Answer
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Hint: $0.\overline {621} $ is a recurring decimal which can be written as $0.621621621......$. Assume it some variable and simplify it to convert in the form of $\dfrac{p}{q}$.
Complete step-by-step answer:
The given decimal is $0.\overline {621} $. Let’s its value is $x$.
Since it’s a recurring decimal, it can be written as $0.621621621......$. So, we have:
$ \Rightarrow x = 0.621621621......$
Multiplying both sides by 1000, we’ll get:
$
\Rightarrow 1000x = 621.621621621...., \\
\Rightarrow 1000x = 621 + 0.621621....., \\
\Rightarrow 1000x = 621 + x. \\
\Rightarrow 999x = 621, \\
\Rightarrow x = \dfrac{{621}}{{999}}, \\
\Rightarrow x = \dfrac{{23}}{{37}} \\
$
Thus, the $\dfrac{p}{q}$ of $0.\overline {621} $ is $\dfrac{{23}}{{37}}$.
Note: Since the above number is converted in $\dfrac{p}{q}$form, it is called rational number. If the decimal is non-recurring, non-terminating (i.e. continuing endlessly without repetition of any group of digits), then it cannot be converted in the form of $\dfrac{p}{q}$. That’s why such a number is called an irrational number.
Complete step-by-step answer:
The given decimal is $0.\overline {621} $. Let’s its value is $x$.
Since it’s a recurring decimal, it can be written as $0.621621621......$. So, we have:
$ \Rightarrow x = 0.621621621......$
Multiplying both sides by 1000, we’ll get:
$
\Rightarrow 1000x = 621.621621621...., \\
\Rightarrow 1000x = 621 + 0.621621....., \\
\Rightarrow 1000x = 621 + x. \\
\Rightarrow 999x = 621, \\
\Rightarrow x = \dfrac{{621}}{{999}}, \\
\Rightarrow x = \dfrac{{23}}{{37}} \\
$
Thus, the $\dfrac{p}{q}$ of $0.\overline {621} $ is $\dfrac{{23}}{{37}}$.
Note: Since the above number is converted in $\dfrac{p}{q}$form, it is called rational number. If the decimal is non-recurring, non-terminating (i.e. continuing endlessly without repetition of any group of digits), then it cannot be converted in the form of $\dfrac{p}{q}$. That’s why such a number is called an irrational number.
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