How to find the square root of a decimal number without a calculator

CuriosityStream: curiositystream...
This video follows on from my previous video where I reviewed an exam from 1866. Finding the square root of a decimal number by hand is not so tricky once you learn this algorithm, but is a skill that is not often taught since we can do it so easily on a calculator these days. My previous video: • This is what a math ex...
More about finding the square root: www.homeschool...
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[@cynodont7391]

I have no doubts that the proposed method (or a similar one) was known at the time of the exam but remember that when an exam question looks difficult then there is probably one or more tricks that can be used to simplify the problem. In that case, the tricks are (1) 0.0043046721 can be rewritten as 43046721 * 10^-10 so the result is the square root of 43046721 multiplied by 10^-5. (2) Look for obvious divisors of 43046721 so 2, 3,, 5 and 9. In that case 9 is a divisor because the sum of the digits 4+3+0+4+6+7+2+1=27 is a multiple of 9 . Compute 4304672/9 = 4782969 (3) Repeat step 2 several times to figure out that 43046721 is actually equal to 9*9*9*9*9*9*9*9 so its square root is 9*9*9*9 = 6561. Remark: After 2 or 3 iterations, a clever student could probably have the intuition that more 9 will follow and then speed up the process by dividing by 81 or any other power of 9. (4) The final result is 6561 * 10^-5 = 0.6561

[@johnpotjewyd5320]

It is simple to calculate the square root without using your method. I cannot show it here because you need a graphic box. The answer, of course, is 0.06561. You take the digits 2 at a time. The demonstration in the video is close to how it must be done. We learned this in the 1960s. The algorithm is not helpful.

[@deweycd23]

CynoDont your answer is off by a factor of ten. John's is the correct answer. This (factoring) is the method that a young, modern North American student would use to solve this question, although a younger student may convert to a fraction first to avoid having to worry about negative exponents.

[@TheZenbudda]

M.Dew he just failed to move the decimal over 5 places. Otherwise it’s correct

[@trueriver1950]

Be careful with this kind of shortcut. The school examiner expects you to display skills taught in class. If classes covered your method that would be great: if not then you would score close to zero. Only at university level do you start to get credit for knowing an easier method. Your method would be great in a maths competition but not in a school exam.

[@boltzmann1688]

@@trueriver1950 I understand... but I think it is stupid to use one and only method taught in class. It makes students less independent. Since it is mathematics, teachers should give more points for alternatives. You need flexible and independently thinking people in mathematics, then and only then you feel the beauty of mathematical problems. Btw, his approach is correct, he just wrote the final answer wrong, although his previous steps are correct

[@stevethecatcouch6532]

Nicely done. I admire that you were able to go from complete ignorance of the algorithm yesterday to sufficient understanding of it today to explain the derivation and confidently go through an example.

[@Ebani]

Since it's elementary math it shouldn't be hard, at all.

[@e4r281]

If you put root beer in a square glass, do you get beer?

[@tibees]

e4r 🍻

[@Escap1st7]

Nop, you'd need a cube root beer, unless you're drinking 2D beer.

[@vk2ig]

@@Escap1st7 I find 2D beer a bit flat ...

[@williammccoll5563]

I liked your joke.

[@satrickptar6265]

Lol

[@dolfinmagikpro]

After seeing you go through this, I now remember having learned this in one of my algebra classes, which we never used again. It was a "Rainy day mini lesson" of sorts. It sure helped having passionate teachers back then! Thank you!

[@mjones207]

Two thoughts... first, my algebra teacher taught me this method in 1976, so cheers to Mrs. Simon. Second, I recognize the 6561 in the answer as 81^2 or 9^4 or 3^8 'cause I'm old school and know my logarithms, but I wonder if advanced arithmeticians in 1866 knew powers up to 3^16, would have automatically seen the 43046721 in the problem, and wrote down the answer very quickly.

[@quantum7401]

My Calculus instructor was Mr. Simon

[@pbierre]

@@quantum7401 We had a substitute teacher one day in 5th grade -- she endeavored to teach the above method. It went over my head at the time, but I never forget that a manual method existed. On your 2nd point, if you know up-front that the square root of n is an integer, does a prime number factorization strategy make sense? If it's a small number (< 13*13*11*11) it will likely work. Where it gets into trouble is numbers embedding squares of larger primes -- you could run out of gas. Whereas, the mechanical algo taught here can handle a problem like 48841 (if you can do error-free arithmetic).

[@peterallam6494]

When were log tables 'invented' ? Who by ?

[@ZipplyZane]

It's usually easiest to get closer to the right number in the (2x+r)r stage by simply diving r into the remaining value. So, when you had 704, you can see that 12*5 = 60 and 12 * 6 = 72, so 5 is the largest the number can be. No need to multiply out 125 * 5 first. Note that this is only an estimation, but it's good to help you find a starting point. It's generally going to be either the number you get here or one lower.

[@pratik3106]

I really enjoy your channel, I gave JEE 3 years ago, did fairly well but my chemistry was very bad. I like how exams abroad aren't very different from each other, especially university ones. Thank you for the great content!

[@joepike1972]

On the cube root question there are two points. 1) Provided the cube is perfect the ending digit will tell what digit is in the ones place; If the ending digit is 9 the ones place will be 9, 2 for 8, 3 for 7, 6 for 6, 5 for 5, 4 for 4, 7 for 3, 8 for 2 1 for 1, 0 for 0. With 389017, the ending digit is 7 so the ones digit will be 3 if this is a perfect cube. 2) Next 10s place multiplied 3 times will be in the thousands so focus on 389. 1^3=1,2^3=8,3^3=27,4^3=64,5^3=125,6^3=217,7^3=343,8^3=512 512>389 so 7 is likely the tens place. 7 in the tens 3 in the ones. Try multiplying 73 three times. It does come out as 389017.

[@SM-pi8up]

This is how your great grandfather used to solve in exams !

[@brunobastos5533]

if he was lucky enough to attend a school

[@malcolm7609]

@@brunobastos5533 Exactly

[@ffggddss]

We were taught this method when I was in junior high school, almost a century after that exam she shows. There weren't yet hand-held electronic calculators, but we had slide rules. There were some places where there were electromechanical desk calculators (Monroe, Friden, etc.), a few of which had square-root-finding built in to them. BTW, a similar method can be devised to find cube roots. And that's even more tedious than this. Fred

[@nirmalasingh5524]

OOO...Your's didn't knew. Send him here our infants shall teach them.

[@ονειρο-μ2ω]

I don't think my great grandfather knew math at all.

[@viditshrestha4054]

I think I have an alternate method to this question that would also work for any square root question. All square numbers have prime factors raised to an even power. You know this because, if root n had prime factors A^a, B^b, C^c....N^n, then n would have prime factors A^2a, B^2b, C^2c etc. By breaking down 0.004304046721 into it's prime factors, you get 10^-10*3^16. To get from root n to n, you multipied the powers by2, so to get from n to root n, you would divide by 2. If you did the same to the number in the question, you would get 10^-5*3^8, which you can quickly work out to be 0.06561. I am very proud to say I found out this method while fiddling around with a difficult question in some maths homework.

[@RandomPerson-xu7rk]

wow ur so smart well done

[@loveof567code.exfitruso8]

This is very interesting.

[@ffggddss]

Thanks for showing us this! It's an ancestral flash-back!! I will add, however, that we were taught this method when I was in junior high; ca 1960. It resembles long division, but the divisor keeps growing, the more digits you work out. A similar method can be developed for finding cube roots, which is, naturally, even more tedious! Curious that the number that the square root of was asked, when the decimal is removed, is just 3¹⁶, and with the decimal, is shifted an even number (10) of places to the right, so that the square root works out to 3⁸, shifted 5 places to the right. I think this tells us something about how the problem was composed. [Someone likely took 9, and squared it 3 times.] Fred

[@unrealworlds5856]

Me: hate math Also me: watching how to find square root in the middle of the night

[@DemoniteBL]

Me: Loves math Also me: Too lazy to learn the stuff showed in this video

[@spyrex3988]

@@DemoniteBL so same lmao

[@SIGSEGV1337]

You don't hate math, you just hate school.

[@black_jack_meghav]

@@SIGSEGV1337 exactly

[@hglundahl]

0:44 In math and science books back then, there was given an algorithm for finding square roots. Works same way before and after decimal comma, but depends on dividing the number into groups of two (decimal comma being at one division, always). (This means, first group of integers or last numeral after a comma may be a sole numeral in its group) If I want to make this for 144, I first divide this into two groups : 1 | 44. Then I figure out square root of first group (or of largest square number below that group), in this case 1. I put a 1 after = sign, and a 1 to the left. THEN I do some interesting stuff. I put a 1 below the 1 to the left. I Then multply them and put product under first group to deduct from it : 1 - 1 = 0. The 0 is put under a line. I then put a line under the two vertical 1, I make an addition. Same height as the 0, ideally. 1+1=2. Now, I put down first numeral in second group next to the zero : 04. I divide 04 by 2 in my head and get 2. Now I need to check it is not excessive. I put a 2 after the 1 after =, I put a 2 after the 2 under the line to the left, I put a 2 under that 2, and multiply: 2 * 22 and as I get 44, I put that under 044 (I now have added the other numeral of second group). 044 - 44 = 0. 12 is the very exact sqrt of 144 ... which we already knew, but I was demonstrating the method. There is a similar one for Cube roots, but it's more complex. BONUS : if the antiquated book you get this from is in Swedish, you are also likely to get an older spelling, previous to 1906!

[@hanniffydinn6019]

Hans-Georg Lundahl what is this called ? Can I look this up somewhere ?

[@albarsutanbaharis.160]

Yea. Where can we find that method?

[@hglundahl]

@@albarsutanbaharis.160 , @Hanniffy Dinn - it si called "extraction of square roots" (kvadratrostutdragning in Swedish), and if you buy an extensive book of mathematics from before the time of pocket calculators, you will probably find it. So, used books' shop.

[@hglundahl]

If you can't find it, simply look up her steps or mine (basically the same, in my version you would have had: 125 005 _______ 130 in one place) on a number of square numbers of which the roots are known. Or use it on 2.00|00|00|00|00|00 and see if this agrees with your pocket calculator's value for sqrt of two to 6 decimals.

[@MarianneExJohnson]

I learned this method from a book called "Mathematics for Statistics," by W. L. Bashaw. It was one of my father's textbooks; I think he used it while working on his Ph.D. thesis. The book was in English, which I had only just started to learn, but the diagrams were so clear I was able to understand the algorithm anyway. www.amazon.com/Mathematics-Statistics-W-Louis-Bashaw/dp/B000V276BS

[@swordofdoom1517]

Fascinating vid! I remember doing these when i was younger. Never understood it then. Looking at the actual algorithm helps a ton

[@andreasholand2828]

This is a nice algorithm to be honest; I can definitely see it being useful before calculators were around. If given this task without a calculator my intuition would be to either use Newton-Raphson's method or to guess on two numbers near what the answer should be, pick the number in the middle between them, square it and then remove the part of the interval the answer must be outside, then repeating the process. In this case that would be, for instance, starting with 0.06 and 0.07, but seeing that the answer we want is greater than 0.065^2, therefore looking at the interval between 0.065 and 0.07, repeating the process with 0.0675. This is more computation heavy, though.

[@PongoXBongo]

Without knowledge of algorithms or special techniques, brute force calculation is the only option. Splitting the difference to narrow a range is a great option. Handy for squaring and cubing as well.

[@edwardhuff4727]

Newton's method applied to square roots, with excellent first guess (a/k/a ansatz). To find S = √A (side = √area), first move the decimal of A two places at a time right or left until you have one or two digits to the left of decimal, and count the steps. At the end, you will move the decimal of the result one place at a time in the opposite direction for the same number of steps. √1234.567 = 10×√12.34567, moved two places right, then one place left. √0.001234 = √12.34 ÷ 100, moved two places right twice, then moved one places left twice. Next, knowing the squares up to 100, guess the square root of the one or two digit whole number. Guessing two digits helps, as in √30 is around 5.5. Repeat the next step up to four or five times. The number of accurate digits doubles each time, so 3 repeats will give around 8 digits. The repeated step: average the guess S with A÷S. If you want the result accurate to N digits, keep at least N+1 m digits of the quotients. S_new = (S + A÷S)÷2 to N+1 digits. When S and S_new are equal to the desired number of decimals, stop. √0.0043046721 = (√43.046721)/100 √43 is between 6 and 7, say 6.5 1) Divide 43.046721 by 6.5, giving 6.6225725. Add 6.5 giving 13.1225725. Divide by 2 giving 6.5612862. 2) Divide 6.5612862 into 43.046721 giving 6.5607138. Add 6.5612862 + 6.5607138 giving 13.1220000 Divide by 2 giving 6.5610000 3) Divide 6.561 into 43.046721 giving 6.561. Stop repeating. √0.0043046721 = 6.561/100

[@fins59]

They had slide rules in the early 1800's that could calculate roots and powers of numbers, the modern version of the slide rule with a runner (the Mannheim) was available from 1850 onwards.

[@sebastianmorataboada9795]

5:06 *_binary search intensifies_*

[@Hydelsius]

In high school, I was very bad at math, so I took the easiest math course possible in order to get that last math credit. I remember doing something very much like this: finding the square root of a number without a calculator. I remember my teacher telling me that we should know this because we'll never have a calculator on us at all times, unaware of the surge of cell phones and smartphones that would happen a few years later in '07 thanks to the iPhone. Once we finished our work for the day he would teach us how to play chess and that's pretty much the only thing I got out of that class!

[@preetamin477]

That was a really interesting method. It is a lot more sophisticated than the method that I used. I solved it by writing it as the square root of a fraction. Because the denominator of the original number was 10 raised to the power of 10, its square root had to have a denominator of 10 raised to the power of 5. To find the numerator of the answer, I split up the numerator of the original number into perfect squares. Because the original numerator was odd, I knew that four was not a factor of it and tried dividing by 9. This worked, so I just kept on dividing the new answer by 9. Once I find out how many times 9 had to be multiplied to get the original numerator of the fraction, I just multiplied that many threes together to get the numerator of the answer. Finally, I divided the numerator by 10 raised to the power of 5 to find the same final answer.

[@Real_Claudy_Focan]

Napier's method

[@yannickdengler]

This method does not only work for decimals by the way. It also works for large roots. Because you can always pull out "2 zeroes" from the root and write a 10 in front the result of the exam is the same as the result of sqrt(43046721) just divided by 10^5. In other words you could have generalized the problem from the exam to the following: sqrt(0.0043046721) = sqrt(43036721) * 10^(-5)

[@danturney5904]

Thanks for this. Many, many years ago my 7th grade math teacher (the most disagreeable teacher I've ever had) did have the one redeeming quality of teaching us this method which I still remember, but I'd never considered its algebraic derivation.

[@mayflowerlash11]

I've been curious about this for some time. And now it has been explained beautifully. Thank you.

[@Alienware173H]

I know I'm late to the party, but I thought I might clarify a minor point at time 2:12 . It seems at first that Tibees is using (x+r) ≤ sqrt(a) and "squaring both sides" to get (x+r)^2 ≤ a. However, inequalities don't quite work the same way as strict equality does, and in general, this is not true. As a counter example, observe that -16 ≤ 2, but clearly we can't just square both sides to get (-16)^2 ≤ 4. Otherwise, we would get 256 ≤ 4 which clearly isn't true. The reason we can get away with what happens at 2:12 is because of the following fact: If x and y are non-negative numbers and x ≤ y, then x^2 ≤ y^2 does, in fact, hold. Because sqrt(a) is a non-negative real number in this context, we can approach sqrt(a) by positive real numbers from below. In other words, we are implicitly assuming the (x+r) is going to be positive. It is ONLY because of this fact that what happens at 2:12 is kosher.

[@gregjhill]

I remember doing this in primary school but of course I’d totally forgotten the method. The method I always remembered was from high school and involved logarithms. In the case of a square root you find the log of the number, divide it by 2, and find the anti log of that number. That is your square root. The beauty of using logs was you just divided the log of your number by the desired root and then worked out the anti log of your result to get an answer.

[@egillandersson1780]

I learned this method, not in 1866 but in 1979 ! We already had scientific calculators (TI, HP, ...), but we still had to learn "by hand" methods like this one, how to use logarithm tables, trigonometric functions tables, ...

[@badchessplyr]

aryabhatta in his book aryabhatiya has mentioned algorithm to find square root and cube root of a given number .he wrote the book in 499 CE if you want to learn the that algorith . go to nptel course of mathematics in india lec 7- 8.. btw my mom taught me this algorith when i was kid

[@heirofslytherin3890]

I forgot this algorithm, now I can use it in my programs

[@padraig4096]

Thanks for posting this. As soon as you started grouping the digits in twos I recognised the method as one I'd learned while browsing in a bookshop in the mid 1960s. I still very occasionally use it as a mental arithmetic aid to calculate square roots to three significant digits.

[@gabor6259]

This is like ASMR. So calming. Also this video is much better than Numberphile's video on the topic.

[@Lumpiaguy]

Thank you so much for this. I had been on a quest for years trying to find a proper demonstration of how to compute the square root of a number by hand.

[@ScoopexUs]

My first thought from your video was that they indeed used an approximation algorithm, but either from a look-up method in books with tables, or by making the number 43046721 instead, and by knowing the root of 1000000 is 1000 and squaring a few times in their heads quickly arrive at 6561.

[@BlueLightningSky]

I remember coming across root 8 in a homework once in my grade school days. So my dad, who studied lived before calculators were common place, showed me by hand how to approximate it. Came to class the next day and teacher said to leave it as root 8.

[@piyushoturkar4535]

😅😅😂😂

[@McRion]

We have this exact algorithm in our 6th grade math textbook. It’s pretty cool that I was able to recall the exact algorithm when you first started writing it down.

[@Kikkerv11]

This is awesome! I have never learned this in school.

[@IvyANguyen]

Thanks so much for this! I imagine most people at this state would have several lower square root approx (& perfect squares) memorised, like square root of 2,3,5,6,7,8,10,11,12,13,14,15.

[@princeofexcess]

if you try 5 then (7 or 8) or (3 or 2) you reduce the number of tries to max 3. So you dont have to calculate 5,6 ,7,8 or 5,4,3,2 of course we can often easily approximate which makes us try 4 attempts very rarely but I thought it's nice to have an algorithm for least number of tries. (This is a binary search for those familiar with CS)

[@trueriver1950]

Thanks for posting this - I knew the technique existed but not the exact details and got stuck trying to figure them out. Also thanks for explaining why this works: now I understand the method I will be able to figure out the details in future. The problem with the way this was taught in the UK in the 1960s was that it was taught by rote without explanation.

[@tsresc]

I love the way she's improvising rapidly. Even if you didn't like her at first, you'll end up liking her. She'd become a great entrepreneur.

[@chakshupunj5943]

Never knew this method of finding square roots. Thanks Tibees!!! 😀

[@willgates32]

It is fine to use this method with minuscule decimal fractions but it is nice to turn the decimal into a fraction with 10^10 as the denominator, then find √43046721 and divide by 10^5 { √10^10 }.

[@condorboss3339]

By inspection: If the number is a perfect square, then it has 5 digits after the decimal place, the first of which must be zero The first non-zero digit must be 6. 66^2 > 4304 > 65^2, so the second non-zero digit must be 5 The final digit must be 1 or 9, as only they can produce a 1 when squared. The final two digits of the product can only be affected by the final two digits of the root. There are only 4 two-digit numbers that can produce 21 as their final two digits: 11, 39, 61, and 89. By elimination, we reach .06561

[@dj-up4nc]

Thank you so much! I remember learning this in middle school, but I couldn’t seem to remember the process. This video helped a lot.

[@wenhanzhou5826]

0.0043046721≈4.3*10^-3 Spilt it into sqrt(43/10)*1/sqrt(1000) =sqrt(43)/sqrt(10000) =sqrt(43)/100 6

[@Arycke]

Love your content, mind, and personality. Lovely video as per usual

[@tibees]

Arycke thanks

[@robertjohnson5838]

On a serious note, to those who say try factoring it, what if it had been a square of a prime number? One insight is to realize the ones place had to be a 1 or a 9, then realize that(ignoring the decimal issues) it was at least 6001 but probably well under 6999, guess 6501 as a start and iterate upwards from there, though if you have all your prime numbers up into the 6000 range (YEAH, RIGHT) you might do the factoring thing that, in this case, was easier perhaps than other methods.

[@misternewoutlook5437]

I think the confusion for some folks comes in where (2xr) enters. It's easy to think of the (x+r)^2 as simply x*x+r*r. But it's not. It becomes clear if we arbitrarily give "a" a value, say 25. Thus we can say x=2 and r=3, which answers the equation. When we get to (x+r)^2=25, we can see the process and rule in action clearly. We want to separate the x and r values out of those brackets. However, x^2=4 and r^2=9 totals just 13. Where's the missing 12? Well, we are actually multiplying this way (x+r)*(x+r)... If you look at the steps here: we really need (x*x)+(x*r)+(r*x)+(r*r)... see? In the middle are those pesky calculations... so she has written those 2xr because you need those twice.

[@user-mg6qo3bw7v]

Thanks for the video. I was having a bit of trouble understanding what the variables stood for like R, and how the exercise tied into the derived formula.

[@ali64fahad]

I thought to use the power series but I finally learn your methods, it is adaptable

[@carl-marvin]

After a day of listening to your teachers speaking into 10$ microphones this asmr math is amazing for finishing the day and calming down, thanks :D

[@Dhalin]

I just wanted to ask you a question.... is this something weird that only I do, with Long Division.... but I came up with a couple methods on my own of how to do single-digit divisor long divisions easy, and how to also make larger ones easier. For example, let's say you have a problem... uh, 151,386 divided by 3. On paper, you'd write the 151386 underneath the division sign, and put the 3 outside like any long division problem. You'd start off by placing a 5 above the 15. Then you move over to the 1... and put a 0 above that and put a tiny little 1 in front of the 3 (You're carrying a tens digit here) and then put a 4 above the 3. Put another tiny 1 beside the 8 (3x4=12, 13-12=1) and then put a 6 above the 8 and then lastly, a 2 above the 6. The answer is 50462, but you used a lot less space on the paper to do this. The second problem, is 65536 divided by 256. The easy way to solve this on paper is to start out by dividing both 65536 and 256 by 2, which gives you a new problem: 32768 divided by 128. These are still even, thus divisible by 2, so you divide by 2 again and get 16384 divided by 64. They are _still_ divisible by 2, so do this again to get 8192/32 .... and again to get 4096/16 .... and again to get 2048/8 .... yet again for 1024/4, again for 512/2 and finally, one last time to get 256/1. Your answer is 256. So basically with this second thing... first thing you ask yourself... are both the divided and divisor even? If so you can chop them both in half. If not, ask yourself if they are divisible by 3. How do you know? If all of the digits added up are divisible by 3, then the whole number is divisible by 3. For example, above, 151,386. This is divisible by three because 1+5+1+3+8+6= 24 which is divisible by 3.

[@crashtech66]

I'm fascinated with the way you draw the letter x, is this common practice where you were taught? I was taught to draw something approximating a sine wave, then bisect it with a small forward slash.

[@walterwilliams6357]

Learned this in a programming logic and design course.

[@mateoslab]

i've never learned this, i love how it looks like a long division problem

[@quill444]

There are several algorithms to compute square roots: I think calculators use logarithms. Another way to test someone in this skill is to ask for something simple, like the square root of five, but to have them calculate the answer to fifteen or twenty places (enough so that a typical calculator won't be able to help). In some of Feynman's lectures, he discusses some brilliant little tricks for finding a few selected cube roots in his head! P.S. I have my New York State Regents Diploma buried in an old box somewhere: in 1976, when I received it, if I remember correctly, these were not only used by college admissions, but could also help a teenage driver get a better rate on his or her auto insurance!

[@jonathonmacbride9951]

Since it was on a test, you knew it was a whole number. The easiest way to simplify it, therefore, is to factor it. As you know that it must be divisible by n^2 if it is divisible by n, this is quicker than you would think. Obviously this won't work if it is a large prime squared or they did not give you a whole number but 43046721 is 3^16.

[@jeffw1267]

There's an old Japanese subtraction method that is very easy and can even be done mentally. Each step is complete in itself and wipes out the previous step, so there's not too much to remember. When applying it to a decimal, you just have to know the rule on how to point off the decimal so it finds the right location in the answer. There is no guessing or rounding. You only determine each digit, one at a time, from left to right. Go to afjarvis.staff.shef.ac.uk/maths/index.html and click the "Square Roots By Subtraction" link at the bottom and download the .pdf file.

[@rogerfurer2273]

Did they have slide rulers back then? My dad was pretty good with a slide rule, but I had a TI calculator, so I never mastered them. Excellent video, thank you.

[@aneophyte1199]

I don't remember the method, but I was taught how to do square roots by hand, when I was in school. What I do remember was it was sort of like long division.

[@saumyagupta4711]

It is possible to find square root using log tables. We used to do that in school. Could it have been possible that they were asking in 1866 paper to find the square/cube roots using log table?

[@Spelterdon]

While playing with bases other than 10, I've discovered that this method works with all bases. This biggest problem, of course is multiplying and subtracting in other bases. This is because very few people (including me) have memorized their multiplication and subtraction tables in any base other than 10. Base 2 wasn't too hard to do. But other bases got pretty hairy. :)

[@ccdsah]

if you had factorized 43046721 you would have found it is 3^16. The whole number is equal to 3^16*10^(-10) thus the square root is 3^8*10(-5)=0.06561

[@VaradMahashabde]

OH. MY. GOD. I finally have an explanation for this method. I was thinking about looking for one but my search is over before it began. This an educational video as in it explains your education THANK YOU

[@VaradMahashabde]

ALSO IS THIS BINARY TOO MAYBE I CAN IMPLEMENT

[@ZipplyZane]

This method will work in any base. Nothing about it requires base ten.

[@josephmartos]

Watching your videos is sooo relaxing..... and the content is perfect obvious xD

[@bmoremike]

I haven't seen this for a long time. My older brother showed me how to do this when I was in grade school, almost 60 years ago, although he didn't show me the algebra part.

[@GomerfromIsaan]

There is another old arithmetic trick that might help. I will skip steps to be brief. First suspect that on a test like this there is probably a shortcut to solving the problem and the answer will come out even. That big 8 digit number is obviously not divisible by 2, 4 or 5. Here is the trick. If the sum of the 8 digits is divisible by 3, the number is divisible by 3. Likewise if the sum of the digits is divisible by 9, the number is divisible by 9. Use long division to divide by 9. Repeat until you find the quotient 6561. solve for the correct position of the decimal point. The square root is .06561.

[@sowmyag5142]

Hiii toby Can you pls do your journey to MIT video? Where did you study before MIT & what courses you took?

[@jaykay6249]

She didn’t go to MIT

[@israel3621]

Great video. Always wanted to know how to do it without a calculator. It does help in understanding.

[@sunnyjoseph558]

You explained it better than my text book. Thanks 👍

[@Shub-e3j]

This was really useful. Thanks tibees 💗😊

[@jonathansobieski2962]

There’s a really cool algorithm that they use at mental calculation World Cup to do these entirely in their heads without paper and it’s surprisingly doable. Regardless, nice video.

[@alexandersanchez9138]

If I were seeing this problem on an exam, I’d want to covert it into a problem about integers: x^2 = .0043046721 (10^5 * x)^2 = 10^10 * x^2 = 43046721 My first guess would be around 6500. So, (6500+r)^2 = 43046721. Solve the quadratic: r^2 +13000r - 796721 = r^2 +13000r [phone]-43046721 = 0 We get (r+13061)(r-61) = 0 by factoring. Hence, x * 10^5 = 6561. Thus, x = 0.06561

[@Sub-lt5yd]

Alexander Sanchez Good

[@renukam5455]

Fantastic explanation , thankyou it helped me a lot❤

[@yvonnetomenga5726]

👍Excellent video. After reviewing your recent videos, playlists and community posts, I decided to subscribe even though I suspect much content will be over my education and intellectual level. Noting Bob Ross in the title of several of your videos, I'm hopeful I'll be able to keep up. Thanks for sharing your knowledge.🥂

[@NataliaBazj]

Here is an algorithm from the arithmometers era. √.00.43.04.67.21 - ? First of all, multiply by 5 43046721 * 5 = 2.15.23.36.05 (care must be taken to ensure that the digits are correctly divided into pairs) then 215 - 05 = 210 - 15 = 190 - 25 = 170 - 35 = 135 - 45 = 90 - 55 = 35 - 65... stop, write next two digits 3523 and replace last 5 by 05 in subtrahend - 605 = 2918 - 615 = 2303 - 625 = 1678 - 635 = 1043 - 645 = 398 - 655... stop, we need two more digits 39836 and subtract - 6505 = 33331 - 6515 = 26816 - 6525 = 20291 - 6535 = 13756 - 6545 = 7211 - 6555 = 656 - 6565... no, two next digits 65605 - 65605 0 yes! zero! we can try to continue 0 - 65615... no, more digits 000 - 656105... no, even more! 00000 - 6561005 00000...00 - 6561000...0005 In other words, 6561 is the precise result. Then just find correct position for decimal point √0.0043046721 = 0.06561 Main idea: 5x² = 5 + 15 + 25 + ... This algorithm allows you to calculate digit by digit the root of any number with any accuracy. A square root is just a series of subtractions.

[@fabfan7939]

you could actually do the prime factorisation as well (though prime factorisation is also quite difficult for certain numbers)

[@mamanikalita7420]

Did this in 8th grade...quite refreshing to see it 3 years later...almost forgot it

[@rivlry1975]

Nicely done. And thank you for the tutorial

[@Pete-Prolly]

My Dad showed me this the same way! He had an "Old-school Greek education" & he helped me with my Math in grade school, (in America.) After I learned "long ÷" he DEMANDED that I learn how to find square roots by hand so he showed me this method. Then, my Dad said "if any of your teachers can find the square root of a number by hand I'll give you $20" so every year I asked until I graduated High School. (I never got the $20 lol.)

[@williamstgeorge7289]

I found that it was a number between 65 and 66 with decimals at appropriate place, and then I quit. Nice work, Tibees, but pretty sophisticated for a high school student. Getting too attached to calculator is unwise. I like doing mental arithmetic because it alerts me to mistakes without resorting to a calculator.

[@jatinramchandani277]

Is there any way to solve this using differential calculus

[@vishvarupa7948]

Maybe newton's method?

[@MukulChopra]

Linearization

[@MarianneExJohnson]

Newton's method works well for square roots. See: en.wikipedia.org/wiki/Newton%27s_method#Examples

[@Davidamp]

Taylor

[@asss808]

Differentiation will give approx answer

[@lagarttemido]

In love with your voice!

[@shantanu925]

I am glad to see that I am not the only one who holds pen in thumb over finger style. People say my pen holding style is wrong.

[@danfg7215]

Shantanu Jain it’s wrong but don’t worry I won’t say anything

[@shantanu925]

dan fg Thanks a lot.

[@robertlozyniak3661]

Does it damage the pen? Does it hurt your finger? Does your handwriting come out legibly? If the answers to those three questions are "no", "no", and "yes", then there is no problem.

[@jameswilson8270]

Great job! Thanks for sharing!

[@TheNameOfJesus]

I just say, aloud, "Hey Siri, what is the square root of 0.043046721?"

[@tutekohe1361]

Brilliant old school "analog"calculating. Thank you tibee. Btw, is that an Australian accent?

[@SIGSEGV1337]

Yeah it is

[@comic4relief]

I actually remember the method from seventh grade. I do like your sort-of derivation of the method. Interesting!

[@lesnyk255]

I used to know this method, having stayed after school one day to ask my math teacher if there were a manual method of extracting roots (we used published tables before we were introduced to logarithms). She must have been taught this method because she was able to explain it to me. (Just for generational reference, this was ca 1963 when I was in the 7th grade, age 13 - no advanced math needed to estimate my current age.) There's another beautiful iterative method supposedly developed by the Babylonians, using only the operations of addition & division - I've put up a YT video explaining it here: kzbin.info/www/bejne/pnfTipiVZ9ykq9U . I've tried without success to derive it as an infinite series, but the terms rapidly expand and become too cumbersome.

[@nirmalasingh5524]

Solve,Simplify,Serenize ... Live Lovely Life

[@maddawgzzzz]

I watch this channel to watch you

[@Omnifarious0]

That's a nifty little algorithm. I would've stumbled around with Newton's method for a few steps, likely made lots of mistakes and probably wouldn't have realized that there was an exact answer.

[@suhashvadlamani1426]

Can you make a video on how to find out square root of any number like 54,53,63,103

[@sebastienberger2890]

Well without knowing how to do it properly I would have used much less sophisticated : devide to conquer. I would first convert the number into a no decimal one and would guest a number to be nearest possible and try to downsize or upsize. By guessing something really close its possible to get close to it.

[@kumargaurav9417]

Learned this in 5th grade..very useful thankful to my classmate

[@TestTest-nf3si]

Hi, You are great! Please keep going on with your channel.

[@draoi99]

I'm not afraid of maths when you are the teacher.

[@sentidude3314]

Hey tibees, I have watched u r videos on 'books for learning mathematics and physics' Which helped me very much. And I am commenting here because this is your latest video. Actually I need help from u on the topic tensors. Please tell me about some books which I can use to learn tensor for the first time. I know scalars , vector and tensor and also known that these all are interrelated, so please help me to choose some books on tensor... 😊😊😊😊😊😊Thanks😊😊😊😊😊😊

[@zoisitemapping]

I finally understand why the square root of the number is a rational number: .0043046721 is 3^16 x 10^-10 and 0.06561 is 3^8 x 10^-5.

[@ceo1OO]

Since 43,046,721= 3^(16) then: Answer = [ (3^16)÷(10^10)]^(0.5) = [(3^8)÷(10^5)] = 6,561÷ 100,000 = 0.06561◾ 📓

[@DarwinianUniversal]

This is off topic, but I wanted a chance that Ti bees would read it rather than posting on an old but more relevant video. Try this deduction................... Clocks are made of quantum mechanical parts, that measure the parameter of general relativity "time". A singular device that corresponds to both of these fundamental theories. If both theories correspond to clocks, then clocks show us how both these theories correspond to each other. Unquestionable and simple deduction. Ok, so what does this tell us? Time dilation corresponds to a property of clock springs which is well termed "force dilation". Clock hands and clock springs are a part of a single system, moving in lock step proportionality with one another. If you log time dilation effects on clock dials, then there is a corresponding shift in position of the spring, and spring positions can be defining in terms of Hookes Law which identifies a varied stored tension, varied force value. Force dilation!. Force dilation is a QM property that corresponds to effect of time dilation, a parameter of relativity. Science hasnt recognized this force dilation effect, but it is an empirical observation, and therefore is not a theory. A literal measurement. If you choose to ignore it, then you are doing so in face of empirical fact. Dont fail such a test if you believe yourself to be scientifically minded. Relativity boiled down to simplest possible terms, is a theory of simultaneity, or point coincidences. Einstein realized that particles dont point coincide, interact, when and where we would expect them too. So he developed a framework which reconciled correctly where and when they do interact. Time dilation is a measurement of when and where they interact, but Force dilation is the cause for this anomalous discrepancy. Force dilation is why particles are late and or early to arrive at a point in space. It is no more mysterious than a ball reaching the other side of the yard quicker, because you throw it harder. Force dilation. Force drives clocks, therefore clocks measure force! Despite this statement adhering to pure logic, people think its nuts. Logic escapes people easily it seams. It is evidently true that clocks measure force, because clock function is entirely a consideration of forces. Dilate the clocks function, you dilate the forces. This all makes perfect sense when forces are considered the cause of changes in the world, and time is but a measure of changes which are caused by force. Therefore force dilation is the cause of time dilation. This demystifies the nature of time as merely being a measurement. Force dilation is a quantum mechanical property which is responsible for the modulation, which is credited to Relativity. This describes relativity in terms of a quantum mechanical effect. It cannot be argued with because it is an empirical observation, a literal measurement. A simplest possible interpretation in terms of cause and effect, which is the goal of science. Do you want to resolve some universal mysteries? Well, keep doing what we did, and keep getting what we got. Somebody with some imagination please, try on a different point of view? Or languish in 100 year old theory which has stubbornly evaded progress.

[@RTYB]

Linearization (Calculus I) is the best method I can think of.

[@briantaylor9266]

Surely students would have had access to log tables and slide rules. These two tools make these problems child's play. (FYI, I'm old enough that I used these as an engineering undergrad.)