Anning's Theorem

From ProofWiki
Jump to navigation Jump to search

Theorem

In any base greater than $1$, the fraction:

$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$

has the property that if the two $1$'s in the center of the numerator and the denominator are replaced by the same odd number of $1$'s, the value of the fraction remains the same.

For example:

$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011} = \dfrac {1 \, 010 \, 111 \, 110 \, 101} {1 \, 100 \, 111 \, 110 \, 011} = \dfrac {9091} {9901}$ (in base $10$).


Proof

Let $b$ be the base in question.

Let $F = \dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$.

Then:

$F = \dfrac {1 + b^2 + b^4 + b^6 + b^8} {1 + b + b^4 + b^7 + b^8}$

It is necessary to prove that for all $k \in \Z_{>0}$:

$F = \dfrac {1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6} } {1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6} }$

This is done by:

multiplying the numerator of one by the denominator of the other

and then:

multiplying the denominator of one by the numerator of the other

and checking that they are equal.


Thus we proceed:

\(\displaystyle \) \(\) \(\displaystyle 1 + b^2 + b^4 + b^6 + b^8\)
\(\displaystyle \) \(\) \(\, \displaystyle \times \, \) \(\displaystyle 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^2 + b^3 + b^6 + b^7 + \cdots + b^{2 k + 4} + b^{2 k + 7} + b^{2 k + 8}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^4 + b^5 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 9} + b^{2 k + 10}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^6 + b^7 + b^{10} + b^{11} + \cdots + b^{2 k + 8} + b^{2 k + 11} + b^{2 k + 12}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^8 + b^9 + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 13} + b^{2 k + 14}\)


and:

\(\displaystyle \) \(\) \(\displaystyle 1 + b + b^4 + b^7 + b^8\)
\(\displaystyle \) \(\) \(\, \displaystyle \times \, \) \(\displaystyle 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b + b^3 + b^5 + b^6 + \cdots + b^{2 k + 3} + b^{2 k + 5} + b^{2 k + 7}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^4 + b^6 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 8} + b^{2 k + 10}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^7 + b^9 + b^{11} + b^{12} + \cdots + b^{2 k + 9} + b^{2 k + 11} + b^{2 k + 13}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle b^8 + b^{10} + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 12} + b^{2 k + 14}\)

Equality can be demonstrated.

$\blacksquare$


Source of Name

This entry was named for Norman Herbert Anning.


Historical Note

Anning's Theorem appears as a problem in The $1970$ publication 250 Problems in Elementary Number Theory by Wacław Sierpiński, who cites N. Anning as its source.

Confusingly, later he refers to P. Anning in the context of the same result. It can be assumed that this is a mistake.

It is apparent that this result was published in volume $22$ of Scripta Mathematica, but it has not been possible to find an online archive to confirm this, or even to determine what the title is of the article it appeared in.


The name Anning's Theorem has been coined by $\mathsf{Pr} \infty \mathsf{fWiki}$ as a convenient way to refer to this result, whose description would otherwise be too unwieldy for convenience.


Sources