# Writing a percent as a decimal and a fraction

I am having a problem understanding some manipulations with recurring decimals. The exercise is

Write each of the following as a decimal and a fraction:

(iii) $66\frac{2}{3}$%

(iv) $16\frac{2}{3}$%

For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$.

For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$.

What I don't understand is a part of the model answer for this exercise. They say

$66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$

and

$16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$

I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator.

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The recurring decimal $0.\overline{a_1\ldots a_n}$ is equal to $$\frac{a_1\cdots a_n}{10^n-1}.$$ E.g., $x=0.\overline{285} = 0.285285285\cdots$, then $$x = \frac{285}{10^3-1} = \frac{285}{999}.$$ That is, you get the periodic portion divided by a number that consists of as many $9$s as the length of the periodic portion.

There are many ways of seeing this; one is using geometric series. Another is to use some manipulations: if $$x = 0.\overline{a_1\ldots a_n}$$ then $$10^nx = a_1\ldots a_n . \overline{a_1\cdots a_n}$$ so $$(10^n-1)x = 10^n x - x = a_1\cdots a_n.$$

The first "model solution" is using this: since $x = 0.\overline{6}$, then $x = \frac{6}{9}$ (the period has length $1$, so you get a single $9$ in the denominator.

When the periodic decimal does not start right after the decimal point, you need to shift it a bit first. So for example, if you had $$x = 0.1\overline{285} = 0.1285285285\ldots,$$ then first we take $10 x = 1.\overline{285}$, then proceed as before: \begin{align*} 10^3(10 x) &= 1285.\overline{285}\\ 10x &= 1.\overline{285}\\ 10x(10^3-1) &= 1284\\ x(9990)&= 1284\\ x &= \frac{1284}{9990}. \end{align*} The second model solution uses this method.

Added. For the series method, in case anyone is interested, suppose that $x$ is of the form $x=0.\overline{a_1\cdots a_n}$. This means, explicitly, that $$x = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{(10^n)^k} = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{10^{nk}} = a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}}.$$ This is a geometric series, with initial term $\frac{1}{10^{n}}$ and common ratio $\frac{1}{10^n}$, so it converges. A geometric series with initial term $a$ and common ratio $r$, $|r|\lt 1$, converges to $$\frac{a}{1 - r},$$ so we have \begin{align*} x &= a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}} \\ &= a_1\cdots a_n\left(\frac{\frac{1}{10^n}}{1 - \frac{1}{10^n}} \right)\\ &= a_1\cdots a_n\left(\frac{\quad\frac{1}{10^n}\quad}{\quad\frac{10^n-1}{10^n}\quad}\right)\\ &= a_1\cdots a_n\left(\frac{1}{10^n-1}\right) = \frac{a_1\cdots a_n}{10^n-1}\\ &= \frac{a_1\cdots a_n}{\underbrace{9\cdots 9}_{n\text{ digits}}}. \end{align*} And similarly if you have to "shift" the decimal before you get to the period; you simply add enough $0$s to the $9$s in the denominator to account for the shift.

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Your answer helped me a lot. Many thanks! – The Gruffalo May 1 '11 at 14:41

Slightly informally, you just multiply to make the repeating decimals subtract out.

\begin{align} 10x &=6.\bar{6} \\x &=0.\bar{6} \\9x &=6 \\ x&=\frac{6}{9} \end{align}

The case for $0.1\bar{6}$ is similar:

\begin{align} 100x &=16.\bar{6} \\ 10x &=1.\bar{6} \\ 90x &=15 \\ x&=\frac{15}{90} \end{align}

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I learn a little more about forum-style Latex very often from reading over your answers. Thanks for that. – mixedmath Apr 28 '11 at 15:31

The model answer for this exercise is too complicated: $66 {2\over 3} = 66 + {2\over 3} = {200 \over 3}$. So $66 {2\over 3} \% = {200 \over 300} = {2 \over 3}$. Similarly, $16 {2\over 3} = {50 \over 3}$ and $16 {2\over 3}\% = {50 \over 300} = {1 \over 6}$.

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The question per se was not difficult for me. What I want to know is how to explain the nines in the denominator. It seems to be some trick to convert the recurrent decimals to fractions but I do not understand it. – The Gruffalo Apr 28 '11 at 15:28

It happens to be well known that a number over 9 (other than 0 or 9) has a repeating decimal. That is: $\dfrac{1}{9} = 0.11 \bar{1}$, $\dfrac{4}{9} = 0.44 \bar{4}$, and so on. Why is this true?

It comes from the fact that if we have the equation $10x = 1.\bar{1}$, then we have $x = 0.\bar{1}$ by division. Subtracting, we get that $9x = 1$, or that $x = \dfrac{1}{9}$.

As happens - I typed this at the same time as Ross.

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