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Here is the exact proof I must solve:

Prove that when the process of long division is used for 2 integers, say m and n, then the resulting decimal fraction is always a repeating one, i.e. m/n=an*an−1...a1*a0 [decimal point] b1...bkc1...cpc1...cpc1...cp... = anan−1... a1*a0 [decimal point] b1...bk(c1...cp).

I don't know where to begin. If m=5 and n=8, then 5/8=0.625 which is terminating. The wording of the question is confusing me.

FoiledIt24
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    I think that in the case of $m = 5$, $n = 8$, that the "repeating" decimal would be $0.625\overline{0}$ (where the $0$ is repeating.) – ETS1331 Jun 19 '19 at 02:24
  • Compare https://math.stackexchange.com/questions/307796/analysis-proof-for-repeating-digits-of-rational-numbers?rq=1 – David K Jun 19 '19 at 02:36
  • Also see https://math.stackexchange.com/questions/61937/how-can-i-prove-that-all-rational-numbers-are-either-terminating-decimal-or-repe – David K Jun 19 '19 at 02:37
  • $5/8$ actually repeats in two ways: it's either $0.625$ followed by an infinite string of $0$s, or $0.624$ followed by an infinite string of $9$s. – David K Jun 19 '19 at 02:38
  • I was wondering about that too... – FoiledIt24 Jun 19 '19 at 03:05

1 Answers1

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A better statement is that the resulting decimal fraction is eventually repeating.

For example when you divide $7$ by $15$ you get $$0.466666...$$ and the repeating part is the $6$ not the $46$

The reason for this eventual repetition is that we only have finitely many possible remainders and when we continue the process of dividing eventually some remainder will happen again which causes a cycle to repeat itself.

Mohammad Riazi-Kermani
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