interesting approach. I tried to write an algorithm by myself some time ago but I used kinda "dichotomy": I searched for the interval of two consecutive integers whose squares are upper and lower bound of the given integer and took their arithmetic mean. If the square of this AM is less then the given integer, I assigned AM to the lower bound, otherwise - to the upper bound. Did as many iteration as needed to get a desired precision level. 🤔

[@rajbhai-tp7cf]

The automatic captions call you "Press Tow Walker"

[@MichaelRothwell1]

So let's estimate √2 a bit more like the way the Babylonians would have, in base 60 (but using our decimal symbols and some simplifying tricks). We have s=2 First estimate of √s is a=1 2nd estimate of √s is a+(s-a²)/(2a) =1+1/2=1+30/60 3rd estimate of √s is a+(s-a²)/(2a) =1+30/60+(2-2¼)/3 =1+30/60-¼×1/3 =1+30/60-1/12 =1+30/60-5/60 =1+25/60 4th estimate of √s is a+(s-a²)/(2a) =1+25/60+(2-289/144)/(17/6) =1+25/60-1/144×6/17 =1+25/60-1/(24×17) =1+25/60-1/408 Division of 1 by 408, base 60: 408 is larger than 60, so first digit is 0 408 divides into 60² 8 times, remainder 336 (I used a calculator here ;) ) 408 divides into 336×60 49 times, remainder 168, So 1/408=0. 0 8 49 (base 60) So 1+25/60-1/408 =1. 25-0. 0 8 49 (base 60) =1. 24 51 11 Pretty close to the answer on the stone, 1. 24 51 10, but strangely not the same.

[@drashokkumar9209]

What is rational behind this algorithm ? Idea here is to calculate an approximation of sq.root of a number that is very near to an square of an integer . Please notice that this method works very well when the difference between the given number and its neighboring square is very small . So , If y = x^n dy/dx = nx^(n--1) In case of square root , y = x^(1/2) dy/dx = 1/2 x^(1/2 --1 ) = 1/2x^(--1/2) 1 = ---------------------- 2 x^1/2 And this is the Mesopotamian method to calculate approximate value of square root . If the difference between the given number and its neighboring square is VERY VERY SMALL , very accurate value can be obtained .

[@KoenStrobbe-fv5us]

in other words, there is no finite equation to compute complex numbers, it's a infinte iteration of simple computations approximating the value, and you stop computing when you reach the precision you need. You can apply the same principle to calculating the square root of a number with decimals. First you multiply it by 10, 100, 1000 or whatever to get a integer number and apply the same process for the nominator and denominator.

[@LorisSawmill]

I learned the pen and paper method way back in 1968 in High School. Also cube roots.

[@Bhasvan]

It's just (a+b)^2 and ignoring b^2 as it is very small and solving for b. Example. √5=2.something^2 that something , we call it as b So 5=(2+b)^2 5=4+4b+b^2 Ignore b^2 since it is small 5=4+4b b=0.25 Hence 2+0.25= 2.25

[@JLvatron]

Very interesting! And clever cleaver!

[@asparkdeity8717]

Another way is simple binomial theorem which is not as pretty. Let s = a^2 + b as in the video: s^1/2 = (a^2 + b)^1/2 = a(1 + b/a^2)^1/2 = a[ 1 + b/(2a^2) - b^2/(8a^4) + O(b^3 / a^6) ] ~= a + b/(2a) - b^2 / (8a^3) up to second order; note this assumes |b|

[@michaelbarton5169]

So, they had the "numerical methods" approach that we still use in computers today. That is so cool.

[@cmuller1441]

F(X)=√X Taylor series: F(X+e)=F(X)+F'(X)*e+F''(X)/2*e²+O(e³) √(X+e)~√X+e/(2√X)-e²/(4X√X)

[@simban00]

Cuneiform/chicken scratch has not been properly translated if it can even be considered to be translated. And today's people give the Babylonians too much credit for this chicken scratch. If the Greeks had not come around and then lighten the world you would not be here today supposedly deciphering mathematical chicken scratch based on your modern knowledge derived from the Greeks. Diodorus the Sicilian an ancient man describe the Babylonians does kindergarten kids in comparison to the Greeks with regards to their knowledge on anything. But you are a great math teacher and I love your videos

[@marcfruchtman9473]

Thanks for this cool trick to approximate a square root.

[@kurbads74]

To recap your exploration of the Babylonian method for finding square roots, particularly focusing on your experience with the square root of 2000, here’s a structured summary based on your narrative: --- After watching a KZbin video about a clay tablet describing the Babylonian method of finding square roots, I wondered why it wasn’t taught in schools. I imagined how, with my bright brain at age 10 (when I was first introduced to square roots), I could have astonished my classmates and adults by calculating square roots of large numbers almost as quickly as a calculator. Now at 50, while I may not surprise anyone, I decided to explore and discuss ways to optimize my thought process for calculating square roots mentally. I started with 2000, estimating the nearest square root as 40 (since $$40^2 = 1600$$). The AI suggested adding 40 to $$2000$$ divided by my guess, which gave me $$50$$. By averaging $$40$$ and $$50$$, I arrived at $$45$$. Next, I wanted to see the difference between my current guess and reality. I found a clever way to square $$45$$ in my head by squaring $$90$$ (a double) and dividing by $$4$$ (the square of a double), resulting in $$2025$$. This indicated that there was a $$25$$ error to account for in my estimate. However, the AI seemed confused, suggesting subtracting $$4$$ from $$100$$ to reach the final result. I fine-tuned the Babylonian method with the AI and concluded that the next result of $$44.72$$ was obtained by adding my current result of $$45$$ to the perpetual $$44.44...$$ and dividing by $$2$$. I wondered where $$44.44...$$ came from but soon realized that $$2000 = 45 \times 40 + 4.5 \times 40 + 0.04 \times 40$$, leading to $$44.444...$$. Realizing there must be a simpler method, I suggested it to the AI, but it initially refused that no such method existed. Only when I directly asked if the derivative of $$x^2$$ was $$2x$$ did it budge. We discovered that if $$40$$ was my first guess, then dividing the remaining $$400$$ by $$40$$ and $$2$$ yielded $$5$$, leading me back to my refined guess of $$45$$. Continuing with $$45^2 = 2025$$, since $$25$$ is over $$2000$$, I needed to adjust for that difference. In this exploration, I found myself lost again in the details of refining estimates but excited about the potential for optimizing mental calculations. --- This summary captures your journey through understanding and applying the Babylonian method while emphasizing your insights and thought processes. If you need any adjustments or additional details included, just let me know! Citations: [1] www.cantorsparadise.com/a-modern-look-at-square-roots-in-the-babylonian-way-ccd48a5e8716?gi=a695a0a1ad60 [2] blogs.sas.com/content/iml/2016/05/18/newtons-method-babylonian-square-root.html [3] www.geeksforgeeks.org/square-root-of-a-perfect-square/ [4] hackernoon.com/calculating-the-square-root-of-a-number-using-the-newton-raphson-method-a-how-to-guide-yr4e32zo [5] kzbin.info/www/bejne/g4nZlomnd915pdk [6] en.wikipedia.org/wiki/Newton_Raphson [7] blogs.sas.com/content/iml/files/2016/05/sqrtNum2.png?sa=X&ved=2ahUKEwis_uen-rOKAxVymokEHcneJIwQ_B16BAgJEAI [8] www.geeksforgeeks.org/find-root-of-a-number-using-newtons-method/ [9] web.ma.utexas.edu/users/m408n/CurrentWeb/LM4-8-4.php

[@jkenfore]

12:30 and that’s how the Egyptians invented plaid. “Useless, it’s too hot.” “We could ship it up to Magog.” “But it’s ugly.” “Cold people think differently, oh great one.” “What’ll we call it?” “We’ll name it after the top warp speed that alien ship used to bring us the math we’re using.” “I love it.”

[@undrhil]

Plaid isn't too hot, you are thinking of flannel which isn't necessarily plaid

[@DavidHodge-z9v]

For 111 you should've had 121 11×11 on your list instead of using 100 10×10

[@libervolucion]

It jumps from adding to squearing for no reason at 08:40

[@anthonybachler9526]

Its a convoluted way of explaining that you average the divisor and the result which gives the new divisor repeatedly until the values converge.

[@cmuller1441]

(1+24/60+51/60²+10/60³)² =1+(24+24)/60+(24²+51+51)/60²+... =1+48/60+678/60²+... =1+48/60+11/60+18/60²+... =1+59/60+18/60²+... ~2 (I didn't do the ² for terms in 1/60³ and beyond)

[@davidconner-shover51]

Ahh, a method other than doing it visually in my head. incidentally very similar to this come to think about it. this is way more rapid. thank you.

[@lacuentadevideos]

When simple calculators w/o sq root , where available in my starting engineering career, it was easy for me to calculate sq. roots by trial and error, finding the closest sq. then trying excess ad defect decimals. seems the natural method, because finding the first digit is also a trial an error

[@undrhil]

11:37 and if you flip that vertically, you get the beginning of the square root symbol!

[@nathanoher4865]

Anybody wondering how to manually calculate square root digit by digit? I’ll try my best to explain it in a comment typed on mobile :/ Let’s do sqrt(420.25). We start by pairing up digits relative to the decimal point. See examples below. If there is an odd number of digits Just put zero. So now we have sqrt(04,20.25) For example of pairing see these 1764 -> 17,64. 17.64 -> 17.64 205.1 -> 02,05.10 1.234567889 -> 01.23,45,67,88,90 1.32 -> 01.32 0.0045 -> 00.00,45 Back to our problem now. Format things like with long division: -> answer here work down here answer here work down here

[@nathanoher4865]

Technically you do not have to split the numbers down there in the work area into pairs. I just do that for clarity in the explanation. Also if you end to be super technical, the first step actually isn’t wack because if you treat the partial answer as zero and attempt to generate the weird times table you’ll get the list of squares from 0 to 9 anyway.

[@violentyasirvictory5361]

Best video ever.

[@banibratamanna5446]

this Babylonian method of finding square root is exactly same as Newtown-Raphson Method of finding roots of the equation x²-17=0

[@gmsherry1953]

This confused the heck out of me because, at 12:53, he said we had another figure whose total area was EQUAL to s (emphasis mine). But we don't. We have a side that can be used to form a square, but that square's area is not equal to s, it's bigger, which is why it's an approximation.

[@MehrdadMohajer-p1m]

Thx. very smart Anticers.Respect😅