A recurring decimal repeats one or more digits forever. A dot above a digit marks the start (and end, if there are two dots) of the repeating block:

Every recurring decimal is secretly a fraction — it's rational. Extended candidates must be able to prove the exact fraction algebraically rather than just quoting it.

1. Let $x$ equal the decimal.
2. Multiply $x$ by whatever power of $10$ shifts one full repeating block to the left of the decimal point.
3. Subtract the original equation from the multiplied one — the infinite tails cancel exactly.
4. Solve for $x$ and simplify.

Worked Example

Show algebraically that $0.\dot{5}\dot{4} = \dfrac{6}{11}$.

1. Let $x = 0.545454\ldots$ — (1)
2. Two digits repeat, so multiply by $100$: $100x = 54.545454\ldots$ — (2)
3. Subtract (1) from (2): $99x = 54$.
4. Solve: $x = \dfrac{54}{99} = \dfrac{6}{11}$.

When digits appear before the repeating block starts (like $0.1\dot{6}$), you need two different multipliers — one that places the decimal point immediately before the repeating block, and one that places it immediately after one full block — so that subtracting them leaves identical, cancelling tails.

Worked Example

Show algebraically that $0.1\dot{6} = \dfrac{1}{6}$.

1. Let $x = 0.16666\ldots$
2. Multiply by $10$ to sit just before the repeating digit: $10x = 1.6666\ldots$ — (1)
3. Multiply by $100$ to move one repeating block past the point: $100x = 16.6666\ldots$ — (2)
4. Subtract (1) from (2): $90x = 15$.
5. Solve and simplify: $x = \dfrac{15}{90} = \dfrac{1}{6}$.

Divide the numerator by the denominator on your calculator (or by long division) and confirm the decimal that comes back matches the original recurring pattern exactly, including where the dot(s) sit.

For $0.1\dot{6} = \dfrac{1}{6}$: dividing $1 \div 6$ gives $0.16666\ldots$, which is exactly a $1$ followed by $6$ repeating forever — matching the original. If your fraction instead produced $0.1666$ terminating, or a different repeating digit, you'd know an arithmetic slip had happened somewhere in the subtraction or simplifying step.

Yes — the method generalises to any block length. The only thing that changes is the power of $10$ you multiply by: for a block of $n$ repeating digits, multiply by $10^n$ to shift exactly one full block across the decimal point before subtracting.

Worked Example

Convert $0.\dot{2}7\dot{6}$ (the block $276$ repeats) to a fraction.

1. Let $x = 0.276276276\ldots$
2. The repeating block has $3$ digits, so multiply by $10^3=1000$: $1000x = 276.276276\ldots$
3. Subtract: $999x = 276$.
4. Simplify: $x = \dfrac{276}{999} = \dfrac{92}{333}$ (dividing top and bottom by their HCF, $3$).