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  2. Standard memberDeepThought
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    03 Dec '14 02:50
    The post that was quoted here has been removed
    Had I just given the original problem then there was a real risk the proof was on the relevant Wikipedia page (I just checked it's not) which would have spoilt it a little. There was no massive problem with the initial statement if anti-commutativity alone was enough to prove associativity. What actually alerted me to the fact there could be a problem was that Soothfast hadn't seen it and since he's a practising mathematician could be expected to know a theorem that simple.
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  4. Standard memberSoothfast
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    03 Dec '14 03:26
    Originally posted by DeepThought
    Had I just given the original problem then there was a real risk the proof was on the relevant Wikipedia page (I just checked it's not) which would have spoilt it a little. There was no massive problem with the initial statement if anti-commutativity alone was enough to prove associativity. What actually alerted me to the fact there could be a problem ...[text shortened]... en it and since he's a practising mathematician could be expected to know a theorem that simple.
    Ah, well, there are a lot of simple theorems in the vast field of abstract algebra that I may not be able to recall with a moment's reflection. For the past couple years I've been focused mostly on the analysis branch of mathematics (and also linear algebra). That'll change next year, finally, and I look forward to getting reacquainted with the finer points of group theory and homological algebra!
  5. Standard memberSoothfast
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    03 Dec '14 03:351 edit
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    Yes, that is not unusual. Someone who does research on partial differential equations may not need any abstract algebra, or only the most basic understanding of it at, say, the junior undergraduate level. An algebraic topologist, on the other hand, may know only as much about differential equations as a sophomore math major at UC Berkeley. Maybe less.

    EDIT: But a good mathematician should be able to quickly pick up the needed tools in an unfamiliar area outside his/her specialty with some "further study." Knowing the "language" of mathematics is the main thing. I'm sure it's the same in the vast realm of physics.
  6. Standard memberDeepThought
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    03 Dec '14 04:29
    Originally posted by Soothfast
    Yes, that is not unusual. Someone who does research on partial differential equations may not need any abstract algebra, or only the most basic understanding of it at, say, the junior undergraduate level. An algebraic topologist, on the other hand, may know only as much about differential equations as a sophomore math major at UC Berkeley. Maybe less.
    ...[text shortened]... anguage" of mathematics is the main thing. I'm sure it's the same in the vast realm of physics.
    Up to a point, there are areas of physics where there's just vast amounts of empirical stuff to know, and semi-empirical results. Something like medial physics (probably I know nothing about it, other than some things about how the kit works) would be difficult to get into because they probably need to know about physiology and stuff. In theoretical physics it's not so much of a problem - it's just being aware of what's been done in the field.
  7. Standard memberDeepThought
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    03 Dec '14 04:32
    Because this was buried under a pile of posts I'd better repeat it. There was not enough information in the OP to solve the problem (probably) the revised problem is:

    Let S be a set which is closed under the product * and has the following properties:

    Anti-commutivity a*b = -b*a
    If a and b are different elements then a*b != 0
    a*b != a and a*b !=b for all a and b.

    Show that S is not associative under '*'
  8. Standard memberDeepThought
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    03 Dec '14 04:382 edits
    Originally posted by Soothfast
    Ah, well, there are a lot of simple theorems in the vast field of abstract algebra that I may not be able to recall with a moment's reflection. For the past couple years I've been focused mostly on the analysis branch of mathematics (and also linear algebra). That'll change next year, finally, and I look forward to getting reacquainted with the finer points of group theory and homological algebra!
    A major factor was that you'd mentioned closure of the set and I started thinking about why you might think that that was necessary. Also Adam Warlock's Edited post was a contributor. He may have solved the problem that I expected to be solved and then spotted the problem. The base problem was the vector cross product in three dimensions (which the revised problem is no longer identical to) which should be a big enough clue. I rehashed it because people could probably solve it from general knowledge, but wasn't careful enough with the rehash.

    I found co-homology easier to follow mostly because you can just use differential geometry - although the only homology I know is from Geometry in Physics by Nakahara - and I don't think he quite realises that other people find these things hard...
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  10. Standard memberadam warlock
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    04 Dec '14 05:53
    Originally posted by DeepThought
    Also Adam Warlock's Edited post was a contributor. He may have solved the problem that I expected to be solved and then spotted the problem.
    That was exactly it I realized that I need an initial assumption that wasn't explicitly stated in the problem.

    Even with the extra assumption now provided I'm not sure that I can prove the solution to the problem.
  11. Standard memberDeepThought
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    04 Dec '14 17:28
    Originally posted by adam warlock
    That was exactly it I realized that I need an initial assumption that wasn't explicitly stated in the problem.

    Even with the extra assumption now provided I'm not sure that I can prove the solution to the problem.
    There's enough information now. Alternatively, show that the vector cross product is not associative, which was the inspiring problem.
  12. Standard memberDeepThought
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    05 Dec '14 16:18
    Originally posted by adam warlock
    That was exactly it I realized that I need an initial assumption that wasn't explicitly stated in the problem.

    Even with the extra assumption now provided I'm not sure that I can prove the solution to the problem.
    Although clearly 0*a = 0 for all a in the set.
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