Consider the following incorrect theorem: [itex]∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)[/itex]
[Translation (not part of the original problem statement): There is at least an [itex]x∈ℝ[/itex] such that, for every [itex]y∈ℝ[/itex], [itex](xy^2 = y-x)[/itex].]
What's wrong with the following proof?
Let [itex]x = y(y^2+1)[/itex], then
[itex]y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2[/itex]
[itex]1. (xy^2 = y-x)[/itex]
[itex]2. x = y(y^2+1)[/itex]
[itex]3. y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2[/itex]
The Attempt at a Solution
Since the first equation is to be proven and the third equation seem to be correct, i think that the problem lies in the second.
I have transformed the theorem as follow:
[itex][∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)] = [∃x(x∈ℝ∧∀y(y∈ℝ→(xy^2=y-x))] [/itex]
From this, i thought that since one of the things to prove is that there is at least an actual x that is true for all y, the substitution done in equation 2 is not correct (since x is substituted not with an actual value but a free variable).
But I'm not sure if this is really the reason for why the proof is incorrect. Any help will be appreciated.