Faith Welch

2022-07-23

Algorithm for converting fraction into (recurring) decimal

I need to determine whether fraction is recurring decimal or not (what are conditions for it?), find period and output it as $\frac{1}{3}=0.\overline{3}$

If it is not recurring, then I already have algorithm.

I need to determine whether fraction is recurring decimal or not (what are conditions for it?), find period and output it as $\frac{1}{3}=0.\overline{3}$

If it is not recurring, then I already have algorithm.

phinny5608tt

Beginner2022-07-24Added 17 answers

Notice that if you have a repeating decimal

$n=0.\overline{{D}_{1}{D}_{2}...{D}_{k}}$

where each ${D}_{i}$ is a digit, you can say that

${10}^{k}n={D}_{1}{D}_{2}...{D}_{k}.\overline{{D}_{1}{D}_{2}...{D}_{k}}$

and so

${10}^{k}n-n={D}_{1}{D}_{2}...{D}_{k}$

and

$n=\frac{{D}_{1}{D}_{2}...{D}_{k}}{{10}^{k}-1}$

This means that $n$ can only be a repeating decimal if $n$ can be expressed in the form

$n=\frac{a}{{10}^{b}-1}$

and so if you are given a number in the form

$I+\frac{h}{j}$

Where $I$ is an integer and $j>h$, then it can only be a repeating decimal if $j$ evenly divides ${10}^{k}-1$ for some $k$. For example, for $j=1,2,5,10$ it will never be a repeating decimal.

Does this help?

$n=0.\overline{{D}_{1}{D}_{2}...{D}_{k}}$

where each ${D}_{i}$ is a digit, you can say that

${10}^{k}n={D}_{1}{D}_{2}...{D}_{k}.\overline{{D}_{1}{D}_{2}...{D}_{k}}$

and so

${10}^{k}n-n={D}_{1}{D}_{2}...{D}_{k}$

and

$n=\frac{{D}_{1}{D}_{2}...{D}_{k}}{{10}^{k}-1}$

This means that $n$ can only be a repeating decimal if $n$ can be expressed in the form

$n=\frac{a}{{10}^{b}-1}$

and so if you are given a number in the form

$I+\frac{h}{j}$

Where $I$ is an integer and $j>h$, then it can only be a repeating decimal if $j$ evenly divides ${10}^{k}-1$ for some $k$. For example, for $j=1,2,5,10$ it will never be a repeating decimal.

Does this help?

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