recurring decimals - base 10 to base 12

Question

I came across duodecimal (base 12) numbers.

In base 10 system 10/3 = 3.333.... i.e repeating decimal. But in base 12 system where "t" represents 10 of base 10 - t/3 = 3.4 which is a non repeating decimal.

I hope have done the above t/3 base 12 division correct.

Now, trying to grasp meaning and significance of recurring decimal. Does a recurring value as a result of 10/3 (in base 10) means we can never divide 10 into 3 parts with 100% accuracy ? But then we change number system to base 12 and t/3 now is not a recurring decimal.So now we can divide t into 3 equal parts accurately. But this is a contradiction.

Answer 1

In base 12: $\frac {10}3 = 3\frac 13 = 3\frac {4}{12}= 3.4_{12}$.

In base 10: $\frac {10} 3 = 3\frac 13 = 3\frac {\frac {10}3}{3\frac {10}3} = 3 \frac {3\frac 13}{10} =3\frac 3{10} + \frac 1{30} = 3\frac 3{10}\frac 3{100} + \frac 1{300} = 3\frac 3{10}+ \frac 3{100} + \frac 3{1000} + ......= 3.333333....$.

In base $b$ a fraction $\frac nm$ will terminate if and only if you can get $\frac nm = \frac k{b^w}$ for some power $w$ of $b$.

In other word $m|b^w$.

In base $10$ to have $m|10^w = 2^w5^w$ we simply need to have $m$ to only have prime factors $2$ and $5$. Example $\frac {17}{40}$ and $40=2^3*5|10^3$ so $\frac {17}{40}*\frac {5^2}{5^2} = \frac {425}{1000} = .425$

In base $12$ to have $m|12^w = 2^{2w}3^w$ we simply need to have $m$ to only have prime factors $2$ and $3$. Example $\frac {23}{36}$ and $36=2^23^2|4^23^2$ so $\frac {23}{36}\frac 44 = \frac {23}{144} = \frac {23}{12^2} = .23$.

Answer 2

It only sounds like a contradiction. Think of the problem of dividing $10$ by $3$ as trying to equally share $10$ cookies among $3$ children. You first give each child $3$ cookies, leaving $1$ left over. Now if you work base $10$, what you in effect do is cut that last cookie into ten (equal) pieces, and then try to mete them out equally: you'll again give each child $3$ pieces, and you'll again have one piece left over, so you'll again cut that piece into $10$ equal pieces, and so on and so forth.

On the other hand, if you work base $12$, you cut the leftover cookie into twelve equal pieces and try to mete them out equally. This time you succeed! Each child gets four of the twelve pieces, and you're done.

The only difference between the two distribution schemes is the arbitrary decision to cut any leftover portions of cookies into a particular number of pieces -- ten in the decimal scheme and twelve in the duodecimal. The most sensible approach here, of course, would be to work base three: cut the leftover cookie into three equal pieces, and give each child a piece.