TITLE: How does proof by contradiction work?
QUESTION [8 upvotes]: I just had a thought from looking at this answer in this thread: For every $\epsilon >0$ , if $a+\epsilon >b$ , can we conclude $a>b$? 
I understand Brian's proof, but I just thought of something. If you prove a hypothesis is untrue, then the logical negation of that hypothesis must be true via proof by contradiction. But why? Can't it be the case that if a hypothesis is false, the negation is also false?

REPLY [3 votes]: Let's say you want to prove $P \implies Q$.
There are some substeps $P \implies P_2 \implies P_3 \implies P_4 \implies Q$.
Assume that $P$ is true and $Q$ is false.
Since $P$ is true, we have $P_2$ is true.
Since $Q$ is false, $P_4$ is false (by contrapositive of the last implication).
Since $P_2$ is true, $P_3$ is also true.
Since $P_4$ is false, $P_3$ is also false (by contrapositive).
So, $P_3$ is both true and false. This is impossible.
So, when $P$ is true, $Q$ cannot be false.