Converting Recurring Decimals and Fractions

What makes a decimal recurring?

A recurring decimal contains a digit or block of digits that repeats for ever. For example,

$0.6666\ldots$

has the single recurring digit 6, whereas

$0.828282\ldots$

has the recurring block 82. The ellipsis, $\ldots$, matters: it says that the pattern continues indefinitely rather than stopping after the digits shown.

In OCR's dot notation, place a dot above a single recurring digit:

$0.\dot{6}=0.6666\ldots$

For a longer block, place dots above its first and last digits. Thus

$0.\dot{8}\dot{2}=0.828282\ldots$

and

$0.\dot{1}4285\dot{7}=0.142857142857\ldots$

The dots identify the entire repeating block, including every digit between them.

From a fraction to a recurring decimal

A fraction bar means division. To convert $\frac{a}{b}$ to a decimal, calculate $a\div b$. If the division does not terminate but settles into a repeating pattern, record that pattern using recurring notation.

For example,

$\frac{13}{3}=13\div3=4.3333\ldots=4.\dot{3}.$

This equality is exact: $4.\dot3$ is not a rounded version of $\frac{13}{3}$. During long division, a remainder that has appeared before must produce the same subsequent digits again, which is why a cycle develops.

A longer cycle can occur. Dividing 1 by 7 gives

$\frac17=0.142857142857\ldots=0.\dot1 4285\dot7.$

When writing the answer, identify the shortest complete block that repeats. Here the block is $142857$, not $142857142857$.

Why algebra turns an infinite decimal into a fraction

Although a recurring decimal has infinitely many digits, its repeating structure allows the infinite tail to be eliminated. The idea is to multiply by a suitable power of 10 so that two copies have identical recurring tails, then subtract them.

Consider

$x=0.828282\ldots$

The recurring block has two digits, so multiply by $100$; this moves the decimal point by two places:

$100x=82.828282\ldots$

Now the decimal tails are identical. Subtracting the first equation from the second cancels them:

$100x-x=82.828282\ldots-0.828282\ldots$

$99x=82$

so

$x=\frac{82}{99}.$

Therefore,

$0.\dot8\dot2=\frac{82}{99}.$

This is an exact fraction: no rounding has been used.

Choosing the correct power of 10

The multiplier is determined by the number of digits in the recurring block:

• one recurring digit: multiply by $10$;
• two recurring digits: multiply by $100$;
• three recurring digits: multiply by $1000$.

For example, let

$x=0.325325325\ldots$

The recurring block $325$ has three digits, so

$1000x=325.325325325\ldots$

Subtracting gives

$999x=325,$

and hence

$x=\frac{325}{999}.$

Always simplify the resulting fraction if the numerator and denominator have a common factor.

When the recurring block does not begin immediately

A decimal may have a non-recurring part before the repeated digits. For example,

$x=0.16666\ldots$

Only the 6 recurs. First move beyond the non-recurring digit:

$10x=1.6666\ldots$

Then move one more place, because the recurring block has one digit:

$100x=16.6666\ldots$

The tails now match. Subtracting the first of these equations from the second gives

$100x-10x=16.6666\ldots-1.6666\ldots$

$90x=15,$

so

$x=\frac{15}{90}=\frac16.$

The essential test is not merely whether a decimal point has moved: the recurring digits after the decimal point must line up before subtraction.

Checking the conversion

A conversion works in both directions. After finding a fraction, divide its numerator by its denominator. For instance,

$82\div99=0.828282\ldots,$

which confirms $\frac{82}{99}=0.\dot8\dot2$. This reverse check can reveal an incorrect multiplier, subtraction error or unsimplified interpretation of the recurring block.

Recognise and write

A recurring decimal repeats indefinitely:

• $0.\dot6=0.6666\ldots$
• $0.\dot8\dot2=0.828282\ldots$

For a multi-digit block, dots mark its first and last digits.

Fraction to decimal

Calculate

$\text{numerator}\div\text{denominator}.$

Identify the shortest repeating block and write it using recurring notation.

Decimal to exact fraction

1. Let $x$ equal the recurring decimal.
2. Multiply by a power of 10 so identical recurring tails align.
3. Subtract the equations to cancel those tails.
4. Solve for $x$ and simplify the fraction.

For an $n$-digit block that starts immediately, multiply by $10^n$. If non-recurring digits come first, use two suitable shifted equations and subtract them.

Final checks

The recurring tails must match before subtraction. The answer must be exact and simplified; verify it by dividing the numerator by the denominator.