Anomalous Cancellation on 2-Digit Numbers/Example/19 over 95

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Theorem

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:

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


So:

\(\displaystyle \) \(\) \(\displaystyle \left({\left({\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i}\right) + 10^n}\right) \Big / \left({5 + \left({\sum_{i \mathop = 1}^n 9 \times 10^i}\right)}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({\sum_{i \mathop = 0}^{n - 1} \left({10^{i + 1} - 10^i}\right)}\right) + 10^n}\right) \Big / \left({5 + \left({\sum_{i \mathop = 1}^n \left({10^{i + 1} - 10^i}\right)}\right)}\right)\) $\quad$ rewritting the $9$'s $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({10^n - 10^0 + 10^n}\right) \Big / \left({5 + 10^{n+1} - 10^1}\right)\) $\quad$ Definition of Telescoping Series $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({2 \times 10^n - 1}\right) \Big / \left({10 \times 10^n - 5}\right)\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 5\) $\quad$ $\quad$

$\blacksquare$


Sources