0.999.../Decimal multiplication by a small number

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.

Assumptions

 * Definition from series
 * Term-by-term operations on series

Theorem
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 $m × A = B$.
 * Statement

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



B = \sum_{n=0}^\infty \frac{b_n}{10^n} = \sum_{n=0}^\infty m \frac{a_n}{10^n}. $$

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



B = m \sum_{n=0}^\infty \frac{a_n}{10^n} = mA. $$