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

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[cynodont]

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

[stevethecatcouch]

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

[er]

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

[viditshrestha]

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

[unrealworlds]

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

[SM-piup]

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

[ZipplyZane]

It's usually easiest to get closer to the right number in the (2x+r)r stage by simply diving r into the remaining value. So, when you had 704, you can see that 12*5 = 60 and 12 * 6 = 72, so 5 is the largest the number can be. No need to multiply out 125 * 5 first.

Note that this is only an estimation, but it's good to help you find a starting point. It's generally going to be either the number you get here or one lower.

[AlienwareH]

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

[mjones]

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

[Hydelsius]

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

[dolfinmagikpro]

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

[pratik]

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

[hglundahl]

0:44 In math and science books back then, there was given an algorithm for finding square roots. Works same way before and after decimal comma, but depends on dividing the number into groups of two (decimal comma being at one division, always). (This means, first group of integers or last numeral after a comma may be a sole numeral in its group)

If I want to make this for 144, I first divide this into two groups : 1 | 44.

Then I figure out square root of first group (or of largest square number below that group), in this case 1. I put a 1 after = sign, and a 1 to the left.

THEN I do some interesting stuff.

I put a 1 below the 1 to the left. I Then multply them and put product under first group to deduct from it : 1 - 1 = 0. The 0 is put under a line.

I then put a line under the two vertical 1, I make an addition. Same height as the 0, ideally.

1+1=2. Now, I put down first numeral in second group next to the zero : 04.

I divide 04 by 2 in my head and get 2. Now I need to check it is not excessive. I put a 2 after the 1 after =, I put a 2 after the 2 under the line to the left, I put a 2 under that 2, and multiply: 2 * 22 and as I get 44, I put that under 044 (I now have added the other numeral of second group). 044 - 44 = 0.

12 is the very exact sqrt of 144 ... which we already knew, but I was demonstrating the method.

There is a similar one for Cube roots, but it's more complex.

BONUS : if the antiquated book you get this from is in Swedish, you are also likely to get an older spelling, previous to 1906!

[joepike]

On the cube root question there are two points.
1) Provided the cube is perfect the ending digit will tell what digit is in the ones place;
If the ending digit is 9 the ones place will be 9, 2 for 8, 3 for 7, 6 for 6, 5 for 5, 4 for 4, 7 for 3, 8 for 2 1 for 1, 0 for 0.

With 389017, the ending digit is 7 so the ones digit will be 3 if this is a perfect cube.

2) Next 10s place multiplied 3 times will be in the thousands
so focus on 389.

1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=217, 7^3=343, 8^3=512
512>389 so 7 is likely the tens place.

7 in the tens 3 in the ones. Try multiplying 73 three times. It does come out as 389017.

[ffggddss]

Thanks for showing us this! It's an ancestral flash-back!!

I will add, however, that we were taught this method when I was in junior high; ca 1960. It resembles long division, but the divisor keeps growing, the more digits you work out.
A similar method can be developed for finding cube roots, which is, naturally, even more tedious!

Curious that the number that the square root of was asked, when the decimal is removed, is just 3¹⁶, and with the decimal, is shifted an even number (10) of places to the right, so that the square root works out to 3⁸, shifted 5 places to the right.
I think this tells us something about how the problem was composed. [Someone likely took 9, and squared it 3 times.]

Fred

[andreasholand]

This is a nice algorithm to be honest; I can definitely see it being useful before calculators were around.

If given this task without a calculator my intuition would be to either use Newton-Raphson's method or to guess on two numbers near what the answer should be, pick the number in the middle between them, square it and then remove the part of the interval the answer must be outside, then repeating the process.

In this case that would be, for instance, starting with 0.06 and 0.07, but seeing that the answer we want is greater than 0.065^2, therefore looking at the interval between 0.065 and 0.07, repeating the process with 0.0675. This is more computation heavy, though.

[swordofdoom]

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

[preetamin]

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

[danturney]

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

[gregjhill]

I remember doing this in primary school but of course I’d totally forgotten the method.

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