Rational numbers in base 6

Question

Express the fractions , for several small values of , in base 6. Determine which rational numbers have terminating expressions in base 6.

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I am unsure how to reduce fractions in base 6. For example, how do you reduce $ {1555 \over 4535} $ ?

I also don't understand how to determine which rational numbers have terminating expressions in base 6. I have found some fractions in base 6 and their corresponding decimal representations (decimal being 0.abcdef, not a number in base 10)

For example $ {2451 \over 2}_6=1223.3_6 , {1 \over 2}_6= 0.3_6 , {43 \over 55}_6=0.4115\overline{15}$ , and so on.

Why is the last term non-terminating? And are the others terminating?

HELP!

Answer 1

(Beware, numbers below are base $6$)

Fractions can be reduced by dividing the greatest common divisor from both the numerator and denominator. And you can find GCD by Euclidean algorithm, but with senary (heximal) multiplication and division. E.g. for $\frac{4535}{1555}$, (it took me some time)

$$\begin{align*} 4535 =& 1555 \times 2 + 541 = 3554 + 541\\ 1555 =& 541 \times 2 + 33= 1522 + 33 \\ 541 =& 33 \times 13 + 24 = 513 + 24 \\ 33 =& 24 \times 1 + 5\\ 24 =& 5\times 3 + 1\\ 5 =& 1\times 5 + 0 \end{align*}$$

Looks like the numerator and denominator are coprime.

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(Appending comments from above)

If a reduced denominator has ONLY prime factors of $2$ and/or $3$, say the denominator is $2^a\cdot3^b$, then one can always multiply both the numerator and denominator by a number $2^{\max(a,b)-a}\cdot3^{\max(a,b)-b}$, so that the denominator becomes $10^{\max(a,b)}$. This fraction obviously has a terminating radix point notation.

The converse, "if a number has a terminating radix point notation in senary base, then the number can be written as a reduced fraction with denominator having only prime factors $2$ and/or $3$", should also be clear.

Answer 2

First, note that if $1/b$ terminates then $a/b$ terminates, so we really only need to consider reciprocals. If $1/b=(0.a_{-1}a_{-2}a_{-3}\ldots)_6$, then $6/b=(a_{-1}.a_{-2}a_{-3}a_{-4}\ldots)_6$. If $6$ is coprime to $b$, then the multiplicative order $n$ of $6\pmod{b}$, which must divide $\phi(b)$, is the least number such that $6^n\equiv1\pmod{b}$. Therefore $\frac{6^n}{b}=\frac{6^n-1}{b}+\frac{1}{b}=(a_{-1}a_{-2}a_{-3}a_{-4}\ldots a_{-n}.a_{-1}a_{-2}\ldots)_6$ So the fraction repeats after $n$.

Meanwhile, if the GCD of $6$ and $b$ is greater than $1$, then you can multiply $\frac{6^k}{b}$ until the simplified fraction has a denominator coprime to $6$ or the fraction is an integer, either of which will determine the nature of the fraction.

Ultimately the only numbers that will have a terminating reciprocal are $d|6^k$ for some $k$.