Common methods for proving: Existence proof For every real number x with  $x\neq -1$ there exists a real number y such that $ \frac{y}{y+1}=x $.
$ \forall x\in \Bbb R\setminus (1) \ \ \exists y \in \Bbb R : \frac{y}{y+1}=x \\, x \neq -1 $
$ So \ \ if \ \ I \ \ choose \ \ e.g. \ x= \frac{1}{2} \ \ then \ \ y=1  \ \ such \ \ that \ \ \frac{y}{y+1}=x $
Is this a correct way of proving this? 

Use the transformation of $y\rightarrow z-1$ to change it to for all $x\neq 1$ there exists $z$ such that $\frac{z-1}{z}=1-\frac{1}{z}=x$, which is easily solvable for $z$ in terms of $x$.
