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 $a, b, c, d$?
Given 4 integers, a, b, c, d > 0, does \frac{a}{b} < \frac{c}{d} imply \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}?
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1You 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. – 2012-10-02
5 Answers
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.
See SIMPLE
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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3Nice way to describe the reason. – 2012-10-01
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 $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.
This is called the mediant. The inequality does indeed hold for all $a,\ b,\ c,\ d$.
That follows from the slightly more general statement:
For $a_1, \ldots, a_n \in \Bbb R$, $b_1, \ldots, b_n > 0$ we have $ \min_i \frac{a_i}{b_i} \le \frac{a_1 + \ldots + a_n}{b_1 + \ldots + b_n} \le \max_i \frac{a_i}{b_i} \, , $ and equality holds if and only if $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots \frac{a_n}{b_n}$.
(Note that the numerators need not be positive.)
Proof: Define $ L =\min_i \frac{a_i}{b_i} \quad, \quad U = \max_i \frac{a_i}{b_i} \, , $ then $ \tag{*} b_i L \le \underbrace{b_i \frac{a_i}{b_i}}_{a_i} \le b_i U \quad \text{for $i=1, \ldots, n$} $ and adding these inequality gives $ L (b_1 + \ldots + b_n) \le a_1 + \ldots + a_n \le U (b_1 + \ldots + b_n) \, . $ Equality holds if and only if equality holds in $(*)$ for all $i$, i.e. if all quotients $a_i/b_i$ are equal.