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is there a function $\gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $\gamma(\frac{a}{b})>\gamma(\frac{a+1}{b+1})$

when you start with $\gamma(\frac{1}{2})$ you get an inequality for all n

$$\gamma(\frac{1}{2})>\gamma(\frac{2}{3})>\gamma(\frac{3}{4})>...>\gamma(\frac{n}{n+1})>\gamma(\frac{n+1}{n+2})>...$$

but if you start with $\frac{1}{3}$ you find no new information because $\frac{1+1}{3+1}$ isn't fully simplified so $3+1$ and $1+1$ isn't coprime so we don't know if $\gamma(\frac{1}{3})>$or$=$ or$<\gamma(\frac{1}{2})$

is there a function that follows this rule for all fractions $\frac{a}{b}$ and that's differentiable everywhere

And if there is a function $\gamma(x)$ then is it made up from elementary functions?

BriggyT
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Assuming that $a, b$ are positive integers: $$ \frac{a}{b} < 1 \implies \frac{a}{b} < \frac{a+1}{b+1} < 1 \\ \frac{a}{b} > 1 \implies \frac{a}{b} > \frac{a+1}{b+1} > 1 \\ $$ (compare also How to prove that adding $n$ to the numerator and denominator will move the resultant fraction close to $1$?).

Therefore any function $\gamma$ which is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, \infty)$ has the property that $$ \gamma(\frac{a}{b})>\gamma(\frac{a+1}{b+1}) $$ for all $a \ne b$, and in particular for co-prime integers.

There are many such functions, a simple example is $\gamma(x) = x + \frac 1x$.

Martin R
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    Just so you comply with all the requirements, make $\gamma$ differentiable. So $$\gamma(\frac ab)=|\frac ab-1|$$ is not good, but$$\gamma(\frac ab)=(\frac ab-1)^2$$ is OK – Andrei Jul 23 '20 at 16:32
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    @Andrei: Yes (I have added a similar example). – Martin R Jul 23 '20 at 16:37