Converting recurring decimals into fractions (Higher)

Type two divided by three into a school calculator. On most of them, the screen fills up: zero point six six six six six six six six six seven. Look at the last digit. A seven? That seven is the calculator giving up. The true answer is sixes forever - but a screen is finite, so it rounds the final digit and shows a number that is very slightly wrong. That is the trouble with recurring decimals. Stop writing digits anywhere, and the number changes. Round, and the number changes. The only way to hold one exactly is as a fraction - two thirds says it perfectly, in one neat package. In this video: the algebraic method that turns any recurring decimal back into its exact fraction. A quick word first - this is a Higher tier topic. On Foundation, it will not be in your exam, so you can skip it guilt-free. On Higher, stay: it is one tidy trick, built entirely from algebra you already know.

First the notation, then the method.

When a digit repeats forever, we mark it with a dot. Zero point five with a dot over the five means zero point five five five, the fives never stopping. When a whole block repeats, dots go over the first and last digits of the block: dots over the four and the five means point four five, four five, four five, and so on.

Now the method, on that first number. We want zero point five five five recurring as a fraction. Step one: give the number a name. Let x equal zero point five five five, and so on forever. Step two: multiply both sides by ten. Every digit shifts one place to the left, so ten x equals five point five five five, and so on. And look at what did not change: the tail. After the decimal point, x and ten x carry exactly the same endless string of fives. Step three: subtract. Ten x minus x is nine x. And on the right, the two identical tails wipe each other out - fives cancelling fives, forever. All that survives is the five in front. So nine x equals five. That is the whole idea. You cannot chop off an infinite tail - but you can line up two copies of it and subtract, and the tails destroy each other. Infinity, handled by subtraction. Solve it: x equals five ninths. And that is exact - not close, not rounded. Five ninths is zero point five five five recurring, precisely. Check it if you like: five divided by nine on a calculator, and the fives march across the screen.

The method in three words: shift, subtract, solve. Shift the digits with a power of ten. Subtract, so the tails cancel. Solve for x. Every recurring decimal at GCSE falls to those three moves.

Your turn, then. What is zero point seven seven seven, recurring, as a fraction? A: seven tenths. B: seven ninths. C: seventy-seven over a hundred.

Have a think. I'll wait.

The answer is B, seven ninths. Multiply by ten, subtract, and nine x equals seven. If you chose A, seven tenths - that is zero point seven exactly, tail chopped off. And the tail is the whole point.

Now the harder version - when the repeating block starts late.

Try this number: zero point two, four five, four five, four five, and so on. The four five block repeats - but that lone two at the front never does. One shift by ten will not line the tails up any more. We need to be a little cleverer. Start the same: let x equal zero point two four five, four five, and so on. Multiply by ten: ten x equals two point four five, four five, recurring. See what that bought us - the awkward two has been pushed out in front of the point, leaving a pure repeating tail: four five, four five, forever. Now shift again, by one whole block. The block is two digits long, so multiply ten x by a hundred: one thousand x equals two hundred and forty-five point four five, four five, recurring. Two lines, ten x and one thousand x - and after the point, their tails are identical. Subtract. One thousand x minus ten x is nine hundred and ninety x. On the right, the matching tails cancel completely, leaving two hundred and forty-five minus two: that is two hundred and forty-three. So nine hundred and ninety x equals two hundred and forty-three. So x is two hundred and forty-three over nine hundred and ninety. But do not stop there - that fraction cancels. Nine goes into both, and the top and the bottom travel together: divide each by nine, and you get twenty-seven over one hundred and ten. That is the exact answer, in simplest form. One thing to notice: how we chose the powers of ten. Shift once to push the non-repeating part clear of the point, then again by one full block so two lines share a tail. Get the tails to line up, and subtraction does the rest - whatever the decimal.

Time for the exam view - how this is marked, and the trap the examiners keep reporting.

This question type appears year after year. Here is the Edexcel examiners' report from 2022, on exactly this topic:

This is a familiar type of question and most students were able to gain the first mark for showing an understanding of the recurring decimal notation. It was pleasing that many students were able to show a complete method leading to a correct fraction.

Notice the two halves of that. Knowing the notation earns credit on its own. But the real reward is the complete method - set up, subtract, solve, and cancel to simplest form. Write it to the end, every single time. And when a recurring decimal appears inside a bigger question, do not round it. Here is what OCR's examiners reported in 2024:

Candidates that used a decimal-based approach were less successful, since they encountered more difficulties both dealing with the recurring decimal numbers and retaining sufficient accuracy within their method. Candidates using terminating decimals such as 1.2 and 1.6 instead of recurring values limited the marks that could be achieved.

That is the calculator lie from the start of this video, wearing exam clothes. One point two is not one point two recurring - chop the tail and your answer drifts. Convert to the exact fraction first, then calculate with the fraction. Fractions never drift. One more line from that same report:

Many candidates demonstrate strong knowledge of how to convert a recurring decimal to a fraction but would benefit from using recurring decimals in other contexts.

In other words: plenty of people can do this as a stand-alone question, then freeze when the same decimal turns up inside an area, a probability, or an equation. Do not freeze. Wherever a recurring decimal appears, the move is the same - convert it to its exact fraction, then carry on.

So, the complete method, start to finish, on a single screen. Name it: let x be the recurring decimal. Shift it: multiply by powers of ten until two lines share the same infinite tail - one shift for a simple repeat, two different shifts when the block starts late. Subtract, and the tails wipe out. Solve for x, then cancel the answer down - top and bottom travelling together - to its simplest form. And the check that never lies: divide your fraction back out, top by bottom, and the original recurring decimal should come marching back. If it does, your answer is exact - no rounding, no drift, ever.

That completes our chapter on fractions - the skills and the conversions, from cancelling all the way to recurring decimals. Everything here feeds the rest of GCSE maths, so it is time well spent.

For more, visit scholafly.com, or watch the next video.