Multiplication of infinite decimals is usually challenging because it involves a great deal of carrying. Fortunately, as in the cases of addition and subtraction, we are interested in identities that involve no carrying at all.

## AssumptionsEdit

## TheoremEdit

- Statement

If there are two decimals *A* = *a*_{0}.*a*_{1}*a*_{2}*a*_{3}… and *B* = *b*_{0}.*b*_{1}*b*_{2}*b*_{3}… and an integer *m* such that for every index *n*, *m* × *a*_{n} = *b*_{n}, then *m* × *A* = *B*.

- Proof

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

Next we apply the fact that a scalar multiple of a series can be computed term-by-term: