I remember hearing years ago that there was a way to figure out the square root of a number with just a pen and paper, no help from a calculator/computer. For some reason, this thought recently came back to me and now I am dying to know how. A quick search of the Dope, Dope message board, and even the sci.math FAQ turned up nothing. Please help! Thanks.
I can tell you an algorithm which will get you close.
Suppose you want the square root of 42. You start with some estimate, say 13/2. Next you take (13/2 + 42/(13/2))/2. you will be reasonably close with this. But you can then take ((13/2 + 42/(13/2))/2 + 42/((13/2 + 42/(13/2))/2))/2 and get an even closer approximation. And so ad infinitum, depending on how close you need to get.
In other words, you take each succeding estimate, and average it with the number whose root you are looking for divided by the estimate.
Hope this is helpful.
This is super-hard to explain in text but here goes.
You put the number in a long-division box and mark off sets of two digits going in either direction from the decimal point. Look at the leftmost set of numbers (1 or 2 digits) and find the square root of them and out it above the box. Subtract its square from the first set of digits. Now, bring down the next two digits to next to the remainder (like in division). Now, double the number on top of the box. Add digit to it that you multiply it plus the digit by to get closest to the remainder. Subtract, and continue the process.
An example, to demonstrate. 12.5 squared is 156.25,
OK, so we mark off 1’56’.25.
The square root of 1 is 1 so that goes at the beginning of our answer. Now, double the 1 to get 2_ where the __ will be filled in with the next digit of the answer. 2__ divides into 56 (the next set of numbers) how many times, where the number of times also fills the blanks? 2, because 22 x 2 goes into 56. So 44 goes below, leaving a remainder of 12, with our answer at 12 so far.
Bring down the next set of numbers, 25 to join the 12 remainder, for 1225. Now, double our answer so far to be 24. How many times does 24_ go into 1225 with the blank being the multiplier as well? 5, as 245 X 5 =1225. It comes out even, and our answer is 12.5
Oh my goodness I’m glad there are calculators for this. And how did I remember that from 15+ years ago??
Fatal Image:
I may be repeating, because I am simply C&P’ing from www.ask.com, but here you go anyway:
When I first saw the OP, I thought of a process I learned long ago that looked like this:
20
/ \
2 10
/ \
2 5
I don’t remember what it was for or what the next step is. Can anyone remind me?
That looks like your standard prime factorization tree. If your looking for a relationship between prime factorization and square roots, the only thing I can think of is that when testing for candidate factors, you only need to test up to the square root of the number…
Dammit, gigi! I thought I was the only one who remembered that!
I think I’m in love!
Dimber, the next step is to write the answer…
The answer is 2<5. Where “<” is the square root symbol. The technique you remember is a way to simplfy square roots. Instead of leaving an answer of <20, you simplify it and write 2<5. This also makes multiply square roots easier.
Trust me, I know. I recently taught this to my Algebra students.
Oooooops.
Actually there is another step… You have to group pairs of numbers. Since you have two 2’s, you circle them and make one 2. Then you are left with just the 5. Basically what you have done is written <20 as <2 * <2 * <5. Since <2*<2=2, the answer is 2<5.
Sorry, I forgot to explain that. Also, this gives an exact answer whereas a calculator may half to cut off a repeating or very long decimal. The same way that 1/3 is exact and .33333 is approximate.
: blush:
And all these threads say math classes don’t pay off… 
I just got finished with my (2nd, unfortunately) course in Algebra. The exact answer to /¯20 is, indeed, /¯2 * /¯2 * /¯5, or 2 * /¯5.
Another, unbearably slow, way to figure out square roots (which I did not learn in my math class, BTW), which I do not recommend doing by hand is to use logarithms. All multiplication and division can be expressed using addition, subtraction, and alog (I think it means “antilogarithm,” it stands for “10^”, which is the opposite of a logarithm.)
So this means in order to find the exact square root, you’d need a huge logarithm table and you’d have to be willing to do complicated equations anyway. I’d recommend using a different method, but just FYI, I believe this is how computers do these equations.
Now, here are the actual equations for multiplication, division, and exponents, which I will also demonstrate by using the variables X and Y, then replacing them with the numbers 3 and 5, to check the results:
Multiplication:
x * y = alog(log(x) + log(y))
example:
3 * 5 = alog(log(3) + log(5))
3 * 5 = alog(0.4771… + 0.6989…)
3 * 5 = alog(1.1760…)
3 * 5 = 15
Division:
x / y = alog(log(x) - log(y))
example:
3 / 5 = alog(log(3) - log(5)
3 / 5 = alog(0.4771… - 0.6989…)
3 / 5 = alog(-0.2218)
3 / 5 = 0.6
Exponent:
x ^ y = alog(y * log(x))
example:
3 ^ 5 = alog(5 * log(3))
3 ^ 5 = alog(5 * 0.4771…)
3 ^ 5 = alog(2.3856…)
3 ^ 5 = 243
Note that alog(y * log(x)) can be futher broken down into alog(alog(log(y) + log(log(x))), but I would definately not recommend doing that equation unless you are programming some low-level calculator using logic gates or whatnot.
So what does this have to do with finding the square root? Well, the square root of any number can also be expressed as x^(1/2) (something else that I did NOT learn in my math class.). Likewise, the cube root of x is x^(1/3). So to find the square root of 20, you would first write it as 20^.5, then do:
alog(.5 * log(20))
alog(.5 * 1.3010)
alog(0.6505)
4.4721
And there you have it. The square root of 20 is approximately 4.4721.
whew
Maybe I’m missing something, but how does reducing /¯20 to /¯5 * 2 help? So now you need to figure out a different square root, but you still need to figure out a square root?!?
At least the averaging method cited above gives you an actual result…
So… 1/3 is exact and .33333 is approximate? Then doesn’t that mean that .99999 is just approximately 1?
:: duckingandrunning ::
