0.999.../Decimal addition and subtraction< 0.999...
Addition and subtraction of infinite decimals includes some easy problems and some hard problems. Even for finite decimals, identities without carrying are easy to verify (123456 + 654321 = 777777), whereas calculations with long runs of carrying are relatively hard to perform (3456 + 6549 = 99915). A similar phenomenon occurs for infinite decimals.
Fortunately, we will not be encountering any addition problems with carrying, so we can concentrate on a few simple proofs of identities without carrying.
Addition by digits is correctEdit
If there are three decimals A = a0.a1a2a3…, B = b0.b1b2b3…, and C = c0.c1c2c3… such that for every index n, an + bn = cn, then A + B = C.
We apply the definition of an infinite decimal as a series:
Next we apply the fact that sums of series can be computed term-by-term:
Subtraction by digits is correctEdit
If there are three decimals A = a0.a1a2a3…, B = b0.b1b2b3…, and C = c0.c1c2c3… such that for every index n, an − bn = cn, then A − B = C.
The proof is almost identical to the previous proof:
The road not takenEdit
If A and B are arbitrary infinite decimals, then it can be tricky to compute the decimal expansion of A + B = C. The problem is caused by the phenomenon of carrying from one digit to the next. To compute any given digit of C, one might need to inspect many more digits of A and B to make sure that their sum doesn't carry into the target digit.
This book does not explore the addition of arbitrary decimals, mostly because it is difficult and unnecessary.