Question

How do you convert $1.19$ ( $9$ being repeated) into a fraction?

Answer 1

Hint: In this number, as $9$ is being repeated, we multiply the number by $100$ and by $10$ separately and then subtract the two multiplied numbers. This gets rid of the repeating digits. We can then easily find out the fraction.

Complete step by step solution:
The given number is
$1.1\overline{9}$
The bar “-“ indicates that the number under the bar is repeating like $1.199999....$ . At first, it may seem quite difficult or impossible as there is no end to the repeating digits after the decimal. But, there is a simple concept or logic that can be implemented to get rid of the repeating digits.
Let the given number be $x$ , or in other words,
$x=1.1\overline{9}.....\left( 1 \right)$
Let us multiply both sides of the above equation by $100$ . The above equation thus becomes,
$\Rightarrow 100x=119.\overline{9}....\left( 2 \right)$
Now, let us multiply both sides of equation $\left( 1 \right)$ by $10$ , The equation thus becomes,
$\Rightarrow 10x=11.\overline{9}....\left( 3 \right)$
Now if we subtract the equation $\left( 2 \right)$ and $\left( 3 \right)$ , we can clearly see that the repeating digit $9$ gets subtracted out completely and we are left out with only non-repeating digits. So, let us subtract equation $\left( 3 \right)$ from equation $\left( 2 \right)$ . This gives,
$\Rightarrow 90x=108$
Dividing both sides of the above equation by $90$ , we get,
$\Rightarrow x=\dfrac{108}{90}=\dfrac{6}{5}$
$x$ is nothing but our assumed given decimal number $1.1\overline{9}$ . This means, we have successfully converted the given decimal into a fraction.
Therefore, we can conclude that the given decimal $1.1\overline{9}$ can be converted to $\dfrac{6}{5}$ as a fraction.

Note: We should have the intuition that upon multiplying x by which number, the repeating digits can be got rid of. This decimal though is easy, there are some recurring decimals out there, which may create a confusion. For them, there is another method. Let the fraction be $\dfrac{p}{q}$ , where,
$p=\left( entire\text{ }number \right)-\left( number\text{ }upto\text{ }no\text{ }bar \right)$ and for $q$ we give slots equal to number of digits after decimal in the number. We will then find the number of repeating digits in the number and insert that many $9$ in the first slots. We then enter $0$ for the remaining slots. Example, for the number $1.1\overline{9}$ , $p=\left( 119 \right)-\left( 11 \right)$ . Number of digits after decimal is two, so we give two slots $\underline{{}}\underline{{}}$ . Number of repeating digits is one, so we insert $9$ in the first slot and $0$ in the second. So, $q=\underline{9}\underline{0}$ .
$1.1\overline{9}=\dfrac{119-11}{90}=\dfrac{108}{90}=\dfrac{6}{5}$