Find Square Root by Hand without Calculator
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- Опубліковано 15 вер 2019
- Learn how to find the square root of a number by hand approximated to at least two decimal places. In this video we approximate the square root of 38 out to two decimal places without a calculator.
Video Link: tinyurl.com/morelligeomvid16
#squareroots
#byhand
#withoutcalculator
My high school teacher insisted there was no way you could do square roots by hand. I insisted there was a way and she challenged me to come up with a method. I never figured it out, but I knew there HAD to be a way. How else could people do them without calculators for all those centuries?
You've shown the way! :D
Same here-it seems like most math teachers like to keep all the cool and interesting math to themselves. My newest math tutorial video shows a *much easier* method for approximating square roots intended to replace this one-I hope it helps you! ua-cam.com/video/xIpoxUPfuRY/v-deo.html
Almost as bad as Math teachers who don’t know there is a quick way to tell if 7 evenly goes into a number. They show 2, 3, 4, 5, 6, 9, 8, 10, but seem to leave out 7 for some reason. 😡
I.e. Double the last digit and subtract that from the remaining digits. Repeat until left with -7, 0, or 7 which means the original number is evenly divisible by 7.
Somehow *good* math textbooks from 70 years taught it but modern ones don’t. 😳
How did she say it was done before the advent of calculators or computers?
how else would they program the calculator?
you should find her on linkedin or facebook and send her this video
Thanks for being the one video that actually explains this process instead of just a “trick” to do rough estimates.
You’re welcome 😊
thats exactly what i was thinking! great video and great explication
@@toplel_96 Me too!
I thought this was my comment for a second 🤦♂️
Actually, the failure of this video is that it only teaches mechanics and does NOT explain why it works. If you can explain why what this video shows works, then you will really understand what the square-root means as a function and will understand a lot more math all at once, including interpolation and principles of calculus. Learning is not about memorizing algorithms, it should be about understanding how they work.
A true teacher!
You don't presume we know but take the trouble to take us through all (not some) of the logical necessary steps to the answer.
Thank you, sir!
Wonder if you got one of these problems if you could do it on your own using these steps and knowing what figures to use as you go along
This is the sort of comment that makes it all worth it...thank you, truly, and I am happy to help. Good luck to you. -Mike
@@AthenianStranger I dropped out of school, but I’m currently learning how to do calculus by myself, thank you for everything you do❤
I was once shown this by a math teacher when I was 13 and have been trying to remember it ever since. Thank you for this video.
Glad to hear it…I actually no longer use this method and will soon upload a much, much easier method which I use to teach students aged 9-12 how to do the whole thing in their head and it is correct to the tenths place.
Keep an eye out for that video. It is so much easier that it makes me cringe to see this video has so many views!
Been about 45 years ago for me, we learned it before the advent of personal use calculators.
@@AthenianStranger I think this video fulfills a different purpose than your new video, because this method makes it possible to calculate square roots to an arbitrary precision
The in your head method is the only useful one in our day and age, because if I need precise square roots I can just use a calculator
My Dad showed it to me when I was about 10. Similarly, I've thought about it from time to time. I only remembered the additional two zeros but forgot the important part of doubling and multiplying by the magic number. Thank you for this video!
HHHGeorge
I hope this not an example of how you have spent your life. Just DO it.
This is how I had learnt to find square roots some 45 years ago in Indian schools. The format and layout were slightly different but the flow was the same. Calculators weren't readily available then. Thanks for this demo.
Did they also teach the theory of this method, the "why" behind the "howto"?
i have never known a time without cellphones and calculators. i have caried a super computer in my pocket, more powerful than the ones used in the moon landings for 20 years.
@@Tubemanjac I need that too. Otherwise this is just a memorized process that makes no sense to me.
🙂 Yes. In Romania we learned this in school as well. And our method was simpler. Same flow, easier logic.
This was first documented by Madhava of Sangamagrama in the 1300s. He also discovered the first recorded formula for calculating pi using infinite series, which is the basis of modern calculus, centuries before Newton or Lebnitz.
What can I really say to explain how grateful I am for your clear explanation? Understanding mathematics does so much for self-confidence not to mention giving you the ability to breakdown important concepts and it is so often so difficult to make the process of learning clear and not something that makes you doubt your abilities. Thank-you very much.
This may be among the kindest and most personally affective compliments I have ever received on this channel. Thank you so much for your elaboration regarding the importance (and rarity) of clear, simple, and direct explanations of math. You have hit upon my central goal as a math teacher. Though I, too, must very often fail in this regard, I try to teach math to “my younger self” the way I wish it had been explained. Perhaps like you, I went from K-12, four years of college loaded with advanced mathematics, and two years of graduate school and managed to emerge without a single coherent dusting of true understanding regarding virtually all mathematics. Then, I became a teacher and never wanted to teach mathematics because I had this secret-my degrees stated that I should by all accounts have the ability to teach everything from rudimentary addition of single digit numbers through advanced calculus but in reality I didn’t have a clue about any of it. Then, one year all my school’s math teachers left-all of them-so my principal asked me if I would be willing to teach Algebra 1 (because in Texas that is the only high school math course with an end of course state assessment). I accepted this challenge and went home and wept into my pillow for trying to impress my boss but it was too late, so here’s what I did: I went on eBay and bought all my old math textbooks in their exact editions from the 1990’s (I was born in ‘86) and then I paid for this online test prep service for the math exam teachers are required to take and pass with a score of 85 or higher. My first attempt at the test earned me a 40 and I went back to weeping in the pillow. Then I just came up with the “keep it simple stupid” plan and worked 100 math problems every day starting with early childhood math (for these I worked 200-300 problems per day) and slowly worked every single problem in every single book all the way through calculus-every day, 7 days a week, for like 6-7 weeks straight with no breaks at all, 12-18 hours a day. The end of summer vacation approached and my principal said I had to take the certification test or he would have to find another solution. Ugh!!!! Pillow weeping ensued, then I just stood up, registered for the first available test which happened to be offered the next day far away on the other side of Texas so I paid for it, hopped in my car, drove straight through the night, and arrived at the testing center just in time. The test was 100 questions and I was given 5 hours. The room was cramped and packed with people so it was hot and humid. I took exactly 4 hours, 59 minutes, and 57 seconds taking the test and obsessively checking over all my work and hit “Submit and End Exam” with literally 3 seconds on the clock. I immediately called my principal and told him I failed for sure. It was the hardest test of my life-NOT A CHANCE that I passed with a 70 much less the required 85. He told me to calm down and I drove home and waited the mandatory three days for the score: I opened the email, clicked the link, logged in, and saw a word I would never have believed I’d see: “PASSED”. I passed the test by just two questions but that was irrelevant-I PASSED and was at that moment forever transformed from a history teacher into a math teacher, spawning the genesis of this channel, and now I have taught every grade mathematics from 7-12. Every time I teach, I learn right along with the kids. I think the kids and perhaps the viewers of this channel sense that-it’s why I always very slowly work each problem and explain every step-and why I use my infamous clear ruler 📏-“straight lines make straight minds.” I am just like all my viewers: Trying to learn math. The journey’s duration is equal to all our respective lifespans and never reaches a conclusion which is what makes this math adventure so interesting and forever awaiting that next newly discovered or mastered skill or rule or hidden internal logical beauty. So thank you for your compliment and for sharing this infinite quest with me 😊. Best wishes and glad tidings, Mike
That’s not a clear explanation at all!
Great explanation, not really a method for fast calculation but it makes the process so easy to understand
Fortunately, my 3rd grade math teacher taught us this. It has turned out useful on many occasions.
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
@@AthenianStranger The second one is called the Bakhshali method, for anyone interested
You learned square roots in 3rd grade??
I've been curious about this for decades! Great instruction, sir. Thanks!
You should also do a video on how to do logarithms by hand. Its an equally interesting process... and gave me a ton of respect for what Napier and Briggs had to go through to make their first log tables. Yikes!
*Very good idea*!
@@AthenianStranger I hope you do go over log by hand; I absolutely abhor the prospect of being forced to use a calculator without understanding the principles at hand, and my senior year, I think ill be in a pre-calc class, something I presume will touch on logarithmic junk.
@@AthenianStranger - Yes! I would be very interested in logarithms.
@@AthenianStranger Yes, please do so! Other YT channels have shown that this can be done, e.g. Numberphile, but not made any effort to actually teach the technique. At least, no where near as clear as I believe you could. You have a desire to teach what you know and not just show off what you know
If I might suggest, stick to base 10, we know the log(10) of 10 is 1 and the log(10) of 100 is 2. We know that any number between 10 and 100 will have a log between 1 and 2. How could we calculate the log of say, 20 or 50 or 66? No more than 2 or 3 decimal places and I think we'd get the idea. You know, lather; rinse and repeat.
Note, you don't have to actually 'double' the number to get the next number in sequence: the pattern shown is this: let's say we have 12_ at the start. You find it to be the number 1. Hence, 121. The next number will be 121+(number you found=1)_. Thus (121+1)_ = 122_. We found in the video that the missing number was 6. So the next number in the sequence will be 1226+(number you found, thus 6)_ = 1332_, etc.
1232_
Wow.
*1232_, but yes.
It is 1232, 1332 is wrong.
Does it work for all equations?
This is way better, and more intuitive, than the method that I was taught in school.
Thank you.
Thank you so much. I understood nothing on my online class when teacher was explaining it. You helped me a lot
If you have to give an answer to 2 decimal places you must go to three because, if the third decimal place is 5 or greater, you would need to increase the second decimal place by one. And, if you go with the convention that even digits before 5 do not get increased, you would have to go even further.
+1 this. Spot on.
I like that you use a transparent protractor as your straightedge. Me too! For only a couple of decimal places, seems like it would be simpler to just guess and check.
This was the best video on square roots that I found, and the easiest for me to keep up with!
Excellent Video!
Thanks for this. My teacher showed me how to use this technique when I was 10 years old to extend me and I had no trouble following it as it was much more interesting than the horrendous long divisions required of us. I am grateful to find the technique so well presented. Yes I had a great teacher!
MOST Middle and Senior High School Math Teachers are terrible at their job.
Thank you! I am 67 and had never had that explained to me ! ( Girls used to be told " you can't do this. ") You are very clear, knowledgeable and good at explaining math. Thank you.
You are so welcome!
I believe you'd have to always calculate an extra digit to determine approximation, for example, if you just calculated the first decimal, you'd get 6.1, however, in the video you can see that this would approximate to 6.2, so calculating that one extra digit every time should increase the consistency of success in this method.
I just recorded a much simpler method for approximating square roots. Here’s the link: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
Unless the digit is 4, in which case you'd need to keep going
@@stoppernz229no
I watched a few videos trying to understand this and yours was the only one that made sense!! Thank you so much! Appreciate you taking it step by step.
Thank you for sharing your kind words.
I just recorded a much simpler method for approximating square roots. Here’s the link: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
This neat method is based on the binomial coefficients. For square root, it's derived from a^2+2ab+a^2 as such: a^2 + b(2a+b). A similar but more complex method gives cube root: a^3 + b(3a^2+3ab+b^2). For fourth root, it's a^4 + b(4a^3+6a^2b+4ab^2+b^3). Fifth root is a^5 + b(5a^4+10a^3*b+10a^2*b^2+5ab^3+b^4), and so forth. You would bring down groups of three 0s, four 0s, and five 0s, respectively.
It's doable by hand for cube root, but rapidly grows to involve gigantic multiplicands when used for higher roots. Nonetheless, it works.
Cool method!
I like developing a Taylor series around the value (it's simpler than most people think, but one does need to know derivatives). I can get to 2 or 3 digits behind the decimal point quite quickly.
Can you explain it further Rony I’m interested
For the example of sqrt(38):
Re-express the expression to be solved as:
sqrt(38)= sqrt(36+2)= 6*sqrt(1+1/18)
We factored the closest perfect square out of the argument of the square root.
Then use the Taylor series of sqrt(1+x), with x=1/18.
The Taylor series is:
sqrt(1+x)= 1+x/2-(x**2)/8+...
So in our example:
sqrt(38) = 6*sqrt(1+1/18)
sqrt(38) = 6*[1+(1/2)*(1/18) +...]
sqrt(38) = 6*[1+1/36+...]
sqrt(38) ~ 6 +1/6
The next term subtracts a less significant amount to correct the result downwards, then the one after that adds an even less significant amount to correct the result upwards, then downwards again, then upwards again, and so on.
You can look at the sizes of successive terms to determine whether you care about continuing on to the next term.
For it to work, the absolute value of x has to be less than 1.
It works better (you need fewer terms for a given desired precision) as the absolute value of x becomes smaller compared to 1. For larger values of |x|, it's not as good of a method. It therefore works better when the argument of the square root is close to a perfect square.
If you are close enough to a perfect square to only need a first order correction in x, then you could use the following formula:
sqrt(N^2 + x) ~ N + (1/2)(x/N)
Taylor series of sqrt(1+x):
www.wolframalpha.com/input/?i=taylor+series+of+sqrt%281%2Bx%29
@@_Longwinded it's a concept introduced late into calc 2, look for taylor series approximations in the comprehensive calc 2 guide on youtube. even if you don't quite understand it it's nice to know how it's performed
Yeah, Taylor series centered at the nearest perfect square number was the first thing to come to my mind, although as for the square root one can use the so called Babylonian method or simply write down a quadratic equation x^2 = a and solve it numerially using the Newton's method (which can be further generalized for an arbitrary nth root problem)
Thanks so much for the great tutorial on square roots! Extremely clear and very helpful!
Glad it was helpful!
That is a neat method if you can’t find your slide rule, which is how I would typically do it for only three significant figures (just transfer from B to C scale). The nice thing is that this method expands farther than three significant figures, a slide rule can’t really do much more than three.
liked and subbed, ive been looking for this all over the net, thank you
Simple, but not precise method is optionally:
1. Estimate the root as well as possible - Lets take the same example (y = 38) and first estimation a = 6
2. Compute the next value as b =(y + a^2)/(2a) e.g. b = (38 + 36)/12 = 6.166666 what is quite good result (delta
A very good explanation - thank you.
I would add that it would be better to show the next decimal place, too. I was taught that, if the answer was to 2 decimal places, you had to work out the third dp. If that figure is 5 - 9, you have to round up.
IF you had to round up, then how would YOU define Pi..3.14157... to a never ending line of numbers. there is NOTHING To" round up". The answer is is a never ending series of numbers.
@@guysabol8743 um... that's the whole point of "rounding". If you wanted to round 3.14157 to the nearest ten-thousandth, it would be 3.1416. Nearest thousandth would be 3.142. Nearest hundredth would be 3.14 (look familiar?)
Very helpful and well explained. Thank you!
You're very welcome!
I haven't done square roots since 1978. Thanks for the refresher.
it is actualy a method that i haven't learned in school. the hand written approximation that we learned was the one where you have a range between two numbers and you half that range and look if the half squared is above or under the root and then you have either the lower number and the mid value or mid value and the higher number as new range and you do that so long until the first few digits don't change anymore and that is then your approximation. a really long way to do it, but the numbers don't get big as fast as here because you can often cut off the new values 3 digits after they start to be differently because of the inaccuracy this method has. and for the squaring the first few digits are enough to say if it is above or below the wanted number so you don't really need to fully calculate that number either.
Fascinating thank you! It’s crazy we were never taught this in grade school
This is one of the coolest mathematical tricks I’ve seen, and it makes intuitive sense. Thank you!
Glad you liked it! I now use a much simpler method-please give it a look 👀 : ua-cam.com/video/xIpoxUPfuRY/v-deo.html
Nice work. I learned how to do square roots on paper more than 60 years ago but had forgotten how. Thank you for reminding me how it is done.
youre a lifesaver, unit final is tomorrow and i had no clue what i was doing
thank you so much! this video deserve more views
Thank you too!
The only and the most amazing video out there on UA-cam that actually explains it.
Thank you for your compliment but I no longer use this method and need to post a new video showing a WAY easier method.
@@AthenianStranger very excited to check it out, pls upload as fast as u could, and thx.
Thanks, the way you explain is just so amazing! I was not able to find any such video for help and seriously you helped me solve my problem.I was stuck in a question and it's now cleared!
you know he's legit when he pulls out the protractor to finish up that square root
I’m obsessed.
I can get 1 decimal pretty quick, which I think is neat if someone wants you to do it on the spot.
for 38 you take the difference between the closest perfect squares 36 and 49. the difference is 13 then you take the remainder from the smaller square so 38-36 which equal 2
and you divide the remainder by the difference so 2/13 = .15.... and add that you 6 and you got a 1 decimal approx and its within a .2% error, not too shabby IMO
obviously workes better for some numbers rather than others
Enjoyed this YT. Have not used this method for years, and am now refreshed on the process. Also triggered my memory for how to do cube roots and 4th roots, etc. by similar methodology.
I just recorded a much simpler method for approximating square roots. Here’s the link: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
I remember doing this trick in 7th grade along with cancelling out nines to check addition. But while I remember doing, I remembered it was somewhat of a convoluted process and could never find a good explanation of the method until this video. As an old engineer, I can still remember using the old ‘slip-stick’ or slide rule to get roughly the same level of precision.
Thank you for this video and verifying that I was not crazy and that there was a means to accomplish this.
I also remember doing matrix calculus to solve optimization problems by hand that are now easily solved by the multiple iterations on a computer. I an now 75 YO and still love remembering how things were done in the old days. Still have my slide rule.
Here's a less convoluted process (I no longer teach or use the method in the video you commented on): ua-cam.com/video/xIpoxUPfuRY/v-deo.htmlsi=Mr3o-MdTBeMC-LvP
This elementary mathematics was taught to our generation in 8th grade math class. Slip sticks were common also for generating educated math replies. H/P made the best slide rules available at the time.
i notice that the double is equal to the sum of the previous two numbers that were multiplied. So, 122 = 121 + 1. And 1232 = 1226 + 6. Could be coincidence of course, but if it's not then that's a slightly easier shortcut.
You are correct. I found another tutorial that did it this "addition" way, rather than "doubling". It appears you get the same numbers either way.
You've really explained it to my satisfaction.
Not the petty petty tricks that ends up confusing you more.
I learned how to do this many years ago when I was in school for drafting. Long since forgotten how to do it. Hopefully this refresher will stay with me, for no other reason than to know the skill.
I just recorded a much simpler method for approximating square roots. Here’s the link: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
My dad taught me this method, I have been looking at solving cube roots this way, it can be done but it's a little bit more complicated than the square root method, just an extension of it, at school I was the only one including my teachers who could calculate these surds, I taught my maths teacher how this was done whilst studying for a levels, that was some 40 years ago, good to see someone else knows this method,
I worked out root 2 to about 22 places and yes the numbers become crazy is 23 digits carried down at each stage, yikes!
Best Math Teacher. Thank you.
Wow, thank you!
For reference, I turn 72 this month. Back when I was in the 5th or 6th grade there was a TV comedy called "My Favorite Martian". In one episode the martian, "Uncle Martin", encountered a 13 year old, I think?, kid who was a genius and he could calculate square root. I asked my father if he could and how to do it. He showed me this very same technique. He didn't use the
Thank you for sharing your story and for your compliments.
My father showed me this 40 years ago - and also a similar method for cube roots (you bring down 3 digits at a time, and the running calculations are more complicated).
FORMULA:
X = the square root answer
Y = the square closest to the number
Z = the square root number
Difference = the Difference you get after Subtracting Z from Y
√Z = X . _____
Z - Y = Difference
Difference / X * 2 + X = X
Example:
√26 = 5 . ______
26 - 25 = 1
1 / 5 * 2 = 1 / 10 = 0.1 + 5 = 5.1
√26 = 5.1
For readability, every time you write "difference" you should replace it with a variable named "D" described as the difference of Z from Y
In most of the parts of India and asia this trick is taught to students at like 8-9 grade, the long division square root method
AWESOME! Thank you. Its just what I have been looking for to do Fire Dept Hydraulic water calculations for Gallons per minute coming out of the fire hose. Without a calculator. Formula is: (GPM=29.7 x d squared x square root of nozzle pressure) (d squared is the nozzle tip size. ie 1 ", 2", etc) (nozzle pressure can be 80psi or 50psi, etc)
I was never taught this in school and while I would never need to use it it's something I was always curious about how it's done
Thanks for explaining it!
You're very welcome! I now use a much simpler method-please give it a look 👀 : ua-cam.com/video/xIpoxUPfuRY/v-deo.html
Why? Im 82. I took square roots by hand for years before calculators could be held in your hand and could be afforded by school kids. But to develp this skill today is VERY low priority. I like your videos even though they are quite verbose. But your recent videos on doing calculator worthy tasks without a calculator seems crazy. I guess your slide-rule series will be next!
to be totally confident on the two decimal places, would you need to check on the third decimal place, in case it were to round up?
I'm 48 years old and never seen someone do this... Thank you.
This is easily the coolest thing I’ve learned today. Thank you!!! 😊
Thank you for the information.
You could also solve this kind of problem using something called Newton method. Say you would like to find the square root of a>0. Then the formula says that the next estimate x(i+1) is a function of the present estimate.
x(i+1) = x(i) - f(x(i))/f'(x(i))
In our case:
f(x) = x^2 - a
f'(x)= 2x
Hence, assuming x != 0
x(i+1) = (x(i)^2+a)/(2x(i))
Hence if we start with x(0)=6
x(1) = (36+38)/(12)=74/12
x(2)=((74/12)^2+38)/(2*74/12)=6.16441441
I agree completely. I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is the Newton-Raphson method, excellent indeed, but this second method is waaaaay faster and still pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
This is really cool I just tried it a few times and double checked with my caculator and I was correct. Thank you.
This could've helped me so much during elementary but better late than never, great video!
Say first approximation is 6. The the second approximation is the linear average of the first approximation and 38 divided by the first approximation: (6 + 38 /6) / 2 = 74 / 12 = 6 1/6. Which is already a slightly better approximation than 6,16.
For the third approximation repeat: ( I use 37/6 because that's more convenient): ( 37/6 + 38 / (37/6))/2 = ( 37/6 + 6x38/37 )/2 = ((37x37 + 36x38)/(6x37))/2 = (1369 + 1368) / (6x37x2) = 2737/444. 2737/444 - sqrt(38) = 0,00000041..
For the fourth approximation I get 14982337 / 2430456 which is only 1,3 x 10^-14 of sqrt(38). But I needed a calculator for the last one, I have to admit.
Yes, I recalled this approach from my school days (many many years ago). Applying this method to binary is truly magical! It's rather easy to create a square-root engine in digital logic! The only choices are between 0 and 1.
Thx for that. Just went through example after converting the radicand 28 to binary. Very cool!!
When I was I kid I asked my math teacher to explain to me how can I calculate the square root of a number by hand. She was unable to give me an answer. After a few decades thanks to you now I know how to do it :) Thanks!
I just recorded a much simpler method for approximating square roots. Here’s the link: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
This was first documented by Madhava of Sangamagrama in Kerala, India, in the 1300s. He also discovered the first recorded formula for calculating pi using infinite series, which is the basis of modern calculus, centuries before Newton or Lebnitz.
Good presentation. I was shown, in grade 7, a style, same method, of square rooting that was much smoother. Check out the square root of 529.4 using this technique. Cheers
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
@@AthenianStranger Newton's method is great because it can work with any continuous function and not just square roots, plus it converges pretty quickly. But, when doing computations by hand, you tend to get rather large divisions, even on the first iteration, which are very bit as difficult as those required using this method, making this method probably easier to do for square roots.
But the reason I commented is because that's a really neat approximation method you just posted, I hadn't encountered that before, do you know what the maximum error is using it? Does the error tend in a direction as you get larger or smaller? You should do a video on that method.
My Brain is the size of a walnut
A step back in time moment for me :D. I haven't done this since school back in the 70's!
This was so helpful 🥰 was literally crying out of stress since I had a math long quiz and I had no time to study 🥺 you explained it very clearly and this will help me so much 🥰
Nice! Please, always use the horizontal middle line in number 7 to distinquish it from 1. You sometimes left it away so it looks a bit confusing.
I've learned alot about squareroots on the way to this video. Here I learned how to really do it!
Thanks!
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
THANK YOU OMG I'VE BEEN FINIDNG A VIDEO FOR THIS FOR A LONG TIME SINCE I HAVE AN UPCOMING MATH TEST TO APPROXIMATE SQUARE ROOTS 🛐🛐🛐🛐
I use a different method I found myself that Is less accurate but much faster.
So using the same example of root of 38. I would say the two closes perfect squares are 36 and 49. Which equate to 6 and 7 after square root. So the sqrt of 38 is between 6 and 7. Next, we find the difference between or two squares which is 13 in this case. Then we subtract the smaller square from the sqrt we want which is 38 so 38-36=2. Now you put 2 as the numerator and 13 as the denominator and the sqrt of 38 is approx 6 2/13. Which would be about 6.15.
Like I said it is easier and faster but less accurate.
Pls review my method thanks.
Yes, except instead of linearly interpolating between 36 and 49 as you did, you should interpolate with a square-root function. You were close enough to the beginning of the interpolation that your answer is almost correct. It will diverge more from the correct answer the closer you get to the other side (49 in this case) and will also get worse the large the numbers are. But… your method is onto something and an excellent way to estimate the answer in your head very quickly without having to all this math rigamarole, which has very little practical use in today’s world.
You don't even need to do the difference between squares as the difference will always be 2 x the lower whole root + 1, so in this case the lower root is 6, so the difference between squares is 2x6+1=13. It helps when you do larger numbers, saves you having to find both squares either side. I also discovered this method by myself. Perhaps we should promote it as we can do these sums in our head in moments instead of the long process shown? You can also "weight" the estimate as it is more accurate closer to the root but will undervalue the root if midway between the lower and upper root. It also becomes more accurate the larger the number you want to square root. So, for example, root 4720 (68.702). As root 100 is 10 you can start with root 47 and get 6 + 11/13 = 6.8. Then root 1470 is 68 + 96/137 (68.700), which is just below 68 + 10/14 (68.714) or 68.7. By taking numbers two at a time you can do very large numbers by this method and have very good accuracy, and likewise could drill down into decimals to improve accuracy. For example root 7. On its own the simple way you would do 2 + 3/(2x2+1) = 2 + 3/5 = 2.6. If you make it root 700 and use the first answer you get 26 + 24/(26x2+1), or 24/53, or just below 0.48. So if root 700 is 26.46, then root 7 is 2.646 (actually 2.645). If you were insane you could then do root 70000 as 264 + 304/264x2+1 = 264 + 304/529 or just below 30/53 or 0.5, so 2.645. As well as being a quick method, each step also gives you the next accurate digit to use for the next step.
I was inspired to discover this by an ad that showed off a smart person when asked the square of 623 answered immediately 24.96. I realised that 623 was 2 below 625 and 0.96 was 0.04 below 25 and became suspicious. When I realised 0.04 was almost 2/2x25 I had the start of a theory.
It might be easier but it is actually NOT accurate - there is no such things as less accurate. Its either accurate or inaccurate. You answer is a whole .01 off.
@@FireMunki63 sqrt(38) is an irrational number, so all calculations can be considered approximations to the actual value. Even the calculator answer show on screen at the end is ONLY accurate up to 14 digits. The 15th decimal place could be rounded thus not accurate. The linear approximation method is a quick approximation method as it requires a lot less computation than what is shown in the video. The linear approximation is better when it is closer to a perfect square number. Try visualize by putting a straight edge through any 2 points on the graph of the sqrt function. The biggest error in linear approximation is in the middle as the sqrt function always concaves down, which James mentioned in his comment above.
I am from India and I only find ur video useful, thanks for ur help 🙃
Thank you for your compliment and good luck to you!
Thank you! Your video helped a lot!
@AthenianStranger Do you have a video or a link to one where it shows why the method is as such? I'm somewhat more interested in why this method works, more so than only knowing how it works.
1. Closest perfect square below √38. That would be √36. So we start with 6.
2. 6 + a fraction. Difference in radicands between √38 and √36 for the numerator. That would be 2.
3. Double 6 and put that in the denominator.
Result is 6 + (2/12) = 6.166. Quite close to 6.1644 . . . .
Note: you can also use this method with the closest perfect square above, but then you subtract the fraction. Try it for √120.
11 - (1/22) = 10.9545. Quite close to 10.95445 . . . .
Or, you can just use your calculator. 😆😆😆
Can you explain why this works?
Thank you for releasing this video! I always wanted to know how to find the square root of a number. Athenian stranger, you know, you can also find the square root of a number by using prime factorization.
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
Thank you so much!! Your video is the only one that made sense and I've watched many. One question though, when choosing to "fill in the blank" is the last number always the same as the number it will be multiplied by? If so, why? (This is at 3:19 and 5:33 in your video). Thank you very much!
Thank you. I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
@@AthenianStranger awesome! I am excited to try both methods. Thanks for reporting back & the new methods. I really appreciate it!
hey thanks for this video, but i have a problem: it doesnt seem to work when the square perfectly multiplys into the number. i wanted to square 48, but the number 6 can go into it without a remainder. I am left with 0 and cannot use this. is there something i am doing wrong or does this not work without remainders?
I don't use this method anymore--lookup the Newton-Raphson method. I should post a follow-up video because there's an even easier way to approximate roots to the tenth's place.
One thing that I have observed is that if a number is not a perfect square then its square root has an infinite number of significant digits. In other words, you can go on forever making the square root more exact. I also want to show you something weird, but it ends up working anyways.
38 - 36 --> R = 6; D00 = 200
200 - 120 * 0 --> R = 60; D00 = 20000 ; Whoops! I picked the wrong digit. I should have picked "1" instead, but there is a reason for this stunt.
20000 - 120[16] * [16] = 544 --> R = 60[16]; D00 = 54400; This is exactly the same remainder that you had after 3 digits; Therefore, we are now back on track. If we were to stop here then you take the quasi-number 60[16] and do a conversion like this --> [16] is too high a digit for base 10. So you subtract 10 from [16] and make that a "6" and carry a "1" (1/10 of the 10 you just subtracted) to the next place on the left. That then causes the "0" to become a "1", and you now have "616" as you are supposed to have.
To calculate 120[16] * [16], I went [16] * [16] = 256; leave the "6" and carry the "25"; I then went 0 * [16] and added "25" to get "25"; leave the "5" and carry the "2"; I then went 2 * [16] + "2" to get "34"; leave the "4" and carry the "3"; I then went 1*[16] + "3" to get "19"; no need to carry since we are at the end; So, we have "19456"; 20000 - 19456 = 544.
Basically, this worked because all numbers in base 10 are the result of plugging the value "10" into a regular polynomial (a polynomial in which all the powers are integers) in which all the constants are integers, too.
You're right about the significant digits thing; in fact, it's possible to prove that perfect squares are the only numbers that can have rational square roots. If the root of some number x can be expressed as a ratio of integers m/n, you can deduce almost immediately that m and n must both have a factor of x. Therefore, the ratio m/n must reduce to an integer in order to avoid an infinite loop, which means that x is a perfect square.
@@Cornix94 I assume you mean to say, "perfect squares are the only integers that can have rational square roots." Small but important difference.
Hello, I have a question. At the part where you said to double six, is it because 38 is being divided by 6, or because the first digit to the answer (of 38 divided by 6) is 6? Which 6 is being doubled: the divisor 6 or the quotient 6?
I am now approaching my 84th birthday and although my memory isn't what it used to me, I cannot remember even once needing to know the square root of any number.... or calculus or trigonometry, for that matter. I let others waste their time learning that stuff and I just called them whenever I needed to know.
However, a little basic algebra has come in useful and I can add, subtract, multiply and divide in my head (which is more than most teenage cashiers can do these days.)
why does this work? it's cool that you can do square roots by hand, but why does this work?
I kind of get the feeling that this should be taught to all grade schoolers by the time they reach the 5th or 6th grade. It is a shame that it isn’t because the Board of Education has failed us.
Preach brother. I couldn't agree more. I teach 7th-12th graders and make them approximate everything possible without a calculator. I don't use the method shown in the video anymore and need to make a follow-up. At first, most kids get frustrated with all the pencil math but they quickly become empowered by their ability to do "big boy" math without a machine.
At least in this case, it would be quite easy to find the answer with a binary-style search, anyway. This weird division stuff seems a lot harder than just "square 6.10, 38 is bigger than result -> square 6.20, smaller -> square 6.15, bigger -> square 6.17, smaller -> square 6.16" and hey presto we've got 2 digits... and if you want to be sure of which way it rounds, you can check 6.165^2 and see that it's bigger than 38, therefore 6.16 < sqrt < 6.165, therefore at two decimal places it rounds to 6.16.
IOW, you can get closer and closer by looking where you "expect" the square root to be based on squares you do know (f.e. 38 is between the squares 36 and 49 so the root has to be between 6 and 7, or even 150 is between the squares 100 and 256) and simply seeing what square you get out. You can even optimize it using your intelligence; rather than doing a true binary search starting at 6.5 (the midpoint of the search range), you can start at 6.1 or 6.25 or something since you know 38 is much closer to 36 than it is to 49.
Agreed 👍 I now use a much simpler method-please give it a look 👀 : ua-cam.com/video/xIpoxUPfuRY/v-deo.html
Thank you, very nice to learn how it's done
It is still better just using a Taylor expansion centered at the closest squared number to your desired value:
√x ≈ √xo + (x-xo)/(2√xo)
Use xo = 36 since this the closest number to x=38 with exact square root:
√38 ≈ √36 + (38-36)/(2√36)
√38 ≈ 6 + 2/12
√38 ≈ 6 + 1/6
√38 ≈ 37/6 = 6,16…
With this method you will always get some excess in the aproximation, this is: √38
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
Ty! I am a 7th grader wanting to learn lol
I AM GRADE 8!!!🥲😅
👌
I invented my own method several decades ago which initialy started with trial and error and then I refined it into more of a system which requires two calculations per decimal place.
Awesome! Thanks for the lesson!
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
Did not see the justification for multiplying by 2 at each stage. The process you presented certainly produces great estimate. Maybe it is something to do with Newton's Method. Thanks for demonstration.
Simple… the square root of N is equal to X + Y DIVIDED BY THE SUM OF X + square root •••
where N is equal to X^2 + Y
The formula for determining the square root is (10X+Y)(10X+Y) which equals 100XX+20XY+YY. 10X is the number already determined and Y is the next digit being determined. X is times 10 because the digits on the left are 10 times that of the new digit being determined as a place holder. After the first digit is determined the 100XX is no longer used and you have left Y(20X+Y) for determining the next digits ad infinitum. He is multiplying the number already determined(X) by 20 then adding the new digit(Y) before multiplying by the new digit Y. The space he is leaving to put the new digit is the where the zero is of the 20. This method is more than an estimate as it is accurate ad infinitum.
This process gives exact decimals and is not based on Newton's method (at least not directly). The 2 comes from the expansion of a binomial square. At each step, consider x as the current estimate (6, 6.1 or 6.16). Then we have 38 = (x + c)^2, where c is the remaining decimals. (x+c)^2 = x^2 + 2xc + c^2. So the remainder is r = 2xc+c^2 = (2x + c)c. This corresponds to the xxx_._
It doesn't produce an estimate, but calculates the real value.
The justification for the two is probably (it's just a guess of mine):
(a+b)^2 = a^2 + 2b + b^2
So, when you square 7.2, it's (7 + 0.2)^2 = (49 + 0.04 + 2 * 1.4) = 50.44
Now, when you want the squareroot, you have to do the inverse process, so you have to get rid of the 2 somehow.
(a+b)²=a²+2ab+b²
The "2" in the "2ab" term is where the multiplication by 2 comes from.
For example, suppose you are trying to find the square root of 1849. You can tell right away that it is between 40 and 50. So, let (a+b)²=1849, and let a=40. Then we try to find b.
1849=(40+b)²=1600+2ab+b²
Therefore, 249=80b+b²
which we can rewrite as 249=(80+b)b
Notice that we *doubled* 40 to get the 80 in this equation.
Then we can use guess-and-check to find b=3
and therefore (a+b)=40+3=43. Therefore, the square root of 1849 is 43.
And what happens when the number is 12 digits?
You start with however many digits you can pull a square root from (pairing two digits at a time to the left of the decimal point. If you have a number like 123456789012, you break it up to 12,34,56,78,90,12. As 3*3 = 9 and 4*4 = 10, the first digit is going to be 3. Repeat the process, bringing the next two digits down, instead of 00. The decimal point in the answer goes where after the digit you get when you bring down the pair of digits immediately to the left of the decimal point. To check, the number of decimal points in the answer is the number of pairs of digits before the decimal in the original number. If there is an odd number of digits before the decimal in the original, just put a zero in front of the front digit. For instance, treat 147 as 01,47.
Your paper magically summons a calculator :)
A year later. Just pair up every 2 digits from right to left. The first number will always be solo or part of a pair (your case).
It would be interesting to solve this also by the Babylonian method, and assess whether that takes more or less hand computation. Or perhaps work it until we feel we've done an equivalent amount of hand computation and see how many digits we got with that effort. When I start with an initial guess of 6, the Babylonian algorithm gives me 6.167 after just one iteration. Granted, though, that was using an exact division of 38 by 6. So, 36/6 = 6.333, (6.333 + 6)/2 = 6.167. It converges very quickly, but I don't have a good feel for the work involved with doing that by hand.
Ask and you shall receive: I wholeheartedly agree so a few months ago I made an updated version of square roots by hand using a modified version of the Babylonian Method (way, way easier):
New Video: Find Square Roots w/o Calculator by Hand Quickly & Easily
ua-cam.com/video/xIpoxUPfuRY/v-deo.html
Nice one!! I was my daughter asked me exactly that just a few days ago! Thank you, I'll show her your video. 👍
I strongly recommend my updated version of this video for your daughter because in the new method I use a much simpler and more intuitive method. Please take a look at the new video and let me know which method was most helpful for your daughter. New video: ua-cam.com/video/xIpoxUPfuRY/v-deo.html
this process can be done to infinity
What would be the process for square root 37 this process can't be applied. I had tried. For 100 121 would be greater🤔
There is a smaller number than one. You can use zero. Same process.
Thanks. Interesting channel. Subscribed. Cheers.
I WANT MORE!!!
Love it!
Wow i got this