# Square Roots by Hand

Ok, so performing tedious calculations by hand that your calculator can do in a millisecond might not be your preferred way to spend a Saturday. However, for the curious mind, or for those who are looking for an unusual math project topic, it is entirely possible to find the square root of any number by hand. And I don't just mean things like $\sqrt{100} = 10$. I'm talking about finding something like $\sqrt{6}$, and to as many decimal places as you could ever want. After all, your calculator has some kind of process built in to figure it out. It doesn't just have every single possible number "memorized" - it uses algorithms (repetitive logic processes) to iteratively crunch and crunch and refine the answer to be more and more accurate until it has a "satisfactory" number of correct decimal places (usually 10 or 12, depending on your calculator model - sideways iPhone calculator has 15 as I'm writing this).We're going to set up our own step-by-step algorithm that we can repeat, repeat, and repeat, until we have as many decimal places as we want. This kind of process is exactly what your calculator does in its circuits, though it probably has you beat for speed.Step 1 - SetupFirst set up a radical and get ready to do some work that looks visually a whole lot like good ol' long division. We'll take the square root of 1411 in this walkthrough example.Also like long division, notice how we've put a decimal after $1411$ and two $0$ placeholder digits for every decimal of accuracy you desire. After all, since $37^2 = 1369$ and $38^2 = 1444$, we know the answer is going to be $37.[\mathrm{something}]$, and since we said we are going to find the answer accurate to three decimal places, we'll need six $0$'s.Next, to make it easier for the algorithm, separate digits into pairs, starting at the decimal place and working up with larger place values. This means you may have one or two numbers in the first block.Step 2 - Draw First BloodTo get the process started, we need the first digit in the answer. Find the left-most "block" (which again, may be either a single digit or a pair), and look at that number. Find the number whose square is less than or equal to it. In this case we're looking at $14$, and the number whose square is less than or equal to it is $3$. Put this number on top.Step 3 - "Division" StepsPart of this repeating algorithm will look a lot like long division. Write the square of this number below, and subtract - just like we do in long division.Then, bring down the next block of digits. Again, this is very similar to long division!Step 4 - Find the Next DigitThis next step is when it gets a little weird and particular. Define $q$ to be our answer so far, ignoring decimals. To get the next number, we need to find the biggest number $d$ such that $d(20q+d)$ is less than $511$, which is the number we were left with from the prior step. Here, $q=3$ since that's all we have up top so far. So, for this problem, we want the biggest number $d$ such that $d (20 \times 3 + d)$ is less than $511$. Hopefully it is easier to digest with the actual example - it's quite difficult to follow theoretically but becomes easier with real examples and practice.$20$ is a special number in this manual square root process. When we repeat steps to get more decimal accuracy, we will always use $20$ times $q$.As we said, here, we want $d$ such that $d \cdot (20 \times 3 +d)$ is less than $511$.$$\mathrm{maximize} \,\,\,\, d \,\,\,\, \mathrm{such} \,\,\,\, \mathrm{that}$$$$d \cdot [60 + d] < 511$$Some quick trial and error will tell you that $d = 7$ is the number we seek. $d(60+d) = 7(67) = 469 < 511$ but if we try $8$ we would get $8(68) = 544 \nless 511$. Trial and error isn't the best "algorithm" practice but recall that's kind of what we do in long division anyway when we seek out the right numbers on for a quotient.Put our magic number $7$ on top.Step 5 - Repeat Steps 3 and 4For as many decimal places of accuracy that you desire, we will continue to subtract, bring down, and find the next digit, using the same ideas we saw in steps 3 and 4. Let's keep going until we have three decimal places of accuracy in this example.Next, we jot down that $d \cdot [20q + d]$ result from the last step and subtract. Then bring down the next two digits.$4200$ will be the next limiting number.Time to do the weird Step 4 stuff again. Grab the whole result so far ignoring decimals (which is now $q=37$) and apply it to our test number expression: $d(20 \times q + d)$. Now we seek the biggest whole number $d$ such that $d(20 \times 37 + d)$ is less than $4200$.$$d(740+d) < 4200$$$$\mathrm{maximum} \,\,\,\, d = 5$$Our maximum whole number $d$ is $5$, yielding $5(20 \times 37 + 5) = 3725$. With this result, we should put the $5$ as the next digit in our answer, and then repeat Step 3 by subtracting and bringing down once again. We will then haveWe now have our answer accurate to one decimal place. We will keep going and continue the same logic in the algorithm. With $q=375$ (again, ignore the decimal), find the number $d$ such that $d(20q + d)$ is less than $47500$.$$d(20 \times 375 + d) < 47500$$$$\mathrm{maximum} \,\,\,\, d = 6$$The next digit in our answer is $6$, and the result $6 (20 \times 375 + 6) = 45036$ will be used to repeat Step 3 with the "long division" steps. Therefore, to two correct decimal places, we are now looking atTo get the last digit (I know it's grueling but we said let's get three decimal place accuracy!), we once again seek the next number in our special expression which will be the next and final digit in our answer. Now, $q = 3756$ and therefore find the biggest whole number $d$ such that we maximize $d(20q+d)$ without exceeding $246400$. That $d$ will turn out to be $3$.We've got what we came to get - the answer correct to three decimal places. Yes we would probably need a calculator to perform the arithmetic with large numbers in this process. Yes that defeats the purpose. But hey, they didn't have calculators at all a few hundred years ago! I guess this is what they did for fun before TV / sports / recorded music existed. #blessed