Let $a,b,c,d \in \mathbb{N}$ and $\epsilon \in \mathbb{R}$ Let $\epsilon < \frac{a}{b} < \frac{c}{d}$ Does this imply that: $\epsilon \leq \frac{a+c}{b+d}$?
$\epsilon < \frac{a}{b} < \frac{c}{d}$ implies that: $\epsilon \leq \frac{a+c}{b+d}$?
4 Answers
More generally, you can show that if $\frac a b < \frac c d$ then:$$\frac{a}{b}\leq\frac{a+c}{b+d}\leq \frac c d$$
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1The last fraction should be $\frac c d$ right? – 2012-12-26
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1I fixed the typo that mathguy pointed out. I hope you don't mind. – 2012-12-26
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0No problem, and thanks! @Nameless – 2012-12-26
Hint: $$\epsilon\le \frac{a+c}{b+d}\iff b\epsilon+d\epsilon \le a+c$$ Because $\epsilon < \frac{a}{b}$ and $\epsilon< \frac{c}{d}$,... (you don't actually need that $\frac ab<\frac cd$ this way)
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3Indeed $\cfrac {a+c}{b+d}$ is symmetrical in the two terms, so the inequality must also work if $\frac c d < \frac a b$ – 2012-12-26
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0Off course, thank you very much! – 2012-12-26
Suppose that $\dfrac{a}{b}\lt \dfrac{c}{d}$, where $b$ and $d$ are positive real numbers. Then $$\frac{a}{b}\lt \frac{a+c}{b+d}\lt \frac{c}{d}.$$
We prove the half of the above result that you do not need, that $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.
A natural approach is to consider the difference $\dfrac{c}{d}-\dfrac{a+c}{b+d}$, which simplifies to $\dfrac{bc-ad}{d(b+d)}$. The denominator is positive. And since $\dfrac{c}{d}-\dfrac{a}{b}\gt 0$, the numerator is positive.
Remark: You might be interested in other properties of the mediant.
More generally, if $\epsilon < a/b$ and $\epsilon < c/d$, where $a, b, c, d$ are positive reals, then $\epsilon < (r a+s c)/(r b + s d)$, where $r$ and $s$ are any positive reals.
Proof: $\epsilon < (r a+s c)/(r b + s d)$ $\iff$ $\epsilon(r b + s d) < r a+s c$ $\iff$ $r(a-\epsilon b) > s(d \epsilon - c)$ which is true because the left side is positive and the right side is negative.
The reverse inequality is also true, but $\epsilon$ likes being less than other values and is uncomfortable when asked to be greater.