Rational Numbers form Prime Field

Theorem
The field of rational numbers $\left({\Q, +, \times}\right)$ is a prime field.

That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\left\{{0}\right\}$, and vacuously $\varnothing$.

Proof
Suppose $F$ is a subfield of $\left({\Q, +, \times}\right)$.

Let $a \in F$.

Then $a^{-1} \in F$ and so $a \times a^{-1} = 1$ and so $1 \in F$.

So, we have that $1 \in F$ and so $-1 \in F$ and thus $1 + \left({-1}\right) = 0 \in F$.

Suppose $n \in \Z$ such that $n \in F$.

Then $n + 1 \in F$ and so by the Principle of Finite Induction that all positive integers are in $F$.

But then if $m \in F$ we have that $-m \in F$ and so all negative integers are in $F$.

Next, as we have seen, $m \in F \implies m^{-1} \in F$.

So:
 * $\forall p \in \Z, q \in \Z_{\ne 0}: p \times q^{-1} \in F$

and it follows by the Subfield Test that $F = \Q$.