0.999.../Decimal multiplication by 10

Multiplying an infinite decimal by 10 is just as simple as multiplying an finite decimal by 10: every digit shifts one space to the left.

Assumptions

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

Theorem
If $A = 0.a_{1}a_{2}a_{3}…$ then $10 × A = a_{1}.a_{2}a_{3}a_{4}…$
 * Statement

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



A = \sum_{n=0}^\infty \frac{a_n}{10^n}. $$

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



10A = \sum_{n=0}^\infty \frac{10a_n}{10^n} = \sum_{n=0}^\infty \frac{a_n}{10^{n-1}}. $$

Next we shift the series:



10A = a_0 + \sum_{n=0}^\infty \frac{a_{n+1}}{10^{(n+1)-1}}. $$

But $a_{0} = 0$ by assumption, so we can simplify:



10A = \sum_{n=0}^\infty \frac{a_{n+1}}{10^n}, $$

which is the desired result.