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

Statement

If there are two decimals $A = a 0 . a 1 a 2 a 3 \dots$ and $B = b 0 . b 1 b 2 b 3 \dots$ and an integer $m$ such that for every index $n$ , $m \times a n = b n$ , then $m \times A = B$ .

Proof

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

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.

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