# Are rational numbers also p-adic numbers?

I’ve recently been learning about p-adic numbers, and I’ve run across a question I can’t seem to find the answer to online...

We talk about some rational number n having a p-adic expansion, and often times I hear people talk as if the p-adic expansion of n IS that rational number. But at the same time, p-adic numbers have different properties than the rationals, so it seems as if the p-adic expansion of n is a new number, and not just a different representation of n.

Further, I’ve heard people say that much like the irrationals, the p-adics “fill in the gaps” between the rationals. Does this mean that there are p-adic numbers which do not correspond to rational numbers? And if so, how do we define “correspond”?

Over all, I suppose I do not understand what we mean by saying there is a p-adic expansion of a rational number, and any guidance on this topic would be much appreciated.

• The $p-$adics are completions of $\mathbb Q$ just like $\mathbb R$ is, see this for example. There's a nice result from Ostrowski that tells us that, up to topological equivalence, any completion of $\mathbb Q$ with respect to an absolute value is one of $\mathbb R$ or $\mathbb Q_p$ for some prime $p$.
– lulu
Oct 22, 2020 at 14:47

Very much like "every rational number has a decimal expansion (but not everything that has a decimal expansion is a rational number)" sums up $$\mathbb Q \subsetneq \mathbb R$$, indeed every rational number also has a (unique!) $$p$$-adic expansion, thereby can be seen as an element of $$\mathbb Q_p$$, but most elements of $$\mathbb Q_p$$ (precisely: the ones whose $$p$$ -adic expansion is not eventually periodic) are not rational numbers. I.e. $$\mathbb Q \subsetneq \mathbb Q_p$$.