Spirit of Math Schools Inc.
Spirit of Math Schools Inc.
Spirit of Math is Canada’s premier after-school math program for high performing students and a global leader in math enrichment, producing students who both love and understand the essence - or spirit - of math. For more than 30 years, Spirit of Math has produced many of the top math students in the nation. Using Spirit of Math methods and materials, our students consistently place on national and international contest honour rolls. Many also qualify for scholarships at private schools and top universities in the world.
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Can you please provide a proof for why this works?
My high school biology teacher (this was 30 years ago and he was near retirement at the time) showed the class this process. I was awestruck, but couldn’t remember the process. I’ve been idly wondering how to do it ever since, and this is exactly it! Thank you so much!
It's so wonderful to read comments like these! People looking for a long time for something they remember (like a song, recipe, or this, or something else), and it emerges for them from the Internet!
Happy to help reel in those (almost) lost memories!
@@alittax One of the many reasons we love being able to share these videos online to people around the globe! Well said.
Really I don't understand how you get that 88 please I want to understand
The first 8 in 88 comes from finding what squared number goes into but not over the first pair of digits (23). 5 squared would be 25, which is over 23, but 4 squared is 16 which is as close as we can get. When you move down to the next line, you have to double that 4 (from 4 squared), which is 8. The second digit comes from looking at the number from the next row (776). You need to find what 2-digit number that starts with 8 and multiplied by the same single digit number equals close to but not over 776. If we use 9 for example, 9 x 89 = 801. If we try 8, 8 x 88 = 704. This is as close as we can get, meaning an 8 goes above 76 and 88 goes to the left, just under 776. Hope this helped!
We did square root problems my senior year but nothing like this!!!
It's never too late to learn a new approach!
successive approximations might be easier.
I learned this almost 45 years ago. Thanks for refreshing my memory! Wonderful.
Happy to help provide you with a blast from the past!
Thanks Stevie Nicks
Right...... 😶
Kalau ade darab dan tambah.. krne selesaikqn darab or bahagi dulu kan
Instead of doubling you can add the left hand side number e.g. instead of calculating 2*48 we can just add 88+8=96 and at next step 967+7=974 etc
We teach this method in India at 7th grade
The method clearly has an international reach
Is not the diameter of a circle a special case of chords of a circle? Have we been short-poured in terms of definitions?
Why wouldnt you complete the equation? Youve done the mathematics equivalent of tearing the last few pages to a book of literature out of the book.
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.)
Or.... use the square root button on your calculator.
Even my dog could push the square root button on a calculator if I would train him to push that particular button that has the square root symbol printed on it. Would it mean he understands advanced mathematics for any high standard? Nope.
How does this work for a cubed root or root of the 4th or etc? This is what breaks my brain with root calculations.
2:10 Also, how would we do this with the (√2)??
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.
My dad developed a method to manually calculate the cubic root as well.
sedqialbaik.blogspot.com/2006/04/blog-post_114434901914567834.html
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.
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?
The square root of 69 is ATE SOMETHING 😂
I was taught this in the 50s and still stretch my brain using this method
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
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.)
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.
What I don’t get is that 9х8 is 72, which is less than 76, obviously, why then you use 8?
Thank you for reminding me why I forgot how to do this.
did I missed something? the last digit: 5 shouldn't that be a 4?
I was taught that in 8th grade in 1970. I'm glad to review that.
I have learned this method, for amusement, some number of times without ever having to memorize it. Calculators are king now. Thanks
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.
brings back memories from grade school
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.
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Would a slide rule come in handy for the middle calculations? 😊
Boooooooooooooooooooo As a mathematician, when you see these bogus methods which simply derail you from proper systematic thinking and methodology. Learn proper methods that work always and leave out these fake that work for corner cases, let alone working at all.
This is stupid
learned this in grade school 1959
Unfortunately, children aren't taught this approach in grade school today and they should be!
Algoritmo di bombelli
Very cool video, and you explained it well. It would definitely take practice and would need math-muscle memory.
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.
This method is based on Binomial Expansion (a+b)squared method. It is very easy. I can show you in a few minutes.
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.
My Dad taught me this method when I was in primary school. Remember it to this day, it has been 16 years.
GREAT VIDEO! Liked and subscribed ❤
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.
Is there anything else you wish you saw earlier? We can help share another video for you.
It's the ×2 that I can't understand
Lucky for you @3Cr15w311 made a comment earlier.
Eventually after many math classes the love of learning was beaten out of me.
Same. Never any explanation or real world examples. Just dreary rote practice out of the textbook.
We're sorry to hear that! We find the best way to learn is in a collaborative, group setting
How can you not double numbers easier
I learned how to calculate square roots nearly 50 years ago. I’m certain they haven’t taught this for probably 30 years
Have you seen our long division video?
I learned how to calculate square roots nearly 50 years ago. I’m certain they haven’t taught this for probably 30 years
It is a lost art, but I am glad that it is in the KZread forever. This is Binomial Expansion Method (BEM).
Ever try converting the number to binary? Still shift by 2. Trial divide always 1 something.
isn't it the 9th row? the first row is index 0 so the 9th row would be 2^8