0.[6]

1x = 0.[6]

10x = 6.[6]

9x = 10x - 1x

9x = 6.[6] - 0.[6] = 6

Therefore the answer is 6/9, or 2/3

When it got to numbers like 0.12[34] (ie 0.12343434343434.....) the lesson was really complicated, but I came up with a more simple approach, so here it is if ever you need it.

My first observation was that we need a large number and a small number. So let's start with the 10x we used above

1x = 0.12[34]

10x = 1.2[34]

therefore 9x = 10x - 1x which is

1.2[34]

-0.12[34]

This is going to give us 1.1(something)

My second observation is that fractions cannot have decimal values in them. In order to get rid of the fraction we need the big number to have the exact same fraction as the small number. e.g.

B.[34]

-S.[34]

=X.[00]

So our small number needs to end with [34]. Given the number 0.12[34] how many places do we need to shift the decimal to the right? The answer is 2, so we need the number 1 and two zeros, which is 100

Small number = 100x

To do a subtraction and get a positive number our big number needs to be bigger than 100x

1x = 0.12[34]

100x = 12.[34]

How many digits are recurring? The answer is 2. This means that the larger factor has two more zeros than the lower factor in order to ensure both numbers end with [34].

1x = 0.12[34]

100x = 12.[34]

10000x = 1234.[34]

1234.[34]

- 12.[34]

=========

1222.[00]

Now we have a whole number!

1222 / (10,000 - 100) is 1222/9900

So the quick way of doing it is this

x = 0.12[34]

0.AA[BB]

Small = x * 10 to the power of the number of non recursive decimal digits [A]

Big = x * 10 to the power of the tge total number of decimal digits [B]

Small = x * 102

Big = x * 104

Big - small = 1222

Big factor (10000) - small factor (100) = 9900

Therefore the answer is 1222/9900

Or like this

1x = 0.12[34]

How many non recurring digits? Two, so our small factor is a 1 with 2 zeros = 100

How many recurring digits? Two, so our large factor has 2 more zeros than the small one = 1 00 00 = 10,000