Is proof by contradiction "same thing" as $A \rightarrow B$ is true when $A$ is false? I encountered earlier today a question "Is the proof by contradiction same as that $A \rightarrow B$ is true when $A$ is false?" continued by "Are they related, then? How?"
I think the answer is "no, they´re not the same thing", and demonstrates some confusion on the subject. However, as I am still more than a little confused about mathematical logic myself, too, I couldn't explain properly why this felt wrong to me. I pointed that proving statement $P$ by contradiction can be formulated symbolically as a $\neg P \rightarrow Q \wedge \neg Q$, and if $P = A \rightarrow B$ (because statements like that were the topic of discussion), we get something ...quite different. (I think I lost my trail of thought here.) However, it (the method of proof by contradiction) does feel like a different thing than the existence of a line in the truth table for $A \rightarrow B$ where $A$ is false and $B$ is true and thus the implication is true. Was I correct in this?
However, afterwards I remembered that $A \rightarrow B = \neg B \rightarrow \neg A$, where the right side is known as proof by contraposition. And the proof by contraposition is "in practice" usually done by assuming that $\neg B$ is true and then showing that $\neg A$ follows. (Because of the equality above, this is same as showing that "$A \rightarrow B$".)
Drawing the truth tables for both $A \rightarrow B$ and $\neg B \rightarrow \neg A$, this line (where $\neg B$ and $\neg A$ are true) does coincide with with $A$ being false and $B$ being true. So maybe one could say that $A \rightarrow B$ is true when $A$ is false is related to (or even "same thing as"?) the proof by contraposition.
Maybe we're both (or just me) confused about the relation of truth tables of statements like $A\rightarrow B$ to actually proving statements like "if A, then B".
Enlighten me, please.
 A: A proof by contradiction is an argument of the form :


*

*assume $A$ and infer $B$;


*then, assume $A$ and infer $\lnot B$,


*and conclude with $\lnot A$.

In words, showing that the assumption leads to a contradiction, licences us to infer the negation of the assumption.
We cam express it in symbolic form (in propositional logic) as :


$(A → B) → ((A → ¬B) → ¬A)$.


We can easily reconcile the two versions noting that, by a "standard" result of a proof system for propositional logic (the Deduction Theorem), we have that :

if $A \vdash B$, then $\vdash A \rightarrow B$

where we have used (as usual) the "turnstile" ($\vdash$) to denote the relation of derivability in a proof system.
Thus, the above argument can be foramlized as follows :


if $\vdash A \rightarrow B$ and $\vdash A \rightarrow \lnot B$, then $\vdash \lnot A$.



How to reconcile it with truth table ?
Having $\vdash A \rightarrow B$ means to have proved the conditional $A \rightarrow B$, and the same with $\vdash A \rightarrow \lnot B$.
Of course we assume that "our logic" is sound, i.e. it does not prove false formulae.
Thus, if both formulae are true, due to the fact that one of $B, \lnot B$ is false, we must conclude that - in order that both conditionals are true - $A$ is false.
A: There is a connection between proving $A \rightarrow B$ and proving "if $A$, then $B$". That's called the deduction theorem, which guarantees that one is true iff the other is. You may want to look that up.
As for your question, I'm a little confused by your wording. A proof by contradiction works by showing, generally, that "if $A$, then $B$" is true, and then showing that $B$ is false, whence $A$ is false. Assuming the deduction theorem, this is the same as showing that $A \rightarrow B$ is true and $B$ is false. So yes, this is related to the idea that $A \rightarrow B$ is true when $A$ is false: given the truth of $A \rightarrow B$ and the falsity of $B$, the only way to jointly maintain this is to assume that $A$ is false.
In case you're unsure, draw the truth table, and then cross-out the row which say that $A \rightarrow B$ is false and the rows which say that $B$ is true. The remaining row will give you that $A$ is false. 
A: A proof by contradiction consists of an argument where you hypothesize something, infer a conclusion which is a contradiction, and then discharge the hypothesis one way or another.
This means either that you infer either two propositions "A" and "$\lnot$A" which qualify as contradictory, or you infer a conjunction which has the form (A$\land$$\lnot$A), or more broadly you infer some formula which is known to consist of the negation of a tautology.
A: Suppose you want to prove statement $X$. If, assuming that $X$ is false leads to a contradiction of the form $Y\land \neg Y$, then, of course, $X$ must be true (by contradiction).
That is certainly the mainstream (i.e. non-instuitionst) view anyway. 
