Counterexample? Other Alternative proofs? There exists real numbers $a$, and all real numbers $b$, such that $a-b^2&lt;0$
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 $a$?
Whenever $a&lt;0$, 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 $a&lt;0$, 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 $a&lt;-b^2$, then the statement is true. But since it is for Natural numbers, $a&gt;-b^2$, and thus it is false.
Whenever $a&lt;0$, 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 $a&lt;0$, 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 $a&lt;-b^2$, then the statement is true. But since it is for Natural numbers, $a&gt;-b^2$, and thus it is false.