Recurring Decimals

Some fractions give terminating decimals — $\tfrac{1}{4} = 0.25$. Others go on forever in a pattern — $\tfrac{1}{3} = 0.333\ldots$

A recurring decimal has at least one digit (or block) that repeats indefinitely. Common examples:

Fraction	Decimal	Notation
$\tfrac{1}{3}$	$0.333\ldots$	$0.\overline{3}$ or $0.\dot{3}$
$\tfrac{2}{3}$	$0.666\ldots$	$0.\overline{6}$
$\tfrac{1}{7}$	$0.142857142857\ldots$	$0.\overline{142857}$
$\tfrac{5}{6}$	$0.8333\ldots$	$0.8\overline{3}$
$\tfrac{1}{11}$	$0.090909\ldots$	$0.\overline{09}$

The vinculum (bar) goes over the digits that repeat. So $0.8\overline{3}$ means $0.8333\ldots$ — the 8 appears once, then the 3 repeats forever.

In some notations, dots are placed above the FIRST and LAST digit of the repeating block: $0.\dot{1}\dot{4}$ means $0.141414\ldots$.

Fact: A fraction $\tfrac{p}{q}$ (in lowest terms) gives a TERMINATING decimal if and only if the denominator $q$ has no prime factors other than 2 or 5. Otherwise it gives a RECURRING decimal.

This is a classic Extended-level technique. Use the variable $x$ to stand for the decimal, then multiply by a power of 10 to SHIFT the repeating part so it lines up.

Worked example 1. Convert $0.\overline{3}$ to a fraction.

Let $x = 0.\overline{3} = 0.3333\ldots$

Multiply both sides by 10 (the period of the repeat is 1): $10x = 3.3333\ldots$

Subtract the first equation from the second — the repeating decimals CANCEL: $10x - x = 3.3333\ldots - 0.3333\ldots$ $9x = 3$ $x = \frac{3}{9} = \frac{1}{3}.$

Worked example 2. Convert $0.\overline{27}$ to a fraction.

Let $x = 0.27272727\ldots$. The repeating block has 2 digits, so multiply by $10^2 = 100$: $100x = 27.272727\ldots$ $100x - x = 27.272727\ldots - 0.272727\ldots$ $99x = 27$ $x = \frac{27}{99} = \frac{3}{11}.$

Worked example 3 (mixed terminating + recurring). Convert $0.1\overline{6}$ to a fraction.

$x = 0.16666\ldots$. The 1 is not part of the recurring block. Multiply by 10 to push past the 1: $10x = 1.6666\ldots$

Multiply by another factor of 10 to shift the repeating block: $100x = 16.6666\ldots$

Subtract: $100x - 10x = 16.6666\ldots - 1.6666\ldots$ $90x = 15$ $x = \frac{15}{90} = \frac{1}{6}.$

General rule: if the repeating block has length $n$, multiply by $10^n$. If there are also some non-repeating digits after the decimal point, use TWO multiplications: one to push past the non-repeating digits, one to shift the repeating block.

Apply the conversion technique to $0.\overline{9}$:

Let $x = 0.\overline{9} = 0.9999\ldots$ $10x = 9.9999\ldots$ $10x - x = 9.9999\ldots - 0.9999\ldots$ $9x = 9$ $x = 1.$

So $0.\overline{9} = 1$, EXACTLY. Not 'very close to'. Equal.

Why is this counter-intuitive? Because we imagine $0.999\ldots$ as a number 'just below' 1. But there is no number between $0.999\ldots$ and 1 — they have ZERO gap, so they must be the same number.

Other ways to see it:

• $\tfrac{1}{3} = 0.\overline{3}$. Multiply both sides by 3: $1 = 0.\overline{9}$.
• Any terminating decimal has an alternative recurring form: $0.5 = 0.4\overline{9}$, $7 = 6.\overline{9}$.

This is a great example for criterion D — discussing what NUMBERS really are, and how different notations can describe the same thing.