1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Putnam Problem 2010 A5

  1. Dec 6, 2013 #1
    I am trying to solve past Putnam Competition problems. I think I have a proof for Problem A5 of the 2010 exam. Here is the set of all problems: http://inside.mines.edu/~dlarue/putnam/exams/2010.pdf

    The solution given required three lemmas to prove, so I wonder whether my proof is flawed (it seems too simple, and the only group property I used was existence of a (right) identity). So can you tell me if my proof is correct?

    Let [itex]\mathbf{e}[/itex] be an identity of the group [itex]G[/itex]. Suppose there exists a vector [itex]\mathbf{a}[/itex] in [itex]G[/itex] such that [itex]\mathbf{a}×\mathbf{e}≠\mathbf{0}[/itex].
    Then [itex]\mathbf{a}×\mathbf{e} = \mathbf{a}\ast\mathbf{e}=\mathbf{a}[/itex], therefore [itex]\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right) = \mathbf{a}×\mathbf{a} = \mathbf{0}[/itex].
    But by properties of the cross product, [itex]\|\mathbf{a}×\left(\mathbf{a}×\mathbf{e}\right)\|=\| \mathbf{a}\|\| \mathbf{a}×\mathbf{e}\| ≠ \mathbf{0}[/itex] since neither [itex]\mathbf{a}[/itex] nor [itex]\mathbf{a}×\mathbf{e}[/itex] are [itex]\mathbf{0}[/itex].

    The obtained contradiction shows that [itex]\mathbf{a}×\mathbf{e} = \mathbf{0}[/itex] for all [itex]\mathbf{a}[/itex] in [itex]G[/itex]. Therefore, every vector in [itex]G[/itex] is proportional to the vector [itex]\mathbf{e}[/itex]. But this means that all vectors lie on the same line, and therefore for every [itex]\mathbf{a},\mathbf{b}[/itex] in [itex]G[/itex], we have [itex]\mathbf{a}×\mathbf{b}=\mathbf{0}[/itex].
    Last edited: Dec 6, 2013
  2. jcsd
  3. Dec 6, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your proof looks sound to me (and very neat).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted