Question

Express \[\dfrac{3}{{40}}\] in decimal form .

Answer 1

Hint: We have to convert the given fractional value $ \dfrac{3}{{40}} $ into its respective decimal form . We solve this by multiplying the numerator and the denominator by suitable terms such that the value of the denominator becomes a number such that it can be written in the power of $ 10 $ . Firstly we find out the prime factors of the numerator and denominator and convert it to terminating decimal expansion by multiplying the numerator and the denominator by the required prime factor .

Complete step-by-step answer:
Given : $ \dfrac{3}{{40}} $ is the given fraction form
Now , we have to find the prime factorisation of numerator and denominator
Prime factorisation of numerator \[ = {\text{ }}3\]
Prime factorisation of denominator \[ = {\text{ }}40\]
 \[ = {\text{ }}2{\text{ }} \times {\text{ }}2{\text{ }} \times {\text{ }}5{\text{ }} \times {\text{ }}2\]
 $ = {2^3} \times {5^1} $
[ Now to make the fraction terminating expansion we have to multiply the numerator and denominator such that the powers of the prime factors become equal ]
So, we have to make the powers of $ 2 $ and $ 5 $ equal
Now, multiplying numerator and denominator by $ {5^2} $
So, the fraction becomes
 $ \dfrac{3}{{40}} = \dfrac{3}{{[{2^3} \times {5^1}] \times (\dfrac{{{5^2}}}{{{5^2}}})}} $
As we know , $ {a^c} \times {b^c} = {(ab)^c} $
 $ \dfrac{3}{{40}} = \dfrac{{(3 \times 25)}}{{({2^3} \times {5^3})}} $
 $ \dfrac{3}{{40}} = (\dfrac{{75}}{{{{10}^3}}}) $
We have acquired the required denominator in the power of $ 10 $
Now , putting decimal points
We get ,
 \[\dfrac{3}{{40}}{\text{ }} = {\text{ }}0.075\]
Thus the decimal expansion of \[\dfrac{3}{{40}}{\text{ }} = {\text{ }}0.075\]

Note: Let \[x{\text{ }} = {\text{ }}\dfrac{p}{q}\] be a rational number , such that the prime factorisation of $ q $ is of the form $ {2^n} \times {5^m} $ where \[n{\text{ }},{\text{ }}m\] are non - negative integers . Then $ x $ has a decimal fraction which terminates .
Let \[x{\text{ }} = {\text{ }}\dfrac{p}{q}\] be a rational number , such that the prime factorisation of $ q $ is not of the form $ {2^n} \times {5^m} $ where \[n{\text{ }},{\text{ }}m\] are non-negative integers . Then $ x $ has a decimal fraction which is non-terminating or non-terminating .