# $\Bbb Q$ is not complete w.r.t. $p$-adic norm

I'm trying to show that $$\Bbb Q$$ is not complete w.r.t. $$p$$-adic absolute value. The answer is already here. But I have two questions in that answer.

1. The answer is valid only for prime $$p>3$$. How can I show it for the case when $$p = 2,3$$?
2. During the proof, it says that using strong triangle inequality, it's easy to show that $$|x-a|_p<1$$. But this is not easy to me. Why is that follow from strong triangle inequality?
• $(x-a)=(x-a^{p^n})+(a^{p^n}-a)$. By FLT $|a^{p^n}-a|_p<1$ and for $n$ large enough $|x-a^{p^n}|_p <1$. By ultrametric inequality (or strong triangle inequality) it implies $|x-a|_p<1$. Commented May 8, 2021 at 10:57
• @Mindlack How $|a^{p^n}-a|_p<1$ follows by FLT? $a^{\varphi(p^n)}\equiv 1\ (\mod p^{n})$ so $a^{p^{n}-p^{n-1}}\equiv 1\ (\mod p^{n})$. I don't get it. Commented May 8, 2021 at 12:01
• $a^{p^n} \equiv a^{p^{n-1}}$ mod $p$ by FLT. By induction $a^{p^n}-a$ is divisible by $p$, QED. Commented May 8, 2021 at 12:34
• @Mindlack Oh, thank you. Commented May 8, 2021 at 12:35

Here’s another explicit idea, different from Dietrich Burde’s very good one (and the other answer that uses the Baire Category Theorem).

Consider the series $$\sum_{n \geq 0}{p^{n!}}$$. It clearly converges in $$\mathbb{Z}_p$$. If its limit were a rational number, then one would have, for finite $$p$$-adic expansions $$\sum_{k=0}^m{a_kp^k}$$ (this one nonzero) and $$\sum_{k=0}^l{b_kp^k}$$, the equality $$\sum_{k=0}^m{a_kp^k}\sum_{k \geq 0}{p^{k!}}=\sum_{k=0}^l{b_kp^k}$$.

Now, it’s easy to write the LHS as $$N_0+\sum_{n\geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$$ and thus $$\sum_{n \geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$$ is the valid (infinite) $$p$$-adic expansion of an integer $$N$$.

If $$N$$ is positive, that’s impossible because positive integers have a finite $$p$$-adic expansion. If $$N$$ is negative, we can write $$N=-p^q+r$$ with $$p^q>r>0$$ and $$-p^q=\sum_{k \geq q}{(p-1)p^k}$$ so the $$p$$-adic expansion of $$-p^q+r$$ has all but finitely many $$p-1$$, which isn’t the case of $$\sum_{n \geq m+1,0 \leq k \leq m}{a_kp^{n!+k}}$$.

• You basically giving an explicit Cauchy sequence in $\Bbb Q$ that does not converge w.r.t $p$-adic norm right? Commented May 8, 2021 at 12:42
• Yes, the sequence $\left(\sum_{k=0}^n{p^{k!}}\right)_n$. Commented May 8, 2021 at 12:57

There is also another way to show that $$\Bbb Q$$ is not complete with respect to the $$p$$-adic metric. For this it suffices to find irrational elements in $$\Bbb Q_p$$ by using Hensel's lemma. Indeed, for all $$p>3$$, $$\Bbb Q_p$$ contains the primitive $$(p-1)$$-th roots of unity. For $$p=2$$ we have $$\sqrt{-7}\in \Bbb Q_2\setminus \Bbb Q$$, and for $$p=3$$ we have $$\sqrt{7}\in \Bbb Q_3\setminus \Bbb Q$$.

The post you have cited has an appendix here, for $$p=2$$ and $$p=3$$:

Showing that Q is not complete with respect to the 2-adic and 3-adic absolute value

If $$\Bbb Q$$ is complete then because it is metric space we can use Baire category theorem.

Because there is no isolated points in $$\Bbb Q$$ (w.r.t. $$p$$-adic absolute value) ($$p^k + q \rightarrow q\ \ \forall q$$) we get $$\forall q\in\Bbb Q\text{ the set }\Bbb Q \setminus \{q\} \text{ is open and dense.}$$

But $$\bigcap_{q\in\Bbb Q}\ \Bbb Q \setminus \{q\} = \varnothing \text{ is not dense}.$$

So $$\Bbb Q$$ is not complete as metric space w.r.t. $$p$$-adic absolute value.