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This question already has an answer here:

Instead of trying to find the square root of square numbers, I try to find the square root of numbers that aren't square numbers like 12 and 35, but it seems hard to do it with pencil and paper, so I use a calculator to figure those out. I wonder how I can find the square root of a not-square number ONLY with paper and pencil.

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marked as duplicate by Ross Millikan, Sasha, user1551, Daniel Robert-Nicoud, egreg Nov 26 '13 at 18:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I hope this isn't a duplicate question like one of my other ones and I really don't know the answer to this, no doubt about it. $\endgroup$ – user111236 Nov 26 '13 at 16:39
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    $\begingroup$ Start with a reasonable approximation, then use the Babylonian method, $x_{n+1} = \frac12\left(x_n + \frac{a}{x_n}\right)$. Depending on the desired accuracy, it can be a lot of paper-and-pencil division. $\endgroup$ – Daniel Fischer Nov 26 '13 at 16:42
  • $\begingroup$ I don't know. That's too complicated and I don't know math that well, but I DO know that the square root of 64 is 8 and that the cube root of that value is 2. Oh, wait! You said it can be A LOT of paper-and-pencil division. Is that really true, or are you faking? $\endgroup$ – user111236 Nov 26 '13 at 16:44
  • $\begingroup$ Yes, but if you want the square root of $61$, say, you can either write $\sqrt{61}$ and call it done, or you have to compute (or you let a machine do the computing). If you compute, the Babylonian method has the advantage that it converges fast. $\endgroup$ – Daniel Fischer Nov 26 '13 at 16:46
  • $\begingroup$ I can't call it done and what does compute mean? $\endgroup$ – user111236 Nov 26 '13 at 16:48
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There is an algorithm that I saw presented in 6th grade, but I never properly learned it. It was standard at one time, but calculators caused it to become neglected.

However, there is the "Babylonian method":

Let $A$ be an approximation to the square root of $N$.

Then $\dfrac{A+\frac NA}{2}$ is a better approximation.

If the first approximation is not absurd, then usually the third is pretty good.

Suppose $3$ is our first approximation to $\sqrt{10}$. Then our second approximation is $$ \frac{3 + \frac{10}{3}}{2} = \frac{19}6 = 3+\frac16=3.16666666\ldots. $$

If $\left(\dfrac{19}{6}\right)^2=10$ then $19^2=6^2\cdot10$, so $361=360$. Not too bad an approximation.

Now let's find the third approximation: $$ \frac{\frac{19}{6}+\frac{10}{19/6}}{2} = \frac{721}{228} = 3.1622807\ldots $$

If $\left(\dfrac{721}{228}\right)^2=10$ then $721^2=228^2\cdot 10$, so $519841=519840$. Off by one part in more than half a million.

All of the above is easily done without a calculator.

A calculator tells me that $\sqrt{10}=3.16227766\ldots$.

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  • $\begingroup$ I didn't say use a calculator, I said only use a paper and a pencil $\endgroup$ – user111236 Nov 26 '13 at 17:32
  • $\begingroup$ All of the above except the last line is quickly and easily done with pencil and paper without a calculator. $\endgroup$ – Michael Hardy Nov 26 '13 at 17:34
  • $\begingroup$ Didn't you hear me? I said no calculator! Oh, wait. You didn't use a calculator! $\endgroup$ – user111236 Nov 26 '13 at 17:37
  • $\begingroup$ Also, how do you use latex like that, Michael Hardy? $\endgroup$ – user111236 Nov 27 '13 at 15:10
  • $\begingroup$ @user111236 : I don't understand your last question. How I coded the mathematical notation can be seen in either of two ways: (1) right-click on mathematical notation in my answer; you'll see something that says "Show Math As", and you choose "TeX Commands", or (2) click on "edit" and look at it. $\endgroup$ – Michael Hardy Nov 27 '13 at 19:15
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Here is the good old method I was taught when I was a kid.

You can quickly find a good approximation of the square root of any number by hand.

The calculations involved are quite easy because you always have to multiply by a one-digit number.

With a bit of practice, you can even do it for small numbers mentally.

Take your number (let's say it's an integer) and add one 0 to its left if it has an odd number of digits.

For example, the number $735$ is written $0735$.

Now consider the two leftmost digits : $\stackrel{\frown}{07}35$ and search for the largest number that is less or equal to $7$ when squared : it's $2$, because $2^2=4<7$ and $3^2=9>7$.

So the square root of $735$, when written from left to right, starts with a $2$.

$2^2=4$, so we have a remainder of $3$.

You can write something like this :

$$\stackrel{\frown}{07}35$$ $$\ \ 3\ 35$$

I've written the two next (and last) digits close to $3$.

Next, I always take the digits I've found so far and multiply them by two, then write a multiplication like this :

$$4\ \color{red}.\times\ \color{red}. \ = \ ...$$

The number I've found so far was $2$, so I doubled it : it makes $4$, hence the line above.

The red dots indicate the same digit. Now the question is : What is the largest digit I can put in place of the red dots such that the multiplication above is less or equal to $335$ ?

$48\times 8=384$ is too big but $47\times 7=329$ is good, so the next number is $7$, i.e. the square root of $735$ is $27,\dots$. The remainder is $6$.

Since there's no digit left, we add a comma to our number and two zeroes to the right of the remainder $6$, so we have so far :

$$\stackrel{\frown}{07}35$$ $$\ \ 3\ 35$$ $$\ \ \ \ \ \ \ \ \ \ \ 6\ 00$$

Now we multiply $27$ by $2$ like before and write :

$$54\ \color{red}.\ \times\ \color{red}.\ =\ ...$$

We quickly see that $1$ is the right digit here : $541\times 1=541$ because a $2$ would make a product over a thousand.

Hence, the square root of $735$ is $27,1\dots$

I'll continue a bit :

$600 - 541 = 59\rightarrow 5900$ and $271\times 2=542\rightarrow 542\color{red}1\ \times\ \color{red}1 =\ 5421 < 5900$ so $\sqrt{735}\approx 27,11$

$5900 - 5421 = 479\rightarrow 47900$ and $2711\times 2=5422\rightarrow 5422\color{red}0\ \times\ \color{red}0 =\ 0 < 47900$ so $\sqrt{735}\approx 27,110$

$47900 - 0 = 47900\rightarrow 4790000$ and $27110\times 2=54220\rightarrow 54220\color{red}8\ \times\ \color{red}8 =\ 4337664 < 4790000$ so $\sqrt{735}\approx 27,1108$

And you can continue until you're satisfied.

Here is how I write it by hand :

Square root of 735

Now let me do it for $12$ and $35$ : (It took me like 30 seconds for each of them)

Square root of 12

Square root of 35

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  • $\begingroup$ That's awesome and how do you make those pictures like that? Also, that's complicated. Too complicated. $\endgroup$ – user111236 Nov 26 '13 at 18:57
  • $\begingroup$ Any camera can do it! I did it with my phone. $\endgroup$ – Philippe Malot Nov 26 '13 at 19:02
  • $\begingroup$ Okay, but how did you get it/them in the writing spot and how can you get the picture(s) to go from your phone or digital camera to the computer? Also, can even a webcam do it? $\endgroup$ – user111236 Nov 26 '13 at 20:16
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    $\begingroup$ I also learned this when I was young. By the way, this method is based on the simple fact that $(10 a + b)^2 = 100 a^2 + (20a + b)b$. Using the expansion $$(10 a + b)^3 = 1000 a^3 + (300 a^2 + 30 a b + b^2)b$$ one has an analogous algorithm to take cube roots. The actual numerical computations however, becomes rather more complicated, since at every step one needs to take the square of the string of digits which we have already found, and that part of the computation becomes somewhat unwieldy after a while. $\endgroup$ – Willie Wong Nov 27 '13 at 8:55
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Here's an alternative method that I actually sometimes use when I do not even have pen and paper. It is less efficient than the Babylonian method, but I can store the required intermediate results in my head, which makes it useful if you only need a few decimal places. Furthermore, it is possible to perform this computation using only addition.

You want the square root of some number $x$.

  1. Pick a number $n_0$ for which you can calculate or perhaps know $y_0 = n_0^2$. Try increasing of decreasing $n_0$ a bit until you are fairly close to $x$, but strictly smaller.
  2. Compute $y_1 = (n_0+1)^2$. This can be done easily by using $y_1 = y_0 + n_0 + n_0 + 1$.
  3. If $y_1 > x$, go back to $y_0$. You are done and $y_0$ is your (temporary) answer.
  4. Otherwise, repeat 2 and 3 until you are done.

This procedure gives you the integer with the closest square that is smaller than $x$. To find more digits, multiply $x$ by 100 and the last $n$ by 10. Then, repeat the procedure. Don't forget to divide by 10 again at the end (simply move the decimal point to the proper place).

An Example

I'll demonstrate this procedure with a simple example. Suppose we want to compute the square root of 11 to 3 decimal points.

We start with $n = 3$, such that $y = n^2 = 9$. Because $4^2 > 11$, this is the closest integer with a smaller square, so this will be the first decimal.

Now, we add one zero to $n$ and two zeroes to $x$ and $y$ to get $n=30$, $y=900$ and $x = 1100$. Find a closer integer by repeating step 2 above:

  1. $31^2 = 30^2 + 30 + 31 = 961$
  2. $32^2 = 31^2 + 31 + 32 = 1024$
  3. $33^2 = 32^2 + 32 + 33 = 1089$
  4. $34^2 = 33^2 + 33 + 34 = 1156$

Oops, we've gone too far. So, apparently $33$ is the integer we want. Recall we multiplied by 10, so this means that actually your answer now is $\sqrt{11} \approx 3.3$. You can repeat this process to find more decimals.

  1. $330^2 = 108900$
  2. $331^2 = 108900 + 330 + 331 = 109561$
  3. $332^2 = 109561 + 331 + 332 = 110224$
  4. Back up to $331$.

So, now we have $\sqrt{11} \approx 3.31$. The actual answer is $3.31662\!\ldots$, so we see that the first 3 decimal places are indeed correct, just like expected.

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