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Definitions: $v_{h,k}=\frac{1}{k(k+k_1)}$ and $w_{h,k}=\frac{1}{k(k+k_2)}$

where $\frac{h_1}{k_1}\lt\frac{h}{k} \lt\frac{h_2}{k_2}$ are sucessive farey fractions.

Proof that: $\frac{h_1}{k_1}-\frac{h}{k}=-w_{{h_1},{k_1}}-v_{h,k}$ and $\frac{h}{k}=\frac{h_1+h_2}{k_1+k_2}$

I know the property of fractions in the farey sequences: $\frac{h_1}{k_1}-\frac{h}{k}=-\frac{1}{k_1k}$

This gives me : $\frac{1}{k_1k}=\frac{1}{k_1(k_1+k_2)}+\frac{1}{k(k_1+k)}$

I get a contradiction. Either I did a simple mistake or the problem I was given is simply wrong (or not stated correctly).

As far as I know the two definitions are quite common and I hope that someone can either correct my mistake or maybe give me the correct problem.

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    $\begingroup$ $\frac{h_1}{k_1}-\frac hk=-\frac{1}{k(k+k_1)}-\frac{1}{k_1(k+k_1)}$ holds, but I don't know if this is what you want. $\endgroup$ – mathlove Jan 9 '18 at 12:35
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I don't see any contradiction in the first part. By definition, $h=h_1+h_2$ and $k=k_1+k_2$, so that $$ \frac{h_1}{k_1}-\frac{h_1+h_2}{k_1+k_2}=\frac{h_1k_2-k_1h_2}{k_1(k_1+k_2)}=\frac{-1}{k_1(k_1+k_2)}, $$ because $k_1h_2-h_1k_2=1$ as Farey neighbours. But in the definition of $v_{h,k}$ and $w_{h,k}$ there is no $h$ appearing.

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  • $\begingroup$ Sry, but I dont see how this explains the statement in the yellow box. I had to edit my question because $\frac{h}{k}=\frac{h_1+h_2}{k_1+k_2}$ $\endgroup$ – XPenguen Jan 6 '18 at 2:02
  • $\begingroup$ and not $h=h_1+h_2, k=k_1+k_2$ $\endgroup$ – XPenguen Jan 6 '18 at 7:31

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