Question

How do you find the fraction of 0.16 bar, or 0.1616...?

I am finding the rational number of this, and I need help!

Answer 1

`16/99`

Explanation:

Note that:

`(100-1) 0.bar(16) = 16.bar(16)-0.bar(16) = 16`

So:

`0.bar(16) = 16/(100-1) = 16/99`

`16` and `99` have no common factors larger than `1` so this is in simplest form.

Why `(100-1)` ?

The multiplier `100` has the effect of shifting the decimal representation `2` places to the left, which is the length of the repeating pattern. Having done that, subtracting the original fraction with the `-1` multiplier cancels out the repeating tail, resulting in an integer.

Another method...

Here's another method potentially useful if you have a calculator to hand...

• Type the approximation of the given fraction into your calculator...

  `0.1616161616`

• Write down the whole number part `color(red)(0)`, subtract it and take the reciprocal to get, approximately:

  `6.187500001`

• Round this to `6.1875` since the `1` on the end is obviously a rounding error.

• Write down the whole number part `color(red)(6)`, subtract it and take the reciprocal to get, approximately:

  `5.333333333`

• Write down the whole number part `color(red)(5)`, subtract it and take the reciprocal to get, approximately:

  `3.000000003`, which rounds to `color(red)(3)`

Hence we can deduce that:

`0.bar(16) = color(red)(0) + 1/(color(red)(6) + 1/(color(red)(5) + 1/color(red)(3))) = 1/(6+3/16) = 16/99`

Why go to all this trouble? For one thing note that we can get good approximate fractions by terminating the continued fraction early.

For example:

`0.bar(16) ~~ 0+1/6 = 1/6 = 0.1bar(6)`

or:

`0.bar(16) ~~ 0+1/(6+1/5) = 5/31 ~~ 0.16129`

If you apply this method to an irrational number like `pi`, you can find good approximations like `22/7` and `355/113`.

Answer 2

`0.161616....->0.16bar(16) = 16/99`

Explanation:

`color(brown)("More in depth explanation of Georges first method.")`

Set `x=0.1616161616......`

One standardised method of writing a cyclical repeat is to put a bar over the top of the repeating part. Which repeating part the bar goes over is down to you. So we can write.:

`x=0.16bar(16)`

Multiply both sides by 100

`100x=16.16bar(16)`

Subtracting

`100x-x=16.16bar(16)`
`color(white)("ddddddddd,")ul(color(white)(1)0.16bar(16)larr" Subtract")`
`color(white)("ddddddddd.")16`

`99x=16`

Divide both sides by 99

`x xx 99/99=16/99`

But `99/99=1 and 1xx x =x` giving:

`x=16/99`