Question

How do you find the repeating decimal 0.231 with 31 repeated as a fraction?

Answer 1

The answer is `1154/4995`

Explanation:

`0.2(31)=0.2+0.0(31)=0.2+0.031+0.00031+0.0000031+...`

Starting from the second number we have an infinite sum of a convergent geometric sequence with:

`a_1=0.031` and `q=0.001`

So we can write the whole decimal as:

`0.2(31)=0.2+0.031/(1-0.001)=0.2+0.031/0.999=2/10+31/999=`

`=(2*999+310)/9990=(1198+310)/9990=2308/9990=1154/4995`

Answer 2

`x=229/990`

Explanation:

let `x=0.2313131.............`

because TWO digits repeat `xx100`

we have therefore:

`100x=23.13131313131313.......`.

&#``x=0.231313131......#

Subtracting the two the repeating 31s will all cancel

`100x=23.1cancel(3131313131313.......)`.

`x=0.2cancel(31313131......)`

we are left with

`110x-x=23.1-0.2`

`99x=22.9`

`x=(22.9)/99=229/990`