Primarygun Posted September 13, 2004 Author Share Posted September 13, 2004 I heard that algebra is a constant for all dimensional space. I know geometry is rejected, and what other are also constants? Link to comment Share on other sites More sharing options...
pulkit Posted September 13, 2004 Share Posted September 13, 2004 nope. i think an average 7th year can solve a simple problem is group theory,. I would definately like to know what you call simple in that case. Even the basic examples of groups such as symmetric groups and quternian groups would be hard to explain. Link to comment Share on other sites More sharing options...
bloodhound Posted September 13, 2004 Share Posted September 13, 2004 i dont know. something like showing that a cyclic group with finite order is Abelian. or something like that. Link to comment Share on other sites More sharing options...
pulkit Posted September 13, 2004 Share Posted September 13, 2004 I agree that the proofs to such results may appear to most to be quite trivial, but they certainly don't constitute any substantial part of group theory. The fact is you won't be able to prove something that you can't fully understand, and to make a school student understand and appreciate fully the beauty of group theory is far from an easy task. As I said earlier, even the simplest examples are beyond the scope of school curiculum. Link to comment Share on other sites More sharing options...
Dave Posted September 13, 2004 Share Posted September 13, 2004 i dont know. something like showing that a cyclic group with finite order is Abelian. or something like that. Think about it: if you'd been given that problem at A-level, you'd probably have gone "wtf?!?" We've had a couple of terms to get used to the idea of university mathematics, and it's a completely different style to what you get at A-level. It took me long enough to get my head around delta-epsilon arguments, and I'm still not sure I completely understand them. Link to comment Share on other sites More sharing options...
Dapthar Posted September 14, 2004 Share Posted September 14, 2004 It took me long enough to get my head around delta-epsilon arguments, and I'm still not sure I completely understand them.When you get into more rigorous proof classes (Real Analysis, Topology, etc.), [math]\epsilon-\delta[/math] arguments will become the easiest type of proof you will encounter. Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 14, 2004 Share Posted September 14, 2004 Maybe if he starts doing that today, he might be done in 20 years or so. It will probably take more time than that Mandrake Link to comment Share on other sites More sharing options...
pulkit Posted September 14, 2004 Share Posted September 14, 2004 Just out of inquistiveness, roughly how many people understand/know the proof of fermat's last theorem ? Link to comment Share on other sites More sharing options...
bloodhound Posted September 14, 2004 Share Posted September 14, 2004 lol. that set would be a null set. Link to comment Share on other sites More sharing options...
pulkit Posted September 14, 2004 Share Posted September 14, 2004 I heard its a 250 page proof. So much for simplicity ! Theres a similar goliath of a proof in group theory for the theorem "all finite simple non-abelian groups are of even order". Link to comment Share on other sites More sharing options...
bloodhound Posted September 14, 2004 Share Posted September 14, 2004 is the contrapositive of that statement true? what about its converse? Link to comment Share on other sites More sharing options...
pulkit Posted September 14, 2004 Share Posted September 14, 2004 Converse is not true. Contra-positive : I don't understand what exactly you mean by that. Link to comment Share on other sites More sharing options...
bloodhound Posted September 14, 2004 Share Posted September 14, 2004 the contrapositive of that statement (in my knowledge) would be "all finite simple Abelian groups are of odd order" Link to comment Share on other sites More sharing options...
pulkit Posted September 14, 2004 Share Posted September 14, 2004 Again i don't think thats true Link to comment Share on other sites More sharing options...
stevem Posted September 14, 2004 Share Posted September 14, 2004 The theorem that Pulkit is referring to is often known as the Odd Order Theorem and is Feit, W. and Thompson, J. G. "Solvability of Groups of Odd Order." Pacific J. Math. 13, 775-1029, 1963 As its title says, the paper proves that all groups of odd order are soluble (the British version of solvable) and as an immediate consequence you get the fact that a finite simple non-cyclic group has even order. Humorous proofs in maths are hard to come by but this impressive theorem is the source of one that, as students, we thought of many years ago. Theorem n! is even for all positive integers n Proof Suppose, for a contradiction, n! is odd. It follows that the symmetric group [math]S_n[/math], which has order n!, is soluble by the Odd Order Theorem. But for n>4 this is well-known to be false, so n! must be even. This leaves the case where n<5 which can be checked individually Link to comment Share on other sites More sharing options...
bloodhound Posted September 14, 2004 Share Posted September 14, 2004 do u know who proved that theorem first? Link to comment Share on other sites More sharing options...
stevem Posted September 14, 2004 Share Posted September 14, 2004 Er, which theorem are you referring to? Link to comment Share on other sites More sharing options...
Dave Posted September 14, 2004 Share Posted September 14, 2004 When you get into more rigorous proof classes (Real Analysis, Topology, etc.), [math]\epsilon-\delta[/math'] arguments will become the easiest type of proof you will encounter. yay Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 15, 2004 Share Posted September 15, 2004 I heard its a 250 page proof. So much for simplicity ! Theres a similar goliath of a proof in group theory for the theorem "all finite simple non-abelian groups are of even order". Yeah but it is whole theory more general than just FMT. The FMT is just a consequence of whole the theory set up by the guy. There are actually people understanding all of this theory since they pointed out an error in the original reasoning and helped to repair it. In any case it is not like on the first page you see proof : and then 250 pages further you see QED or something like that. It is an entire theory with many results. Mandrake If i remember correctly it was Andrew Wiles who proved FMT in 1994 and corrected his proof in 1995. Link to comment Share on other sites More sharing options...
Primarygun Posted September 15, 2004 Author Share Posted September 15, 2004 What do you do(think) first when you need to prove. Think? Or just applied the knowledge that seems to be the most optimum to the question?> Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 15, 2004 Share Posted September 15, 2004 To whom are you wondering that ? Mandrake Link to comment Share on other sites More sharing options...
bloodhound Posted September 15, 2004 Share Posted September 15, 2004 If i remember correctly it was Andrew Wiles who proved FMT in 1994 and corrected his proof in 1995. i meant who proved that theorem about groups first. Link to comment Share on other sites More sharing options...
pulkit Posted September 15, 2004 Share Posted September 15, 2004 Walter Feit and John Thompson Link to comment Share on other sites More sharing options...
matt grime Posted September 15, 2004 Share Posted September 15, 2004 The theorem that Pulkit is referring to is often known as the Odd Order Theorem and is Feit' date=' W. and Thompson, J. G. "Solvability of Groups of Odd Order." Pacific J. Math. 13, 775-1029, 1963 As its title says, the paper proves that all groups of odd order are soluble ([i']the British version of solvable[/i]) and as an immediate consequence you get the fact that a finite simple non-cyclic group has even order. Humorous proofs in maths are hard to come by but this impressive theorem is the source of one that, as students, we thought of many years ago. Theorem n! is even for all positive integers n Proof Suppose, for a contradiction, n! is odd. It follows that the symmetric group [math]S_n[/math], which has order n!, is soluble by the Odd Order Theorem. But for n>4 this is well-known to be false, so n! must be even. This leaves the case where n<5 which can be checked individually n=1 anybody? Link to comment Share on other sites More sharing options...
stevem Posted September 15, 2004 Share Posted September 15, 2004 n=1 anybody? Link to comment Share on other sites More sharing options...
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