Logic of the implication in $ε$-$δ$ proofs Im confused by why epsilon delta proofs logically work.
An example is 
Proof: 
Given $ε>0$, choose $δ = {ε\over3}$. For all $x$, if $0<|x−2|<δ$ then $|(3x−1)−5| < ε$.
That last part  if $0<|x−2|<δ$ then $|(3x−1)−5| < ε$ LOOKS A LOT LIKE $P\to Q$ because of the "if then" but yet the proof in the book solves it like as if its $Q\to P$?:
$$\begin{align}|(3x−1)−5| &= |3x−6|\\ &= |3(x−2)|\\ &= 3|x−2|\\ &<3δ\\ &= 3\left({ε\over3}\right) \\ &= ε\end{align}$$
So my question is how come it looks like a $P\to Q$ proof but yet we start with $Q$ to show $P$?

Let's just run through the proof.
We want to prove that, for any $\epsilon > 0$, there exists $\delta > 0$ such that for all $x$, $0 < |x-2| < \delta \implies |(3x-1)-5|<\epsilon$.
Hence, we take an arbitrary $\epsilon > 0$. We now want to prove the existence of a $\delta > 0$ such that the statement "for all $x$ (...)" holds.
Note that $|(3x−1)−5| = |3x−6| = |3(x−2)| = 3|x−2|$. Now we want to construct a $\delta > 0$ such that IF $|x-2| < \delta$, THEN $3|x-2| < \epsilon$. Hence, if we take $\delta = \frac{\epsilon}{3}$, we have reached our goal: $|x-2| < \delta = \frac{\epsilon}{3}$, hence $3|x-2| < \epsilon$.
