# Quick math question **URL:** \ **Category:** Factual Questions\ **Created:** [August 4, 2000, 9:04pm UTC](https://boards.straightdope.com/t/quick-math-question/28173 "2000-08-04T21:04:54Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![Fatal\_Image](https://avatars.discourse-cdn.com/v4/letter/f/73ab20/32.png) [@Fatal\_Image](https://boards.straightdope.com/u/Fatal_Image)\ **Post date:** [August 4, 2000, 9:04pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/1 "2000-08-04T21:04:54Z")
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. ---
**Author:** ![IzzyR](https://avatars.discourse-cdn.com/v4/letter/i/2acd7d/32.png) [@IzzyR](https://boards.straightdope.com/u/IzzyR)\ **Post date:** [August 4, 2000, 9:14pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/2 "2000-08-04T21:14:02Z")
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. ---
**Author:** ![gigi](https://avatars.discourse-cdn.com/v4/letter/g/a587f6/32.png) [@gigi](https://boards.straightdope.com/u/gigi)\ **Post date:** [August 4, 2000, 9:28pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/3 "2000-08-04T21:28:03Z")
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?? ---
**Author:** ![sdimbert](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/sdimbert/32/20204_2.png) [@sdimbert](https://boards.straightdope.com/u/sdimbert)\ **Post date:** [August 4, 2000, 9:34pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/4 "2000-08-04T21:34:49Z")
**Fatal Image** : I may be repeating, because I am simply C&P’ing from [www.ask.com](http://www.ask.com/main/followup.asp?qCategory=SCI_&ask=how+do+you+figure+a+square+root&qSource=0&origin=0&frames=yes&site_name=Jeeves&metasearch=yes&ads=&aj_ques=snapshot%3DJeeves%26kbid%3D270971&aj_logid=64AE5039526AD4118AD6009027737DE1&aj_rank=3&aj_score=0.8&x=21&y=11), but here you go anyway: > [@](#): > > How do you find the square root of a number by hand? > > The square root of a number is just the number which when multiplied by itself gives the first number. So 2 is the square root of 4 because 2 \* 2 = 4. > > Start with the number you want to find the square root of. Let’s use 12. There are three steps: > Guess > Divide > Average. > … and then just keep repeating steps 2 and 3. > > First, start by guessing a square root value. It helps if your guess is a good one but it will work even if it is a terrible guess. We will guess that 2 is the square root of 12. > > In step two, we divide 12 by our guess of 2 and we get 6. > > In step three, we average 6 and 2: (6+2)/2 = 4 > > Now we repeat step two with the new guess of 4. So 12/4 = 3 > > Now average 4 and 3: (4+3)/2 = 3.5 > > Repeat step two: 12/3.5 = 3.43 > > Average: (3.5 + 3.43)/2 = 3.465 > > We could keep going forever, getting a better and better approximation but let’s stop here to see how we are doing. > > ``` > 3.465 * 3.465 = 12.006225 > > ``` > > That is quite close to 12, so we are doing pretty well. ---
**Author:** ![sdimbert](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/sdimbert/32/20204_2.png) [@sdimbert](https://boards.straightdope.com/u/sdimbert)\ **Post date:** [August 4, 2000, 9:39pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/5 "2000-08-04T21:39:36Z")
When I first saw the OP, I thought of a process I learned long ago that looked like this: ```auto 20 / \ 2 10 / \ 2 5 ``` I don’t remember what it was for or what the next step is. Can anyone remind me? ---
**Author:** ![ren](https://avatars.discourse-cdn.com/v4/letter/r/ed655f/32.png) [@ren](https://boards.straightdope.com/u/ren)\ **Post date:** [August 4, 2000, 9:54pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/6 "2000-08-04T21:54:49Z")
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… ---
**Author:** ![rjk](https://avatars.discourse-cdn.com/v4/letter/r/ed655f/32.png) [@rjk](https://boards.straightdope.com/u/rjk)\ **Post date:** [August 5, 2000, 12:19am UTC](https://boards.straightdope.com/t/quick-math-question/28173/7 "2000-08-05T00:19:39Z")
Dammit, gigi! I thought I was the only one who remembered that! I think I’m in love! ---
**Author:** ![Bear\_Nenno](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bear_nenno/32/3358_2.png) [@Bear\_Nenno](https://boards.straightdope.com/u/Bear_Nenno)\ **Post date:** [August 5, 2000, 12:37am UTC](https://boards.straightdope.com/t/quick-math-question/28173/8 "2000-08-05T00:37:38Z")
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. ---
**Author:** ![Bear\_Nenno](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bear_nenno/32/3358_2.png) [@Bear\_Nenno](https://boards.straightdope.com/u/Bear_Nenno)\ **Post date:** [August 5, 2000, 12:43am UTC](https://boards.straightdope.com/t/quick-math-question/28173/9 "2000-08-05T00:43:03Z")
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. ---
**Author:** ![gigi](https://avatars.discourse-cdn.com/v4/letter/g/a587f6/32.png) [@gigi](https://boards.straightdope.com/u/gigi)\ **Post date:** [August 7, 2000, 7:41pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/10 "2000-08-07T19:41:46Z")
> [@](#): > > \*Originally posted by rjk \* > \*\*Dammit, gigi! I thought I was the only one who remembered that! > > I think I’m in love! > \*\* : blush: And all these threads say math classes don’t pay off… 😉 ---
**Author:** ![KJ](https://avatars.discourse-cdn.com/v4/letter/k/b2d939/32.png) [@KJ](https://boards.straightdope.com/u/KJ)\ **Post date:** [August 7, 2000, 9:56pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/11 "2000-08-07T21:56:24Z")
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_ ---
**Author:** ![ren](https://avatars.discourse-cdn.com/v4/letter/r/ed655f/32.png) [@ren](https://boards.straightdope.com/u/ren)\ **Post date:** [August 7, 2000, 10:14pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/12 "2000-08-07T22:14:04Z")
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… ---
**Author:** ![sdimbert](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/sdimbert/32/20204_2.png) [@sdimbert](https://boards.straightdope.com/u/sdimbert)\ **Post date:** [August 7, 2000, 10:22pm UTC](https://boards.straightdope.com/t/quick-math-question/28173/13 "2000-08-07T22:22:40Z")
> [@](#): > > \*Originally posted by Bear\_Nenno \* > \*\*Oooooops. > > …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. \*\* > > _Originally posted by KJ_ > \*\* > The exact answer to /¯20 is, indeed, /¯2 \* /¯2 \* /¯5, or 2 \* /¯5. So… 1/3 is exact and .33333 is approximate? Then doesn’t that mean that .99999 is just _approximately_ 1? _:: duckingandrunning ::_ 😃