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 A = 0.a₁a₂a₃… then 10 × A = a₁.a₂a₃a₄…

Proof

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

${\displaystyle A=\sum _{n=0}^{\infty }{\frac {a_{n}}{10^{n}}}.}$

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

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

Next we Shifting a series|shift the series:

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

But a₀ = 0 by assumption, so we can simplify:

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

which is the desired result.