Question

Express \[1.\overline {143} \] in the form \[\dfrac{p}{q}\] where p and q are integers \[q \ne 0\].

Answer 1

Hint: A repeating decimal or recurring decimal is a decimal number whose digits are repeating its value at regular intervals and the infinitely repeated portion is not zero. Decimals are equivalent to fractions with denominators that are a power of ten.
Generally, to write a decimal number into fraction \[\dfrac{p}{q}\] form we divide the decimal number with \[{10^n}\] where n is the total number of digits after decimal and then we reduce it but in the case when the value of n if very high then it becomes difficult to reduce the fraction \[\dfrac{p}{q}\]. In such a case we multiply the decimal number with a very small \[{10^n}\]number and its difference from the original number is found.
Here in the question, we need to transform (or express) \[1.\overline {143} \] in \[\dfrac{p}{q}\] form for which first multiply \[1.\overline {143} \] with 10 and again with 100 and subtract the equation obtained to get the final result.

Complete step by step answer:
Given the number \[1.\overline {143} \] a bar on a given number represents the number to be a recurring number, hence we can write the number as \[1.143143\].
Let \[1.143143 = x - - - (i)\]
Now multiply the number by 10 on both the sides, we will have
\[
1.143143 = x \\
10 \times \left( {1.143143} \right) = 10x \\
10x = 11.43143 - - - (ii) \\
\]
Again multiply identity (i) with 1000, hence we have
\[
1.143143 = x \\
1000 \times \left( {1.143143} \right) = 1000x \\
1000x = 1143.143 - - - (iii) \\
\]
Now subtract equation (ii) from (iii), we have
\[1000x - 10x = 1143.143 - 11.43143\]
Hence by solving we get
\[
990x = 1132 \\
x = \dfrac{{1132}}{{990}} \\
= \dfrac{{566}}{{495}} \\
\]
Hence the recurring decimal number in \[\dfrac{p}{q} = \dfrac{{566}}{{495}}\]
Where \[q = 495 \ne 0\]

Note: It is easy to find the fraction form of a given decimal when it is non-recurring or non-repeating since the digits after the decimal are very less but in case of recurring number of digits are more hence it get difficult to reduce the fraction.