# Calculating square roots without a calculator [closed]

Can anyone help me calculate $\large{\sqrt{\frac{4}{11}}=\sqrt{0.\overline{36}}}$ using the digit by digit method?

## closed as off-topic by wythagoras, Did, zz20s, user228113, FutureMay 2 '16 at 13:28

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• 'Help' means you've tried something yourself. Can you please share it? – barak manos May 2 '16 at 9:30
• My dad taught me this when I was in 5th-6th grade, and I find it excessively difficult to explain it by this means. Perhaps you need a mentor or at least someone to talk by telephone/facebook/whatsapp with. – DonAntonio May 2 '16 at 10:09
• Do you mean $\sqrt{0.3636363636}$ or $\sqrt{0.\overline{36} }$ ? – callculus May 2 '16 at 10:22
• The first option which is the equivalent to 4/11 – Mary May 2 '16 at 10:53
• No, $0.\overline{36}=\frac4{11}$. This is the second option. – callculus May 2 '16 at 11:48

Let's start with the approximation $\dfrac 4{11}\approx 0.36363636$ that we multiply by $100^{\,4}$ to get an integer (this is not an obligation but somewhat simplifies the discussion).

If we want the square root of $N$ and know that $a^2 \le N <(a+1)^2$ then the next digit $d$ must verify $$(10a+d)^2 \le 100 N <(10a+d+1)^2$$ Since we want $\;100a^2+20ad+d^2\le 100 N\;$ we are searching the largest digit $d$ such that $$(20a+d)\,d\le 100\,(N-a^2)$$ At the next iteration $a$ will be given by $\,a'=10a+d\,$ and we may continue the operations.

Here we start with $\,a=6\,$ and get :

\begin{array} {r|ll} \color{#ff00ff}{36}.36.36.36 & 6\\ \hline \\ \underline{-\color{#ff00ff}{36}.00.00.00\;} & \color{#ff0000}{6}&\text{the largest integer which square is smaller than}\ \color{#ff00ff}{36}\\ \color{#ff00ff}{36}.36.36 & 12d\times d&\text{the largest $d$ such that this is smaller than $\color{#ff00ff}{36}$ is $\ d=\color{#ff0000}{0}$}\\ \underline{-\color{#ff00ff}{00}.00.00\;} & \color{#ff0000}{60}&\text{(partial result)}\\ \color{#ff00ff}{36.36}.36 & 120d\times d&\text{the largest $d$ such that this is smaller than $\color{#ff00ff}{3636}$ is $\ d=\color{#ff0000}{3}$}\\ \underline{-\color{#ff00ff}{36.09}.00\;} & \color{#ff0000}{603}&\text{(partial result)}\\ \color{#ff00ff}{27.36} & 1206d\times d&\text{the largest $d$ such that this is smaller than $\color{#ff00ff}{2736}$ is $\ d=\color{#ff0000}{0}$}\\ \underline{-\color{#ff00ff}{00.00}\;}&\color{#ff0000}{6030}\\ \color{#ff00ff}{} \end{array} If we want more precision we could continue with $\;\color{#ff00ff}{27.36.36}\;$ at the left to get the next digit $d=2$ and the solution $\sqrt{\dfrac 4{11}}\approx 0.60302$.

• I believe this is the definitive answer to the question. A similar method that was once taught in French schools is in math.stackexchange.com/a/916263; an equivalent method I learned in US schools (and used in math.stackexchange.com/a/1495943) is at xlinux.nist.gov/dads/HTML/squareRoot.html. I think the only differences among these methods are in how much one explicitly writes at each step and how the steps are formatted on the page. – David K May 2 '16 at 13:41
• Thanks to explain all this @David K! This is indeed the French method as I learned it. The same method works for cubic root as explained here but the work becomes more tedious (Newton iterations considered there converge faster). Cheers, – Raymond Manzoni May 2 '16 at 14:22

$x=0.3636\ldots$

$100x=36.3636\ldots$

This implies that

$99x=36$

$x=\frac{36}{99}$

$\sqrt{x}=\frac{6}{3\sqrt{11}}$

$\sqrt{x}=\frac{2}{\sqrt{11}}$

As for calculating $\sqrt{11}$, there are two ways I have in mind right now:

1) Successive approximations:

$3^2<11<4^2$, so you pick 3.

Then you try $3.5^2$ and see it's above $11$, so you try $3.4^2$ which is again above $11$.

However, $3.3^2<11$ so you got $\sqrt{11}$ to be accurate by one decimal digit ($3.3$). You continue similarly for as much as you want.

2) You use the taylor series of $\sqrt{x}$ about $x=9$

$\sqrt{11}=3+\frac{11-9}{6}-\frac{(11-9)^2}{216}+\frac{(11-9)^3}{3888}\ldots$

• Not helpful. The asker specifically asked for a certain method. This is not that method. – Anon May 2 '16 at 9:40
• Thank you for your response but I'm just wondering if you could outline how to use the digit by digit method on a square root, even root 11? My answer is only working out accurate by 2 decimal places – Mary May 2 '16 at 10:03
• I think this can be helpful as calculating by hand the square root of a positive integer is simpler. +1 – DonAntonio May 2 '16 at 10:11
• $0.3636...=0.\overline{36}$ is not equal to $0.3636363636$ – callculus May 2 '16 at 10:18
• @callculus, I know but I thought he made a mistake or so... – Hasan Saad May 2 '16 at 11:37

In base 10 it goes like this:

Look at the first two digits. Find the greatest digit $x$ with $x^2 \leq 36$, this is $6$. That is the first digit. There is no remainder to carry in this case, because $6^2=36$.

Because there is no remainder we just look at the next two digits, but now we need to find the greatest digit $x$ with $x(20 \cdot 6+x) \leq 36$. (This $6$ is the part of the root that we have already found, ignoring the decimal point.) This is $0$. That is the second digit. Now there is a remainder of $36$.

Carrying the remainder and throwing in two more digits to the right of it, you get $3636$, and you need the greatest digit $x$ with $x(20 \cdot 60+x) \leq 3636$. This is $3$. That is the third digit. You have a remainder of $27$, you carry that and continue.

Finding where the decimal point goes is a separate, but easier, part of the problem. Do you know how to do that? Once you've taken care of that, what we've done so far says that the first three significant digits are given by $0.603$. For comparison the answer from Windows calculator is $0.60302268915552724529362413394012$.

You can see more details at https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Decimal_.28base_10.29 One thing that might not be obvious is where this seemingly magic "20" came from. It is not magic: it is the "2" in the formula $(a+b)^2=a^2+2ab+b^2$, multiplied by the base of the number system that we choose to use.

• I once saw a function $f(x,s) = \frac{1}{2} \big ({x + \frac{s}{x}} \big )$ for generating approximations for a square root, and the function works really well, generating an approximation of $\sqrt 2 = \frac{577}{408}$ after about 4 iterations. What would be the equivalent function for cube roots, et al? My first guess was $f(x,s) = \frac{1}{3} \big ({x + \frac{s}{x}} \big )$, but I didn't generate an accurate approximation when I tried it out. Disclaimer: I'm not educated in calculus. – Braden Best Nov 4 '18 at 6:42
• I realized it was silly to ask a question in the comments, so I migrated it – Braden Best Nov 4 '18 at 7:11