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A theorem of Erdos states:

"There exists an absolute constant c such that, if n>ck, and if a1/b1, a2/b2, ... are the Farey fractions of order n, then ax/bx and ax+k/bx+k are similarly ordered."

Can someone provide a definition of "similarly ordered" as used here?

Thanks for any insight.

Cheers, Scott

@ARTICLE{Erdos:1943, author={Erd{\"o}s, Paul}, title={A note on {F}arey series}, journal={Quart. J. Math., Oxford Ser.}, fjournal={The Quarterly Journal of Mathematics. Oxford. Second Series}, volume={14}, year={1943}, pages={82--85}, issn={0033-5606}, mrclass={40.0X}, mrnumber={MR0009999 (5,236b)}, mrreviewer={G. Szeg{\"o}}

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  • $\begingroup$ A comment, not an answer: I assume that $a_{x+k}b_{x+k}$ is a typo for $a_{x+k}/b_{x+k}$. Can you give us the reference for the paper so we can have a look at the context? $\endgroup$
    – Gerry Myerson
    Feb 20, 2010 at 11:15

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Here is a link that gives the definition: http://qjmath.oxfordjournals.org/cgi/pdf_extract/os-13/1/185

It's to the paper by Mayer that Erdos refers to. (The link gives you just the first page of the paper, but fortunately the definition is on that page.)

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Most likely it means that for any $x,y$, we have $a_x/b_x < a_y/b_y$ iff $a_{x+k}b_{x+k} < a_{y+k}b_{y+k}$.

Edit: On second thought, this seems unlikely, because with the definition I gave the theorem doesn't make sense for $k=1$, $n > 5$: we have $a_1/b_1 = 1/n, a_2/b_2 = 1/(n-1), a_3/b_3 = 1/(n-2)$, and $a_1/b_1 < a_2/b_2$ but $a_2b_2 > a_3b_3$. Maybe it means that for more than half of all pairs the order will be the same?

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