How To Write Repeating Decimal As Fraction

Converting a repeating decimal into a fraction is a fundamental skill in algebra that bridges the gap between decimal approximations and exact rational values. Day to day, while terminating decimals convert easily using place value, repeating decimals—numbers where a digit or block of digits repeats infinitely—require a specific algebraic technique to express them as a ratio of two integers. Mastering this process not only strengthens number sense but also prepares students for more advanced topics in sequences, series, and calculus Simple, but easy to overlook. No workaround needed..

Worth pausing on this one.

Understanding Repeating Decimals and Rational Numbers

Before diving into the conversion method, it is essential to understand why this works. Day to day, a repeating decimal (also called a recurring decimal) is a decimal representation of a number whose digits are periodic. Here's the thing — the infinitely repeated portion is called the repetend or repeating block. Take this: in $0.Also, \overline{3}$, the digit $3$ repeats forever. Here's the thing — in $0. \overline{142857}$, the block $142857$ repeats.

It sounds simple, but the gap is usually here.

A critical theorem in number theory states that a number is rational if and only if its decimal expansion is either terminating or repeating. Day to day, since every repeating decimal represents a rational number, it must be expressible as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. The algebraic method we use exploits the properties of equality and the base-10 number system to isolate the repeating part.

The Standard Algebraic Method: Step-by-Step

The most reliable way to convert a repeating decimal to a fraction involves setting up an equation, multiplying by a power of 10 to shift the decimal point, and subtracting to eliminate the repeating tail. Here is the universal workflow:

1. Assign a Variable

Let $x$ equal the repeating decimal. Example: Let $x = 0.\overline{7}$

2. Identify the Length of the Repeating Block

Count how many digits are in the repeating pattern.

• $0.\overline{7}$ $\rightarrow$ 1 digit ($7$)
• $0.\overline{27}$ $\rightarrow$ 2 digits ($27$)
• $0.\overline{123}$ $\rightarrow$ 3 digits ($123$)

3. Multiply by the Appropriate Power of 10

Multiply both sides of the equation by $10^n$, where $n$ is the number of repeating digits. This shifts the decimal point $n$ places to the right, aligning the repeating parts perfectly.

• For 1 repeating digit: Multiply by $10$.
• For 2 repeating digits: Multiply by $100$.
• For 3 repeating digits: Multiply by $1000$.

Example: $x = 0.\overline{7}$ has 1 repeating digit. Multiply by $10$: $10x = 7.\overline{7}$

4. Subtract the Original Equation from the New Equation

Subtract $x$ from $10x$ (or the respective multiples). Because the decimal parts are identical ($.\overline{7}$), they cancel out completely, leaving an integer on the right side. $ \begin{align*} 10x &= 7.\overline{7} \ -\quad x &= 0.\overline{7} \ \hline 9x &= 7 \end{align*} $

5. Solve for $x$ and Simplify

Divide by the coefficient of $x$ to isolate the variable. Reduce the resulting fraction to its lowest terms. $x = \frac{7}{9}$

Worked Examples: From Simple to Complex

Example 1: Single Digit Repeating ($0.\overline{3}$)

1. $x = 0.\overline{3}$
2. Repeating block length = 1. Multiply by 10: $10x = 3.\overline{3}$
3. Subtract: $10x - x = 3.\overline{3} - 0.\overline{3} \rightarrow 9x = 3$
4. Solve: $x = \frac{3}{9} = \frac{1}{3}$

Example 2: Multi-Digit Repeating Block ($0.\overline{45}$)

1. $x = 0.\overline{45}$
2. Repeating block length = 2 ($45$). Multiply by 100: $100x = 45.\overline{45}$
3. Subtract: $ \begin{align*} 100x &= 45.\overline{45} \ -\quad x &= 0.\overline{45} \ \hline 99x &= 45 \end{align*} $
4. Solve: $x = \frac{45}{99}$. Simplify by dividing by 9: $x = \frac{5}{11}$.

Example 3: Mixed Recurring Decimals (Non-Repeating + Repeating)

This is where many students struggle. Consider $0.2\overline{6}$ (which is $0.2666...$). The digit $2$ does not repeat; only the $6$ repeats.

1. $x = 0.2\overline{6}$
2. We need the repeating part ($6$) to align immediately after the decimal point in our multiplied equation. Currently, the $2$ is in the way.
3. Step A: Multiply by 10 to move the non-repeating digit to the left of the decimal. $10x = 2.\overline{6}$
4. Step B: Now the repeating part is isolated. Since the repeating block is 1 digit, multiply this new equation by 10 (effectively multiplying original $x$ by 100). $100x = 26.\overline{6}$
5. Step C: Subtract the equation from Step A from the equation in Step B. $ \begin{align*} 100x &= 26.\overline{6} \ -\quad 10x &= 2.\overline{6} \ \hline 90x &= 24 \end{align*} $
6. Solve: $x = \frac{24}{90} = \frac{4}{15}$.

Alternative Shortcut for Mixed Decimals: For a decimal like $0.a_1a_2...a_k\overline{b_1b_2...b_n}$ (where $a$ are non-repeating and $b$ are repeating):

• Numerator: (Entire number without decimal point) $-$ (Non-repeating part without decimal point).
• Denominator: As many $9$s as repeating digits followed by as many $0$s as non-repeating digits.

Applying to $0.2\overline{6}$:

• Numerator: $26 - 2 = 24$.
• Denominator: One $9$ (for the one repeating digit $6$) and one $0$ (for the one non-repeating digit $2$) $\rightarrow 90$.
• Fraction: $\frac{24}{90} = \frac{4}{15}$.

Example 4: Repeating Decimal Greater Than 1 ($2.\overline{3}$)

The process is identical; the integer part simply carries through the subtraction Easy to understand, harder to ignore..

1. $x = 2.\overline{3}$
2. $10x = 23.\overline{3