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I get the idea about duodecimals from what I read till I reach the fractions point where:

$\frac{1}{8}=0.16$ instead of $0.15$

$\frac{1}{9}=0.14$ instead of $0.13333333$

$\frac{1}{5}=0.249797979797$ instead of $0.24$

Why is that happening I tried to convert the decimal representation into duodecimal and it came to prove me right on every case.

Thank you very much.

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  • $\begingroup$ sorry about the gibberish at the end but it wouldn't let me post till i wrote a few more lines $\endgroup$
    – user3566929
    Jun 19, 2015 at 4:12
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    $\begingroup$ What is your actual question? $\endgroup$
    – James
    Jun 19, 2015 at 4:22
  • $\begingroup$ I'm not sure what you mean by "$x$ instead of $y$" In decimal $\frac{1}{8} = 0.125$ and in dozenal $\frac{1}{8}=\frac{16}{100} = 0.16$ (with arithmetic done in dozenal). Where does $0.15$ come from? Perhaps you are thinking that since it is "one and a half dozenths"... but remember that half of a dozen is in fact $6$, not $5$, whereas in decimal half of ten is $5$. Likewise, a third of a dozen is $4$, whereas a third of ten is $3.\overline{3}$. $\endgroup$
    – JMoravitz
    Jun 19, 2015 at 4:27

3 Answers 3

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Note:

In every representation, you have: $\frac{1+1+1+1+1+1+1+1+1+1+1+1}{1+1}= 1+1+1+1+1+1$

Likewise, $\frac{\text{"one dozen"}}{\text{"two"}}=\text{"six"}$

In decimal, you have $\frac{12}{2}=6$

Notice though, in dozenal, the representation for "twelve" is not $12$, but is instead $10$.

In dozenal, you have $\frac{10}{2}=6$, just like all of the previous as this is in fact representing the same expression, and so similarly $\frac{1}{2}=0.6$ in dozenal.

You seem to have confused the fact that $\frac{1}{2}=0.5$ in decimal with the fact that $\frac{1}{2}=0.6$ in dozenal. That is to say "one half of one is five tenths" and "one half of one is six dozenths." It appears that you stopped thinking in dozenal halfway through the process of computing the arithmetic once you saw a fraction you thought you were familiar with. Indeed, $\frac{1}{8}=0.16$ in dozenal, i.e. "one eighth of one is a dozen and six gross-th's" or rather "one eighth of one is one and a half dozenths"

With $\frac{1}{8} = $" one and a half dozenths" you jumped to $1.5$ as meaning "one and a half" but again, $1.6$ is one and a half in this context. As such, we intend to use $0.16$ instead of $0.15$ to mean this quantity.

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  • $\begingroup$ i chose your answer for the reason that you insisted on doing it intuitively instead of mathematical derivation, with that said i got it from you insisting on that a half is 0.6 and not 0.5 cause the one means 12 now and not 10 and you are right i stopped thinking in dozenal halfway ,thank you $\endgroup$
    – user3566929
    Jun 19, 2015 at 8:49
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$\frac{1}{8} = 0.16_{12}$ is correct. From the definition of base 12, $0.16_{12} = \frac{1}{12} + \frac{6}{144} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$.

Please give some working on how you obtained $0.15$ etc. so that we can help identify the mistake.

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  • $\begingroup$ well, i always assumed that 1 in the numerator could be replaced by the base number and then divided by 10 to solve it right for instance 1/2(base10) = (10/2)/10 =0.5 ,1/2(base16) = (16/2)/10=0.8 so for 1/2(base 12) it must be 0.15 $\endgroup$
    – user3566929
    Jun 19, 2015 at 8:20
  • $\begingroup$ Your method is in general not correct. I suspect that it converts only the first digit between bases because you are only multiplying and dividing by 12 and 10. If 0.1 is interpreted in duodecimal and 0.05 is interpreted in decimal (as in half of 0.1) you would be correct. But half of 0.1 in base 12 is in fact 0.06. $\endgroup$
    – user242594
    Jun 19, 2015 at 10:34
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If you convert $1/8_{10}$ to base $12$, you might realize that $8$ divides $144_{10}=12_{10}^2$, so it will have a two digit terminating representation. $\frac 18=\frac {18}{144}_{10}=0.16_{12}$ because $16_{12}=18_{10}$. The others are similar. As you did not show where you got your values, we cannot see where the errors are.

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  • $\begingroup$ very nice way of derivation, but i just wanted to understand the underlying cause $\endgroup$
    – user3566929
    Jun 19, 2015 at 8:56

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