Quantifier question $!\exists x ! \exists y \forall w(w^2>x-y)$

Question

I have a question about the following quantified sentence if it is true or false. $!\exists x ! \exists y \forall w(w^2>x-y)$ for the real numbers

I think this this is true because if take two negative numbers $x<y$ then it will work Like if you have $-8-(-7)$ $x=-8$ $y=-7$

They exist but they're not unique. For example $x=0$ and $y=1$ works for all $w$. — CyclotomicField, May 23 '20 at 22:17

score 2 · Accepted Answer · answered May 23 '20 at 22:36

2

The question asks whether there is a unique pair $x, y$ such that the square of any real number is greater than the difference $x - y$. This is clearly false: there are infinitely many pairs $x, y$ such that $x - y < 0$ (just choose any pair such that $x < y$) so that, for every real number $w$, we have $w^2 \geq 0 > x - y$.

answered May 23 '20 at 22:36

twosigma

3,244

Ok thank you I get it – Fernando Martinez May 24 '20 at 23:00

Quantifier question $!\exists x ! \exists y \forall w(w^2>x-y)$

1 Answers1