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 edit

Theorem edit

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.