# Finding an example of a non-rational p-adic number

We know that every rational number can be written as a $p$-adic integer with expansion $\sum\limits_{n=-m}^\infty a_n p^n$, where $a_n\in\{0,\dots,p-1\}$ and $m\in\mathbb{N}$; therefore there exists an injection $\mathbb{Q} \hookrightarrow \mathbb{Q}_p$.

But how do I show that $\mathbb{Q}_p$ is bigger? How do I find an example of a $p$-adic number which is not rational?

I heard $p$-adic numbers and even $p$-adic integers being uncountable, is the proof easy?

First of all, it is every integer (not every rational) that can be written as you have presented it as a sequence $a_n\in\{0,\ldots, p-1\}$. The number $1/p$ cannot be so represented.

To answer your cardinality question, there are at least two arguments that $\mathbb Z_p$ is uncountable, and thus that $\mathbb Q_p$ is uncountable.

From your presentation of the $p$-adics as the series $\sum a_n p^n$, we see that the $\mathbb Z_p$ has the cardinality of the set of maps $\mathbb N^+\rightarrow \{0,\ldots, p-1\}$, (i.e., $n\mapsto a_n$) and hence uncountable.

Another argument is to start with the fact that $\mathbb Z_p$ is infinite and (Hausdorff) compact, e.g., because $\mathbb Z_p= \lim \mathbb Z/p^{n}$, so a closed set of a compact set, hence compact (and infinite). By Baire's theorem, in a compact Hausdorff space, the countable union of nowhere dense closed sets cannot contain a non-empty open set. In particular, an infinite countable set cannot be (Hausdorff and) compact.

• Sorry, I'll correct that integer/rational thing. The "set of maps" argument seems to work nicely! – Ottavio Bartenor Dec 20 '15 at 16:51

HINT: A rational number has a periodic $p$-adic expansion. So, an example would be $$\sum_{n\ge 0} p^{n^2}$$

• Nice example, but I'm afraid I'm not aware of the reason why rationals expansion is periodic... Can you enlighten me? – Ottavio Bartenor Dec 20 '15 at 16:58
• @Ottavio: The argument is very much like the one showing that fractions have periodically repeating decimal expansions. Save for the cases when the denominator has common factors with ten, when you get terminating or eventually periodic expansions. Here $p$ takes the role of ten (even though the expansion goes a bit differently). – Jyrki Lahtonen Dec 20 '15 at 17:04

Take a prime $p>2$, and $m$ an integer with $m\equiv1\pmod p$. Then $\sqrt m\in\Bbb Z_p$.

For $p=2$, if $m\equiv1\pmod8$, then $\sqrt m\in\Bbb Z_2$.

Finally, $\Bbb Z_p$ contains all $p-1$ of the $(p-1)$-th roots of unity. Uninteresting for $p=2$ and $p=3$, but very interesting for bigger primes.

• [earlier comments deleted... just to alert you] – paul garrett Dec 26 '15 at 23:35