3 ms·
I feel for the guy. At the same time one can argue that if len ( Proof1 ) < len ( Proof2 ), where Proof1 and Proof2 are of the same theorem, then Proof1 impos
by udev 12y ago
I feel for the guy.
At the same time one can argue that if len ( Proof1 ) < len ( Proof2 ), where Proof1 and Proof2 are of the same theorem, then Proof1 imposes less cost, and hence is more valuable, since it will be easier to teach, use. etc.
- trhway 12y agoa frequent situation is that the first proof is really long and convoluted. Later, after building up of much of a theory construction blocks around, the proof becomes "simpler" (i.e. simpler combination of well established in that area "blocks") and that is what frequently taught later to students and to "outsiders". For such first long proofs among the best approaches is to throw it to the pack of hungry wolves ... err ... students and first years PhD-s and let them gnaw on it (with presenting their progress on weekly department seminars - thus saving time to more valuable members of the department and providing the students with real-life experience of a mathematician) Well, at least this is how it was done back then at our University (in Russia :).
- wetmore 12y agoThis would be relevant if there was another proof of the ABC conjecture available.
- udev 12y agojust take len ( Proof2 ) = infinity