Fail in the reasoning? I'm trying to prove that this is a true statement:
$$\forall x\in\Bbb R:2<x<3\implies3<\frac{2x}{x-1}<4$$
My reasoning is if $3<\frac{2x}{x-1}$ adding to that the restriction $1<x-1$ we have $3(x-1)<2x$ so $3x-3<2x$ thus $x<3$, since the restriction $1<x-1$ doesn't violate $x<3$ the conclusion is true.
The same procedure can be done to $\frac{2x}{x-1}<4$
My justification is that if we read the proof backwards it will be 100% true, so there's enough hope for it to be true.
However my friend violently dismissed this reasoning without given me a reason.
So who's true?

I wont solve your simple algebra question here, only the reasoning question:
Your friend seems to be right here. If you want to prove a proposition like $p \rightarrow q$ you can assume $p$ to be true and then deduce $q$, but you cannot assume $q$ to be true and then deduce $p$.
If by "read the proof backwards" you mean that all arguments of your proof are "if and only if" then you managed to prove $q \leftrightarrow p$ and of course that implies $p \rightarrow q$. But it doesnt appear to be the case in your example.
