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Finding fault in proof [contains logical quantifiers]

  • Thread starter iopz
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  • #1
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Homework Statement



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]​

Homework Equations



[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.
 

Answers and Replies

  • #2
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What's wrong with the following proof?

Let [itex]x = y(y^2+1)[/itex]


This is already wrong. x is a single number and can't depend on y
 
  • #3
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This is already wrong. x is a single number and can't depend on y
I see, thank you!
 

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