Question

$answer \: the \: questio \: which \: is \: in \: attachment$
•Here back with an challenging question.
•Answer only if u know.
[Hoping great answers from @Sparklingboy,@Mathsdude500,@Taken name]​

Answer 1

$\large{\text{\underline{Question:-}}}$

Express $\sqrt{19}$ as a continued fraction, hence find the approximation.

$\large{\text{\underline{Method:-}}}$

View the attachment for explanation. There might be latex errors, so please view in desktop mode.

First, let the real value be $r$. We are going to isolate the integer part $i$ and the fractional part $f$.

Then, the following equation is made.

$\hookrightarrow r=i+f$

Since fractional part is an irrational number,

$\hookrightarrow \boxed{0<f<1\implies \dfrac{1}{f} >1}$

This means, we can always isolate the integer part since it is always greater than 1.

The value of $i$ is the integer part that comes in place of the boxes.

$\hookrightarrow \dfrac{19}{111} =\boxed{0}+\dfrac{1}{\boxed{5}+\dfrac{1}{\boxed{1} +\dfrac{1}{\boxed{5}+\dfrac{1}{\boxed{3}} } } }$

$\large{\text{\underline{Solution:-}}}$

We have,

$\hookrightarrow \sqrt{19}= 4+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\ddots}}}}}}$

This is the first approximation, without the integer parts recurring.

$\hookrightarrow \dfrac{1421}{326} =4+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8}}}}}}$

which is,

$\hookrightarrow \sqrt{19} \approx \dfrac{1421}{326} \text{ (1st approximation.)}$

We could find the closer rational number if we simplify further. Note that the denominator is repeating, so we will simplify further by manipulating the 1st approximation.

$\hookrightarrow \dfrac{117}{326} =\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8}}}}}}$

Taking inverse, we find the value that comes in the next denominator.

$\hookrightarrow \dfrac{326}{117} ={2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8}}}}}$

Putting the result, we get the result where the integer parts recurring twice.

$\hookrightarrow \sqrt{19}\approx 4+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\dfrac{1}{\dfrac{326}{117} }}}}}}}$

which is,

$\hookrightarrow \sqrt{19} \approx \dfrac{483136}{110839}\text{ (2nd approximation.)}$

Again, to manipulate the 2nd approximation, subtract 4 to get,

$\hookrightarrow \dfrac{483136}{110839} -4=\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\dfrac{1}{\dfrac{326}{117} }}}}}}}$

So,

$\hookrightarrow \dfrac{39780}{110839} =\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\dfrac{1}{\dfrac{326}{117} }}}}}}}$

Taking inverse, we find the value that comes in the next denominator.

$\hookrightarrow \dfrac{110839}{39780} =2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\dfrac{1}{\dfrac{326}{117} }}}}}}}$

Putting the result, we get the result where the integer parts recurring thrice.

$\hookrightarrow \sqrt{19} \approx\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{8+\dfrac{1}{\dfrac{110839}{39780} } }}}}}}$

$\hookrightarrow \sqrt{19} \approx \dfrac{164264819}{37684934}\text{ (3rd approximation.)}$

$\large{\text{\underline{The result:-}}}$

Now let's compare.

$\hookrightarrow \sqrt{19} \approx 4.35889894354067355223$

$\hookrightarrow \dfrac{1421}{326} \approx\underline{4.358}9153688\text{ (Correct up to 3 decimal places.)}$

$\hookrightarrow \dfrac{483136}{110839}\approx \underline{4.3588989435}12662\text{ (Correct up to 10 decimal places.)}$

$\hookrightarrow \dfrac{164264819}{37684934}\approx \underline{4.358898943540673}30992\text{ (Correct up to 15 decimal places.)}$

Turns out, the method is really accurate.

$\large{\text{\underline{Note:-}}}$

There are many ways to approximate the square root. This is one of the ways, along with the Babylonian method.

In the continued fraction notation $[4;\overline{2,1,3,1,2,8}]$, the line represents the recurring integer parts. So, we can represent the continued fraction of the irrational number $\sqrt{19}$ as $[4;\overline{2,1,3,1,2,8}]$. For another example, $[0;5,1,5,3]$ represents the continued fraction of the rational number $\dfrac{11}{119}$.