Simple Proof - ∃n∈ Z: ∀k ∈ Z: n < k How to prove these 2 (one is true, the other one is false):

*

*∃n∈ Z: ∀k ∈ Z: n < k


*∀n∈ Z: ∃k ∈ Z: k < n
where Z = {0,±1,±2,...}
EDIT1: What I have so far:
.
∀n ∈ Z: ∃k ∈ Z: k < n
Pose n = x et k = x - 1,
For any x ∈ Z, there will always be a term inferior to x such as:
if n = -665, k = -665 - 1 =-666 where k < n
Indeed, ∀n ∈ Z: ∃k ∈ Z: k < n is true.
.
∃n ∈ Z: ∀k ∈ Z: n < k
Pose n = x and k = y for any x,
There will exist x < y
But, limite of Z is +-∞
Then, for any x, there will always exist y = x-1 < x.
This proposition is false.
.
I know this is not working, i'm new to this
EDIT2:
To prove the second part, i would have to prove the opposite.
∃n ∈ Z: ∀k ∈ Z: n < k would be ¬(∃n ∈ Z: ∀k ∈ Z: n < k) = ∀n∃k: k < n

Observe that $$t\in \Bbb Z \iff \{t+1 \in \Bbb Z\} \cap \{t-1 \in \Bbb Z\}\tag 1$$
An implication of this is that $\Bbb Z$ has no greatest or least element.
We can see this by assuming $t$ is the greatest (*least) element of $\Bbb Z$, and noticing that $(1)\implies  t+1(^*t-1)\in\Bbb Z$ and that $t+1>t \ (^*t-1<t)$ and so $t$ is not the greatest (*least) element. By contradiction, there must be no such element.

The first statement asks whether there exists an $n\in\Bbb Z$ such that every $k\in\Bbb Z$ is greater than it. Such an $n$ would be the least element of $\Bbb Z$, but as stated earlier, it doesn't have one, and this statement is false!

The second statement asks whether for every $n\in \Bbb Z$, there exists a $k\in\Bbb Z$ less than it. Indeed, since there is no least element of $\Bbb Z$,  we may easily find such an element. $(1)$ identifies that picking $k=n-1$ works fine here.
