eli5: How were square roots calculated in the first place, before calculators?
In: Mathematics
You would look them up in a table. The tables were populated by hand calculation.
For example, what’s the square root of 20? Well, you know it’s going to be more than 4 and less than 5, because 16 is 4² and 25 is 5². So you might try 4.5. So then you would square 4.5 and get 20.25. This tells you that the square root of 20 must be a bit less than 4.5. So you could try 4.4. You square 4.4 and you get 19.36.
So now you know that the square root of 20 must be 4.4xxx. So you might try 4.45. 4.45² is 19.8025, so that’s too low. 4.46² is 19.8916, so that’s too low. 4.47² is 19.9809.
That gives us a second decimal place. Now let’s add another decimal place. 4.475² is 20.025625. 4.474² is 20.016676. 4.472² is 19.998784…
So now by essentially guessing and checking, we’ve gotten three decimal places on the square root of 20. 4.472. We could keep going and keep adding decimal places through the exact same method.
There isn’t really an ELI5, since you don’t learn this until Calc 2 at most schools, but a common way to calculate square roots, exponential functions, or log functions is with a Taylor series expansion. A Taylor series takes a function that’s difficult to calculate by hand and expresses it as the infinite sum of simpler functions. You keep adding together smaller and smaller numbers until they are close enough to the exact number for your practical use.
This article does a decent job showing you how: https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Series_and_Expansions/Taylor_Expansion_II
Quick edit: many modern calculators actually do something like this for finding square roots and logs. They can calculate 8 decimal places in less than a second.
You used a format that looks like long division but isn’t, to get the answer one digit at a time. Using (a+b)^2 = a^2 + 2ab + b^2, but backwards, guessing each new b as you went.
1 . 4 1 4
__________________________________________________
1 | 2 . 00 00 00
1 . 00 = 1*1
. ——
1 . 00
4
. 96 = 2(4 *) 4
——
. 04 00
1
. 02 81 = 28(1 *)1
——–
. 01 19 00
4
. 01 12 96 = 282(4 *)4
———–
. 00 06 04 00
, etc. At each layer you take what you’ve got, multiply by 2, stick on your guess for the next digit, then multiply by that guess, shifted over to the right 2 more places; find the guess that gets you closest to the current remainder without going over. Subtract, put that guess digit on the end, repeat.
–Dave, cube roots can be done this way too, but not fourth roots or higher
Before calculators there were books of these mathematical functions like square roots, logarithms, antilogarithm, trigonometric functions, powers of e, etc.
Was in high school in late 1970s and used these. Godfrey and Siddons ‘Four Figure Tables’. The math teacher had an electronic calculator the size of a house brick and he would show it off.
First, think about what a square root actually is: the length of a side of a square that has the given area.
Let’s say you want to find the square root of 10005. Draw a square to help visualize this, and assume that that square’s area is 10005. You know that the square root of 10000 is 100, so let’s draw that square inside the first one with the upper left corners of the two squares on top of each other. You now have a square that you know has an area of 10000, and the square surrounding it shares the same area, plus a strip at the bottom and a strip at the right. Those two strips are the same size: 100 (the length of the sides of the inner square) times some unknown additional width, which we’ll call x. (There’s also a small square at the bottom right that’s x by x, but let’s ignore it for the moment.) That extra area must be equal to the difference of the areas of the two squares: 10005-10000, or 5. So let’s assume that the two strips together equal 5. Since we have two, let’s say each is 2.5. Now we have an area, 2.5, and the length of one of the sides, 100, so it’s easy to find the length of the other side: 2.5 ÷ 100, or 0.025. Well, the length of that additional unknown side is how much bigger the square of area 10005 is than the square of area 10000. So that means the square’s sides are 100.025, and that’s what the square root would be. Except we ignored that x by x sized square at the bottom right, so this is just an estimate, but it’s very close.
If it’s not close enough, though, you can repeat this process with the new 100.025-sided square we just made. It’s actually a little bigger than 10005, so you’ll have to deal with subtracting strips along the edge instead of adding them, but that shouldn’t be too big a deal. Then you can continue repeating the process as many times as you like. (It’s important to note that square roots that are not whole numbers are definitively irrational numbers, so you will never find the exact value. Estimates are the best you can hope for.)
[Matt Parker has a video about this](https://youtu.be/Bwt5EZEb1Ns), and I stole his example values. His drawings may help.
I was at school before calculators were widely available, and we used [log tables](https://en.wikipedia.org/wiki/Mathematical_table) – books of logarithm values to do large multiplications and divisions, and to calculate roots.
To find a square root of a number: look up its logarithm in the table, divide it by 2, then look up the related exponential value.
Here’s a trivial example to illustrate the method…
To get the square root of 25:
log(10) of 25 is 1.39794 (looked up in the base-10 log table)
1.39794 / 2 = 0.69897
10^0.69897 = 5 (looked up by working backwards in the same table, or by using an “anti-log” table)
There are tricks to how to scale and combine values to make it possible to apply the table to various large and small numbers, not just the values that can be directly looked up in the table.
Here is a link that describes the process in detail?
Stepping through the hand calculations required.
[how to calculate square root by hand. ](https://www.freecodecamp.org/news/find-square-root-of-number-calculate-by-hand/#:~:text=How%20to%20find%20the%20square%20root%20of%20a,The%20Right%20Match.%20…%206%20Subtract%20Again.%20)
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