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Given two fractions $\frac{h}{k}$ and $\frac{h^{'}}{k^{'}}$ both in reduced form. I am unable to find a case when $\frac{h+h^{'}}{k+k^{'}}$ does not lie in the interval $\big[ \frac{h}{k},\frac{h^{'}}{k^{'}} \big]$. Is there such a case ?

PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.

Bill Dubuque
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advocateofnone
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2 Answers2

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Let $\dfrac hk=a$ and $\dfrac {h'}{k'}=b$.

Then, we have $h=ak$ and $h'=ak'$.

Then, $\dfrac{h+h'}{k+k'}=\dfrac{ak+bk'}{k+k'}$.

WLOG, let $a<b$.

$\dfrac{ak+bk'}{k+k'}>\dfrac{ak+ak'}{k+k'}=a$

$\dfrac{ak+bk'}{k+k'}<\dfrac{bk+bk'}{k+k'}=b$

$a<\dfrac{ak+bk'}{k+k'}<b$

$\dfrac hk<\dfrac{h+h'}{k+k'}<\dfrac{h'}{k'}$

The case where $a=b$ and where $a>b$ is left to the reader as an exercise.

Kenny Lau
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Let's prove that, for positive $a,b,c,d$, with $\frac{a}{b}\le\frac{c}{d}$, it holds $$ \frac{a}{b}\le\frac{a+c}{b+d}\le\frac{c}{d} $$ The inequality on the left is equivalent to $$ ab+ad\le ab+bc $$ that is, $ad\le bc$, which is true.

The inequality on the right is equivalent to $$ ad+cd\le bc+cd $$ that is, $ad\le bc$, which is true.

Note that, if $a/b<c/d$, the two inferred inequalities are strict too. The hypothesis about the fractions being in reduced form is redundant.

egreg
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