The extension $\mathbb{Q}_3 / \mathbb{Q}$ is not algebraic I'll start from the beginning: the definition of $\mathbb{Z}_p$ (the ring of $p$-adic integers) we were taught is the set of infinite series $(a_1, a_2, \ldots)$ such that $a_{i+1}\equiv a_i \pmod{p^i}$ (I'm omitting some formal details, this is just a clarification). Then we defined $\mathbb{Q}_p$ as the field of fractions of $\mathbb{Z}_p$ (equivalence classes of pairs of $p$-adic integers...).
I'm saying all this because this is truly all I know about $p$-adic numbers. 
Oh, and also that there's a natural embedding of $\mathbb{Q}$ into $\mathbb{Q}_p$
(derived from viewing the $p$-adic integer $(k, k, k,\ldots)$ as the integer $k$).
So now I'm requested to show that the extension $\mathbb{Q}_3 / \mathbb{Q}$ is not algebraic, that is, there exists a $3$-adic rational that is not a root of any polynomial in $\mathbb{Q}[x].$
I can think of some $3$-adic rationals that are not in the embedded $\mathbb{Q}$, but showing one of them is not algebraic is quite harder.
Could someone please enlighten me?
 A: If you just want to show that there exist transcendental elements of $\mathbb{Q}_p$, Franz Lemmermeyer's argument from the comments is a good way to go: $\mathbb{Q}_p$ is uncountable, but only countably many elements can be algebraic over $\mathbb{Q}$.
It is possible to write down an transcendental element $\mathbb{Q}_p$ using a $p$-adic version of Liouville's constant.
Claim: The $p$-adic number
$$
\alpha:=\sum_{n=0}^\infty p^{n!}
$$
is transcendental over $\mathbb{Q}$.
First we need a lemma.
Lemma: Let $f(T)\in\mathbb{Z}[T]$ be a non-constant square-free polynomial of degree $d$, and suppose $x_0\in\mathbb{Z}_p\backslash\mathbb{Z}$ is a root of $f$. Then there is a constant $C$ such that for all $x\in\mathbb{Z}$ sufficiently close to $x_0$,
$$
v_p(x-x_0)\leq d\cdot\log_p(|x|) + C.
$$
Proof: We can write $f(x)$ as a polynomial in $(x-x_0)$, say
$$
\tag{1}f(T)=\sum_{n=1}^d a_d(T-x_0)^n.
$$
The constant term vanishes because $x_0$ is a root of $f$, and $a_1\neq 0$ because $f(T)$ is square-free. For $x$ sufficiently close to $x_0$, the biggest term in $(1)$ is the first, so
$$
v_p(f(x)) = v_p(a_1) + v_p(x-x_0).
$$
Now $f(x)$ is a non-zero integer, so $v_p(f(x))\leq \log_p(|f(x)|)$. We also have $|f(x)|\leq A |x|^d$, where $A$ is the sum of the absolute values of the coefficieints of $f$. Thus we have shown
$$
v_p(x-x_0)=v_p(f(x))-v_p(a_1)\leq \log_p(A|x|^d)-v_p(a_1)=d\log_p(|x|)+\log_p(A)-v_p(a_1).
$$
This completes the proof.
Proof of the claim: For the sake of contradiction, suppose $f(T)\in\mathbb{Z}[T]$ is irreducible of degree $d$ and has $\alpha$ as a root. For $N=0,1,2,\ldots,$ define
$$
x_N:=\sum_{n=0}^N p^{n!},
$$
so that $x_N\to\alpha$ as $N\to\infty$. Let $C$ be be the constant guaranteed to exist by the lemma ($f(T)$ is square-free because it is irreducible). Then
$$
(N+1)!=v_p(\alpha-x_N)\leq d \log_p(x_N)+C\leq d (N!+1) + C.
$$
This is a contradiction once $N$ is larger than $d+|C|$.
