# Anomalous Cancellation on 2-Digit Numbers/Examples/49 over 98

## Example of Anomalous Cancellation on 2-Digit Numbers

The fraction $\dfrac {49} {98}$ exhibits the phenomenon of anomalous cancellation:

$\dfrac {49} {98} = \dfrac 4 8$

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

This is part of a longer pattern:

$\dfrac 4 8 = \dfrac {49} {98} = \dfrac {499} {998} = \dfrac {4999} {9998} = \cdots$

## Proof

 $\displaystyle \frac {499 \cdots 99} {999 \cdots 98}$ $=$ $\displaystyle \paren {4 \times 10^n + \paren {\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i} } \Big / \paren {\paren {\sum_{i \mathop = 1}^n 9 \times 10^i} + 8}$ $\displaystyle$ $=$ $\displaystyle \paren {4 \times 10^n + 9 \times \paren {\frac {10^n - 1} {10 - 1} } } \Big / \paren {9 \times 10 \times \paren {\frac {10^n - 1} {10 - 1} } + 8}$ Sum of Geometric Progression $\displaystyle$ $=$ $\displaystyle \dfrac {4 \times 10^n + \paren {10^n - 1} } {10 \times \paren {10^n - 1} + 8}$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac {5 \times 10^n - 1} {10 \times 10^n - 2}$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac {5 \times 10^n - 1} {2 \times \paren {5 \times 10^n - 1} }$ factoring $\displaystyle$ $=$ $\displaystyle \dfrac 1 2$ $\displaystyle$ $=$ $\displaystyle \dfrac 4 8$

$\blacksquare$