Anomalous Cancellation on 2-Digit Numbers/Examples/49 over 98
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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
\(\ds \frac {499 \cdots 99} {999 \cdots 98}\) | \(=\) | \(\ds \paren {4 \times 10^n + \paren {\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i} } \Big / \paren {\paren {\sum_{i \mathop = 1}^n 9 \times 10^i} + 8}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {4 \times 10^n + 9 \times \paren {\frac {10^n - 1} {10 - 1} } } \Big / \paren {9 \times 10 \times \paren {\frac {10^n - 1} {10 - 1} } + 8}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 \times 10^n + \paren {10^n - 1} } {10 \times \paren {10^n - 1} + 8}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 \times 10^n - 1} {10 \times 10^n - 2}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 \times 10^n - 1} {2 \times \paren {5 \times 10^n - 1} }\) | factoring | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 4 8\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $16 /64$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $16/64$