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 = ???). 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.

## Theorems

### Addition by digits is correct

Statement

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.

Proof

We apply the definition of an infinite decimal as a series:

${\displaystyle C=\sum _{n=0}^{\infty }{\frac {c_{n}}{10^{n}}}=\sum _{n=0}^{\infty }{\frac {a_{n}+b_{n}}{10^{n}}}.}$

Next we apply the fact that sums of series can be computed term-by-term:

${\displaystyle C=\sum _{n=0}^{\infty }{\frac {a_{n}}{10^{n}}}+\sum _{n=0}^{\infty }{\frac {b_{n}}{10^{n}}}=A+B.}$

### Subtraction by digits is correct

Statement

If there are three decimals A = a0.a1a2a3, B = b0.b1b2b3, and C = c0.c1c2c3 such that for every index n, anbn = cn, then AB = C.

Proof

The proof is almost identical to the previous proof:

${\displaystyle C=\sum _{n=0}^{\infty }{\frac {c_{n}}{10^{n}}}=\sum _{n=0}^{\infty }{\frac {a_{n}-b_{n}}{10^{n}}}=\sum _{n=0}^{\infty }{\frac {a_{n}}{10^{n}}}-\sum _{n=0}^{\infty }{\frac {b_{n}}{10^{n}}}=A-B.}$

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.