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I have a question.

As you know, $\mathbb{Q}$ is the set of finite decimals and circulating decimals.

I want to prove this.

Here rational numbers $\mathbb{Q}$ is the set of all fractions.

The set of finite decimals and circulating decimals is a subset of $\mathbb{Q}$.

This part is clear.But I do not know how we prove the remaining part.

Thank you in advance.

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    The heart of the matter is that the decimal expansion of every rational number is [ultimately periodic](http://www.encyclopediaofmath.org/index.php/Ultimately_periodic_sequence) (a more canonical term than your *circulating*...).2012-09-15

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Let $\frac{m}n$ be a rational number, i.e., $m,n\in\Bbb Z$ and $n\ne 0$. There’s no harm in assuming that $0: once we handle the rational numbers between $0$ and $1$, the rest are easy.

Imagine doing a long division of $m$ by $n$ to expand $\frac{m}n$ as a decimal. At each stage you get a quotient, the next digit of the decimal expansion, and a remainder. The only possible remainders are the numbers $0,1,2,\dots,n-1$. Either you get a remainder of $0$ at some point, in which case the decimal terminates, or you eventually repeat a remainder. Say that you get the same remainder after finding the $\ell$-th digit as you got after finding the $k$-th digit; then you’re doing the same division to get the $(\ell+1)$-st digit that you did to get the $(k+1)$-st digit, and so on. That is, the block of $\ell-k$ digits starting with the $\ell$-th digit exactly repeats the block starting with the $k$-th digit, and so on.

I’ve left this very informal; it would be a good exercise to try to make formalize it a bit.

Added: Here is an example of what I’m talking about. We wish to expand $\frac27$ as a decimal:

                 0.2857142...                   -------------                7)2.0000000...                    0                    ---                    2 0  <-- Remainder of 2                    1 4                    ----                      60                      56                      ---                       40                       35                       ---                        50                        49                        ---                         10                          7                         ---                          30                          28                          ---                           2 <-- Remainder of 2 

At this point you’re going to get the same sequence of remainders ($2,6,4,5,1,3$) and the same sequence of digits in the quotient ($2,8,5,7,1,4$) all over again. And then again. Dividing the same thing, in this case $7$, into the same quantity must yield the same result every time.

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    The remainder you get after step $l$ completely determines all remaining digts (and remainders). Therefore if the remainder after step $k$ is the same, it determines that the remaining digits from position $k$ then on are the same as from position $l$ on. Hence as sson as a remainder repeats (not necessarily already when a digit repeats), the digits become periodic2012-09-15
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This is another way of looking at the long division method with successive remainders.

The essence of long division is to find an expression $10^sm=k_sn+r_s$ for successive integers $s$, and integers $k_s$ and $0\leq r_s\leq n-1$. If we ever find a $t>s$ with $r_t=r_s$ we can subtract the expressions to eliminate the remainder term, obtaining $10^s(10^{t-s}-1)m=(k_t-k_s)n$ or$\frac m n=\frac{k_t-k_s}{10^s(10^{t-s}-1)}$

Then use that the right-hand side is a multiple of $\frac 1 {10^{t-s}-1}$, which has a particularly simple form as a recurring decimal to show that $\frac mn$ is also ultimately recurring.