Is there a way to define addition and multiplication of natural numbers using the language of category theory? Like, one could say that "Addition is the unique functor that satisfies…" and "Multiplication is the unique functor that satisfies…". I would be very interested in such a definition.
Category Theory – How to Define Addition and Multiplication of Natural Numbers?
category-theorynatural numbers
Related Solutions
1) Define $n+0=n$. Assume $n+m$ has been defined. Then define $n+\sigma(m)=\sigma(n+m)$. This defines addition on all ordered pairs of naturals. Suppose there was some alternate addition $+'$ satisfying the above properties. We notice that for pairs $(n, 0)$, $n+0=n+'0$. Assume that for each $n$, we have shown that $n+k=n+'k$ for all $0 \leq k \leq m$. Then $n+\sigma(k)=\sigma(n+k)=\sigma(n+'k)=n+'k$ and so we see that $+, +'$ are the same for all pairs. This gives uniqueness. You will want to prove distributivity of addition for later parts. Try to get this via induction.
2) Pick any $n \in \mathbb{N}$. We proceed by induction on $m$. If $m=0$, then $p=m$. If there exists $p$ so that $n+p=m$, then we observe that $n+ \sigma(p)=\sigma(n+p)=\sigma(m)$. If there exists $p$ so that $n=p+m$, then if $p=0$, we have $n+1=\sigma(m)$. If $p \neq 0$, then the number we need is the natural whose successor is $p$. Try proving that if $p \neq 0$, it has a predecessor. (Hint: induction!)
3) This relation is clearly reflexive. For anti-symmetry, if there exists $p$ so that $a+p=b$ and $p'$ so that $b+p'=a$, then $b+p+p'=b$. You should be able to show this implies $p+p'=0$. If $p \neq 0$ or $p' \neq 0$, this should give you a predecessor of $0$, which is a contradiction and shows that $p=p'=0$ and so $a=a+p=b$. Transitivity is easy, and totality is part 2).
4) This is not hard. Try to work it out for yourself.
5) Your intuition is correct. You can easily show that $\sigma (n) \geq n$ but $n \neq \sigma(n)$.
6) If this were not the case, you would have an infinite descending chain in $\mathbb{N}$. To get a contradiction, you need to show that for each $m \in \mathbb{N}$, there are only finitely many naturals less than $m$. You can do this by induction.
7) You are correct.
8) This is somewhat similar to addition. Define inductively and use your definition to prove what you need.
9) You can do this by induction.
Every preorder $(P,\leq)$ can be regarded as a category. The object set is $P$, and a morphism $x \to y$ exists (and is unique) iff $x \leq y$. These are precisely those categories in which every diagram commutes. But still, it is interesting to apply category theory to preorders. A limit is just an infimum, a colimit is a supremum. In particular, initial (terminal) objects are least (largest) elements. Monotonic maps are just functors. Galois connections are just adjunctions between preorders.
We can define ideals and prime ideals in $P$. If $p \in P$ is an element, one says that $p$ is a meet prime iff the generated ideal ${\downarrow}p$ is a prime ideal, i.e. iff $p$ is not the largest element and $\inf(x,y) \leq p$ implies $x \leq p$ or $y \leq p$.
If $P = \mathbb{N}^+$ and $\leq$ is the relation of divisibility reversed(!), then prime elements are precisely the usual prime numbers. You can omit this reversion by looking at join prime elements.
More generally, if $P$ is the preorder of ideals of a ring $R$, ordered by inclusion, then prime elements are precisely the prime ideals of $R$ in the sense of ring theory.
One can generalize many notions of number theory / ring theory to preorders resp. lattices (see Wikipedia for a start; I have also found many papers by a quick google research). In order to answer your interesting question "That is, can one explain (or at least motivate) deeper ideas in elementary number theory in terms categorical language?" I would like to see an explicit "deeper idea in elementary number theory" first - then we may try to explain it in category-theoretic terms (although I doubt that we will gain anything from that).
Best Answer
"Using category theory" is a bit too general of a question. So there are multiple answers.
There is a notion of a natural numbers object in any suitable category $\mathcal{C}$. One can use the universal property to define addition and multiplication and verify the Peano-axioms internal to $\mathcal{C}$.
The natural numbers with addition are the free monoid on one element. Using plain category theory one can show that the forgetful functor $\mathsf{Mon} \rightarrow \mathsf{Set}$ admits a left adjoint, giving you access to $(\mathbb{N},+)$. Multiplication comes from this universal property by using that $(\operatorname{Map}(\Bbb N, \Bbb N),\circ,\operatorname{id})$ is a monoid.
The natural numbers are the set of isomorphism classes of the category of finite sets. Addition and multiplication come from the symmetric monoidal structures given by disjoint union and cartesian product.
The natural numbers with addition can be regarded as the free living endomorphism, ie. the category generated by the graph with one vertex and one loop, as suggested by @rschwieb. If you also want multiplication, I think you should regard the natural numbers as the free commutative-monoid-enriched category on one object and one loop.