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That'd be great! I'm planning on spending a few weeks full-time on this (8 hours a day, or more if I can handle it). That should be a minimum of one section a d

That'd be great! I'm planning on spending a few weeks full-time on this (8 hours a day, or more if I can handle it). That should be a minimum of one section a day, but hopefully more, even a whole chapter on some days. It doesn't matter how fast, though, as if either falls behind we can just compare those solutions. Email me at technoguyrob[at]gmail[dot]com.

brl 18y ago
At that pace I think I would definitely be holding you back. Maybe I'll try to tackle something a bit less steep, like Mount Fraleigh. Have you already studied some abstract algebra, or would you mainly be learning it through this project?
- technoguyrob 18y ago
  You can still read through my solutions and point out all my errors if you like, though. ;) I've already had a couple classes, but I've never felt intimately comfortable with it. With several programming languages, I don't even need the occasional googling, nearly all the libraries (and of course basic syntax) are already in my head. When doing commutative algebra and Noetherian rings, I found myself looking back at ring theory for various properties about rings. Same goes for homological algebra: I had to keep going back to properties of modules. I understand it in more in terms of theorems and facts rather than mathematical intuition and maturity, and my goal is to change that around.
  - Xlp-Thlplylp 18y ago
    Note on the problem that a group G of even order contains an element of order 2. Partition G into classes {{x,x^{-1}|x\in G}. This is a partition since inverses are unique. For x\in G, call {x, x^{-1}} the class of x. At least one other element x besides the identity has a class of size one. Otherwise, the order of G would be 1 + 2*(# of classes of size 2) which is odd. Hence there is an x with x != 1 and x = x^{-1}; i.e., an element of order 2.
    - technoguyrob 18y ago
      Thanks, that's exactly the kind of comments I was looking for. That was exactly what I thought in my head, but it came out very distorted and ugly on paper. I'll update the solution later. Do you want credit?
      - Xlp-Thlplylp 18y ago
        Credit for this is optional. Another way to see obtain a partition of G is to observe that the relation x ~ y if and only if x = y or x = y^{-1} is an equivalence relation on G. The equivalence classes induce the partition above. The point is that any two equivalence classes {x, x^{-1}},{y,y^{-1}} are either disjoint or equal, and their union is G. But this is obvious (the word 'obvious' means "I thought of it.")

brl 18y ago

At that pace I think I would definitely be holding you back. Maybe I'll try to tackle something a bit less steep, like Mount Fraleigh. Have you already studied some abstract algebra, or would you mainly be learning it through this project?

technoguyrob 18y ago
You can still read through my solutions and point out all my errors if you like, though. ;) I've already had a couple classes, but I've never felt intimately comfortable with it. With several programming languages, I don't even need the occasional googling, nearly all the libraries (and of course basic syntax) are already in my head. When doing commutative algebra and Noetherian rings, I found myself looking back at ring theory for various properties about rings. Same goes for homological algebra: I had to keep going back to properties of modules. I understand it in more in terms of theorems and facts rather than mathematical intuition and maturity, and my goal is to change that around.
- Xlp-Thlplylp 18y ago
  Note on the problem that a group G of even order contains an element of order 2. Partition G into classes {{x,x^{-1}|x\in G}. This is a partition since inverses are unique. For x\in G, call {x, x^{-1}} the class of x. At least one other element x besides the identity has a class of size one. Otherwise, the order of G would be 1 + 2*(# of classes of size 2) which is odd. Hence there is an x with x != 1 and x = x^{-1}; i.e., an element of order 2.
  - technoguyrob 18y ago
    Thanks, that's exactly the kind of comments I was looking for. That was exactly what I thought in my head, but it came out very distorted and ugly on paper. I'll update the solution later. Do you want credit?
    - Xlp-Thlplylp 18y ago
      Credit for this is optional. Another way to see obtain a partition of G is to observe that the relation x ~ y if and only if x = y or x = y^{-1} is an equivalence relation on G. The equivalence classes induce the partition above. The point is that any two equivalence classes {x, x^{-1}},{y,y^{-1}} are either disjoint or equal, and their union is G. But this is obvious (the word 'obvious' means "I thought of it.")