Repeating decimals
A fraction to a repeating decimal and back: 0.(54) = 6/11, 0.91(6) = 11/12. Long division and the rule “the repeating block over nines”.
How to solve it
In long division the remainders can only be $0, 1, \dots, b - 1$, where $b$ is the denominator. So sooner or later a remainder repeats and the digits go round in a circle: that is the repeating block. The way back rests on one observation: $0.(1) = \frac{1}{9}$, $0.(01) = \frac{1}{99}$ and so on.
Step by step
- Fraction to decimal: do long division and write down the remainders. As soon as a remainder repeats, the digits between its two appearances are the repeating block.
- A pure repeating decimal to a fraction: divide the block by as many nines as it has digits: $0.(54) = \frac{54}{99}$.
- With a non-repeating part: multiply by $10^k$ to move that part to the left of the point, convert, then divide by $10^k$.
- Reduce the fraction.
Common mistakes
- Dividing the block by $10$, $100$, … instead of $9$, $99$, …: that gives a terminating decimal, not a repeating one.
- Putting the brackets in the wrong place: $\frac{13}{22} = 0.5909\ldots$ has the non-repeating part $5$ and the block $90$, so the answer is $0.5(90)$.
- Stopping the division too early: $\frac{1}{7}$ has a six-digit block, $0.(142857)$.
- Not reducing: $\frac{54}{99} = \frac{6}{11}$.