# Is it possible to turn a recurring decimal into a terminating decimal?(examples too please :D)

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Nicarolina-

that is usually true, with the following exception:

In base 10, the number .999999... has two representations

similarly, in base 2, the number .11111...

in base 3, the number .222222...

etc, have multiple representations.

To see this, what is .999... ?

It is 9/10 + 9/100+ 9/1000+...

which is (9/10)(1 + 1/10 + ...)= (9/10)*(1/ (1-1/10) ) = (9/10)*(10/9) = 1

Perhaps more intuitively:

1/3 = .333333...

but .3333... *3 = .999999...

.999... = 1

On a side note: every repeating decimal can be turned into a fraction. Which means you can put the number in a different base and it will terminate. For example, .142857142857... = 1/7

If you put that in base 7, that number is just .1

So another way to write it as a terminating decimal is to switch the base to the denominator of the equivalent fraction

If you stay in base 10 (our standard ones-tens-hundreds system), then for the most part you can't turn a repeating decimal into a terminating decimal. The only exception is a number that ends in 9999999...

example:

.9999999 ... = 1

2.345999999999... = 2.346

99.9999999... = 100

You *can* turn a repeating decimal into a fraction:

`.15151515...`

`=15 (.01010101...)`

`=15 ( (1)/(100) + (1)/(10000) + (1)/(1000000)+ ... )`

`=15 ( (1)/(100) + (1)/(100)^2 + (1)/(100)^3 + ... )`

`=(15)/(100) (1 + (1)/(100) + (1)/(100)^2 + ...)`

`=(15)/(100) (1)/(1-(1)/(100)) `

`=(15) * (1)/(100-1) `

`=(15)/(99)`

`=(5)/(33)`

It is impossible to do so, as a recurring decimal has a never ending numeral or chain of numerals, and if you take away a terminating numeral or a recurring numeral from a recurring numeral, it will always have the recurring numbers as the last digits of the decimal.

e.g.

0.(8recurring) - 0.4 = 0.4(8 recurring)

0.(8recurring) - 0.(4recurring) = 0.(4recurring)