Multiplication by 2 over 3 in Egyptian Fractions

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Theorem

Let $\dfrac 1 n$ be an Egyptian fraction not equal to $\dfrac 2 3$.


In order to multiply $\dfrac 1 n$ by $\dfrac 2 3$ and have it that $\dfrac 1 n \times \dfrac 2 3$ is also expressed in Egyptian form, we have:

$\dfrac 1 n \times \dfrac 2 3 = \dfrac 1 {2 n} + \dfrac 1 {6 n}$


Proof

\(\ds \dfrac 1 {2 n} + \dfrac 1 {6 n}\) \(=\) \(\ds \dfrac 3 {6 n} + \dfrac 1 {6 n}\)
\(\ds \) \(=\) \(\ds \dfrac {3 + 1} {6 n}\)
\(\ds \) \(=\) \(\ds \dfrac 2 {3 n}\)
\(\ds \) \(=\) \(\ds \dfrac 1 n \times \dfrac 2 3\)


Note the case where we multiply $\dfrac 2 3$ by $\dfrac 2 3$ itself:

\(\ds \dfrac 2 3 \times \dfrac 2 3\) \(=\) \(\ds \dfrac 4 9\)
\(\ds \) \(=\) \(\ds \dfrac 3 9 + \dfrac 1 9\)
\(\ds \) \(=\) \(\ds \dfrac 1 3 + \dfrac 1 9\) which is in Egyptian form

$\blacksquare$


Examples

Example: $\dfrac 2 3 \times \dfrac 1 5$

$\dfrac 2 3 \times \dfrac 1 5 = \dfrac 1 {10} + \dfrac 1 {30}$


Example: $\dfrac 2 3 \times \dfrac 1 8$

$\dfrac 2 3 \times \dfrac 1 8 = \dfrac 1 {16} + \dfrac 1 {48}$

despite the fact that it is apparent that:

$\dfrac 2 3 \times \dfrac 1 8 = \dfrac 1 {12}$


Sources

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