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Maybe I'm just a sucker for group theory but that strikes me as too trivial for any mathematician. In that same vein, I'd suggest proving that a group G is abe

Maybe I'm just a sucker for group theory but that strikes me as too trivial for any mathematician. In that same vein, I'd suggest proving that a group G is abelian if and only if (ab)^2 = a^2.b^2, where a, b are in G. It's not as straightforward as your question in the sense that proving it requires a little idea that you have to come up with.

stephencanon 14y ago
I don't really see this as being any less trivial than proving inverses are unique. Both require writing down the statements, and a couple applications of the group axioms.
- Locke1689 14y ago
  I like my phrasing a little bit more because it at least requires knowledge of the definition of a group hom. Fizzbuzz also requires little more than an understanding of for loops and if statements, so I'd say they're kind of similar.
  - stephencanon 14y ago
    On the other hand, I know a couple extremely good analysts who would stare blankly at you until you reminded them of the definition of a homomorphism =) To be clear, however: I think these are pretty good analogues of FizzBuzz.
dllthomas 14y ago
That seems too easy. (ab)^2 = aabb a^2 b^2 = abab ab = ba iff G is commutative, so for an Abelian group we can substitute for the middle bit aabb = a(ab)b = a(ba)b = abab Which won't hold if G is not commutative. QED I recognize that FizzBuzz is supposed to be easy, but it's supposed to recognize programmers with basic competence; I am not a mathematician. (But maybe I underestimate myself or overestimate some of those with advanced degrees in mathematics?)
- stephencanon 14y ago
  In fairness, if you handed this in for homework in a sophomore algebra course, you likely wouldn't get credit (you've definitely proven the "only if"; the "if" is a bit murky). However, it's not too much of a stretch to clean it up into a proper proof.
  - dllthomas 14y ago
    In what way did the aabb = a(ab)b = a(ba)b = abab fail to prove the if? edit: I did fail to re-state it, but figured it was obvious in the not-really-formal-proof setting. If that's all you meant by "a bit murky" then nevermind.
    - stephencanon 14y ago
      You have very clearly established the following: ab = ba --> (ab)^2 = a^2b^2 It is less obvious that you have proven that: (ab)^2 = a^2b^2 --> ab = ba If I were grading a sophomore algebra class, I would expect to see something along the lines of: Suppose (ab)^2 = a^2b^2. Re-associating gives us a(ba)b = a(ab)b; multiplying on the left and right by the inverses of a and b gives the result. In any domain outside of a sophomore algebra class, I happily accept much briefer and more hand-wavy proofs.
      - dllthomas 14y ago
        Ah, when I read your response I flipped the order of the problem around in recalling it, so had the if and only-if backwards. Yes, I was handwavy there but it seemed clear enough for the setting (which you seem to be granting anyway) - just wanted to be sure I wasn't misunderstanding something. Thanks :)
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Locke1689 14y ago
Wow, I actually edited my comment without reading yours and came up with the same problem.
- hypeibole 14y ago
  Gotta love HN :)

stephencanon 14y ago

I don't really see this as being any less trivial than proving inverses are unique. Both require writing down the statements, and a couple applications of the group axioms.

Locke1689 14y ago
I like my phrasing a little bit more because it at least requires knowledge of the definition of a group hom. Fizzbuzz also requires little more than an understanding of for loops and if statements, so I'd say they're kind of similar.
- stephencanon 14y ago
  On the other hand, I know a couple extremely good analysts who would stare blankly at you until you reminded them of the definition of a homomorphism =) To be clear, however: I think these are pretty good analogues of FizzBuzz.

dllthomas 14y ago

That seems too easy. (ab)^2 = aabb a^2 b^2 = abab ab = ba iff G is commutative, so for an Abelian group we can substitute for the middle bit aabb = a(ab)b = a(ba)b = abab Which won't hold if G is not commutative. QED I recognize that FizzBuzz is supposed to be easy, but it's supposed to recognize programmers with basic competence; I am not a mathematician. (But maybe I underestimate myself or overestimate some of those with advanced degrees in mathematics?)

stephencanon 14y ago
In fairness, if you handed this in for homework in a sophomore algebra course, you likely wouldn't get credit (you've definitely proven the "only if"; the "if" is a bit murky). However, it's not too much of a stretch to clean it up into a proper proof.
- dllthomas 14y ago
  In what way did the aabb = a(ab)b = a(ba)b = abab fail to prove the if? edit: I did fail to re-state it, but figured it was obvious in the not-really-formal-proof setting. If that's all you meant by "a bit murky" then nevermind.
  - stephencanon 14y ago
    You have very clearly established the following: ab = ba --> (ab)^2 = a^2b^2 It is less obvious that you have proven that: (ab)^2 = a^2b^2 --> ab = ba If I were grading a sophomore algebra class, I would expect to see something along the lines of: Suppose (ab)^2 = a^2b^2. Re-associating gives us a(ba)b = a(ab)b; multiplying on the left and right by the inverses of a and b gives the result. In any domain outside of a sophomore algebra class, I happily accept much briefer and more hand-wavy proofs.
    - dllthomas 14y ago
      Ah, when I read your response I flipped the order of the problem around in recalling it, so had the if and only-if backwards. Yes, I was handwavy there but it seemed clear enough for the setting (which you seem to be granting anyway) - just wanted to be sure I wasn't misunderstanding something. Thanks :)
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Locke1689 14y ago

Wow, I actually edited my comment without reading yours and came up with the same problem.

hypeibole 14y ago
Gotta love HN :)