Chandler tells Aubrey that the decimal value of -1/7 is not a repeating decimal. Should Aubrey believe him?

Guest Nov 26, 2014

#3**+10 **

A fraction will have a terminating decimal only if we can write the denominator as (2^{n} x 5^{n}) where n ≥ 0

To see why this is true......note that all terminating decimals can be written as A / 10^{n} = A / (2 * 5)^{n} = A / (2^{n} x 5^{n}) where A is the integer formed by moving the decinal point in the non-repeating decimal n places to the right.

For instance

1/4 = 1/(2^{2} x 5^{0}) = .25 = 25 / (2^{2} x 5^{2}) = 25 / 100

And

1/5 = 1/(2^{0} x 5^{1}) = .20 = 20 / (2^{2} x 5^{2}) = 20 / 100

But, note that fractions such as 1/6, 1/13, 1/23 will repeat because the denominators cannot be factored solely in terms of 2 and 5.

CPhill
Nov 26, 2014

#1**+10 **

- 1/7 = - 0.142857(142857)...

In fact, all fractions have repeating decimals (or a finite number of decimals).

Guest Nov 26, 2014

#2**+10 **

Hi anon

Not all fractions have repeating decimals. What about 1/2 that is just 0.5 nothing is repeating :)

$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}} = -{\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}$$ but this is a repeating decimal just like you said

Melody
Nov 26, 2014

#3**+10 **

Best Answer

A fraction will have a terminating decimal only if we can write the denominator as (2^{n} x 5^{n}) where n ≥ 0

To see why this is true......note that all terminating decimals can be written as A / 10^{n} = A / (2 * 5)^{n} = A / (2^{n} x 5^{n}) where A is the integer formed by moving the decinal point in the non-repeating decimal n places to the right.

For instance

1/4 = 1/(2^{2} x 5^{0}) = .25 = 25 / (2^{2} x 5^{2}) = 25 / 100

And

1/5 = 1/(2^{0} x 5^{1}) = .20 = 20 / (2^{2} x 5^{2}) = 20 / 100

But, note that fractions such as 1/6, 1/13, 1/23 will repeat because the denominators cannot be factored solely in terms of 2 and 5.

CPhill
Nov 26, 2014