If a statement has direct proof, it can not be proved by contradiction. Is it true? I want to know if a question has direct proof: can we once again prove by contradiction method too?
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Take for example this direct proof of the irrationality of $\sqrt{2}$: http://mathoverflow.net/questions/32011/direct-proof-of-irrationality. I'm sure you know the proof by contradiction. – forallepsilon Aug 26 '14 at 04:22
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Suppose that $P$ is a statement and $\sigma$ is the proof of proving $P$ directly. Now consider the proof which begins by assuming $\lnot P$, then repeating $\sigma$ to obtain $P$. Contradiction.
The other direction is the impossible one: if you have a proof by contradiction for $P$, it might not be possible to obtain a direct proof for $P$.
Asaf Karagila
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Proposition (Contraposition)
In order to proof a statement $B$ from $A$ it is (classically) the same if you have a direct proof of $B$ starting from $A$ or you proof $\neg A$ starting from $\neg B$ - in the sense of:
$A\longrightarrow B \equiv \neg A \longleftarrow \neg B $
(where for $F$ and $G$ sentences, $F \equiv G$ means that $F$ and $G$ are logically equivalent - i.e. $F$ is true if and only if $G$ is true).
Proof:
Say $A\longrightarrow B$ is true. (*)
We want to show, that then also $\neg A \longleftarrow \neg B $ must be true.
So we assume $\neg B$ is true and show $\neg A$ must be true:
If $\neg A$ would not be true, we would (classically) have $A$ is true. But then by (*) it would follow (using modus ponens) that $B$ is true - contradicitng that $\neg B$ is true.
The opposite direction is shown exactly analogue..
Say $\neg A \longleftarrow \neg B $ is true. (**)
We want to show, that then also $A \longrightarrow B $ must be true.
So we assume $A$ is true and show $B$ must be true:
If $B$ would not be true, we would have $\neg B$ is true. But then by (**) it would follow (using modus ponens) that $\neg A$ is true - contradicitng that $A$ is true. $\diamond$
FWE
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