 # 3.16 repeating as a fraction. Convert 3.16 to a fraction. This question aims to find the repeating number $3.16$ as a fraction. Fraction is any number written in the form of a quotient. In the quotient, any integer written above is called the numerator and the integer written below is called the denominator. An integer can be any real number or complex number.

If the integer written in the numerator is less than the denominator, then it is called a proper fraction. Similarly, if the integer written in the numerator is greater than the denominator, then it is called an improper fraction.

Repeating fractions are those numbers that have infinite digits after the decimal point. The digits do not stop and they keep on repeating. These types of fractions are also called recurring fractions. They can be written in the form of:

$\dfrac { 17 } { 9 } = 1 . 8888889 . . . .$

If we have to convert the repeating decimal into fractions then we have to take two equations. Assume:

$x = 3 . 1666 . . . eq . 1$

To eliminate the decimal point, we will multiply $eq.1$ with $10$.

$10 x = 31 . 666 . . . eq . 2$

By subtracting $eq.2$ from $eq.1$ we get:

$10 x – x = 31 . 666 . . . – 3 . 1666 . . .$

$9 x = 28 . 5$

$x = \dfrac { 28 . 5 } { 9 }$

$x = \dfrac { 285 } { 90 }$

$x = \dfrac { 19 } { 6 }$

$x = 3 \dfrac { 1 } { 6 }$

## Numerical Solution

The fraction of repeating number $3 . 16 . . .$ is $3 \dfrac { 1 } { 6 }$ .

## Example

Convert $1.888$ into a fraction.

Let us assume:

$x = 1 . 888 . . . eq . 1$

To eliminate the decimal point, we will multiply $eq.1$ with $10$.

$10 x = 18 . 888 . . . eq . 2$

By subtracting $eq.2$ from $eq.1$ we get:

$10 x – x = 18 . 888 . . . – 1 . 888 . . .$

$9 x = 17$

$x = \dfrac { 17 } { 9 }$

The fraction of repeating number $1 . 888$ is $\dfrac { 17 } { 9 }$ .

$2$ ) Convert $0 . 414141 . . .$ into the fraction.

Let us assume:

$a = 0 . 414141 . . . eq . 1$

To eliminate the decimal point, we will multiply $eq.1$ with $100$.

$100 a = 41 . 414141 . . . eq . 2$

By subtracting $eq.2$ from $eq.1$ we get:

$100 a – a = 41 . 4141 . . . – 0 . 414141 . . .$

$99 a = 41$

$a = \dfrac { 41 } { 99 }$

The fraction of repeating number $0 . 414141 . . .$ is $\dfrac {41}{99}$ .

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