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:

$C=\sum _{n=0}^{\infty }{\frac {c_{n}}{10^{n}}}=\sum _{n=0}^{\infty }{\frac {a_{n}+b_{n}}{10^{n}}}.$ [ఉ 1]

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

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

### 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:

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