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In an excellent post several years ago, we learn that the period of the decimal expansion of a rational number $\frac{p}{q}$ must divide the multiplicative order of $10\pmod q$ assuming that there are no factors of $2$ or $5$ in $q$.

Length of period of decimal expansion of a fraction

How does the period of the decimal expansion of $\frac{q}{p}$ relate to that of $\frac{p}{q}$? Let us assume that $p, q$ are coprime integers. (I have added this in response to an answer given.)

DDS
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  • Why should they be related? – user10354138 Jun 26 '19 at 16:28
  • If they are not, can we prove that? – DDS Jun 26 '19 at 16:33
  • Pick $p,q$ unrelated, with 10 is a primitive root both mod p and mod q, for example. – user10354138 Jun 26 '19 at 16:36
  • I think you answered your own question in the post. The period of $\frac pq$ must divide the multiplicative order of $10 \pmod q$ and the period of $\frac qp$ must divide the multiplicative order of $10\pmod q$. As these multiplicative orders are completely unrelated the periods are unrelated. Basically the numerator does very little to affect the period. – fleablood Jun 26 '19 at 16:43

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I'm not sure how to prove this, but it seems like the periods of $\frac1q$ and $\frac pq$ are identical. So what you are asking is if for any primes $p$ $q$ how the periods of $\frac1p$ and $\frac1q$ are related. I would say they are not.

Ruben
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  • It's not necessary that $p, q$ be prime, but coprime. $p/q$ is presumed to be a fraction in lowest terms. – DDS Jun 26 '19 at 18:13