Recurring Decimals – Secrets of the Mathematical Ninja

— Thanks to Rosalind for showing me this trick.

It’s one of the questions in the GCSE that looks like it ought to be easy: What is $0.1\dot{4}3\dot{6}$ as a fraction?

But it’s a lot less easy than it seems at first. I’ve taught the longwinded way for years. It turns out there’s a ninja shortcut you can do.

What you do is, starting from the right hand end, put a nine under each recurring digit (under or beneath the dots) and a 0 under anything else. In this case, you’d get 9990 — this will be the bottom of your fraction.

Now, temporarily ignore all of the dots and think about what number you have: 1436 here. If you have anything that’s not recurring — in this case, that’s the 1 at the start — away from it, to get 1435. That’s your top.

So, $0.1\dot{4}3\dot{6} = \frac{1435}{9990}$.

Being ninjas, we choose to cancel that down — there’s clearly a factor of 5, making it $\frac{287}{1998}$, which doesn’t cancel.

This is nastier than anything you’ll see in a GCSE (at worst, you’ll have two numbers under dots and possibly one in front).

You can also use this to show that $0.\dot{9} = 1$.

I haven't figured out why this trick works -- I'd love to hear your theories!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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One comment on “Recurring Decimals – Secrets of the Mathematical Ninja

  • Mark Ritchings

    Hi Colin, I like that. My example below shows why it works?

    $0.23\dot{1}04\dot{1}=\frac{23}{100}+\frac{1041}{999900}=\frac{23(10000-1)+1041}{999900}=\frac{231041-23}{999900}=\frac{231018}{999900}$

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