Implications of field $\mathbf{F}$ with characteristic $0$.

Question

I am required to show that given any field $\mathbf{F}$ of characteristic zero the set $\mathbf{Q}$ of all rational numbers is a subset of $\mathbf{F}$.

The following is my attempt at the problem is it correct?

Proof. Since $\mathbf{F}$ is a field of characteristic zero it follows that $\forall n\in\mathbf{N}\left(\sum_{j=1}^{n}1\neq 0\right)$ consequently given any arbitrary rational number $\phi = \frac{p}{q}$ where $q\ne 0$ it follows that $p = (\sum_{j=1}^{p}1)\in\mathbf{F}$ and $q = (\sum_{j=1}^{q}1)\in\mathbf{F}$ since $q\neq 0$ we can deduce using the axioms of a field that $\frac{1}{q}\in\mathbf{F}$ and by extension reason that $\frac{p}{q}\in\mathbf{F}$.

$\blacksquare$

Answer 1

There are several problems (some of which were mentioned in the comments):

1. The statement is false. The best you can do is to prove that $\mathbf F$ contains a subfield which is isomorphic to $\mathbb Q$.
2. You assert that you found a way of associating each element of $\mathbb Q$ with an element of $\mathbf F$. No, you did not. To each representation of a positive rational number as $\frac pq$ (with $p,q\in\mathbb N$) you associated an element of $\mathbf F$. But you did not prove that, say $\frac23$ and $\frac69$ are associated with the same number (they should, since they are the same rational number).
3. What about the negative rationals?
4. You still have to proved that this association is a field homomorphism (it is).