# 0.999.../Decimal addition and subtraction

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.

## Contents

## Assumptions[edit]

## Theorems[edit]

### Addition by digits is correct[edit]

- Statement

If there are three decimals *A* = *a*_{0}.*a*_{1}*a*_{2}*a*_{3}…, *B* = *b*_{0}.*b*_{1}*b*_{2}*b*_{3}…, and *C* = *c*_{0}.*c*_{1}*c*_{2}*c*_{3}… such that for every index *n*, *a*_{n} + *b*_{n} = *c*_{n}, then *A* + *B* = *C*.

- Proof

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 correct[edit]

- Statement

If there are three decimals *A* = *a*_{0}.*a*_{1}*a*_{2}*a*_{3}…, *B* = *b*_{0}.*b*_{1}*b*_{2}*b*_{3}…, and *C* = *c*_{0}.*c*_{1}*c*_{2}*c*_{3}… such that for every index *n*, *a*_{n} − *b*_{n} = *c*_{n}, then *A* − *B* = *C*.

- Proof

The proof is almost identical to the previous proof:

## The road not taken[edit]

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.