Rational number with more than ten digits Could we have a repeating rational decimal number with more than 10 repeating digits (something like $0.0123456789801234567898...$) after the decimal point?
What is the maximum number of repeating digits after the decimal point in a number?
Could the answer be generalized to state that we could / couldn’t have a repeating rational number in base $b$ with more than $b$ repeating digits?
 A: This also happens for "ordinary" rational numbers.  For instance, $\frac1{17}$ has a period of 16 digits, and $\frac1{983}$ has a period of 982 digits.  Look up "full reptend prime" or "long prime" for more information about them.
A: Short answer:
$$\dfrac1{17}=0.\color{green}{0588235294117647}0588235294117647\color{green}{0588235294117647}0588235294117647\cdots$$

Also think that you are free to take any finite sequence of digits and repeat it forever, you will always get a rational number.
A: What determines the period length of a repeating decimal number?
I would like to add a intuitive answer.
If you calculate the decimal representation of a fraction $1/n$ by hand, by long division on a sheet of paper, then in each step, you perform a division by $n$ and write the integer part of the quotient to the result. You also write down the remainder, append a zero and use that as input for the next step. 
When you get a remainder that you already had in a previous step, the result will start to repeat.
So the number of possible remainders limits the length of the period. When dividing by 17, there are 16 possible remainders different from 0. (A remainder of 0 means that the calculation is complete and you get a terminating decimal number.)
On the other hand, the number of available digits in the number system used does not limit the period length. The result does not start to repeat when you get twice the same integer quotient, but with different remainders.
A: The period of a periodic sequence of digits can be as large as you like. To see this, multiply the number by $10^T$, where $T$ is the period, and then subtract the original number. Since this is definitely a whole number $n$ – the repeating parts of the sequence cancel out – the original must have been a rational number, specifically $n/(10^T-1)$.
