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

Statement

If Template:Math then Template:Math

Proof

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

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

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

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

Next we shift the series:

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

But Template:Math by assumption, so we can simplify:

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

which is the desired result.