Can You Assume the Converse in Mathematical Proofs?

[Kamataat]

I didn't want to post this in the logic subforum, because it's really basic...

I know that when proving mathematical statements, one can't assume what he/she is trying to prove. But can one assume its converse?

- Kamataat

 

[dextercioby]

You mean like reductio ad absurdum...?It's by far the most elegant method of mathematics...

Daniel.

 

[matt grime]

The constructivists may disagree that it is the most elegant, and indeed proofs by contradiction are to be frowned upon if they are unnecessary - often doing a proof by contradiction merely shows what we first started with to be true directly, and we've jist added pointless length to it.

You may assume not(X) is true and show that if so then we derive a logical contradiction, hence not(X) is false, X is true. Is that what you're getting at?

 

[honestrosewater]

Do you mean converse or negation? I've only seen converse applied to conditionals: (Q -> P) is the converse of (P -> Q). Note that the reason you "can't" assume the proposition you're trying to prove is only that such an argument is circular; Circular arguments are still valid- they just don't tell you anything new. With one exception- when P and Q are equivalent- you can assume the converse of a proposition without being circular, since [(Q -> P) -> (P -> Q)] is a contingent proposition, i.e., [(Q -> P) therefore (P -> Q)] is an invalid argument; Its counterexample is [Q is false, P is true]).

 

[Kamataat]

...

You may assume not(X) is true and show that if so then we derive a logical contradiction, hence not(X) is false, X is true. Is that what you're getting at?

Yeah, that's what I meant. Thanks!

- Kamataat