Addition of irrational numbers algorithmically How to perform addition of irrational numbers algorithmically, in the standard $b$-adic expansion I do not know how to start the "standard" digit-by-digit way of doing this cause irrational numbers are non-periodic in their expansion?
 A: So, the problem is that standard algorithm starts from the right-most digit, which, for all irrational numbers (and also periodic ones) does not exist.
But you can also start from the left-most digit. When the sum of digits exceeds 9, you must update the digits you have already computed on the left. This way your algorithm gets a sequence of digits (i.e. rational numbers) which approximate the exact result.
Of course this algorithm never terminates. And there are cases when you cannot be sure that the digits you have found so far are the exact expansion of the result.
A: The problem is carries. If I add $4.31\ldots$ to $1.64\ldots$, I can start in the one's place and see that I get $4+1=5$: so in the sum, the digit in the one's place is either $5$ or $6$, depending on whether or not there is a carry from the right.
Looking at the tenth's place, we get $6+3=9$: in the sum, this digit is either a $9$ (and there is no carry to the left), or it is a $0$ (and there is a carry to the left).
In the hundredths place, we get $1+4=5$. While this might turn out to be a $6$ due to carries, there cannot be any carries to the left from this position.
Thus, we can guarantee that the first two digits of the sum are $4.9\ldots$.
Usually, if you go out far enough, you will eventually come to a point where there is no question about whether there is or is not a carry going to the left, and you can be sure of all of the digits up to that point. The only exception is when a sum ends in all $9$'s with unknown carry. Either the true sum does end in all $9$'s without a carry in the next place before the $9$'s, or it ends in all $0$'s and there is a carry.
Fortunately, both choices give the same value: e.g. $1.\overline{0} = 0.\overline{9}$.
However, depending on precisely what you mean by "algorithmically", this ambiguity can mean that you can't do addition algorithmically in a radix representation: if you ever run into a long sequence of digits that are either $9$ without carry or $0$ with carry, then you are stuck and can never compute any more digits of the sum until you reach the end of such a sequence, or can prove it continues forever.
