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We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$?

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Yes; you’re getting the mediant of the original fractions. You may also find the Stern-Brocot tree interesting in this connection, not to mention Farey sequences. – Brian M. Scott Oct 1 '12 at 22:27
You can think of the mediant as a weighted average: if you have a bag of $b$ balls, $a$ of which are white, and you combine it with a bag of $d$ balls, $c$ of which are white, then you get a bag of $b + d$ balls, $a + c$ of which are white. – Qiaochu Yuan Oct 2 '12 at 1:15
up vote 10 down vote accepted

Yes, the inequality holds. One standard approach to proving that $x\lt y\,$ is to show that $y-x\gt 0$.

Apply this to $x=\dfrac{a}{b}$ and $y=\dfrac{a+c}{b+d}$.

The difference is $\dfrac{a+c}{b+d}-\dfrac{a}{b}$, which simplifies to $\dfrac{bc-ad}{(b+d)b}$. But $bc\gt ad$ follows from our initial inequality.

The same method works for showing that $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.

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Hint $\ $ The middle term, known as the mediant, is the slope of the diagonal of the parallelogram with sides being the vectors $(b,a),\ (d,c).\:$ Clearly the slope of the diagonal lies between the slopes of the sides.

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Nice way to describe the reason. – André Nicolas Oct 1 '12 at 23:45

This is called the mediant. The inequality does indeed hold for all $a,\ b,\ c,\ d$.

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The setting where this occurs as a matter of course is simple continued fractions for, in this case, some positive quantity. If the "partial quotient" is some $k$ that is not necessarily equal to $1,$ the two related cases of the next "convergent" are $$\frac{a}{b} < \frac{a+kc}{b+kd} < \frac{c}{d},$$ which is what you get if the two convergents happen to be in increasing order, otherwise $$ \frac{c}{d} > \frac{c + ka}{d + kb} > \frac{a}{b}$$ where the inequality signs need to be massaged as the two convergents happen to be in decreasing order. Anyway, both displayed inequalities are true.


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