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 Template:Math and Template:Math and an integer Template:Math such that for every index Template:Math, Template:Math, then Template:Math.

Proof

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

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

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

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