Mediant is Between/Corollary 2/Proof 4
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Corollary to Mediant is Between
Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.
Let $\dfrac a b = \dfrac c d$.
Then:
- $\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$
Proof
Let $\epsilon \in \R_{>0}$.
Then:
- $\dfrac a b < \dfrac {c + \epsilon} d$
- $\dfrac a b < \dfrac {a + c + \epsilon} {b + d} < \dfrac {c + \epsilon} d$
By Inequality of Sequences Preserved in Limit, letting $\epsilon \to 0$:
- $\dfrac a b \le \dfrac {a + c} {b + d} \le \dfrac c d$
Since by hypothesis:
- $\dfrac a b = \dfrac c d$
we have:
- $\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$
$\blacksquare$