Does function: $ f: \mathbb{Z} \to \mathbb{Z} $ exist, such that this statement is true: $$(\forall{x} \in \mathbb{Z}:f(x) \geq 2)\Rightarrow(\forall{x}\in \mathbb{Z}:f(x)<10)$$ and this statement is false: $$ \forall{x} \in \mathbb{Z}:(f(x) \geq 2\Rightarrow f(x)<10) $$ Now I understand that the second statement is stronger. But I am not sure how I would go about proving whether the function exists or not on paper.
Any help would be greatly appreciated.
Recall the following truth table for the conditional $p\Rightarrow q$:
p q | ( p => q )
T T | T
T F | F
F T | T
F F | T
You see that you can make an implication of the form $p\Rightarrow q$ true by choosing $p$ to be false (and $q$ to have any truth value). To make the claim $$(\forall{x} \in \mathbb{Z}:f(x) \geq 2)\Rightarrow(\forall{x}\in \mathbb{Z}:f(x)<10)$$ true, I can thus choose a function $f$, such that $\exists x\in\Bbb{Z}\colon f(x)<2$ is a true statement, i.e. the function should take a value less than $2$ at at least one integer point. You also want $$\forall{x} \in \mathbb{Z}:(f(x) \geq 2\Rightarrow f(x)<10)$$ to be false. I can make the conditional $p\Rightarrow q$ false by choosing $p$ to be true and $q$ to be false, so I need at least one integer point $x$, such that $f(x)\geq 2$ is true, but $f(x)<10$ is false (i.e. $f(x)\geq 10$ is true). This can be done by simply choosing a point $x$, such that $f(x)\geq 10$.
Summa summarum, I need a point $x_1$, such that $f(x_1)<2$, and another point $x_2$, such that $f(x_2)\geq 10$. Any function with these properties works.