TITLE: Implications of field $\mathbf{F}$ with characteristic $0$.
QUESTION [2 upvotes]: 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 $\frac{p}{q}\in\mathbf{F}$.
$\blacksquare$

REPLY [3 votes]: There are several problems (some of which were mentioned in the comments):

The statement is false. The best you can do is to prove that $\mathbf F$ contains a subfield which is isomorphic to $\mathbb Q$.
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).
What about the negative rationals?
You still have to proved that this association is a field homomorphism (it is).
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