Take Any Square Root by Hand - Easy to Learn!
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- čas přidán 13. 07. 2024
- In just three easy steps, learn how to take the square root of any positive number, regardless of whether or not the result is rational!
The long division method is fast, efficient, and not too much different from dividing numbers normally.
If you have any questions, feel free to leave a comment and I'll get back to you as soon as possible!
I am approaching 80 and this is the first time someone has had the ability to explain it to me. Thank you!
Almost 9! ;-)
@@tmst2199 8.944
When I was in elementary school in the 50s, having a teacher that was really competent in Maths was rare. I rather doubt that they are much, if any, more common today.
I am currently in London next weekend for a subscription to work on bed and bed for I am currently in London next weekend for a subscriptions to work on bed and bed for
@@carolinahenao3330 Is this posted to the wrong thread?
That was exactly the method I've been shown in school 35 years ago. I have forgotten the exact algorithm, and now that my daughter has asked me if it is possible to calculate a sqrt by hand, I am happy to share that with her. Thank you, sir!
43 years, 1979, for me. It was just before calculators became affordable and compact on a wide-scale basis.
There’s nothing like helping your kids understand math. (Makes us seem pretty smart. Kids are proud of smart parents.)
It would have been much easier to remember if you simply had been taught to multiply the currently completed root by 20 in each iteration. Thats about it! 20a
Use that as a mental divider to "estimate" your next digit, b. Add b to the 20a figure, multiply by the b. Put that below the remainder in your LD configuration. Subtract. You now have your new currant remainder.
The material in this vid is a whole rote of rote and routine which she will easily forget, just as you did. She will have forgotten the vid material in a month. She will never forget just multiplying the currant remainder by 20, adding the new digit b, and multiplying by the new digit.
The method I outline above is the old tried and true method, the one in the vid isnt.
This method is really old . My primary school taught me this over sixty two years ago when I was preparing for Common Entrance Exam. Her name was Miss Valerie Retemeyer. God bless this dear lady
So?
@@andrewwallace1218 show some respect
Isn’t it good to refresh good ideas?
I’m in India and I learnt it last year ;-;
@@stak321 Did u knoe that literally almost everything u study has been there for centuries
I always remember this method as Horner’s method. Your description of the different stages is how I remember it, except that at each stage you double the number above the line, I.e ‘2’ becomes ‘4’ then ‘29’ becomes ‘58’ then ‘291’ becomes ‘582’ etc and then determine the red digit as you describe.
Same for me. Thank you to point it out.
Yes. That is how I was taught. We also put the 'divisor' on the same line as the 'dividend'.
2 questions:
1) Who is credited with discovering this method?
2) Why does it work and where can I read about that?
Did anyone find/know what the answer to this comment is? I really want to find the book or paper that this method came from
@@thomasbates9189 well.. what about this? en.wikipedia.org/wiki/Long_division :-D
@@thomasbates9189 The Muslim mathematician Jamshid al-Kashi developed this algorithm in his book Miftah al-Hisab (The Calculator's Key), in 1427. Albeit $90, the first English translation has recently been published of al-Kashi's Calculator's Key by my professor. I've been studying this algorithm (and others by al-Kashi) as a part of our history of Islamic mathematics class. If you can find access to J.L. Berggren's episodes in the mathematics of medieval Islam, the justification is laid out in chapters 2.6-2.7.
@@gildoringlorion3484 do you also have a Geek and Egyptian math history class?
It is based on the binomial theorem: a²+2ab+b²
At 5:12 you get the 58. A much simpler way to get this is to DOUBLE the EXISTING answer gotten SO FAR. Double 29 = 58. Do the same at 6:41, doubling existing answer (29) to get 582. At 7:08, you talk about decimal point. This is much simpler if you write the answer digit over the last pair dripped down (spacing answer digits out). This way the decimal points will be vertically aligned. At 8:24 you double 2911 to get 5822. To illustrate the alignment, I show the final result calculations aligned as I suggest.
2 9 1. 1 4
4 8 47 65.23 47
4
58 4 47
4 41
582 6 65
5 81
5822 84 23
58 21
58228 26 02 47
thanks
Brian, I agree with you. Double the 'quotation' (refer to the square formula (a+b)*(a+b) = a*a + 2ab + b*b.
Brian, I learned your way over 60 years ago! I think it's simpler than the method in the video.
@@bobward2322 You ought to see what happens if you try this method in binary. It becomes totally awesome, as calculating the next text value reduces to appending "0 1" to the end of the existing answer!
@@paulc3960 You ought to see what happens if you try this method in binary. It becomes totally awesome, as calculating the next text value reduces to appending "0 1" to the end of the existing answer!
Most useful - I had learned this in school about eighty years ago, and later after learning the easier method of Log table-use did not have to use it! However, delighted to be reminded of this method again that should not be forgotten and remains useful for teaching youngsters that creates their interest in learning maths comfortably. - Ashit Sarkar (89 yrs young)
BEWARE THE LOG TABLE! There is a distinction between INTEGER SQUARE ROOT (method presented in this video) & "approximation" methods such as Newton's method, & log/antilog tables. Sometimes it is ABSOLUTELY important to get the largest possible number whose square is equal to of less than the number to be rooted. All of the approximations can't GUARANTEE that this will b true, NO MATTER HOW MANY PLACES YOU CALCULATE beyond the precision you need. Calculating more places gives you a BETTER CHANCE of being correct, but no GUARANTEE.
I tried to learn this 230 years ago but never got the hang of counting :)
@@tim40gabby25 bro, do you even know how long that is? That's like a decade!
@@noon1117 Yes, it is a long time. It's like 23 decades, or Two (2) centuries and three (3) decades. I do not mean to be insulting, having fun.
@@pher5179 that was the continuation of your joke; in this case, of not knowing what a decade is ;)
I am the class of 1957 (yeah, that song) I was taught this by one Mr. Lynn Mauntner (amazed I remember his name). He also taught us logarithms, slide rule using a log log antilog slide rule (a four foot long one hanging on the wall) and after class he would show the few of us who hung around how to do things not in the syllabus for the class, like taking the cube root. I remembered (mostly) how to do the square roots and after you did the second step I was back in the rhythm (thanks) . Taking Cube roots I can sorta remember but I will need a bunch of hand holding (never used it after the precalculus course in 1956)
learnt this method in 1972. I guess I was a less student than you are. You have my respect, Sir
I remember slide rules too. This square root procedure I don't remember ever learning until last year. I wanted to know how it worked, so I had to derive the steps to the procedure starting from the symbolic assumption that the root has the form of (a + b); then expanded it into the square (a² + 2ab + b²). The '2' in the '2ab' is the doubling. Great fun deriving it. I even used a spreadsheet to run the procedure for me; of course, it finishes instantly, but every digit result and sub-result is in the spreadsheet. Even expanded it to do cube roots too.
@@alterherrentspannt You have more ambition than I. As soon as an RPN hand-held calculator was available from
HP I went to a calculator and never looked back. My compliments to you.
I never learned anything like this, and math was my strong suit and I studied engineering. I guess when calculators came in things got lazy.
@@alterherrentspannt I can't remember if I ever learned how to use a slide rule.
The last time I saw this was when I was in high school-1956 through 1959. We did it slightly different, with the writing of the numbers to the right of the system. We also left the square root sign in place instead of drawing a division box. We did need to remember to put a +/- in front of the answer, however. Thanks for the memories.
The root is always positive! (By definition). Of course,if you take the negative of the root and square it, you will also get the original number.
@@torstenbroeer1797 ... I think you're (i)magining things! ...I'll show my self out... ;p
I remember my Dad patiently, and carefully explaining this to me in Grade 3; while I sat in astonishment, shock and fascination thinking: "But I just wanted the square root of 25!"
I did remember the details well enough to put it all together some years later in high school, when my math skills were somewhat more mature.
Thank you for prompting a fond memory.
One point my Dad made was to place each digit of the result above the corresponding group that generated it; making placement of the decimal point quite obvious.
Love you Dad. Miss you every day.
Your father's advice makes sense.
the square root of 25! is probably a very big number
@@DOROnoDORO ;-)
Thank you, man. I cannot believe that the best explanation for this is coming from such a small channel. Keep up the good work.
Great stuff, I've been a math nerd for a long time always happy to see things like this out there for the world to learn.
brilliant! i remember learning a method of taking a square root from a book but never mastered that method, especially since I learned it on my own (they didn't teach taking square roots from all sorts of numbers, only the obvious ones). Thank you for making this video! Very clearly taught, too. :)
I'm 40, an engineer and part time math and science tutor, and since high school I've been pretty good with being to do everything mathematically by hand. And I've never learned this method lmao. Love it!
I'm 70 and learned this at school. I met professors of mathematics who didn't know this method.
@@xbia1 Im 71, did excellent at maths, but this was never taught in any class I took. I soon taught it to myself from an old encyclopedia. My method is better -- simpler, easier to remember. And more likely the ancient one.
In each iteration, I simply multiply the current root by 20, estimate the new digit, add that to the 20x figure, multiply by the same digit. Done! Subtract from the current remainder and proceed to the next iteration.
Sometimes people just like to make things sexy, rather than simple. As in this vid. 95% of the kids learning this from the vid will have forgotten it in 2 months.
That is because the vid method is rote and routine, with little understanding of why one is doing what one is doing.
I used to teach this all the time in Ontario. Past the initial step, I used to say " Double the top, write it on the side leaving a space. Whatever fills that space must also go on top" Seems easier to do than your adding procedure for the side number.
Two details.
I would describe what you do as multiplying by the last digit.
I would also write the digits of the result over each GROUP. That way the decimal point automatically comes in the right place.
Yes, thank you, these small things were on my mind as well. Not complaining! But would be a slight improvement in clarity.
@@EmpyreanLightASMR Came to say write the digit over the group (how I learned it as a kid.)
Thanks! I learned this in 5th grade (1960) from Mrs. Simms, but then forgot it. I looked around a couple of years ago and everyone on the internet wants to do a much more complicated and barely more accurate method. But this is more than adequate. Most of the time I want to use it is to estimate the Square root of a hypotenuse to the nearest unit value. Basically, "How much would we save if we cut across instead of going around?". Unlike the other methods, I can see doing this in your head.
Nothing is more accurate... Each digit is exact, and you can perform iterations indefinitely
Thank you. Very well done.
I am a retired QC mechanical inspector.
Always did my math with pencil and paper.
I am old enough to have gone through grade school pre-calculator. Never taught this, but, now, I'm going to watch this video again, and do a few square roots by hand. This is amazing. (To the nay-sayers - if you get a wrong answer, you messed up a step. As a math teacher, I'm very careful before declaring a book has a typo, or a fellow teacher, a wrong answer.)
Math teacher who didn't learn how to extract the square root "by hand" ... WOW
"I teach, therefore I learn"!
@@shaguna -- I think that's entirely possible. This was never taught while I was in school. I saw it in books on my own, but if I wasn't a reader seeking out information because I like knowledge, school certainly would not have been sufficient for me to be anything other than a worker drone.
This takes me back to my first year of secondary school. So aged 11. Afterwards we were taught to read off the log of a number from a book of tables and divide it by 2 and then take the antilog. This was also back in the days of slide rules. What was great about them is you had to have an idea of what the answer should be so could catch errors quickly and get the decimal point in the right place. Even at University we had a lecture on fast estimation. Look at a problem, simplify the numbers (ie. PI is 3), do a rough calculation and then find you have gustimated the solution accurate to two decimal places. Calculators are great but they have killed people when they give rubbish answers and are believed.
I remember using slide rules! My dad was Air Force, so I went to DoD schools abroad. I guess they were more advanced than in the states, because when we transferred to Texas when I was in grade 6, I was like a year or two ahead of everyone. I had to sit in a cubicle off to the side of the class by myself and study separately from the rest of the class.
The great thing was I got to go to math contests, which were actually pretty fun (I was competitive back then). Like you said with your log tables and having to have a good guesstimate of what you were looking for beforehand (like 5800 is *about* 6000, so I know I need a 4 digit number, etc), using slide rules made you actively think ahead while calculating. And the competitive environment made you think quick on your feet. Seems like I was a lot smarter when I was younger... now I have only vague concepts of the methods. My math skills are 2-3 times slower than way back when.
I was in high school in the late 1970’s. This was the beginning of the era of hand held calculators.
My high school algebra teacher didn’t want us to use calculators, in class.
He always said “Garbage in - Garbage out”.
Did all of the above in community college. It does kinda give away our age group, doesn't it, lol ?
Although slide rules have been out of use for more than 40 years, schools still teach children how to do rough calculations, mental arithmetic, and pencil and paper calculations, at least in Germany.
@@suryahitam3588 Good for them. Being able to do quick mental math can be very important at times.
I learned this method in elementary algebra class in junior high school. Most kids didn't have calculators back then, so the teacher taught us how to compute square roots with pencil and paper. But I totally forgot how to do it afterward. It was one of those things you learned to pass your exam and then dropped.
Well back to Jr hs with you then. I don't think there any more Jr hs. Everything is middle school. I know one suburban city that built a new school and called it Central Middle School. I guess a lot of thought went into that one. I think I graduated from the last 3 yr hs in the country. It finally went to 4 yrs as # of students dropped.
OMG! I had learned this in the 7th or 8th grade and back then this was easy to me and I liked it. Then over time I forgot how to do and spent the next 10 or 15 years trying to find someone that knew how to do square root by long hand (as we called it) and never did find out how to do it again. And then you come along and retaught an old fart how to do it again. I was spell bound when you started then little by little it all came back, amazing. Thank you very much, although I don’t really use it or need it but just seeing it one more time was fun. Thank you for making, now a 73 year old man happy for a little bit. Many times over the years I would tell people that I once could do square root on any number by long hand and they either just looked at me like I had three heads or they just laughed their asses off. Ahhhhhh, now I can go in peace. 😂
I'd like to see if you could do this in your head or on paper without taking much much longer than using a calculator. Lol
I learned this method in my high school algebra class, in 1979.
This was the beginning of the era of hand held calculators.
One day, during class, I went to my teacher’s desk and asked, “Before we had calculators, how did people figure out the square roots of numbers?”
He showed me this same method.
All these years later, I have never forgotten it.
Thank you for sharing a video about this.
This method of solving square roots reminds me of my childhood 70 years ago. This is what we call it mental math. I still remember this method and used it my whole life when required without a calculator or computer. thanks for refreshing my knowledge. There was no calculator those days.
Great explanation ! When teaching a subject that you are very familiar with, and your students are not at all ... NEVER use words like "simple" and/or "easy" in your explanation. Rather, show enough examples such that the students will make a connection and understand the concept. Not everyone will grasp it no mater how many times you show it. That is just the bell curve in action.
"such that the students will understand the concept": No "concept" was presented here. If it had been, we would know how (why) the "process" works. Do you know "why" it works?
@@toddmarshall7573 you're going a little beyond the point abz's trying to make, but I'll humor you: in what question is "why" a demonstrably sensible word, and not "how"? ;P
Fully agree, though, somewhat technically, with your analysis. "Concept" was used in the wrong way.
Baloney. I tell my students when something is easy or simple... because it is. He said "easy to learn" not "easy" and unless you're at the extreme left end of your bell curve, severely brain-damaged or have a sub-70 I.Q., this *is* easy to learn. I learned to do it in the 5th grade.
I can’t believe that I was never taught how to do this. Thank you so much.
By the comments here i guess, it isn't taught any more now.
This is really great. I am 65 and I never heard this ever. Thank you very much !!
I learned this square-root process long ago when I was a kid, but I had forgotten it... just the other day I was remembering that I used to know how to do this... so I was delighted to stumble across your video today (thanks CZcams for listing it), and be reminded of some fun times in distant long-ago mathland.
Me too,
We did it in high school in class 8th and I can still do any number by hand.
Just want to add that instead of adding 9 into 49 the simplist way is just double the 29 at the top in answer place. Which still give you same thing. Similarly in first place double the 2 in black color instead the red one. Anyway your way is also correct!!
Yes, I learnt this in school in the !960s
It was taught to us as the " plonk" method !
Without paper or pencil, the sqr(6) must be between sqr(4) and sqr(9), that is, 2 and 3. Linear interpolation of 6 between 4 and 9 gives (6-4)/(9-4) = 2/5 = 0.4, which is the offset of 2. So, the result should be approximately 2+0.4 = 2.4.
then square 2.4 (take 24*24 and then divide by 100)... to find 5.76 and you know a) your answer is quite close and b) the actual answer is more than 2.4 You can also continue this: Try 2.5*2.5 = 6.25 - 6.0 is almost halfway in between 5.76 and 6.25 so you could guess just below 2.45.
Actual answer is 2.4494897, so you get pretty close quite quickly this way too :).
If you're not familiar with the Newton-Raphson method, you can quickly get a bunch of digits using some simple arithmetic.
Use your method to come up with an initial guess.
x0=2.4
x1=(2.4 + 6/2.4)/2=2.45
x2=(2.45 + 6/2.45)/2=2.4494898
You have the answer to 6 decimal places in just 3 steps.
I'm 63. When I was in 9th grade, I learned to do this in algebra class. By the time I took chemistry in 11th grade, we had calculators that would take square roots, so I never used it, and forgot it after a few years. It's amusing that it has "come back around." Nice job of teaching it. The reason you add the last digit to the number on the left is because you are doubling the number on the top (the estimated square root). This is because (x+y)^2= x^2+2xy+y^2. If you untangle that expression, you can see what's going on and why this method works. (x is your estimate, y is the additional part of the square root you are trying to find). I enjoyed your video, thanks!
My 89 year old dad was showing my teenage daughter this the other day. He said he learned this math when he was a kid. He says they don't teach kids how to do math today...everything is done with calculators. So true.
Things in school are eliminated, when technology makes them not needed anymore. For instance: Math of course, because of calculators. Cursive handwriting, because who actually writes formal letters anymore, when there are computer word processors (on a side note kids can't ever sign their name cause of this). Probably many more that I can't think of.
So underrated!
This is soooooo cool! So I made at least 20 other square-roots! It’s just too much fun! What a great method! Thanks!
lot of people are saying this is underrated but it has been used in india from a lot of time
Lots of people assume that everyone learns and understands things the same way they do. The best teachers know that everyone learns and understands differently.
I saw this technique one time in junior high over 40 years ago and never again until just now. Thanks for posting!
Excellent explanation. We were made to do this in year 9. Used log tables and sin and cos tables, mean difference tables and both types of root tables. No calculator until year 10 (1978). Gave a very good grounding in arithmetic!
If only kids today would be as lucky! Basically the education system is misguided. Basically they have given up. As for maths ability among the young, the US, the people who put a man on the moon, has become a scandal among the nations of the world. It has been like this for decades
I think educators today dont even realize what (teaching things like this to young minds), does for the development of those minds. You have a chance when they are young, and then it is gone
Kind of sad, really
Thanks for showing this. I had learned it back in grade school, but this refresher is great. I do have a few clarifying suggestions. 1) When the tutorial says take the number (out to the left) and add it to the "first" digit of itself, it should say add it to the "last" digit of itself. 2) In the answer at the very top, the number placed there should be "centered" over the group of numbers to which it pertains. In the example, the 9 in the answer should be centered over the group of 47; and the 1 in the answer should be centered over the group of 65. 3) Keep the decimal in the answer at the top in the same exact position as the decimal place in the subject number. That is, the two decimals should be exactly one on top of the other. This keeps everything a lot clearer.
The alignment you described is how I learned this as well. Keeps you from getting confused about where you are,
I personally think that both "First" and "Last" are equally confusing, since they're both relative terms... Perhaps the ideal solution would be to refer to it in a more absolute sense as the "Ones" digit instead.
I remember my second daughter in 6th grade doing this, and she had taken the number out to 6 digits, but got frustrated because the calculator she used to test "her" number didn't match the tables in the calculator. I used mine at the time to shew her that her number was correct; it was a cheap calculator she was using, and it didn't give all of the decimal places.
No school like the "no tech" old school, i.e., Pencil (with a good eraser) and Paper!
Thanks, Tet, for the excellent video!
I saw my Algebra teacher in 7th grade doing this for a student at the end of the class before me. Always wondered what she was doing. Now I know!
A couple of comments: another way to do the step of finding the next digit by guesswork is simply to pretend you're doing long division and cover the last digit of the "dividend" (in the last step completed, it was 260247, and simply cover the final digit with your thumb and consider how many times the full running "divisor" (in the example, 5822) goes into 26024, and to me at least, that seems to be a simpler way to do that step. Next comment: it's helpful to know that each time he's added the last obtained digit to the running "divisor", what you're actually doing is simply doubling the entire result obtained so far. In the example, when he adds 1 to 5821 to get 5822, that number is simply twice the answer so far obtained at that point (2911*2=5822). This last observation makes it more clear how and why this whole method works. Consider what you get when you square the expression a+b...you get a**2 + 2ab + b**2. (I hope everyone knows I'm using ** to mean the operation of squaring). Every time a new digit is obtained in the result, all the previous digits form the "a" value, and the newest digit obtained is the "b" value. The 2ab is the one that's harder to explain, so pretend in his example that the number having its square root taken is simply 841, which is a perfect square. You'd obtain the same results as he did up to the point where the 9 in the answer is multiplied by 49 to get 441. Now with 841 being the number having its square root taken, the number produced when the next group of 2 digits was brought down will be 441, NOT 447, and you'll be subtracting 441 from 441 to get 0, so with no remainder, the original number was a perfect square, no need for any further steps. In this simpler case a is 20, b is 9, and 2ab is 2*20*9=360, so a**2 + 2ab + b**2 is 400 (20*20) + 360 (2ab) + 81 (9*9) = 841. The hard part to explain is why I'm working with 20 and not 2, the first digit obtained. But what's happening with the numbers he produces in red includes as a hard-to-see step a multiplication of the running "divisor" by 10 before adding on the next digit obtained. I'm sorry, that may be as clear as mud. I tried to explain it better, but the fact remains that the formula (a+b)**2 = a**2+2ab+b**2 forms the foundation for this method, with "a" representing all the digits obtained so far and b representing the next new digit in the answer.
Exactly. See my comments elsewhere. I've done all the steps in this procedure and even used a spreadsheet to do all the steps instantly. (a² + 2ab + b²) = (a + b)² is the starting point to derive the procedure where 'a' is the first guess and 'b' is the next digit approximation to the root.
I _think_ the common symbol for the exponentiation operator is ^, so squaring "a" would be "a^2" rather than "a*2". That's been my understanding. If anyone wants to correct me, please do so. Just trying to elevate confusion. ;-)
@@dienekes4364 Sorry, I was using keyboard codes for exponentiation. ALT-0178 gives the symbol for the high-two. Like here a². Here are some other special characters: ü õ ß é ä
@@alterherrentspannt Oh, cool. My response was to liesulin, not you. I was wondering how you got the high-two. I didn't realize there was an ascii code for it, but I'm not surprised. Good to know! :-)
@@dienekes4364 I see that you're right. I picked up the ** thing somewhere, not sure where. I also see that in my initial comment I typed "a*2" one time when I meant "a**2". Anyway, you're right.
Chad, thanks mate this helped a lot
My father (who finished school in 1927 at the age of 14) taught me this method when I was about 10 years old. A couple of years later in grammar school the maths teacher (Mr Hodges) was teaching us square roots by prime factor which only worked for perfect squares. He claimed the only way to calculate the square root of a number that was not a perfect square was to use log tables. When I told him I could calculate square roots without log tables he basically said I was wrong. So I went to the blackboard and showed him the method described in this video. He was amazed and, to his credit, he apologised to me.
I then showed him how to calculate cube roots by a similar method.
Subbed and adding a comment to help boost your channel with the algorithm. Legit love me some math channels. Keep up the educational videos.
Cool! I learned this method (or something very similar) in 8th grade (thanks, Lynne!).
In high school I came across a book that explained how it works, based on the fact that (a+b)^2 = a^2 +2ab +b^2, and it even showed a similar method for cube roots, which could probably be extended to fourth, fifth, etc. roots.
Do you Know what book it is? Or where can i find lhe explanation? Thanks
@@ptrexy This is going back to a book I borrowed about 40 years ago, so I'm sorry, I don't remember much about the book other than that algorithm and that the cover (paperback) may have been red... :/
But this page in Wikipedia:
en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation
describes it pretty well (dig down to the heading: "Decimal (base 10)" if you want to cut to the chase...).
As a method for calculating square roots, it's slower than many others, but it is interesting. Have fun!
It gets VERY cumbersome for roots above cube roots. For 4th roots it is way easier to just find the sqrt of the sqrt. For the 6th root, the cube root of the sqrt, etc.
@@herbertlong3981 True. For practical use, you are probably better off using logarithms, etc. (Or a calculator... ;) ) What I like about these methods is the way they show you what's happening, even if they aren't very practical for day-to-day use..
@@zevfarkas5120 It is very common for people who make videos of the cuberoot "long division method" to declare after they have done it to 3 digits : "it becomes so tediously time consuming and cumbersome that it is impractical"
As for cube roots, and of course sqrts, that is not really true. If you apply yourself, you can teach yourself to find the CR of any number to 7 place accuracy in less than 10 minutes on a 4 x 6 inch piece of paper. That is, pencil only, no calculators before or during, no separate worksheets, no erasing, and w/o any "savant ability" involved. To 10 digits in less than 20 min, on 2/3rds of a standard sheet. I have even shown that I can do it to 25 digit accuracy on one side of one sheet, taking roughly 2 hrs.
I have never seen a vid of my improvement on utube. (Although I believe that anyone interested in this should very shortly and instinctively be led there). Even sent a page to a maths teacher who had made an LD method vid, but he returned with -- "I cannot figure out what you did there, but it seems very clean and concise. But I will study it and return to you". So far I havent heard from him. I was kind of mystified as to why he couldnt figure out what I did, because it was just the LD method with several very obvious improvements
I guess I am the only person in the world who has developed a practical method to evolve cube roots by hand!! (-;
I was taught this long ago in elementary school. The difference was instead of adding the last number to the previous divisor, simple double the result above the line. (Quotient) I impressed my boss by using this to calculate the length of a hip rafter when the battery of his carpenter calculator died.
Most people calling themselves carpenters these days don't even know how many degrees are in a triangle.
I was taught this in a 1960s rural grade school that had recess 3 times a day. We also just doubled the on-going value above the line.
@Ryan Brown - Exactly same situation here in rural midwest. I still (very) occasionally use it. We also doubled the value above the line. (I'll have to see which method I prefer after some trial and error.)
I always wondered what practical applications would be. Can anyone share others?
@@ChrisW228 Just as a carpenter I can tell you that in order to square a building and make rafters you must use the Pythagorean Theorem. Finding the square root is essential. Todays calculators do all the work for us, but some us want to learn how it was done in the old days before electricity.
For whatever it's worth, I was taught this method in the sixth grade in 1960 in private school in New York City. Everyone in the class got it even the poor students.
I learned this method in Jr. High, back in 1966 and have taught it to many GED students. I also learned to do cube roots long hand but don't recall the method. I told my GED students that this is the method that you use when you have no calculator, such as when you are on a deserted desert island where it's an important skill.
Nicely done. I had to teach this to Grade 7 and 8 students and the hardest thing for them was keeping the position of the numbers in the quotient. The fix I had for this made the work seem more difficult than it was but only seemed as they began the way I taught them and later ditched the part which made it look messy. Usually, only the ones who had a real problem with the times' tables had any problem at all.
Then again the pocket computer came out and made everything actually harder to get across. Why learn how when all you had to do was press the square root key?
A very fine explanation.
However, the first stage is not doubling the 2, it's ADDING 2.
What difference does it make?
Well, the progression is purely one of addition, which makes it much easier to remember, long term,
Doubling AND adding during a continuous process has no logic to it.
This seems to imply that the video is wrong, so I will ignore it. Confuses me, anyway.
@@david203 The video is correct David.
I'm merely saying that it is better to THINK of the first stage on the left as adding the number to itself, rather than doubling it.
Ie, that way, you are adding the black two on the top line to the red two, then adding the black 9 on top to the red nine, etc, etc.
It looks like that both way work, but the video mixes them.
You can either add the red digit to the number itself, or see all red digits as a number(which is the current approximation result) and double it to get the next black number.
@@artsmith1347 I always step the left side sensibly, eg the 49 alongside the 447, the 581 alongside the 665 and so on, (as you suggest).
However, I thought the mixture of doubling and adding almost guaranteed a failure of memory, and hence was more important?
In effect, all we're doing is adding each red number to itself to start the next new line underneath on the left.
I deleted my other comments because someone deleted a comment I put a lot of work into. It could have saved others the effort of going down the same disappointing rabbit hole. Social media sucks when it is antisocial and feels free to delete substantive comments.
A refresher--for a step by step that I would have NEVER recalled but was SO familiar. This is a toll that Hewlett Packard/Texas Instruments obscured since I often need the square root in the years since school.
I've loved math since 9th grade and even tutored it for 6.5 years as an adult and never saw this before. Thanks for teaching me something new.
My equation is as follows. I have written down straight forward !
SQRT X ~ (X-A)/(B-A) + SQRT A
A: Biggest perfect square less than X
B: Smallest perfect square bigger than X
SQRT B = 1 + SQRT A
It works better for bigger numbers !
SQRT 111 ~ (111-100)/(121-100) + SQRT 100 = 10.523 (actual is 10.535)
For square roots use(10x+y)
To do sqrts just multiply your current x by 20, add your new y, multiply by your new y. Done! Subtract from your current remainder, bring down the next group, and repeat
Fore cuberoots square your current x, multiply by 300. 300x^2
Also multiply your current x by 30. 30x
Employing those as divisors, mentally determine your next y
Add that y to the 30x figure.
Multiply that by your y.
Add that figure to your 300x^2 figure.
Multiply by your y.
Done! This algorithm satisfies everything you said in your post
Subtract that figure from your currant remainder.
Bring down the next group of 3
Proceed as many times as desired
Btw, if you are interested!
The above algorithm for CR gets tediously cumbersome as I think you know by now. Think of trying to use that to find the CR of a number to 10 digit accuracy. Yet if you think about it, you can improve that method to be able to find CR to 7digit accuracy in less than 10 minutes on a piece of paper of 4 x 6 inches. We are talking no erasing, no calculator assistance befor or during, no tiny writing, no savant abilities, no separate worksheets.
You can find the CR to 10 digits on 2/3rds of a standard sheet in less than 20 min
I even proved one can find to 25 digit accuracy on one side of one sheet of paper. Takes roughly an hour
I have never seen a vid of my method, but I have faith you can figure out what I do.
Math teachers/professors put vids about the LD method on utube and they ALWAYS capitulate after 3 digits. "Too tediously long and cumbersome". That is not so. They just have never figured it out
@@37rainman I showed my H.S. math teacher how to manually extract cube roots on the blackboard in front of the class. He used logarithms to get the cube root and said I was wrong because his last digit was one off from mine. I left it alone so as not to embarrass the teacher because he didn't believe it was possible.
@@bobs182 To how many digits had you evolved your root?
Was he using a calculator to find the root. If he was ............ !!??? ........ I dont understand how he could be wrong........
How long ago was this?
@@37rainman Are you familiar with logarithms? He found it to the 4th digit. The last digit I got was like 7 and he got 8 so the difference was likely his number was rounded. This was in the early 1960s before calculators. Logarithms and slide rules were used to calculate back then.
THANK YOU. I have been wanting to learn this forever but I never knew what to look up. We never learned this in school.
Thanks! i remember learning this method in 6th grade, and in the intervening 36 years i've (somehow :) completely forgotten it. Good to be reminded
At 4:00, referring to the 49 you said add 49 to the first digit of itself. For clarity, the phrase "first digit" is relative to perspective. I write left to right so the first digit (from a layman's point of view) would be the 4, not the 9. When you said 49x9 you shut my brain down.... perhaps you should say "add it to the 1's column digit. just sayin... Great video!
I totally agree. He should have said the units digit or the ones digit... Not the first digit.
I learnt this method at the age of nine by a teacher who saw I had an interest in mathematics, I later developed a method for cubic roots but I can't remember it now, anyway getting a slide rule stopped me using "long" methods, and the calculator stopped me using log tables, usually two decimal places was acceptable as an answer...
Thank you I did this in HS, years ago, very well done and fastinating...
I found this method in an antique encyclopedia when I was young. Helped me fly thru square roots in high school.
In this particular example, a good approximate answer can be obtained in four steps:
Step1: 84765 is close to 90,000 so sqrt(90,000)=300 with difference of -5235.
Step2: 300 x 2 + 1 = 601 ~ 600
Step3: -5235/600 ~ -8.7
Step4: 300 - 8.7 = 291.3
its accurate if you round up the number to a no-decimal number, but if you keep the decimals, its different by ~0.15.
That's Newton's method basically, and it's what I would have naturally done, but I love how this proceeds directly forward without a start-over and square step.
I learned this as a "Horse and Goggle" method from a H.S. math teacher who wore really wide paisley ties and was kind and funny.
We used to learn this method in school thirty years ago, but had forgotten it very quick thanks to pocket calculators. Now this video refreshed my memory. Thank you!
This is amazing! I'm almost 58 yo and, beleive it or not, this method used to be taught in Greek High Schools in the late 70s! Not taught any more….Thanks for reminding me!
This is an interesting explanation, but is different from the method I was taught (other than starting by grouping the number in pairs from the decimal point). We were taught to multiply the answer thus far by 20 and then add the next (estimated) digit to that to become the next divisor. So, in your example it would be 2 (first digit of answer) x 20 = 40 and then say to yourself 40 something divided into 447. That something will be 9 so divide 49 into 447, then write that product of 441 under 447 and subtract. the next dividend becomes 665. Multiply the answer thus far (29) by 20 and get 580. Say to yourself that five hundred eighty something is to be divided into 665 so it has to be 581. At its basics, this is the same as your method, but just explains it slightly differently. I always said to my students that this was a pencil - not a pen - procedure!
I learned this in the USA and taught it there. Perhaps somewhat surprisingly, when I taught in the UK, none of my fellow math/s teachers had ever learned how to calculate square roots by hand. The only colleague who knew how to do it was a French teacher who was French and had been taught it as a schoolchild in France.
That is the method I learned. I think it would be much easier to remember the algorithm, than his method because you know exactly what you are doing and why you are doing it. It is not so rote. I am 71. Nothing like this was taught in schools in my time. This method would have been more from my fathers time..
Btw, way back then I immediately examined the same method for cube roots. I improved the method to make it practical. Today I can still actually find cube roots to the 7th digit, on a 4 x 6 inch scrap, in less than 10 minutes. No erasing, no tiny writing, no calculator help before or during, and no "savant abilities" involved. Just a lot of addition subtraction, multiplication, and a bit of mental division to estimate the next digit. I can find to 10 digits on 2/3 a standard sheet in under 20 min. I even demod that I can find to 25 digit accuracy on one side of one sheet of paper' in about 100 minutes or so
Have never found a vid on utube showing my improvement, (guess I am the only person in the world who has figured it out!!!), and the standard thing for maths professors (who do vids on utube) is to go to 3 digits, and declare "it becomes too long and cumbersome to be practical". But it is not! It kinda proves that some maths teachers also have a bit of trouble thinking outside the box
Never learned this method but did learn (and sometimes use) the divide and average method. To find sqrt(x), divide x by a rough guess at sqrt(x). Average that with the original guess. This is your new guess for the next iteration, Just two iterations gets you pretty damm close. Example sqrt(5): Guess is 2. 5/2 2.5. Average 2 and 2.5 = 2.25. This is your new guess. 2nd iteration: 5/2.25 = 2,2222. Average this with 2.25 gives 2.2361 which is pretty close to the real sqrt(5) which is 2.236068. Of course, more iterations will refine the result.
Thank you Mark.
This seems much simpler. Going to compare the two methods when it isn’t 3:39am. LOL
This seems much simpler. Going to compare the two methods when it isn’t 3:39am. LOL
This is called "Newton's method". It is OK with hand 4-function calculator, but cumbersome by hand with all the full-width divides. Once the iteration becomes close, the convergence becomes insanely fast. But it difficult to determine the number of iterations required for error less than "x". it is even more difficult for computer program to determine when to stop looping. Sometimes program can hang up & loop forever due to "bobble" in the calculations.
Thank you, man !
Thank you for the upload. I was taught this by my 4th grade math teacher about 30 years ago, and have always wanted to refresh my memory of it. It is surprising difficult to find any type of explanation for how to use this method.
I beat you.
I was taught this exact method by Sisters Of The Incarnate Word (Nuns) in the 4th grade over 63 years ago.
@@frankgonzalez607 Usually, it's the Sisters who beat YOU....
Thank you for the video. As an advanced math teacher, I have two suggestions. First, start with a much easier example like √2 or √3(numbers whose value they might already know to 3 or 4 decimal places) and then something like √50. And secondly, then show what your solution squared equals so the viewer can see how precise your method is. Don't make the viewer have to write your answer down, then find a calculator, and then come back and compare it to the original number.
And you as "an advanced math teacher" then tell your students the "concept" behind the process...right? As an "advanced math teacher" what do you think about Trachtenberg's methods?
This was a great explanation. I definitely needed that info like how to do the groupings, if there is an odd # of digits, and where that decimal goes.
It's definitely a tricky little bit of calculation and would take me a lot longer than just plugging it into a calc... but with a bit of practice I feel like I could crank out the figures by hand if I needed to... like if I was helping check my kid's math
great, I learned this in school,& now at the age of 75, I cannot think I've ever used or needed it. Paul Lucas.
Ray Seto
Ray Seto
3 days ago
brilliant! i remember learning a method of taking a square root from a book but never mastered that method, especially since I learned it on my own (they didn't teach taking square roots from all sorts of numbers, only the obvious ones). Thank you for making this video! Very clearly taught, too. :)
9
Yes indeed. Like most of the garbage taught in schools.
I learned this in grammar school, and then learned to find cube roots, which is similar, only you start by sequestering groups of three rather than groups of two.
Excellent explanation! I learned this when I was in 3rd year of junior school (in UK) so about 10 years old. We had a very progressive young teacher who also taught us logarithms and it has been a party-piece of mine ever since.
I honestly have never met anyone who could do this.
The only difference we were taught was to double the answer, ignoring decimal points, and bring that down as the next divisor instead of what you showed but it comes to the same thing anyway.
Do you know how to calculate cube roots? I've never been able to figure that out.
We were taught to double the answer and bring it down which is easier. I’m trying to work out why both things work out the same!
@@lindamanas6735 I always just multiplied the current answer by 20. There is our new divisor. When you determine your new digit, simply add that, and multiply by the new digit. Done! Subtract from your current rmainder, and proceed.
Any x can equal a + b (actually an infinite number of them! But we employ one where in each iteration a is the current finished part and b is the remainder, or in practice the next digit)
So x^2 = (a + b)^2 = a^2 + 2ab + b^2. At each iteration, the a^2 part has already been satisfied. and the remainder is always the 2ab + b^2 part, or (2a + b)b. Any of the various methods used satisfies this, you just must look closely and figure it out.
In my particular method, you are going to immediately ask why I multiply by 20 instead of the 2 in the formula. Look closely, you will see why
But certainly his method is very clean, concise, and interesting, although mine is probably the easiest to remember, simply because there is nothing rote about it
One can also learn to find the cube root of any number with this method, and you may not believe it, but one can perfect the process so that you can find the CR to 7 digit accuracy on a scrap on paper 4 inches x 6 inches, in less than 10 minutes. (Note, finding a CR by hand is immensely more involved than finding a sqrt.) You will not find my method on any utube vid, but it is simply a logical perfection of the Long Division Method
I knew such a method existed but never saw it explained, thanks for the video.
Now, for a challenge, can you make a video explaing how to calculate (by hand) the base 10 Log of a number? Of the few methods I've seen (on YT) it was either: A) a contrive i.e. easy example calculate for 10 or 100 or, B) a number whose log was know is some other base, e.g Log (base 3) of 27.
How would one calculate the Log of 20? I know it's between 1 and 2 but that's as far as I can get.
You are right it might be hard to find the log of 20 with base of 1 or 2. I used to find any member's log by hand with any base 24yrs ago. But still remember there used to be table at the end of maths book in which we used to see the first number in column and after decimal in a row that was called Mantissa I think but those books are long been changed.
Taking natural log (log base e)by hand i only know two ways. Approximating the integral of 1/x from 1 to n, or Newton Rathson method iterating x' := x - 1 + n/e^x, though that requires approximating e^x with its power series
Basically, you mean that if the bases don't come equal then how we can do it. Let's, log20^100=x
20^x = 10^10, since bases are different we can't can't say x = 10.
Therefore we can take Ln of both sides. So, Ln 20^x = Ln 10
XLn20 = Ln 10
X = Ln10/ Ln 20. So we can get x value. But as I said in early comments that you will need a table to see the ln10 and ln20 then divide it you can get exact answer.
Coming to your example, Log20?
First find the references number ( first non zero number) that is 2 in this case. So 1 is your charactric and now use the table and search for fist 2 numbers ( in this case is 20).
Since there is no other number after 0 therefore look in the table in 20 row and 0 column. So your charactric was 1 therefore your answer will be 1. (.......). After 1put a point and after point write the number you find in 20 row and 0 column which is called Mantissa.
Hope this should be your answer.
If your number is log 0.00235
Then bring the point in the front of non zero number which is 2 and you are going back 2 digit back therefore your characteristic is -2. Looking for Mantissa now, look in table in row 23 and 5th column your will get Mantissa. So your answer will be
-2. (......) meaning that -2 then point then your Mantissa which you found in column 5.
Adding to this if there is another digit after 5. Let after 5 is 8 then you will look in the same row and 8th column the difference which will be added in Mantissa.
Lastly, if the number number has 0s before a non zero number then you will bring point to the front of non zero number( in the above case is 2)
If a number is like 234 then bring the point after first non zero.
This way if you go back your answer will be in minus if you go forward your answer will be in positive.
Hope this will be clear.
@@davidcivil1651 My apologies for not seeing your comment earlier. I used such a a book and still have it, "Standard Mathematical Tables" 14th edition. It was new when I used it. We just called it the "CRC" book, the book was published by the Chemical Rubber Company. This was the mid 60's, it was a few years later that I encountered my first 4 function pocket calculator. I guess there's no need for these books any more, still glad I have it. No batteries required.
@@SlaveToMyStomach No problem, Richard, To be honest I don't remember the book writer or edition, because I was in school that time and was just focused on learning to be ready for exams. I remember I achieved 98/100 in maths. Pretty much I had learned by heart all the answers. Teachers were saying I will be a mathematician in future but unfortunately I lost my track and wasted around 10 yrs of my life and remained away from studies. Thankx God that I was blessed once again and completed my degree in structural Engineering where I studied applied mathemtics too. When I flesh back and remember my those 10 yrs makes me upset alot however, I learned from life that there is no friend in adversity and lots of friends in success.
I'm 65 and was never taught this. Slide rule was the first tool we were taught to find square roots.
I googled how to do this a few months ago. Got unsatisfying results. Simply put, there is a formula to find rafter lengths but the square root of the rise and span is required and sometimes it's hard to figure out. I don't remember my teachers in school ever explaining how to do it and I'm sure whatever they taught wasn't this... easy to learn.
I've always held the belief that if you can't explain something, you don't truly understand it yourself. Your tutorial was easy to understand.
I use a much faster, but imprecise method in my head, just for quicky analyses, and for this I got 281 in no time.
Here's how I do it.
84,000 is 2 times 42,000
or 4 x 21,000
8 x 10,500
16 x 5,250
32 x 2500
64 x 1250
128 x 625
256 x 312
At this point you go to the halfway point between the two numbers, and decrease it a little because the graph y=x^2 is curved such that the halfway point on the x axis corresponds to a point on the y axis that is less than halfway.
The halfway point is 284. I guess I should have stuck with that because I neglected all the digits after the 84.
The halfway point of what is 284?
Oh, I see nvm.
Your "method" is clear as mud.
Double one number, halve the pfber, repeat until they are about the same. Then split the difference
84,000 equals 84 times 1000
168 times 500
336 times 250
Halfway point is 293.
I have not needed square roots for a long time, but watching this was fascinating. 🤗
I was shown this method in school exactly once but the teachers didn't test you on it. Later on I was shown how to interpolate, which was resonably accurate and simpler to do.
Brilliant, not done this for ages.
Good organization of the data! Thank you.
Thank you! I was wondering if you have a visual proof that indicates why that rule works. My students would appreciate a visual proof in addition to memorizing the procedural steps.
It's based on adding something, then squaring it. Lets call the root abcde... where a, b and so on are the digits. You start by finding a. Now you wanna calculate (a+b)² = a² + 2ab + b² = a² + (2a + b)b and out of that we already have a² so you only need to look for the latter. You can visualise this with rectangles. The next steps go the same way, except your first term is the number you already have.
My 5th grade teacher Mr. Biot taught this to us in 5th grade.
Was really magical to me .. and still is.
I learned this in grade school before calculators had square root on them. I have been looking for this solution for a while now. Thanks.
An interesting method. Very curious as to how/why this works. If I needed to do this by hand I think I would use Newton-Raphson method. In the second itteration I already got to 3 decimal places.
it is just (x + y)^2 = x^2 + 2xy + y^2 iterated.
Zach E, Thank you for the clarification (x+y)^2= x^2 + 2xy + y^2.
The differences between this technique and N-R iteration are:
1. You know in advance how many iterations you need to get a given precision
2. This is a process that inherently uses integer arithmetic to model rational numbers. Thus you are finding successively more accurate rational approximations each time.
3. Also you can predict how many digits would be needed at any given step in the worst case for that step. You need at most 2n digits for each nth step.
4. In contrast N-R iteration can diverge in some cases (tho that can be avoided for a function like sqrt where the second derivative does not change sign)
5. It is harder to predict exactly how many iterations will be needed, and harder still to predict what precision is needed in intermediate calculations to get a specified precision in the answer. Therefore the usual approach is to keep going till successive iterations are closer than some limit. But you don't really know how close you are.
6. Noting the finite precision of numbers that computer engineers call "real" (actually they are just rationals in another guise) a N-R iteration can get "stuck" at a value that is not quite the best estimate. So even when successive iterations keep the same value you still don't know how many digits to trust.
In the eighties there were debates about whether "real" numbers made sense. Languages like Forth were invented that did everything in integer arithmetic (in binary you would group the bits in pairs).
The needs of engineers (who are very comfortable with floating point notation) won out over the number theory purists. Commercially that was the right decision.
But I do sometimes wonder if we could now do better with whatever Forth would have become given four decades of Moore's Law...
@@trueriver1950 Thanks for the elaborate explanation, predictability is certainly a benefit. In the case of taking a square root I am not too worried about diverging iterations of N-R.
You got me interested in the Forth language. (And, yes, I am one of those engineers that is very comfortable with floating point numbers. :-) ) As for real numbers, in my opinion, they only make sense in maths, everything else does have a finite precision and uncertainty.
@@trueriver1950 I think in the case of finding square roots you can also put some pretty hard limits on how many NR iterations you need to achieve a given precision. There really are no problems at all using NR for finding square roots (unless you're daft and make your first guess zero). y = x² - N is really perhaps the most well-behaved function you could ask for, where something more advanced than just linear intersections is actually necessary.
There's probably some crossover in how much effort is involved in finding a square root of a given number of digits with the two methods. For just a handful of digits this "long division" style method is probably faster. But past maybe about 20 digits I think you'll cover fewer pages in ink with the NR method due to its doubling the number of correct digits (for this problem in particular) with each iteration.
Heck, I was taught this exact method by Sisters Of The Incarnate Word (Nuns) in the 4th grade over 63 years ago. And I still remember it and occasionally use it to refresh my memory.
When my step daughter was in the 4th or 5th grade, I asked her if she was or had been taught this method, and her answer was "the deer in the headlights look." Very disappointing.
Today's education systems are abysmal. It's why I would venture that the overwhelming majority reading this don't know the differences among "there", "their" and "they're."
Oh, we were also taught to (and did in fact) diagram sentences! Subject-Verb-Object.
I could continue, but I would be up past midnight.
In fairness, adults don't need to compute square roots by hand when a $10 calculator or a $5 smartphone app can do it for them. Times change, and the skillsets have to change. Driver's ed classes don't teach students how to ride a horse either.
@@stevenlitvintchouk3131 Maybe they can calculate, but they often lack number sense.
I've learned this when I was 10 flipping a book published circa 1920, and yes, a great full explanation!
From the recent comments here, clearly some quirk in youtube's algorithm is suddenly putting this old video on a tiny channel in front of people. Fascinating.
I would have thought that the way to do something like this would be to work out Newton's method by hand. But probably this is totally different since it gives one digit at a time. It might be fun to work out the algebra of why it works, and i love how the "guess the number" part generally doesn't get more difficult even as the numbers get bigger.
In fact the "guess the number" part becomes easier and easier the further you go because it becomes more and more like you're simply doing long division when you make your "guess".
Someone worked that out a few days before your reply... scroll up to leinsulin’s comment. The rest shared it in replies.
"easy and fast"
all lies
Compared to having to use the bisectional search method (which is what I was using for quite a while before I saw thus video), yes it is. Compared to a calculator, of course not, but for a by-hand method, it's relatively super fast
Lol
@@samk2407 LOL imagine complaining when this guy is helping you. SMH
He’s a fraud
big capper right here, ladies n' gentlemen
I was always told to check my work. So squaring 291.14 gives us 84,762.4996 which is very close to the original number. Performing more steps will obviously improve the accuracy. Great video!
My, my, that unnecessary squaring of big numbers is going to HUGELY add to the time this takes!
For a fast estimate, Just know or calculate the perfect squares 1, 4, 9, 16, 25, etc., and estimate for the fill in of decimal places. Folks thought I was Einstein as I could say these right off the top of my head. Approximately 4.1 for the sq root of 17. 4 squared is 16 and because 17 is close to 16 (1) the decimal part is numerically small also. Then would proof it on paper. Its actually 4.12 at 2 places. This is a great presentation.
That is sooooo cool. Our little school never excited us like this.
Thank you for the very nice explanation! I remember learning a much more difficult method in school some 50 years ago.
Great method! Thank you.
I am in my 50's and was never taught this method. It is amazing. Who invented it ? When ? Finding square roots was one of the reasons Calculators were invented.
Thank you for this video. My father tried to teach that to me 50+ years ago. I think I finally get it.
Amazing! Love the way you explain it.
Wow, that's pretty ingenious
I learned this method way back in the 80s when I was in my 8th grade in Eritrea. In step 3, after you bring the pair down on right, you were supposed to leave out the last digit of 7 (44 instead of 447) and divide the number by the 'doubled' 4 (or 44/4) to get the rough factor estimate of 9 (so 49x9=441). You were supposed to do this all along every time you drop down (leave out the last digit then divide and multiply - helps you take out the guesswork). --- Another thing, instead of doubling the single digits on left, you also could simply have doubled the entire number sitting on top of the bar each time. Still, great video presentation though.
The best way to proceed is to simply multiply your currently completed root by 20. Use that to mentally determine your next digit. Add that new digit to the aforementioned product, multiply by that new digit. Subtract from the current remainder, bring down the next 2 digit group. Done!! Repeat as many times as desired
His method is nifty and clean, but will not be remembered as well, because you are simply doing something by rote