Terminating Decimals

In this wiki we will be talking only about rational numbers. There are 2 types of rational decimal numbers: repeating and terminating.

Terminating decimals are rational decimal numbers which can be written in the decimal form by using a finite number of digits.

A terminating decimal can be easily identified. If the number doesn't have a bar over it, it is a terminating decimal.

A fraction can be identified as a terminating decimal if its lowest term is in the form $\frac { k }{ \prod _{ i=1 }^{ n }{ { p }_{ i }^{ { a }_{ i } } } },$ where ${ \left\{ { p }_{ i } \right\} }_{ i=1 }^{ n }$ is the set of prime factors of the base being used. In base 10, the form is $\frac { k }{ { 2 }^{ a }{ 5 }^{ b } } .$

Which of the following is a terminating decimal?

$\begin{array}{c}&(a)~ 0.2312 &&&&(b)~ 0.\overline { 232 }\end{array}$

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The answer is $(a)$ because $(b)$ has a bar on it, representing a repeating decimal. $_\square$

Is $\displaystyle \frac { 14 }{ 270 }$ a terminating decimal?

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The answer is no because

$\frac { 14 }{ 270 } =\frac { 7 }{ 135 } =\frac { 7 }{ { 3 }^{ 3 }{ 5 }^{ 1 } } ,$ which is not in the form $\displaystyle \frac { k }{ { 2 }^{ a }{ 5 }^{ b } }$. $_\square$

The number of decimal places in $\displaystyle \frac { k }{ { 2 }^{ a }{ 5 }^{ b } }$ is $\displaystyle \frac { a+b+\left| a-b \right| }{ 2 }$. In other words, it is the greater of the two numbers $a$ and $b$.

Find the number of decimal places in $\displaystyle \frac { 14 }{ 12500 }$.

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We have

$\frac { 14 }{ 12500 } =\frac { 7 }{ 6250 } =\frac { 7 }{ { 5 }^{ 5 }{ 2 }^{ 1 } }.$

So, the answer is $5$, because $5>1$. $_\square$