Counterexample? Other Alternative proofs?

Question

There exists real numbers $a$, and all real numbers $b$, such that $a-b^2<0$

I know you can use a counterexample of $a=5$, and $b=0$, but are there alternative proofs?

What about if there exists all natural numbers $a$?

Answer 1

Whenever $a<0$, then the statement is true. It is asking for a value $a$ that exists to prove this statement right for the very first part.

since $-b^2\leq0$, with $a<0$, we can always get a value that is lower than $0$. For example let $a=-0.1$ and $b=0$, then for the first statement is true.

There exists an $a$, for all $b$s.

As for the second statement, it's obviously false as it asks for $a\in\mathbb{N}$, and we just proved that if $a<-b^2$, then the statement is true. But since it is for Natural numbers, $a>-b^2$, and thus it is false.