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Is this the correct way to convert fractions to duodecimals and duodecimals to fractions?

Write $\left(\frac{7}{13}\right)_{12}$ as a duodecimal.

$\left (\frac{7}{13}\right)_{12}=\left(\frac{7}{15}\right)_{10}=.4666..._{10}$

$4.666666 \cdot 12 = 5.5999992$

$.5999992 \cdot 12 = 7.1999904$

$.1999904 \cdot 12 = 2.3998848$

$.3998848 \cdot 12 = 4.7986176$

$.7986176 \cdot 12 = 9.5834112$

$.5834112 \cdot 12 = 7.0009344$

Then $\left(\frac{7}{13}\right)_{12}= .5729497_{12}$

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Write $\left(.2929...\right)_{12}$ as a fraction.

$\left(.2929...\right)_{12}$ =$\frac{2}{12} + \frac{9}{144}+\frac{2}{1728}+\frac{9}{20736}...$

On this one, I am not clear on how to write this in lowest terms or even properly represent the repeating part?

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Let us observe that for a "basis" $b$, e.g. $b=10$ or $b=12$, we have $$ \begin{aligned} \frac 1{b-1} &= \frac 1b+\frac 1{b^2}+\frac 1{b^3}+\dots &&=0,111\dots_{(b)} \\ \frac 1{b^2-1} &= \frac 1{b^2}+\frac 1{b^4}+\frac 1{b^6}+\dots &&=0,010101\dots_{(b)} \\ \frac 1{b^3-1} &= \frac 1{b^3}+\frac 1{b^6}+\frac 1{b^9}+\dots &&=0,001001001\dots_{(b)} \\ \frac 1{b^4-1} &= \frac 1{b^4}+\frac 1{b^8}+\frac 1{b^{12}}+\dots &&=0,000100010001\dots_{(b)} \ , \end{aligned} $$ and so on. For instance:

sage: a = 1/(10^5-1)
sage: a.n(200)
0.000010000100001000010000100001000010000100001000010000100001000

Now we will use this to compute the "decimals" of $7/15$ for the basis $12$. (If no basis is specified, we will use truly decimal digits.)

First, $15$ and $12$ are not relatively prime, so we use a small "dance around" to put us in position. This is similar to computing for instance

sage: (1/28).n(200)
0.035714285714285714285714285714285714285714285714285714285714

where we have a "specific beginning", $0.03$, and only after this there is a period repetition. (Ten is $2\cdot 5$, and $2^2$ is in the denominator, so we expect two atypical places, after that we have a periodic repetion of decimals.) In the above case example, one could proceed as follows, rewriting $1/28=25/(28\cdot 25)=25/700=(21+3)/700 = 3/100+3/700$. The first term contributes to $0.03$, the second term corresponds to a shifted representation for $3/7$. In our case, we rewrite correspondingly. $$ \begin{aligned} \frac {7}{15} &= \frac {7\cdot 4}{15\cdot 4} = \frac {28}{12\cdot 5} = \frac 1{12}\cdot\frac {28}{5} \\ &= \frac 1{12}\cdot\left(5+\frac {3}{5}\right) \\ &\qquad\text{... now use the fact that $5$ divides $12^4-1=20735=5\cdot 4147$, $4$ being $\phi(12)$,} \\ &= \frac 1{12}\cdot\left(5+\frac {3\cdot 4147}{5\cdot 4147}\right) \\ &= 0,1_{(12)}\cdot\left(5_{(12)}+7249_{(12)}\cdot 0,0001\,0001\,0001\,\dots_{(12)}\right) \\ &= 0,\ 5\, 7249\,7249\,7249\dots_{(12)} \\ &= 0,\ 5\, (7249)_{(12)} \end{aligned} $$ The other direction is arguably simpler: $$ \begin{aligned} 0,(29)_{(12)} &= \frac{2}{12} + \frac{9}{12^2}+\frac{2}{12^3}+\frac{9}{12^4}+\dots \\ &=(2\cdot 12+9)\cdot\left(\frac 1{12^2}+\frac 1{12^4}+\frac 1{12^6}+\dots\right) \\ &=33\cdot \frac 1{12^2}\cdot\frac 1{1-\frac 1{12^2}} \\ &\qquad\text{sum of the geometric series $1+q+q^2+\dots$ is $1/(1-q)$, here $q=1/12^2$,} \\ &=33\cdot \frac 1{12^2-1} =\frac {33}{143}=\frac 3{13}\ . \end{aligned} $$ Computer check:

sage: ZZ(int((3/13)*12^20)).digits(base=12)
[9, 2, 9, 2, 9, 2, 9, 2, 9, 2, 9, 2, 9, 2, 9, 2, 9, 2, 9, 2]

(We took the decimally written fraction $3/13$, multiplied with a big power of $b=12$, converted to an integer, then asked for the digits, written in converse order.)

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  • $\begingroup$ I am trying to follow each step above, but I am confused, at the end...is $\frac {3}{13}$ in base 12 or 10? You referenced it as the "decimally written fraction". That seems base 10, meaning $.2929..._{12}$ = $\left( \frac{3}{13} \right)_{10}$=$ \left ( \frac{3}{11} \right)_{12}$. Am I understanding correctly? Thanks in advance for clarifying. $\endgroup$ – PerpetualStudent Jun 22 '18 at 22:34
  • $\begingroup$ Yes, $3/13$ (without any indices) means that numbers are written w.r.t. the usual decimal system. The problem only wanted to write it as a fraction, so we only need to know the numerator, and the denominator (in some working basis). Since all computations were done using decimal writing, i was letting the final fraction also w.r.t. base $10$. Note that - as a parallel to what we do in base ten - e.g. by writing $0,(141143)=141143/999999$, we could also have written directly $0,(29)_{12}=29_{(12)}/BB_{(12)}$, and we can simplify... Here $A=10$, $B=11$, in base $12$. $\endgroup$ – dan_fulea Jun 26 '18 at 21:20
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Doing everything base twelve, with $T$ and $E$ for the extra two digits, we have

$$100\times.292929\ldots=29.292929\ldots$$

(just as in ordinary base ten), hence

$$29=29.292929\ldots-.292929\ldots=(100-1)\times.292929\ldots=EE\times.292929\ldots$$

and thus

$$.292929\ldots={29\over EE}$$

We could leave it like this. However, this fraction reduces: Since $10=3\times4$, we find $29=3\times E$, which gives us

$$.292929\ldots={29\over EE}={3\over11}$$

As for writing $7/13$ in duodecimal form, note that $13=3\times5$, so we have

$${7\over13}={7\over3\times5}={4\over10}\times\left(1+{2\over5}\right)={1\over10}\times\left(4+{8\over5}\right)={1\over10}\times\left(6-{2\over5}\right)$$

Now

$$\begin{align} 20&=4\times5+4\\ 40&=9\times5+3\\ 30&=7\times5+1\\ 10&=2\times5+2 \end{align}$$

so long division says

$${2\over5}=.497249724972\ldots$$

It follows (using $EEEE-4972=7249$) that

$${7\over13}={1\over10}\times(6-.497249724972…)={1\over10}\times5.724972497249\ldots=.5724972497249\ldots$$

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