Proof Techniques of Subsets

Question

How many ways could one proof that the natural numbers are a subset of the integers, and what is the most simple proof technique?

Answer 1

The standard method for showing $A \subseteq B$ is to take $a \in A$ and show that $a \in B$ using the definition or properties of the sets.

If $A = \Bbb N$ is the set of natural numbers (and is defined as the set of all non-negative integers) and $B = \Bbb Z$ is the set of integers, then if $a \in A$, we know that $a$ is a non-negative integer. Therefore it is an integer and $a \in B = \Bbb Z$.

Answer 2

Depending on how integers are constructed versus how natural numbers are constructed in your system, there may be many ways to prove this. In the end, all constructions use the reasoning in Alexis Olson's response.

Answer 3

Well... for kicks and giggles I went to wikipedia to see how the integers are constructed and read:

Therefore, in modern set-theoretic mathematics a more abstract construction, which allows one to define the arithmetical operations without any case distinction, is often used instead. The integers can thus be formally constructed as the equivalence classes of ordered pairs of natural numbers $(a,b)$.

i.e. $(a,b) \equiv (c,d) \iff a+d = b+c$

sooo..... $(n,0) \equiv (m,0) \iff n+0 = m+ 0 \iff n = m$ so the map $f:\{(n,0)\}\subset \mathbb Z \rightarrow \mathbb N; f((n,0)) = n$ is injective. If $m \in \mathbb N$ then $(m,0) \in \mathbb Z$ so $f$ is surjective. So $\mathbb N \equiv \{(n,0)\} \subset \mathbb Z$.

.... which I suppose is a far cry from simple....