Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A rational number can be represented in the form p/q. prove that the period of the the repeating decimal should at the most q-1.

share|improve this question
add comment

2 Answers

Hint $ $ If a fraction is writable with denominator $\rm\:10^f\!-1,\:$ and also with denominator $\rm\:10^{q-1}\!-1\:$ then it is writable with their gcd $\rm\:10^{(f,q-1)}\!-1\:$ as denominator, i.e. $\rm\: mq,nq\in\Bbb Z\:\Rightarrow\:(m,n)q\in\Bbb Z.$

Remark $\ $ Recall that this properties of denominators is the basis of one conceptual fractional proof of unique factorization of integers. See also this answer for a derivation of the (pre)period formula.

share|improve this answer
    
@Farrukh12 The last linked post shows the connection with periodicity and writable with denominator $\rm 10^f\!-\!1.\:$ Was that where you needed help, or elsewhere? You may find it helpful to work through a specific example. –  Bill Dubuque Dec 11 '12 at 15:32
add comment

If, for some $m>n$, $10^np\equiv10^mp\pmod{q}$, then $(10^n-10^m)p=kq$. Therefore, for some $k_1,k_2$ where $0\le k_2\lt10^{n-m}-1$, we have $$ \begin{align} \frac pq &=\frac k{10^n-10^m}\\ &=\frac{k_1(10^{n-m}-1)+k_2}{10^m(10^{n-m}-1)}\\ &=\frac{k_1}{10^m}+\frac{k_2}{10^n}(1+10^{m-n}+10^{2(m-n)}+10^{3(m-n)}+\dots)\tag{1} \end{align} $$ Obviously, $(1)$ represents a decimal number that eventually repeats with period $n-m$.

To see that we always have some $m>n$, so that $10^np\equiv10^mp\pmod{q}$ and $n-m\lt q$, we simply need to note that there are only $q$ possibilities for $10^np\bmod q$ and if $10^np\equiv0\pmod{q}$, then $\frac pq=\frac k{10^n}$, and therefore, the decimal terminates.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.