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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: $\epsilon \leq \frac{a+c}{b+d}$?

4 Answers 4

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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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    No problem, a$n$d tha$n$$k$s! @Nameless2012-12-26
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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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    Off course, thank you very much!2012-12-26
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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.

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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.