Q:

Check my logical argument for this proof.

if x is a real number $x \not =\ 1 $, then  there exists y which is also a real number $  ((y+1)  \div ( y-2) ) = x .$
Prove it's converse also.
Logical Argument:
given: $x \not = 1$
Goal:  $  \exists y ((y+1)  \div ( y-2) ) = x .$
Proof by contradiction 
$ \forall y ((y+1)  \div ( y-2) ) \not = x  --------(1) $
Take $y =-1$
which make $(1) $ to  $0$ but $x$ can be zero bcoz the given is  $x \not = 1 $ leads to $0 \not = 0$. Hence a contradiction.
Converse: 
  Take $ x=1 $ then $ (y+1) \div (y-2) = 1 $  For this to be true, $ y+1 = y-2$ which means $+1=-2$ but that's not possible. Hence it's contradiction.
Suggestion of better way is recommended.

A:

Your second argument is fine. Your first one is not. I'll leave it for you to contemplate your solution a little bit. Meanwhile, here's the solution:
In order to prove the existence of $y$ here, you'll have to construct such one. Assuming the contrary isn't constructive in such situations. To construct $y$, scratch on the draft:
$$\frac{y + 1}{y - 2} = x \implies y + 1 = xy - 2x \implies y - xy = -(1 + 2x) \implies y(1-x) = -(1 + 2x)$$ 
And since we are allowed to divide by $1 - x$:
$$y = \frac{2x + 1}{x - 1}$$
Now because we illegally found $y$, it remains is to show that this $y$ verifies the requirement, and verify that $y = 2$ can't happen. Then we are done.

