Question

∀x∈R, x²≥0.
What is wrong with this proof?
Suppose not. Then for every real number x, x² < 0. In particular, plugging in x = 3, we would get 9 < 0, which is clearly false. This contradiction shows that for every number x, x²≥0.

Answer 1

the negation of "for all" is "there exists"

Answer 2

the negation u r using is wrong:
\[ \large \neg(\forall x)(P(x))\equiv(\exists x)(\neg P(x)) \]

Answer 3

ooh nice