Proof of the Simple Theorem: x^2 >= 0 for Real x

[Howers]

Theorem: Suppose that x is real, then x^2 > or = 0.
Incorrect Proof: Suppose not. Then x is real and x^2<0. Consider x=3. 9<0, a contradiction. Therefore x^2 >=0.

 

[Werg22]

I'm sorry, but what exactly is the brain teaser at hand here?

 

[Howers]

I'm sorry, but what exactly is the brain teaser at hand here?

Why the proof is incorrect, even though seemingly it looks like it is.

 

[Werg22]

I don't see how could this fool anyone... Given the assumption P -> -Q , --Q(3) does not give rise to P -> Q. One needs to show that the assumption P -> -Q gives rise to --Q (a statement for all x being considered, not just 3) and therefore conclude P -> Q by reductio ad absurdum.