The domain for x, y, z is real numbers.
i) $\forall x \exists y (y^2<x)$ FALSE counterexample: $x=0$
ii) $\forall x \exists y (y^3<x)$ TRUE
iii) $\forall x\exists y \forall z ((y>0) \land ((z^2<y) \rightarrow (z^2+1<x^4)))$ (Not sure how exactly x,y,z relate to each other but from my current understanding it is False) Update: It's false because the second part of the if-then statement, $(z^2+1<x^4)$ does not always hold true for all $x$ and for all $z$ in the reals.
iv) $\exists y \forall x (y^2<x)$ FALSE counterexample: all $x \lt 0$
please let me know if my answers are right or wrong, if wrong, please provide an explanation. For iii) please provide an explanation in either case.
Thanks!
i) false — correct; ii) true – correct.
iii) says: for every $x$, there's $y > 0$ such that $$\forall z (z^2 < y \to z^2 + 1 < x^4) \tag{*} $$ Let $x = 1/2$, the statement claims there is a $yy > 0$ such that (*) is true. But then for some/any $z$ with $z^2 < y$ (and there are such $z$), we would have $1 \le z^2 + 1 < \frac 1 {16}$. So iii) is false.
iv) is false, and your counterexample works.