Question

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
a). $\dfrac{13}{3125}$
b). $\dfrac{17}{8}$
c). $\dfrac{64}{455}$
d). $\dfrac{15}{1600}$
e). $\dfrac{29}{343}$
f). $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$
g). $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$
h). $\dfrac{6}{15}$
i). $\dfrac{35}{50}$
j). $\dfrac{77}{210}$

Answer 1

Hint: Here, to check whether the rational number $\dfrac{p}{q}$ will have a terminating decimal expansion or not we have to reduce the denominator, $q$ into prime factors and identify whether it has prime factors other than 2 and 5. If $q={{2}^{m}}{{5}^{n}}$ where m and n are integers, we can say that $\dfrac{p}{q}$ will have a terminating decimal expansion. If q can be written as $q={{2}^{m}}{{5}^{n}}k$ where m and n integers and k is any prime number other than 2 and 5, we can say that $\dfrac{p}{q}$ will have a non-terminating repeating decimal expansion.

Complete step-by-step answer:
Here, without performing the long division, we have to check whether the rational numbers have a terminating decimal expansion or a non-terminating decimal expansion.
To check whether the rational number $\dfrac{p}{q}$ is terminating or non- terminating, we have to reduce q into its prime factors.
If q can be written as $q={{2}^{m}}{{5}^{n}}$ where m and n are integers, we can say that $\dfrac{p}{q}$ will have a terminating decimal expansion.
Else, if q can be written as $q={{2}^{m}}{{5}^{n}}k$ where m and n integers and k is any prime number other than 2 and 5, we can say that $\dfrac{p}{q}$ will have a non-terminating repeating decimal expansion.
Now, we need to check the same for each rational number above.

a). Consider the rational number $\dfrac{13}{3125}$ which is of the form $\dfrac{p}{q}$ .
Here, q = 3125, now we have to reduce q = 3125 into its prime factors.
Here, to reduce 3125 into its prime factors first do the prime factorisation:
$\begin{align}
  & 5\left| \!{\underline {\,
  3125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  625 \,}} \right. \\
 & 5\left| \!{\underline {\,
  125 \,}} \right. \\
 & 5\left| \!{\underline {\,
  25 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \,\,\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
From above prime factorisation, we will get:
 $\begin{align}
  & 3125=5\times 5\times 5\times 5\times 5 \\
 & 3125={{5}^{5}} \\
\end{align}$
Here, 3125 can be written as a power of 5 and it contains only a prime number 5. Therefore, we can say that $\dfrac{13}{3125}$ will have a terminating decimal expansion.

b). Consider the rational number $\dfrac{17}{8}$ where q = 8
Now, we can write:
 $\begin{align}
  & 8=2\times 2\times 2 \\
 & 8={{2}^{3}} \\
\end{align}$
Here, 8 can be written as a power of 2 alone and it contains only a prime number 2. Therefore, we can say that $\dfrac{17}{8}$ will have a terminating decimal expansion.

c). Next, we can consider the rational number $\dfrac{64}{455}$ where q = 455.
Now, we have to do the prime factorisation:
$\begin{align}
  & \,\,5\left| \!{\underline {\,
  455 \,}} \right. \\
 & \,\,7\left| \!{\underline {\,
  91 \,}} \right. \\
 & 13\left| \!{\underline {\,
  13 \,}} \right. \\
 & \,\,\,\,\,\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
From the above prime factorisation, we can write 455 as,
$455=5\times 7\times 13$
The prime factors of 455 includes prime numbers 7 and 13 other than 5. Therefore, we can say that $\dfrac{64}{455}$ will have a non- terminating decimal expansion.

d). Consider the rational number $\dfrac{15}{1600}$ where q = 1600.
First, we have to do the prime factorisation of 1600.
$\begin{align}
  & 5\left| \!{\underline {\,
  1600 \,}} \right. \\
 & 5\left| \!{\underline {\,
  320 \,}} \right. \\
 & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & \,\,\,\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Now, we can write 1600 as:
$\begin{align}
  & 1600=5\times 5\times 2\times 2\times 2\times 2\times 2\times 2 \\
 & 1600={{5}^{2}}{{2}^{6}} \\
\end{align}$
The prime factors of 1600 include only two prime numbers 2 and 5. Therefore, we can say that$\dfrac{15}{1600}$ will have a terminating decimal expansion.

e). Next, consider the rational number $\dfrac{29}{343}$ where q = 343.
First, we have to find the prime factorisation of 343.
$\begin{align}
  & 7\left| \!{\underline {\,
  343 \,}} \right. \\
 & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \,\,\,\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
That is, we can write 343 as:
$\begin{align}
  & 343=7\times 7\times 7 \\
 & 343={{7}^{3}} \\
\end{align}$
The prime factors of 343 contains only a prime number 7.Therefore, we can say that $\dfrac{29}{343}$ will be a non - terminating repeating decimal expansion.

f). Next, consider the rational number $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ where q = ${{2}^{3}}{{5}^{2}}$.
Here, we don’t have to do the prime factorisation since q is already reduced into its prime factors and the prime factors contains only two prime numbers 2 and 5, we can say that $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ will have a terminating decimal expansion.

g). Now consider the rational number $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$ where q = ${{2}^{2}}{{5}^{7}}{{7}^{5}}$.
Here, we don’t have to do the prime factorisation since q is already reduced into its prime factors.
Here, the prime factors contain the prime numbers 2, 5 and 7. That is, apart from 2 and 5, q contains another prime number 7. Therefore, we can say that the rational number $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$ will have a non-terminating decimal expansion.

h).Consider the next rational number $\dfrac{6}{15}$ where q = 15.
Now, 15 can be reduced into its prime factors. i.e. 15 can be written as:
$15=3\times 5$
From above, we can say that 15 contains a prime number other than 2 and 5, i.e. 3. Therefore, we can say that the rational number $\dfrac{6}{15}$ will have a non-terminating repeating decimal expansion.

i). Now, consider the next rational number $\dfrac{35}{50}$ where q = 50.
Here, first we have to do the prime factorisation of 50.
That is, 50 can be written as:
$\begin{align}
  & 50=2\times 5\times 5 \\
 & 50=2\times {{5}^{2}} \\
\end{align}$
From above we can say that the prime factors of 50 contain only two prime numbers 2 and 5. Therefore, we can say that the rational number $\dfrac{35}{50}$ will have a terminating decimal expansion.

j).Consider the last rational number $\dfrac{77}{210}$ where q = 210.
Here, first we have to do the prime factorisation.
$\begin{align}
  & 5\left| \!{\underline {\,
  210 \,}} \right. \\
 & 7\left| \!{\underline {\,
  42 \,}} \right. \\
 & 2\left| \!{\underline {\,
  6 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \,\,\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Now, we can write 210 as:
$210=5\times 3\times 7\times 2$
The prime factors of 210 contain prime numbers 3 and 7 other than 2 and 5. Therefore, we can say that the rational number $\dfrac{77}{210}$ will have a non-terminating decimal expansion.

Note: Here, this method is applicable only for non-terminating recurring decimal expansion not for non-terminating non-recurring decimal expansion. Also here, instead of doing separately you can also create a table and write the prime factors of the denominators. From the table you can identify whether the rational number will have terminating decimal expansion or a non-terminating repeating decimal expansion.
.