Question

Write decimal expansion of $\dfrac{7}{{125}}$.

Answer 1

Hint: We will first see which fractions have a terminal expansion. Then, we will modify the given fraction by multiplying and dividing by some number so that we have powers of 10 in the denominator and thus have a fraction.

Complete step-by-step answer:
We know that if we have a fraction then it has a terminal expansion if the denominator is of the form ${2^m} \times {5^n}$, where m and n are non-negative integers.
Now, we see that $\dfrac{7}{{125}}$ has a denominator of 125.
$ \Rightarrow 125 = 5 \times 5 \times 5 = {5^3}$
if we compare it to ${2^m} \times {5^n}$, we get m = 0 and n = 3.
Therefore, the fraction given to us has a terminal expansion.
Now, we have $\dfrac{7}{{125}}$.
We can write:-
$ \Rightarrow \dfrac{7}{{125}} = \dfrac{7}{{{5^3}}} \times 1$
Now, this can be written as:
$ \Rightarrow \dfrac{7}{{125}} = \dfrac{7}{{{5^3}}} \times \dfrac{{{2^3}}}{{{2^3}}}$
Simplifying the RHS, we will get:-
$ \Rightarrow \dfrac{7}{{125}} = \dfrac{{7 \times 8}}{{{{(5 \times 2)}^3}}}$
Simplifying the RHS further, we will get:-
$ \Rightarrow \dfrac{7}{{125}} = \dfrac{{56}}{{{{10}^3}}}$
$\therefore \dfrac{7}{{125}} = \dfrac{{56}}{{1000}} = 0.056$

$\therefore $ The required answer is 0.056

Note: The students must not forget to check if the fraction has a terminal expansion or not before carrying on with the procedure of division (if preferred by the students).
The students must wonder that they have used the condition for a fraction to be terminal but how does this condition work? Let us ponder over it.
If we have a fraction with a denominator in the form of ${2^m} \times {5^n}$. We can always somehow multiply and divide that fraction with 2 or 5 depending upon whether n > m or m > n to make the denominator a power with base 10 and thus easily replicable into a decimal.