Recurring Decimals to Fractions

You already know the awkward truth that 0.333\ldots is secretly just \tfrac{1}{3} — a tidy fraction hiding behind an endless string of threes. But that raises a cheeky question: given any recurring decimal — say 0.727272\ldots, or 0.8\dot{3}, or something nastier — can you work out the exact fraction it came from?

You can, and the method is a small piece of algebra that feels like a magic trick the first time you see it. The endless tail looks like the hard part, but it's actually the key: we're going to make two copies of the decimal, line their tails up, and subtract so the infinite part vanishes in a puff of smoke — leaving an ordinary equation we can solve.

The trick: shift, then subtract

Take x = 0.\dot{3} = 0.333\ldots. Multiply by 10 to slide the decimal point one place along. Here's the crucial thing: because the tail repeats forever, the digits after the point are exactly the same in both lines — 333\ldots either way:

10x = 3.333\ldots \qquad x = 0.333\ldots

Now subtract the second line from the first. Every single digit after the decimal point cancels against its twin, and the endless tail simply disappears:

10x - x = 3.333\ldots - 0.333\ldots = 3 9x = 3 \quad\Longrightarrow\quad x = \frac{3}{9} = \frac{1}{3}

There it is: 0.\dot{3} = \tfrac{1}{3}, proved, no hand-waving. And notice a pattern already: a single recurring digit always lands over 9.

When a longer block repeats

Example — a two-digit block, 0.\dot{7}\dot{2} = 0.727272\ldots. One shift isn't enough now: multiplying by 10 would leave the tails out of step. We need to jump the whole repeating block past the point, so we multiply by 100:

100x = 72.727272\ldots \qquad x = 0.727272\ldots 100x - x = 72 \quad\Longrightarrow\quad 99x = 72 x = \frac{72}{99} = \frac{8}{11}

(Don't forget the last step — simplify. Here 72 and 99 share a factor of 9, giving \tfrac{8}{11}.)

The pattern is now unmistakable: one recurring digit gives a denominator of 9, a two-digit block gives 99, a three-digit block gives 999 — the number of 9s matches the length of the repeating block. So 0.\dot{2}5\dot{1} = \tfrac{251}{999} straight off the bat.

The tricky case: a non-repeating start

Example — 0.1\dot{6} = 0.1666\ldots. The 1 after the point does not repeat; only the 6 does. That leading digit gets in the way, so we shift twice — once to clear the non-repeating part, and once more to line up the tail.

Multiply by 10 (to step past the lone 1) and by 100 (one block further), so both lines end in the same 666\ldots tail:

100x = 16.666\ldots \qquad 10x = 1.666\ldots 100x - 10x = 16 - 1 = 15 \quad\Longrightarrow\quad 90x = 15 x = \frac{15}{90} = \frac{1}{6}

Sure enough, \tfrac{1}{6} = 0.1\dot{6}. The idea never changed — pick two multiples of 10 whose tails match, then subtract to kill them.

To turn a recurring decimal into a fraction: A single recurring digit gives a denominator of 9; a two-digit block gives 99; an n-digit block gives n nines.

The whole trick rests on the two tails being identical so they cancel. Two ways to wreck it:

This is the internet's favourite maths argument — and our trick settles it in three lines. Let x = 0.\dot{9} = 0.9999\ldots and run the method:

10x = 9.999\ldots \qquad x = 0.999\ldots 10x - x = 9 \quad\Longrightarrow\quad 9x = 9 \quad\Longrightarrow\quad x = 1

No sleight of hand, no rounding, no "getting closer and closer". It comes out exactly 1. So 0.9999\ldots isn't almost 1 — it is 1, just written in a needlessly dramatic way. (Makes sense, too: \tfrac{1}{3} = 0.\dot{3}, so 3 \times \tfrac{1}{3} = 0.\dot{9} and also = 1.) It's the perfect punchline to the big idea of this page: every recurring decimal is a fraction in disguise — even the ones that look like they can't be.