# Show that a decimal with a repeating pattern represents a rational number?

Exercise 1.3.1 (Guide to Analysis, Hart, p.4) Consider a decimal of the form $$x=0.a_1a_2...a_n$$ with a repeating pattern of $$n$$ digits. Write $$x = 0.a_1a_2 ... a_n$$. Express $$10^nx$$ as a decimal. Then subtract $$x$$ and check that $$10^nx - x$$ is an integer. Deduce that $$x$$ is rational. Now extend the method to prove that all decimals with a repeating pattern represent rational numbers.

First question: let $$x= 0.33$$, then $$10^2*.33 - 0.33=33-0.33=32.67$$ is not an integer. Now, it seems that I miss understood something. Can someone explains to me what he is saying.

• Let $x=0.14141414...$. Then $100x=14.141414...$, so $100x-x=14,$ so $99x=14,$ so $x=14/99$ Commented Mar 16, 2020 at 19:20
• @J.W.Tanner Thank you! this is an answer to my question. I miss that I write .33333333 for ever. I see the point! write your comment to accept it as an answer. Commented Mar 16, 2020 at 19:25
• 1) How did you go from $x=0.25$ to $0.33$. 2) $0.25000000....$ and $0.3300000....$ terminate and the repeating terms are $0$. If you do this by the letter for $0.33$ then the repeating parts are the $0$ at the third third position so you do $10^4*0.33 - 10^2*0.33 = 3300 - 33 = 2367$ is and integer and $x =0.33 = \frac {2367}{9900}$. Commented Mar 16, 2020 at 19:29
• I was also wondering how you went from $0.25$ to $.33$; I took the liberty of changing $10^x-x$ to $10^nx-x$, which I think you intended; note: repeating decimals can also be thought of as geometric series; @fleablood: I think you meant $\color{red}{32}67$ Commented Mar 16, 2020 at 19:31
• If you mean $x=0.333333......$ then you have the repeating pattern of length 1 starting at position 1 so you multiply by $10^1$ so $10x = x = 3.3333..... - 0.33333... = 3$ is and intger and $x = \frac 3{9} = \frac 13$. Commented Mar 16, 2020 at 19:34

For example, let $$x=0.14141414...$$.
Then $$100x=14.141414...$$, so $$100x-x=14,$$ so $$99x=14$$ so $$x=14/99$$.
In general, if $$x=0.\overline {a_1a_2\dots a_n}$$, where the overline indicates it's repeated,
then $$10^nx=a_1a_2...a_n.\overline{a_1a_2\dots a_n}$$, so $$10^nx-x=a_1a_2\dots a_n$$, so $$x=\dfrac{a_1a_2\dots a_n}{10^n-1}$$ is rational.
• This could be further generalized to show that $b_1b_2\dots b_k.c_1c_2\dots c_m\overline{a_1a_2\dots a_n}$ is rational as well Commented Mar 16, 2020 at 19:27