Question on "Proof by contradiction" I am a little confused whether the following set of arguments can constitute a proof by contradiction:
Required to prove: Statement A is true.
Assumption: Suppose statement A is false.
We then show that if A is false then a (different) statement B is true.
But, statement B is true implies A is true.
Hence, A must be true.
These set of arguments feel a bit bizzare to me and perhaps might be incorrect. I am unable find a mistake.
 A: You start your proof from "let A be false". Now, using "true tools" you prove a particular statement which implies that A has to be true. So to sum up you prove that A is false implies A is true. This is a contradiction. For example:A: there isn't a smallest rational number greater than zero.
Assumption: there is a smallest rational number greater than zero.
So, call him $r$.
Now you can take $r/2$.
So B: given a rational number $r$ greater than zero, there exists another rational number $r/2$ that is greater than zero and smaller than $r$.
So A.
A: This technique is actually quite common.  The answer above from LuckyS is a nice one.  Here's my favorite simple example.

*

*We have three beans, each one either white or black.

*We will prove that there must a pair of beans of the same color.

*Suppose this is not so.

*Then the first two beans are not a pair of the same color, so one is white and the other is black.

*Then if the third bean is white, there is a pair of white beans, and if the third bean is black, there is a pair of black beans.

A: Yes, this is a proof by contradiction, but written in a different way.
You have two statements: $B \implies A$ and $\neg A \implies B$, both of which are true.
If you have come across proofs by contrapositive before, you will know that $B \implies A$ and $\neg A \implies \neg B$ are equivalent. This means that what you really have are the two statements
$$\neg A \implies B \\ \neg A \implies \neg B$$
This is a contradiction, and this is how one typically learns proofs by contradiction. So what you have is essentially the same thing, but with the second statement $\neg A \implies \neg B$ written as the equivalent $B \implies A$.
A: (Posted after a previous answer was accepted.)
We have the tautology:

We can verify this using a truth table. Or we can can prove it using a simplified form of natural deduction as below. We take the negation of proposition $A$ on line 4 to mean that "$A$ is assumed to be false" (a premise). Lines 4-9 are a proof by contradiction that $A$ must be true given the premise on line 1. (Screenshot from my proof checker)

A: The way proof by contradiction works is that by assuming A is false, we can find some logical reasoning that shows that some known false statement B is true.  Sometimes the key to a proof is that the false statement B is "A ^ ¬A", but it doesn't need to be that: B could be "1 = 2" or any other false statement.
Then, having shown that assuming ¬A means that the false statement B actually is true, we state that this is absurd, and conclude that ¬A must actually be false.  Since ¬A is false, A must be true.
