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$
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 $\text T$).
