Anomalous Cancellation on 2-Digit Numbers/Examples/16 over 64

From ProofWiki
Jump to navigation Jump to search

Example of Anomalous Cancellation on 2-Digit Numbers

The fraction $\dfrac {16} {64}$ exhibits the phenomenon of anomalous cancellation:

$\dfrac {16} {64} = \dfrac 1 4$

as can be seen by deleting the $6$ from both numerator and denominator.


This is part of a longer pattern:

$\dfrac 1 4 = \dfrac {16} {64} = \dfrac {166} {664} = \dfrac {1666} {6664} = \cdots$


Proof

\(\displaystyle \frac {166 \cdots 66} {666 \cdots 64}\) \(=\) \(\displaystyle \paren {10^n + \paren {\sum_{i \mathop = 0}^{n - 1} 6 \times 10^i} } \Big / \paren {\paren {\sum_{i \mathop = 1}^n 6 \times 10^i} + 4}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {10^n + 6 \times \paren {\frac {10^n - 1} {10 - 1} } } \Big / \paren {6 \times 10 \times \paren {\frac {10^n - 1} {10 - 1} } + 4}\) Sum of Geometric Progression
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\paren {10 - 1} 10^n + 6 \times \paren {10^n - 1} } {60 \times \paren {10^n - 1} + 4 \paren {10 - 1} }\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {9 \times 10^n + 6 \times 10^n - 6} {60 \times 10^n - 60 + 36}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {15 \times 10^n - 6} {60 \times 10^n - 24}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {15 \times 10^n - 6} {4 \times \paren {15 \times 10^n - 6} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 4\)

$\blacksquare$


Sources