How to handle rounding in uneven number bases? Is there any formal rule for handling rounding operations for the middle element in an uneven base? For example, in ternary {0,1,2}, what would I round a number ending in 1 to?
 A: I was about to ask a similar question when I encountered yours. I've considered this quite a bit and come to realize that the rule for handling the middle element is up to the person doing the rounding, regardless of the base. But I think your understanding of "the middle element" is incorrect. Mine was, too. I'll explain.
In base ten, we're used to rounding to the nearest thousand pesos or tenth of a second or whatever. But ten is an arbitrary radix, so rounding to the nearest multiple of a power of that radix is also arbitrary.
If you're rounding to "the nearest multiple of one thousand", for example, the typical rule is to round down when $0$ through $4$ appears in the hundreds digit and round up for $5$ through $9$. That seems fair because each direction has an equal number of digits. But it's not! It's only fair if the $5$ is followed by another non-zero digit, however tiny that fraction might be. The actual midpoint ($500$) requires an uncomfortable compromise because the two closest multiples of $1000$ are equally near. So, at your discretion, you can apply different rounding techniques. In away-from-zero rounding, $500$ rounds up to $1000$. But in nearest-even-digit rounding, $500$ rounds down to $0$ while $1500$ rounds up to $2000$. This strives to minimize accumulated rounding errors by rounding half the midpoints up and half down. Away-from-zero achieves the same end if you're rounding both positive and negative numbers. The point is, this is a decision that depends on the application. And often, it's arbitrary.
By defining a round number as a multiple of a power of the radix, we may make our lives easy. But is that really the definition of a "round" number? If so, every integer is a round number if you only but change the radix!
On the contrary, I think a round number is whatever you've defined it to be. You can round to the nearest multiple of three. The comment by @froggie demonstrates this in base three, but it works even with decimals. $1$ rounds down to $0$, $2$ rounds up to $3$, $3$ stays $3$, and so on.
Herein lies where you went astray, and I did, too. The number one is not "the middle element", as you call it; one and a half is.
What happens with $1.5$? It rounds up to $3$ if you're doing away-from-zero rounding. If you look at this case in ternary, you've got $1.\overline{1}$, which, if you follow the same midpoint rule, rounds to $10$. (Doubling ternary $1.\overline{1}$ gives you $2.\overline{2}$, which equals $10$.)
So is there any formal rule? No, the decision lies with the person (or algorithm) doing the rounding.
