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:

\[ \dfrac { 17 } { 9 } = 1 . 8888889 . . . .\]

## Expert Answer

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}$ .**

*Image/Mathematical drawings are created in Geogebra. *