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Lets generalize this.Consider a decimal $0.\overline{a_1a_2a_3\dots a_n}$ .Now lets prove that this is rational.So let

$A=0.\overline{a_1a_2a_3\dots a_n} \dots (1)$

If we multiply $A$ by $10^{n}$ , the decimal point gets shifted to left by $n$ places which gives us :

$10^n \times A = \overline{a_1a_2a_3\dots a_n} . \overline{a_1a_2a_3\dots a_n} \dots (2)$

Now lets subtract $1$ from $2$ because that will make the right side of decimal as $0$ . Thus we get:

$10^n A - A = \overline{a_1a_2a_3\dots a_n} \\ \Rightarrow A(10^n - 1) = \overline{a_1a_2a_3\dots a_n} \\ \Rightarrow A = \dfrac{ \overline{a_1a_2a_3\dots a_n}}{10^n - 1}$

Since $A$ can be clearly seen to be equal to $\dfrac{p}{q}$ for $p,q \in \mathbb{Z}$ and $q\neq 0$ , we conclude that $A$ is rational.

In this problem $a_1=2 \ , \ a_2 = 8 \ , \ a_3 = 5 \ , \ a_4 = 7 \ , \ a_5 = 1 \ , \ a_6 = 4$ and by our generalization we have $0.\overline{285714}$ as rational number :)

If it is rational number, let's check:

$A= 0. \overline{285714}$

$1000000A= 285714.\overline{285714}$

$1000000A-A= 285714$

$999999A=285714$

$\frac{285714}{999999}=\boxed{\frac{2}{7}}$

This is simply $\frac{285714}{999999}$ which implies it is rational.

If a number has finite recurring decimals, it is rational.

$0.\overline {285714}=0.285714285714.......$

A number IS a rational number if it repeats a sequence of finite digits over and over.

$\huge \therefore 0.\overline {285714} \in \mathbb{Q}$

1/7=0.142857142857... 2/7=0.285714285714... 3/7=0.428571428571... 也就是说以7为分母的分数的循环结构由1、4、2、8、5、7构成 It's interesting.

All repeating decimals are all rational numbers.

$\displaystyle 0.285714285714\cdots = 0.\dot 2 8571 \dot 4 = \frac { 285714 } { 999999 } = \frac { 2 \times 142857 } { 7 \times 142857 } = \frac { 2 \times \cancel { 142857 } ^ { 1 } } { 7 \times \cancel { 142857 } _ { 1 } } = \frac { 2 } { 7 }$ .

If a decimal repeats, it is rational.