$\mathbb{Z}\times\mathbb{Z}\not\subseteq\mathbb{Z}\times\mathbb{N}$ I was asked to prove the following statement:
$\mathbb{Z}\times\mathbb{Z} \not\subseteq \mathbb{Z}\times\mathbb{N}$.
Intuitively seems simple enough, but I have not been trained in rigorously proving inclusion cases with cartesian products.
Your help will be greatly appreciated and thank you beforehand for you welcome assistance.
This is as much as I have been able to do:

 A: For any three sets $A$, $B$, and $C$ with $A \neq \emptyset$, $A \times B \subseteq A \times C$ if and only if $B \subseteq C$. So, since $\mathbb{Z} \neq \emptyset$ and $\mathbb{Z} \not\subseteq \mathbb{N}$, $\mathbb{Z} \times \mathbb{Z} \not\subseteq \mathbb{Z} \times \mathbb{N}$.
A: Hint: $$(1,-1)\in\Bbb Z\times \Bbb Z.$$

 But $$(1,-1)\notin \Bbb Z\times \Bbb N.$$

A: All you need is a single counter example.
$A \subset B$ means for all $x \in A$ then $x \in B$.  So $A\not \subset B$ means there exists at least one $a\in A$ but $a \not \in B$.
So to show $\mathbb Z \times \mathbb Z \not \subset \mathbb N \times \mathbb N$ all you have to do is find one $(a,b) \in \mathbb Z\times \mathbb Z$ where $(a,b)\not \in \mathbb N \times \mathbb Z$.
That is find just one $a,b$ pair where $a \in \mathbb Z, b \in \mathbb Z$ but either $a \not \in \mathbb N$ or $b\not \in \mathbb Z$.
Well, obviously $b\in \mathbb Z$ and $b\not \in \mathbb Z$ is impossible, but surely finding an $a\in \mathbb Z$ where $a \not \in \mathbb N$ is possible, isn't it?
You just need ONE example.
So how about $(-27, 534)$?   $-27\in \mathbb Z$ and $534 \in \mathbb Z$ so $(-27,534)\in \mathbb Z \times \mathbb Z$.  But $-27 \not \in \mathbb N$ so $(-27, 534) \not \in \mathbb N \times \mathbb Z$.  ANd that's it.  It is NOT the case then every $(a,b)\in \mathbb Z \times \mathbb Z$ is an element of $\mathbb N \times \mathbb Z$ because we find one where that wasn't true.
So $\mathbb Z\times \mathbb Z \not \subset \mathbb N \times \mathbb Z$
That's all there is to it.
