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See if you can prove this:

  1. Dec 12, 2009 #1
    I've got a proof but I'll wait a couple days to post mine to give you guys a chance to take a crack at it.

    Prove that:

    For all equations E(A1, ... , An), solving for Ai is equivalent to finding the function
    f(A1..A(i-1),A(i+1)...An), such that when f is substituted in E in place of Ai, E reduces to an identity (i.e. 0 = 0, 1 = 1, a^2 + b^2 = c^2, etc.)

    Use whatever axioms you'd like.
  2. jcsd
  3. Dec 13, 2009 #2
    Seems like implicit function theorem proof.
  4. Dec 14, 2009 #3
    Not necessary.
  5. Dec 14, 2009 #4
    I don't think this is true. If you consider the equation [tex]x_1=x_1^2[/tex] and the function [tex]f(x_2)=0[/tex] reduces to the identity 0 = 0, but misses the solution [tex]x_1=1[/tex].
  6. Dec 15, 2009 #5
    My wording of the problem was poor, I forgot to account for multiple independent solutions. So, here it is, better worded:
    Finding a solution of Q in the equation is equivalent to finding a function f, such that when f is substituted for Q in the equation, the equation will reduce to an identity.

    Both of your solutions satisfy this property, no other number satisfies this property and no other number is a solution.
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