# Given 4 integers, $a, b, c, d > 0$, does $\frac{a}{b} < \frac{c}{d}$ imply $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$?

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 $a, b, c, d$?

• 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

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 $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.

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.

• 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 $a,\ b,\ c,\ d$.

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

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.