Quantifier difference

Question

What s the difference between $ n \in Z \implies n(n+1) =2k $ such that $k \in Z$ and $ \forall n \in Z \implies n(n+1) =2k $ such that $k \in Z$

Is this true: $ (n \in Z \implies n(n+1) =2k $ such that $k \in Z) \implies \forall n \in Z; n(n+1) =2k $ such that$ k \in Z$

Answer 1

Is this true that :

$(n ∈ \mathbb Z ⟹ \dfrac {n}{(n+1)} = 2k ) ⟹ ∀n ∈ \mathbb Z \dfrac {n}{(n+1)} = 2k \text { , for } k ∈ \mathbb Z$ ?

The formula :

$\dfrac {n}{(n+1)} = 2k$

is about two unspecified numbers $n$ and $k$; it can be either true or false, according to the values we assign to them.

Specifically, the formula is true only for $n=k=0$.

This means that the consequent : $∀n ∈ \mathbb Z \dfrac {n}{(n+1)} = 2k$ is false.

Thus the original formula is false for $n=k=0$ (because in that case we have $\text T \to \text F$, which is $\text F$) and true in all other cases (because $\text F \to \text F$ is $\text T$).