Recurring decimal to fraction calculator

Example: $0.\dot{2}\dot{7} = 0.272727\ldots$

1. Let $x = 0.272727\ldots$
2. Two digits repeat, so multiply by 100: $100x = 27.272727\ldots$
3. Subtract to remove the recurring part: $99x = 27$
4. $x = \dfrac{27}{99} = \dfrac{3}{11}$

Example: $0.1\dot{6} = 0.1666\ldots$

1. $x = 0.1666\ldots$
2. $100x = 16.666\ldots$ and $10x = 1.666\ldots$
3. Subtract: $90x = 15$
4. $x = \dfrac{15}{90} = \dfrac{1}{6}$

Multiply by two different powers of 10 so that both numbers have exactly the same recurring tail, then subtract. The same method gives $2.1\dot{3} = \tfrac{32}{15}$.

A dot over a digit means it repeats: $0.\dot{3} = 0.333\ldots$ Dots over the first and last digits of a block mean the whole block repeats: $0.\dot{1}2\dot{3} = 0.123123\ldots$ Some books use a bar over the repeating digits instead.

A single repeating digit $d$ is $\tfrac{d}{9}$; a repeating pair is $\tfrac{\text{pair}}{99}$.

• Multiplying by 10 when two digits repeat (it must be 100).
• Not lining up the recurring parts before subtracting.
• Forgetting to simplify the final fraction.

Practise with our recurring decimals questions, and use the fraction calculator to check simplifying.

• "Prove that $0.\dot{2}\dot{7} = \tfrac{3}{11}$" needs the full algebra, not just a calculator check.
• Show the two equations lined up, then the subtraction.
• This is a Higher-tier GCSE topic.

Every recurring decimal is a rational number: it can be written as a fraction of two whole numbers. The algebra method always works because subtracting removes the endless tail, leaving a whole number on the right.

The reverse is true too: fractions whose denominator (in simplest form) has prime factors other than 2 and 5 give recurring decimals. So $\tfrac{1}{8} = 0.125$ terminates (8 = 2³), but $\tfrac{1}{6} = 0.1\dot{6}$ recurs because 6 has a factor of 3. Numbers like $\pi$ and $\sqrt{2}$ never recur, which is what makes them irrational.