
How do you write the repeating decimal $0.73$ where $73$ is repeated as a fraction?
Answer
548.4k+ views
Hint: In the given question we have, $0.737373...$ , here there is a two-digit repeating number which is $0.73$, thus we need to form an equation such as $n = 0.7373$ and then multiply both the sides of the equation with the power of $10$which will bring the repeating number to the left of the decimal point. We further subtract the new equation with the given equation and get our required answer.
Complete step-by-step solution:
The given number is $0.737373$ where the repeating number is $73$
According to the question; we need to bring this decimal form into the fraction form.
In order to do so, we first form an original equation which is:
$n = 0.73$ ………….. $\left( 1 \right)$
Now we need to multiply both the sides of the above equation with the power of $10$, so that one repeating part gets shifted to the left of the decimal point.
Here since the repeating number has two – digits, therefore we multiply equation (1) with ${10^2}$ which is equal to $100$:
$100\left( {n = 0.7373} \right)$
$ \Rightarrow 100n = 73.73$ ⟶ this is equation $\left( 2 \right)$
Now, in order to solve further we need to subtract equation $\left( 1 \right)$ from equation $\left( 2 \right)$
\[\begin{array}{*{20}{c}}
{100n = 73.73} \\
{\underline {{\text{ }} - n = 0.73{\text{ }}} } \\
{99n = 73} \\
{\_\_\_\_\_\_\_\_\_\_\_}
\end{array}\]
Thus we are only left with the integer part which is $73$
On further solving, we get:
$99n = 73$
On dividing both sides with $99$, we get:
$n = \dfrac{{73}}{{99}}$
Hence we get the required answer.
The required answer is $\dfrac{{73}}{{99}}$.
Note: Repeating decimals are numbers which have a recurring decimal part which keeps on repeating itself over and over again with no end, which means that it follows an infinite loop. They are periodic in nature, which means the numbers reoccur after a fixed period of time. The repeating portion of a decimal number is denoted by a vinculum, for example:
$n = 0.585858........ = 0.\overline {58} $. Rational numbers either have a finite expansion such as : $\left( {\dfrac{2}{5} = 0.4} \right)$ or they can have infinite decimal expansions like: $\left( {\dfrac{1}{3} = 0.3333333......} \right)$ but irrational numbers like $\pi = 3.14159....$ are not repeating nor are they periodical.
Complete step-by-step solution:
The given number is $0.737373$ where the repeating number is $73$
According to the question; we need to bring this decimal form into the fraction form.
In order to do so, we first form an original equation which is:
$n = 0.73$ ………….. $\left( 1 \right)$
Now we need to multiply both the sides of the above equation with the power of $10$, so that one repeating part gets shifted to the left of the decimal point.
Here since the repeating number has two – digits, therefore we multiply equation (1) with ${10^2}$ which is equal to $100$:
$100\left( {n = 0.7373} \right)$
$ \Rightarrow 100n = 73.73$ ⟶ this is equation $\left( 2 \right)$
Now, in order to solve further we need to subtract equation $\left( 1 \right)$ from equation $\left( 2 \right)$
\[\begin{array}{*{20}{c}}
{100n = 73.73} \\
{\underline {{\text{ }} - n = 0.73{\text{ }}} } \\
{99n = 73} \\
{\_\_\_\_\_\_\_\_\_\_\_}
\end{array}\]
Thus we are only left with the integer part which is $73$
On further solving, we get:
$99n = 73$
On dividing both sides with $99$, we get:
$n = \dfrac{{73}}{{99}}$
Hence we get the required answer.
The required answer is $\dfrac{{73}}{{99}}$.
Note: Repeating decimals are numbers which have a recurring decimal part which keeps on repeating itself over and over again with no end, which means that it follows an infinite loop. They are periodic in nature, which means the numbers reoccur after a fixed period of time. The repeating portion of a decimal number is denoted by a vinculum, for example:
$n = 0.585858........ = 0.\overline {58} $. Rational numbers either have a finite expansion such as : $\left( {\dfrac{2}{5} = 0.4} \right)$ or they can have infinite decimal expansions like: $\left( {\dfrac{1}{3} = 0.3333333......} \right)$ but irrational numbers like $\pi = 3.14159....$ are not repeating nor are they periodical.
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