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Numbers in the Farey sequence are expressed as fractions e.g $F_5$:

$0\over1$ $1\over5$ $1\over4$ $1\over3$ $2\over5$ $1\over2$ $3\over5$ $2\over3$ $3\over4$ $4\over5$ $1\over1$

All of the $n\over5$ terms can be calculated by adding the numerators and denominators of the fractions immediately to either side of them, e.g. ${3\over5} = {1+2\over2+3}$ but obviously ${1\over2}+{2\over3}\neq{3\over5}$. Is there an operator to express this? ${a\over b} ? {c\over d} = {a+c\over b+d}$

I should point out that this is not for standard equalities, but for generating sequences, such as the above mentioned Farey sequence. It seems cumbersome to separate the numerators and denominators, perform calculations on them and then recombine them to get the next term.

CJ Dennis
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  • Thanks @Integrator! I think I'll make it fractal so it takes an infinite amount of time to write it! – CJ Dennis Nov 08 '14 at 04:02
  • On the wiki page you cite, there is the term "mediant" to denote the operation. Of course that isn't a symbol to place between the two fractions to mean the result, maybe there isn't such a symbol since it isn't a common enough operation. (On the other hand for all I know there may be a symbol in use fror it.) – coffeemath Nov 08 '14 at 04:06

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Yes and no. On one hand, given the fact that we usually identify the fractions $\frac{a}b$, $\frac{2a}{2b}$, $\frac{3a}{3b}$, $\ldots$ - that is, since there are many ways to write a given rational number as the quotient of two integers, it is not clear that this is a particularly natural operation; for instance, keeping $?$ as the operation of adding numerators and denominators gives $$\frac{1}2?\frac{1}1=\frac{2}{3}$$ $$\frac{1}2?\frac{2}2=\frac{3}4$$ and yet, typically, we would consider the left-hand expressions to be equal, so it makes little sense for the results to be different. We could (and this suffices for the Farey sequence, since no fractions in it ever reduce) define $?$ to sum the numerator and denominator of the fraction in lowest terms, but that operator doesn't have a lot of nice properties - it is not even associative since $$\left(\frac{1}2?\frac{1}2\right)?\frac{1}3\neq\frac{1}2?\left(\frac{1}2?\frac{1}3\right)$$ $$\frac{2}4?\frac{1}3\neq\frac{1}2?\frac{2}5$$ $$\frac{1}2?\frac{1}3\neq\frac{1}2?\frac{2}5$$ $$\frac{2}5\neq\frac{3}7$$

However, we could consider that the notation $\frac{a}b$ actually refers to an ordered pair $(a,b)$ - so we make none of the usual identifications - and that, essentially, we perform this coordinate-wise addition to these pairs, and only afterwards do we consider it as an actual fraction. So, here, we define our operation in $\mathbb{Z}^2$ as the entirely sensible vector addition $(a,b)+(c,d)=(a+c,b+d)$ and consider the map $(a,b)\mapsto \frac{a}b$ only afterwards, recognizing that our operation is not well-defined in $\mathbb{Q}$.

Milo Brandt
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  • Interesting concept. So the angle of the vector would give you an idea of the magnitude of the term when reduced to a fraction, however, the magnitude of the vector itself would in most cases be misleading. – CJ Dennis Nov 08 '14 at 04:10
  • Exactly; in particular, the slope of the line through the origin to $(a,b)$ would be the fraction it reduces to. – Milo Brandt Nov 08 '14 at 04:13
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The Wikipedia article on Farey sequences may be of interest to you. This question may be interesting as well.

The quantity $\frac{a + c}{b + d}$ is the mediant of the fractions $\frac ab$ and $\frac cd.$ I don't know of an operator symbol for mediant (like $+$ for sum or $\cdot$ for product) but there is a long tradition of making up symbols when it makes mathematical manipulations easier, so if you had reason to embed a "mediant operation" in some mathematical formula it would make sense to create a symbol for it.

David K
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    Thanks for the mediant information. As you can see from my question I referred to the same article. Henceforth, the operator for the mediant shall be $+\over+$. – CJ Dennis Nov 08 '14 at 04:22
  • Oh, that's where I got that link. :-) After digging around on a particular topic for a while, it's easy to forget why I opened a particular tab. – David K Nov 08 '14 at 14:44