Calculating square roots without a calculator

Question

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

Answer 1

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 $\sqrt{\dfrac 4{11}}\approx 0.60302$.

Answer 2

$x=0.3636\ldots$

$100x=36.3636\ldots$

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 $x=9$

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

Answer 3

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.