Mediant is Between/Corollary 2/Proof 4

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:

$\dfrac a b < \dfrac {c + \epsilon} d$

By Mediant is Between:

$\dfrac a b < \dfrac {a + c + \epsilon} {b + d} < \dfrac {c + \epsilon} d$

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:

$\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$

$\blacksquare$