How to write a recurring decimal 0.17(7 is recurring) as fraction ?

Hint:Here the given question need to find a fraction in which the result of the fraction gives the value in which “7” on the tenth place of decimal make recurring decimal, for solving this question we need to apply the algebraic approach in which we will assume a variable for the decimal number given and then solve for the fraction.

Complete step by step solution:
The question needs to find the fraction in which the number seven on the tenth place of decimal comes as recurring number, to solve this question we are assuming the variable “x” for the given decimal number, on solving we get: on constructing the mathematical equation for variable we get:
\[ x = 0.1777777.....\]
Because tenth digit is recurring so multiply by ten on both side of equation, and then on solving we get:
\[10x = 1.777777.....\, \to eq(1)\]
\[\Rightarrow \dfrac{{10x}}{{10}} = \dfrac{{1.777777.....}}{{10}} \\
\Rightarrow x = 0.1777777...\, \to eq(2) \\ \]
Solving both the equation one and two by subtracting them we get:
\[eq(2) - eq(1),we\,get: \\
\Rightarrow 10x - x = 1.777777... - 0.1777777... \\
\Rightarrow 9x = 1.6 \\
\Rightarrow x = \dfrac{{1.6}}{9} \\
\Rightarrow x = \dfrac{{16}}{{90}} \\
\therefore x = \dfrac{8}{{45}} \\ \]
Here we got the final fraction as per instruction given in the question, after getting the fraction with the help of variable we have multiplied by ten to get the better fraction without decimal, but all the fraction obtained are answer and even multiples of these fraction can also be the required answer.

Hence,a recurring decimal 0.17 as fraction is $\dfrac{8}{{45}}$.

Note: Here in this question we get the fraction be using algebra rule, In which we form an equation with the help of variable then by removing the recurring terms, by making two equations using math’s, we get our desired required fraction.