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

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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
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 5\)

$\blacksquare$


Sources