Irrational P-adics $\mathbb{Q}_p$ is completion of $\mathbb{Q}$ by defining a new metric. So, with respect to this new metric they are complete.
I just want to be sure, are there p-adic rationals? If there are P-adic irrationals, what sense they make. I mean, how are they constructed?
Edit My question was if P-adic irrationals exist. I made some type rendering a little confusing question. Sorry. I did not intend to ask if p-adic rationals are existant. 
Thank You.

Ittay Weiss already answered the main question, so I will only add an example construction of an irrational $5$ adic number. Let's do a $5$-adic $\sqrt{-1}$.
We begin by noticing that $2$ is a good $5$-adic approximation to $\sqrt{-1}$, because $2^2=4=-1+5$ is pretty close to $-1$ as $5$ is a small number in this context. Can we find a better one? After a little bit of searching we notice that
$$
7^2=49=-1+50=-1+2\cdot5^2
$$
is even closer to $-1$ as $2\cdot 5^2$ is a smaller error than $5$ (the higher the power of $5$ dividing a number, the smaller it is $5$-adically). Observe that $7=2+1\cdot5$, so I left the "most significant digit" $2$ in tact, and only tuned up the next digit the best I could.
Proceed with this turning our attention to the next digit. How could we find it? We are looking for an even better approximation $q=7+a\cdot 25$. Given this we can calculate
$$
q^2=7^2+14a \cdot 25+ a^2\cdot 5^4.
$$
Aiming at an error divisible by $5^3=125$ we can ignore that last term. We want to select $a$ in such a way that $7^2+14a\cdot25\equiv -1\pmod{125}$. This is equivalent to $$14 a\cdot25\equiv -50=-2\cdot25\pmod{125},$$ and thus also to $14a\equiv -2\pmod5.$ We see that $a=2$ is a solution to this congruence. Therefore we end up with $q=7+2\cdot25=57$. Let's check
$$
57^2=3249=-1+3250=-1+26\cdot 125.
$$
The error in the square has, indeed, shrunk to a multiple of $125$.
It is not hard to see (once your studies reach Hensel's Lemma, you'll learn why)
that we can go on forever, and find numbers $a_i\in\{0,1,2,3,4\}$, $i=3,4,\ldots,$ such that the partial sums of the series
$$
2+1\cdot5+2\cdot5^2+a_3\cdot5^3+a_4\cdot5^4+\cdots
$$
have squares differing from $-1$ by multiples of everincreasing powers of five.
This justifies calling the sum of this series (to get it to converge we need to go the $5$-adic completion)
$$
"\sqrt{-1}"=\lim_{n\to\infty}\left(57+\sum_{k=3}^na_k5^k\right)
$$
A caveat: there is another $5$-adic square root of $-1$, not unexpexctedly, the negative of the one above. It is a bit dangerous to pick one of them and call it "the square root of $-1$" so I really shouldn't. As an exercise you can do
a few iterations of this process starting with another good initial approximation
$\sqrt{-1}\approx3$ ($3^2=9=-1+2\cdot5$), and check that you get the "negative" of the series I was calculating. The verification will be easier, if you observe that
$$
-1=\frac{4}{1-5}=4\frac1{1-5}=4\sum_{k=0}^\infty5^k=4+4\cdot5+4\cdot5^2+\cdots
$$
as a consequence of the sum formula of the geometric series.
