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?
 A: 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.
A: Taking into account that $\;x\,,\,\,x-1>0\;$ , we have
$$2<x<3\implies 1<x-1<2\implies 1>\frac1{x-1}>\frac12$$
And now you can work it :
$$\frac{2x}{x-1}=2+\frac2{x-1}\begin{cases}<2+2\cdot1=4\\{}\\>2+2\cdot\frac12=3\end{cases}$$
A: The way you approached this is a good way to figure out how a proof might work.
It's not the same thing as doing the proof, however.
You do have a good reason to expect that if you work all the steps backward, 
each step will justify the next and you will have proved that $3 < \frac{2x}{x-1},$
but for this to be sufficient as a proof, 
you'd have to make sure that all the steps really do work both ways.
The simplest way to approach this is, now that you think you know what the steps are,
actually do write them in reverse, proving each step using the previous steps.
So you start with $2<x<3.$
From $2<x$ you get $1<x-1,$ and from  $x<3$ you get $3x-3<2x,$
therefore $3(x-1)<2x.$  Now, since we already have $1<x-1,$ we can divide
both sides by $x-1$ and we have $3<\frac{2x}{x-1}.$
(Actually it's was sufficient that $x - 1 > 0$ in order for this step to be correct.)
And there you are! (At least for the first half of the inequality.)
If you were paying close attention, you may have noticed that I essentially just
copied all your expressions in the opposite order you wrote them, with reasons why
the expressions I wrote first justify the ones I wrote later.
So you were correct that you could do this, and that it would be a valid proof,
but since  you had not actually done it, your friend was technically
correct that you had not proved the inequality.
