Mediant is Between/Corollary 2/Proof 2

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

\(\ds \dfrac {a + c} {b + d}\)

\(=\)

\(\ds \dfrac a b \times \dfrac {b \paren {a + c} } {a \paren {b + d} }\)

multiplying by $\dfrac a b \times \dfrac b a$

\(\ds \)

\(=\)

\(\ds \dfrac a b \times \dfrac {1 + c / a} {1 + d / b}\)

dividing top and bottom by $a b$

\(\ds \dfrac {a + c} {b + d}\)

\(=\)

\(\ds \dfrac c d \times \dfrac {d \paren {a + c} } {c \paren {b + d} }\)

multiplying by $\dfrac c d \times \dfrac d c$

\(\ds \)

\(=\)

\(\ds \dfrac c d \times \dfrac {1 + a / c} {1 + b / d}\)

dividing top and bottom by $c d$

\(\ds \leadsto \ \ \)

\(\ds \dfrac a b = \dfrac c d\)

\(=\)

\(\ds \dfrac {a + c} {b + d}\)

This article, or a section of it, needs explaining. In particular: How does $\dfrac {a + c} {b + d} = \dfrac a b \times \dfrac {1 + c / a} {1 + d / b} = \dfrac c d \times \dfrac {1 + a / c} {1 + b / d}$ imply that $\dfrac a b = \dfrac c d$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

$\blacksquare$