# See if you can prove this:

1. Dec 12, 2009

### tickle_monste

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. Dec 13, 2009

### trambolin

Seems like implicit function theorem proof.

3. Dec 14, 2009

### tickle_monste

Not necessary.

4. Dec 14, 2009

### rochfor1

I don't think this is true. If you consider the equation $$x_1=x_1^2$$ and the function $$f(x_2)=0$$ reduces to the identity 0 = 0, but misses the solution $$x_1=1$$.

5. Dec 15, 2009

### tickle_monste

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.