Proof of an $\frac{a}b=\frac{c}d⟺\frac{a+b}{a−b}=\frac{c+d}{c−d}$.

Question

I'm trying to prove this : $\frac{a}b=\frac{c}d⟺\frac{a+b}{a−b}=\frac{c+d}{c−d}$

but going from $\frac{a}b=\frac{c}d$

I already proved it backwards

$\frac{a+b}{a−b}=\frac{c+d}{c−d}$ I multiply each side by (a-b)(c-d)

$ac-ad+bc-bd = ac - bc + ad - bd$ so $2bc=2ad$ which by dividing both sides by $2bd$ gives us : $$\frac{a}b=\frac{c}d$$

any ideas how to proceed ?

Edit : this equivalence only works when $a\neq b $ and $c \neq d$.

Answer 1

Suppose that $a \neq b, c\neq d$, and $b,d \neq 0$. Then, $\frac ab \neq 1$. We are asked to assume that $\frac ab = \frac cd$.

Now, we can do the following:

$$ \frac ab = \frac cd \implies \frac ab + 1 = \frac cd + 1 \implies \frac{a+b}b = \frac{c+d}d\\ \frac ab = \frac cd \implies \frac ab - 1 = \frac cd - 1 \implies \frac{a-b}{b} = \frac{c-d}{d}\\ $$

Divide the first equation by the second: $$ \frac{\frac{a+b}{b}}{\frac{a-b}b} = \frac{\frac{c+d}d}{\frac{c-d}d} \implies \frac{a+b}{a-b} = \frac{c+d}{c-d} $$

This is a proof, provided the conditions that are given in the starting hold true.

On a side note, I think this is called the componendo-dividendo equality (I may be wrong).

Answer 2

I do not think the two statements are equivalent. Take $a=b=c=d=1$ then your first equality holds while the second on is undefined.