# Anomalous Cancellation on 2-Digit Numbers/Examples/19 over 95

## Example of Anomalous Cancellation on 2-Digit Numbers

The fraction $\dfrac {19} {95}$ exhibits the phenomenon of anomalous cancellation:

$\dfrac {19} {95} = \dfrac 1 5$

as can be seen by deleting the $9$ from both numerator and denominator.

This is part of a longer pattern:

$\dfrac 1 5 = \dfrac {19} {95} = \dfrac {199} {995} = \dfrac {1999} {9995} = \cdots$

## Proof

Formally written, we have to show that:

$\displaystyle \paren {\paren {\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i} + 10^n} \Big / \paren {5 + \paren {\sum_{i \mathop = 1}^n 9 \times 10^i} } = \frac 1 5$ for integers $n > 1$.

So:

 $\displaystyle$  $\displaystyle \paren {\paren {\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i} + 10^n} \Big / \paren {5 + \paren {\sum_{i \mathop = 1}^n 9 \times 10^i} }$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {\sum_{i \mathop = 0}^{n - 1} \paren {10^{i + 1} - 10^i} } + 10^n} \Big / \paren {5 + \paren {\sum_{i \mathop = 1}^n \paren {10^{i + 1} - 10^i} } }$ rewriting the $9$'s $\displaystyle$ $=$ $\displaystyle \paren {10^n - 10^0 + 10^n} \Big / \paren {5 + 10^{n + 1} - 10^1}$ Definition of Telescoping Series $\displaystyle$ $=$ $\displaystyle \paren {2 \times 10^n - 1} \Big / \paren {10 \times 10^n - 5}$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac 1 5$

$\blacksquare$