Why introduce the $p$-adic numbers? My current intuition about the p-adic numbers comes from the following three facts:


*

*You can describe $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ with the $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ groups.

*Hensel's lemma

*We find that the p-adic integers are formal neighborhoods of closed points in $\mathbb{Z}$, so they naturally show up in the deformation theory of arithmetic objects (Schemes over $\mathbb{Z}$)


What are other reasons to consider p-adic integers/numbers?
 A: About 40 years ago, a fellow who was doing C-star Algebras was teasing me about how none of my Number Theory would ever find applications in "the real world". I told him to just wait and see – some day, there would be crucial applications of $p$-adic numbers in Physics. I was kidding, but apparently there is a serious pursuit of $p$-adic quantum mechanics. See here. 
A: Quoting Wikipedia, "the Skolem–Mahler–Lech theorem states that if a sequence of numbers is generated by a linear recurrence relation, then with finitely many exceptions the positions at which the sequence is zero form a regularly repeating pattern. More precisely, this set of positions can be decomposed into the union of a finite set and finitely many full arithmetic progressions.... Its proofs use p-adic analysis." 
A: On one hand, the $p$-adic numbers are extremely natural objects of study: by Ostrowski's theorem every nontrivial absolute value on $\mathbf Q$ is equivalent to either the usual absolute value or the $p$-adic absolute value for some $p$. So the $p$-adic numbers, together with the real numbers, give all the posible completions of $\mathbf Q$. 
On the other hand, the $p$-adic numbers are also extremely useful, even if you only care about $\mathbf Q$. A great example of this is the Hasse principle, that says (for example) that a homogeneous quadratic equation has a nontrivial solution over $\mathbf Q$ if and only if it does over $\mathbf R$ and $\mathbf Q_p$ for each $p$, and the latter question turns out to be straightforward to answer. 
