Watch this educational video from the Spirit of Math Curriculum, presented by Spirit of Math founder and CEO Kim Langen teaching methods of calculating square root by hand. to find out more about Spirit of Math visit spiritofmath.com/
I too learned and later forgot this method years ago. I was amused that the presenter used chalk, which broke, while working the problem. That really brought back the 60s tome.
I learned this method 50 years ago from my Chemistry teacher. I later found out that this method is based on Binomial Expansion (a+b)squared. Not many knew this long division method in school these days. Thanks to RU-vid, this method has been revealed. I love this method.
@@pbworld7858 I even had a teacher who taught me the formula for the Nth Fibonacci number with the phi in it. A friend was verifying that by hand in Chess club! And it worked!
We learned this when I was in 6th grade. Years later, I used Newton's method: start with any reasonable guess, then iterate: new guess = 1/2( guess + number/guess). It converges quite rapidly.
The advantage of the "divide and average method" is if you make a mistake, it will work out if you don't make more mistakes. With the way presented in the video (the way I first learned square root), any mistake will ruin the result from there on.
My dad showed me this method many years ago, and I've never met anybody else that knows this method. This is the first video that I've run across that explains it step by step.
Back when i was in grade six (in Canada), i went to Ecuador for the summer. I was bored as everyone was in school. So my mom enrolled me in school there for a couple of months. In that short period, my math skills jumped to a Canadian grade 8 level. I learned how to do square root by hand. When i got back to Canada, i went back to learning long division, and in grade eight, we learned to use calculators.
The lost art of doing mental math or calculating solutions to challenging problems by hand is one of the reasons our parents say they keep coming back!
Our math teacher showed us this method in extra classes. Everything was almost the same, except that she said that you can not only multiply by 2, but also add. For example, 48 * 2 = 96. But you can get 96 by adding 8x + x (88+8 = 96), which was usually intuitive, since we put two dots when we were guessing the number for multiplication. Exactly the same in the second case: 487 * 2 = 974, but you can get the same thing if you add 7 to 967. Thus, 967 + 7 = 974. It always works. That is, once again. When you have decided on a digit, multiplied, calculated the difference, and you need to multiply the top number by 2, we don’t have to do this. You can take the number that was the last one on your left and add it with the digit that you put the last (its own last digit).
I learned this from my 3rd grade (4th year) math teacher. He made math fun. Subsequent math teachers varied in quality, but I didn't have another who was that good until university. Even if you pick a number that is too high for the next step, the algorithm is self-correcting.
I too leaned this in Grade 3, at age 9, from my Dad. His explanation wasn't quite as tight as one now finds on the internet, but was sufficient for me to have some fun.
I learned this method in 7th grade, back in 1962 or -63. It wasn't part of the curriculum, but I asked our teacher, Mrs Galloway, if there were such a manual method, and she showed me after class. I'd long since forgotten it when I stumbled across this video. The Babylonian method is another way - much simpler to flowchart, but involves ever more lengthy long divisions.
@@johnchristian7788 Consumer-grade electronic calculators wouldn't be invented for another ten years. We were probably shown where to look up tabulated values in a handbook. Use of log tables wasn't introduced until high school (grade 9-10). My dad showed me how to use a slide rule at some point, but I don't remember when. Geez, this was over sixty years ago - I don't remember when they taught what.
@@lesnyk255 It's funny to think that even before calculators became popular, they didn't teach square root by pen and paper. They should really include in the curriculum in all countries. I used to love using log tables.
@@johnchristian7788 Well, personally, I wouldn't go back to using log tables, slide rules, or manual typewriters except maybe at gunpoint. There are easier ways to get rough manual estimates of square roots if you've left your calculator or iPhone at home - polynomial approximation, for example, or the Babylonian method. This video was a bit of a nostalgia rush - 7th grade, Walpole NH JHS... long time ago....
Feel same. Been a bit and I feel as you, just a little reminder to do elementary problems! Want a nice square concrete pad and although few concrete workers remember and quiet likely never did by the fun of me when I mentioned hypotenuse they get a big laugh at there 10th grade drop out lol. He who laughs first laughs last right lol. Bless their hearts lol. I always like the 3,4 and 5 or even double the number helps. What I really love is running a say three foot diameter pipe through a floor system lol. Usually take my measurements home lay out on piece of cardboard then bring in to work and always fits so nice nothing even gets mentioned lol but that’s fine huh. I will give anyone who may be interested the Pipe fitters hand book is small and like anything the more you do it you get even noticed less but who wants noticed if it all works nicely. I was a Union Ironworker and modest. Again thank you for the refresher, very nice ❤️. Calculators are very handy lol.. Left 4 men to form up for a metal building and wanted the exterior sheets to run down the side of the concrete pad to eliminate water 💦 running inside the building. Many ways of laying out and having one nice square corner sure simplifies ✌🏼. Sincerely Grateful, HB
This method is based on Binomial Expansion (a+b)squared method. It is very easy. I can show it to you in a few minutes. This teach makes longer than it is.
Never learned this when I was in school in the 80s and 90s, likely by then they already just assumed everyone had calculators. I appreciate the method, it's very interesting! (Whenever I've wanted to do this without a calculator I've just basically made educated guesses and worked my way to something close, I have squares memorized to about 25 which helps.)
I went through five bad videos before I found yours. One guy even helpfully blocked the view of the whiteboard while he explained what was on it. It took about a minute to catch on watching you. Thank you!
try stretching your brain doing the method for cube roots No one taught this in grades 1 thru 12. But I got interested on my own When the stress closes in, I often find myself evolving the cube root of a number looks like you are about 5 to 10 years older than I
I was never taught how to calculate square roots. When I was in grade school, I tried a few different ways on my own, and they ended up being very much trial and error. This is a more refined approach that improves on what I figured out, but I see that it still involves some. Thanks for showing it!
This method is based on Binomial Expansion (a+b)squared method. It is very easy. i can show you in a few minutes. This teacher makes it look longer than it really is.
An over-reliance on calculators makes your math muscles weak. We always encourage our students to learn the core concepts and do the arithmetic mentally or by hand whenever possible
@@SpiritofMathSchools I tend to agree. Teachers can choose values that can be computed in the head or simple multiplication and long division on paper. Real world math rarely has those convenient numbers. Calculators, I would argue do not make one's math weak as doing the calculations is only the lowest skill on the math "tree." Knowing how to set up the problem is where the math skills shine. I suspect that those who do real world math will rarely use hand calculations, and they will quickly notice when their calculator have given faulty inputs. It is important that students learn the basic arithmetical calculation techniques and practice them in the classroom.
it’s neat to see a long division style algorithm for the square root! what makes long division not too bad is that the subcomputations for each digit (guessing the closest multiple below a given number) all involve numbers around the same magnitude, whereas here it seems getting another digit involves a subcomputation with numbers around a magnitude larger than those on the previous step. i wonder if there’s another long division like algorithm where the subcomputations don’t inevitably grow in magnitude? i also wonder if doing this in base 2 would feel simpler?
Your way of manually doing square roots is the way my 8th math teacher Mrs Wilker taught us how to do it . I will study this problem and do more problems like it. Lot of WHACK out ways of finding the square roots . They work, but very CONFUSING You is worth your weight in gold raised to 20^20 power . (HUNDRED QUINTILLION) Thank you.
I haven’t started the video yet and I am interested to see it, but I always like to try things before I watch the video. I mean when it comes to math. So in a few seconds, I came up with an estimate that the answer is just shy of 50, since the number is shy of 2500 and then in under three minutes, I came up with a slightly better approximation of 48.77, which I got from interpolation between 48^2 and 49^2 (having already rounded to 2377^1/2, and rounding 103 to 100…and rounding 2401 to 2400.
I grew up learning how to do square roots manually . Kids today do not learn how to do sq. rts. manually. They press the magic button on the calculator.
And we really appreciate the positive feedback! Perhaps you could check out some of our other videos and let us know if there's any other topics you'd like to see in the future?
People underestimate muscle memory, especially when it comes to mathematics! That's part of our approach with our students that we notice makes such a difference.
It's obvious that the square root lies between 48 and 49, because 48^2 is 2304 and 49^2 is 2401. I can use a linear approximation to determine additional digits. 2376.6 - 2304 is 72.6, and the difference between 48^2 and 49^2 is 97, so 72.6/97 is my linear approximation, which gives me the next digit of 7. So far I have 48.7, and I can use linear approximations to double the number of significant digits with each iteration. But everyone knows this.
You ought to see what happens if you apply this on binary numbers! You start as usual, grouping the numbers, etc. On the first digit, it is one for the first pair of non-zero digits (there are only 00, 01, 10, 11 cases). To generate the next test number to subtract, you take the answer you have so far, & append to the right of it 0 1. Why? Appending the 0 to the right doubles the number. Appending the 1 is the test digit. Multiplying by 1 is trivial case, just copy the number! If it "fits", write "1" for the next digit of the answer. If not, write "0" & discard the subtract. (You do not cover the case where even "1" is too large. In that case you need to write "0" in the answer & discard the result of the subtract, leaving the partial remainder intact. Then you being the next 2 digits down alongside the existing remainder & proceed from there.)
I got 48.75 in about 20 seconds. Divide 2376.592 by an approximate square root ie 50. That gets you 47.53184. Average 47.5 and 50 and you get 48.75 Trial and error can get you 3 significant figures very quickly by hand.
The Cube Root: A Practical Method to Find It from Any Number The Cube Root A Practical Method to Find It from Any Number Sidqi Mohammed Al-Baik In the Abbasid era, Arabs excelled in mathematics, enriching the facts of arithmetic, establishing algebra and logarithms, dealing with exponents (powers) and roots, and organizing tables. It is not unlikely that they devised practical methods to find the square root or cube root, other than the method of prime factorization, but these were not known to modern mathematics scholars or were not published. However, students following the French curriculum recently learned a practical method to find the square root (as in Syria and Lebanon) while those who studied according to the English curriculum did not. I have not come across a practical method to find the cube root, nor have I found any mathematics specialists who know a practical method for the cube root. Therefore, I worked hard and for a long time, spanning several years, fluctuating between despair and hope, until I discovered this practical method to find the cube root of any large number, other than the prime factorization method. Many may now find it unnecessary to use this method and others by using calculators, which also spared them from many calculations. However, people, especially students, still need to learn different methods. This method may be an intellectual effort added to other mathematical information and facts. Here is this method, which requires knowing the cubes of small numbers from one to nine, which are (1, 8, 27, 64, 125, 216, 343, 512, 729). Method and Steps Divide the number into groups of three digits, starting from the right, after writing the number in the correct format. Start the first stage with the leftmost group, approximate its cube root, and place it above the group. Place the cube of this number under the leftmost group and subtract it. Bring down the second group next to the previous subtraction result and start the second stage. Prepare the root factor according to the following steps in the left section: A. Square the root obtained in the first stage and place a zero before it. B. Mentally divide the number obtained in step (4) by three times the squared root (from step A) by underestimating, and assume this result as the second digit of the root and place it above the second group. C. Multiply this assumed number by the previously obtained root with a zero before it. D. Add steps A and C. E. Multiply this sum by three. F. Add the previous multiplication result to the square of the assumed number. G. Multiply the sum in step (F) by the assumed number, place the product under the number obtained from bringing down the group (step 4), and subtract it. Bring down the third group to the right of the previous subtraction result, start the third stage, and repeat the steps in (5) as follows: A. Square the previous root (both digits) with a zero before it. B. Mentally divide the number obtained from bringing down the group (in step 6) by three times the squared root (from step A). C. Multiply the assumed number (from step B) by both digits of the root with zeros before them. D. Add steps (A) and (C). E. Multiply this sum by three. F. Add the previous multiplication result to the square of the assumed number. G. Multiply the previous sum (from step F) by the assumed number, place the product under the number obtained from bringing down the group (step 6), and subtract it. Continue this process. If a remainder remains after subtraction and no groups are left, add a group of three zeros and repeat the previous steps, placing a decimal point in the root as the result will have decimal parts. Practical Example Cube Root of (77854483) Divide the number: 7 2 4 77,854,483 Approximate the cube root: The approximate cube root of 77 is 4, place 4 above the first group. Subtract the cube: The cube of 4 is 64, place it under the first group and subtract it. 77 - 64 = 13 Bring down the second group: Bring down the second group: 13,854 Prepare the factor: Square the root with a zero before it: 40 × 40 = 1600 Mentally divide 13,854 by 1600 × 3 = 2 approximately Multiply 2 by 40: 2 × 40 = 80 Add 1600 and 80: 1680 Multiply 1680 by 3: 1680 × 3 = 5040 Add the square of the assumed number: 5040 + 4 = 5044 Multiply 5044 by 2: 5044 × 2 = 10,088 Subtract 10,088 from 13,854: 13,854 - 10,088 = 3,766 Bring down the third group: Bring down the third group: 3,766,483 Repeat the previous steps: Another Example: Cube Root of (12895213625) Divide the number: 5 4 3 2 12,895,213,625 Approximate the cube root: The approximate cube root of 12 is 2. Subtract the cube: The cube of 2 is 8, place it under the first group and subtract it. 12 - 8 = 4 Bring down the second group: Bring down the second group: 4,895 Prepare the factor: Square the root with a zero before it: 20 × 20 = 400 Mentally divide 4,895 by 400 × 3 = 1 approximately Multiply 1 by 20: 1 × 20 = 20 Add 400 and 20: 420 Multiply 420 by 3: 420 × 3 = 1,260 Add the square of the assumed number: 1,260 + 1 = 1,261 Multiply 1,261 by 1: 1,261 × 1 = 1,261 Subtract 1,261 from 4,895: 4,895 - 1,261 = 3,634 Bring down the third group: Bring down the third group: 3,634,213 Repeat the previous steps.
Just use Newton's method x=(x+a/x)/2 where a is the number whose sq root is to be found and x is the current approximation to the sq root. And iterate.
The first rule of optimization is to identify the operations that take the most total time, and work on making those faster. If this is an infrequently used procedure, i.e. it won't represent a significant portion of a student's life, then why not teach the conceptually simpler approach of progressively refining an initial guess using a binary search?
I am pretty ticked off that I was never shown this in any year of schooling. Yeah it might have been rough at a young age, but the mental workout it would be if all kids had to learn this stuff. People would be way better thinkers as grown up as well as following rules for things and how to solve problems, in life not just math as the problem solving skills are applicable everywhere.
I learned this from a high school classmate but I didn't get what he did. He wrote on paper so quickly. I didn't have time in class. I think if you're in an east or as Southeast Asian country or somewhere from South America they might have taught this. Asian countries taught tough stuff forvyoung kids that's not taught in the USA or Canada.
I learned this method long long time ago when there were no electronic calculators ,am now 70. y/o ,but instead of multipying by 2 we multiply by 20.Now a day they don't do this method any more.
Yes. I have always simply multiplied the currently completed root by 20, (20a). then estimate how many times that divided into the current remainder . That is your tentative next digit (b). Add the b to the 20a figure and multiply by b. (20a + b)b Subtract from current remainder, bring down the next group of two, for your next current remainder This simple method can be remembered forever, because you know why you are doing what you are doing It is never taught on utube, because it doesn't appear as sexy. But in our father's time, my method was used, because I eventually saw it in a very old encyclopedia
2:10 Really? I was really hoping this was gonna be the universal equation that solves any square root, or cubed root, or etc. I've never understood roots because there is no reverse calculation for it like division is for multiplication. I also watched a video a few days ago where I was introduced to n⁰=1 and 0⁰=1. Math is suppose to be about logic, but I feel the more advanced maths are just number manipulation to get a desired answer.... Basically arbitrary like language and to me, arbitration is not based on logic.
Question for viewers Can you derive such method for cube roots ? If you really understand why this method works you will be able to derive method for cube root yourself I was taught this method in high school once we were solving quadratic equation (to determine if discriminant is perfect square or to approximate roots) and derived method for cube root myself
What country did you go to school that they just told you to find the method yourself? I'm suspecting that instead of multiplying by 2 we should multiply by 3 and use cubes instead of squares in the same method. Not sure if I should group by 3 digits 🤔
@@johnchristian7788 In Poland I derived method for cube root for myself and it was not homework As soon as I understood why method for square root works I was able to derive method for cube root Yes you group 3 digits Yes you multiply by three but square of actual approximation not just actual approximation Instead of appending last digit of next approximation you append square of last digit of next approximation To number created in this way you add triple product of current approximation and last digit of next approximation shifted one position to the left (10a+b)^3 = 1000a^3+300a^2b+30ab^2+b^3 (10a+b)^3 - 1000a^3 = 300a^2b+30ab^2+b^3 (10a+b)^3 - 1000a^3 = (300a^2 + 30ab + b^2)b (10a+b)^3 - 1000a^3 = ((300a^2 + b^2) + 30ab)b
Crystal Clear Maths has a vid on utube where he examines cube roots by the LD method But he concludes that it is not practical beyond a few digits. This is not true. I have demonstrated that with pen/paper I can find the CR of any number to 25 digit accuracy on one side of one sheet. No calculators involved, no separate worksheets, no erasing, no savant ability, just plain old addition subtraction, multiplication.
Step 1: Convert to binary. This avoids any need to guess. Step 2: Apply the algorithm for binary numbers. Very fast. Step 3: ( Optional ) Convert to base ten.
Looks like a neat method, but frankly you lost me and I have a strong background in mathematics. May I suggest you redo this video? Writing out a script with queue cards may help. Citing a published source for this trick would be great. Other commentators suggest it is a reorganized Binomial expansion....I tend to agree, though more background would be nice .
The decimal will never end since the square root of non perfect square is non terminating as well as non repeating. In otherwords they are irrational numbers.
@@cbruata5198 Well, it depends on when you multiply the answer by itself how close you get to the original number. In this case, you can round the answer off to two decimal places, but as it stands the answer is not an approximation.
Calculators are great when you have it on hand. What if you don't have one? You mind is the best calculator. This method is based on Binomial Expansion (a+b)squared method.
No. You can convince yourself by looking at the problem from the opposite direction: if you take a rational number and square it, will you always get a square number? If it's an integer, yes (2*2 = 4; 3*3 = 9; etc), but if it's not an integer, then no: 0.5*0.5 = 0.25, so there exist non square numbers with rational square roots.
@@Merione thank you for taking my question seriously. I appreciate your response. Just like everything that is explained it seems obvious in hindsight and I probably should have just thought about it harder. That was a very satisfying and simple explanation.
By "as close as possible" I assume it is, as you say in the first case, as close to but less than. And the amazing statement at the end about square roots never repeat. Well, some certainly do, e.g. a square of a rational, such as 2.25, repeats with infinite 0s. So the divisor changing doesn't guarantee non-termination
Final top digit , following YOUR rules, is 4 ,NOT 5. WHATEVER YOU DO IN MATHEMATICS , YOU MUST EXPLAIN YOUR DEPARTURE FROM YOUR RULE! {ESPECIALLY IF YOU REFUSE TO EXPLAIN EITHER YOUR RULE, OR YOUR UNEXPLAINED DEPARTURE. YOU CANNOT CALL IT MATHEMATICS WHEN YOU ARBITRARILY MODIFY YOUR RULE WITHOUT EXPLANATION.
I agree with why you're complaining (although lay off the ALL CAPS next time perhaps, eh?). But I think I have a plausible explanation for this departure from her previously stated rule... She added a zero after the original number, so that she had a pair of digits at the end (instead of a single digit) to match the rest of her method. Given that the zero is technically not significant in relation to the original number, the use of 5 (instead of 4, as one would expect from her original rule) kind of acts as a rounding factor for the final answer. This is only an explanation I've come up with, after the fact. Interestingly, using 4 (going by her original rule) gives a resulting divisor of 9744, which ends up giving 38976 for the drop-down subtraction, and then an answer of 10044 ... implying a very very scary iteration of the algorithm if there was another pair of digits in the original number!! 😨
Ugh! Aside from all the comments thanking the presenter for a trip down nostalgia lane, this is a dreadful use of 7 minutes and 23 seconds. The algorithm is VERY complicated, and there is no explanation of why these particular steps are taken. A guess-and-check method would at least reinforce what a square root is.
Evidently you never paid attention when multiplication was being taught in school, or paid attention when she was explaining the very step you're complaining about. The 6 comes from the first 8*8 (=64), she put the 4 down and carried the 6 to the tens column for the next step. Pay attention in future, champ. 🤡
@@robertveith6383 Indeed. However, you would think that the whole point is not the mechanics, which a calculator will do right quickly, but to explain why the calculator and this method come up with (approximately) the same answer. She might also show us what 48.75 x 48.75 equals. It's 2376.5625. She says you could go on and on, but she doesn't say that you would get closer and closer to the target square, 2376.592 if you did. It's weird, I know, but this mechanics for the sake of mechanics reminds me of filling out the capital gains page of an IRS form. You know, you've put in the amount you paid for 200 shares of Zockman Birtwistle Corp. stock and the amount you sold it for and subtracted the first number from the amount you sold it for. Then the instructions to me just get silly. Something like: Take the total on line 3 and multiply by .15 Take the total on line 1 and multiply by .35 if you bought the shares more than two years ago. If less than two years ago, multiply by .28. Write this number in on line 4. Take the total on line 2 and add it to the number on line 4. Write this number on line 5. Subtract line 5 from line 3. I expect it to continue with: then sing the 4th verse of The Star Spangled Banner and write the number of words in it on line 6. Count contractions as one word. Every time I had to fill out such a form, I had no idea why I was merrily multiplying by, adding to, subtracting from those numbers from the top to the bottom of the page, and somewhere in the middle, I would start giggling because I had no idea WHY I was doing those particular calculations with those particular numbers. It was like being given a set of 10 algebra problems that had no relationship to the real world, just to practice the mechanical steps to the solution. OK, now that I got the solution, what's the point? There seems to be no point. You got seven correct, so you get a 70% score. Oh, now I get it. The point of learning math is to learn math. You don't actually use it for anything. Well, at least tell students that. The idea is to train your brain to think in a variety of ways so that it is functional to its full potential by the time you're 18. Well, that's what it seems like in too many math classes. I'm not against getting the right answer. That is, of course, important. I'm not against showing the steps to the teacher, so she knows you didn't come up with a lucky guess. But not often enough do we hear why we would want the right answer in the first place except to please the teacher. PS I cannot believe that this teacher cannot subtract 7259 minus 6769 in her head. I'm hoping she's putting on a little act for young students, who might be struggling to remember what you do to subtract 6 from 5 and 7 from 2. But how young a student would be trying to find the square root of 2376.592 and why on earth would they want to?
@@SpiritofMathSchools thanks for engaging, More assurance in expression and more fluidity in idea into idea I’m sure sis is a dope mathematician It just didn’t feel like she knew and has to look elsewhere, Whoever she is looking to should have done the expo
@@WEBLY12121 Thanks for the feedback! We're happy to take these notes down to improve our future content. Have you checked out any of our other videos? Have you noticed the same things there?
Not a "waste of a video" for people curious enough about maths. Somehow I doubt that students who already think "that maths is confusing and opaque" are her target audience, champ. Waste of a comment. 🤡
I learned this algorithm 50 years ago, and I still remember we had to find the root of 5 in a test back then using this " pedestrian" method. It's not bad to know that this method exists and what it is based on.
Few times she says to use calculator to find the number you put on right hand side of two digit, three digit, .... to multiply the whole thing by ...... my question is if I'm using a calculator to find the next digit, why not just use it to find the square root to begin with. I wonder.
We place importance on the process, which is why we challenge our students to complete these kinds of problems by hand, even if they're using a calculator for pieces of it. The priority is understanding each step.
