Prove that each subfield of the field of complex numbers contains every rational number.
Solution
Let \(F\subseteq\CC\) be a subfield.
1. Every integer belongs to \(F\).
Because \(F\) is a subfield, \(0,1\in F\). Closure under addition gives \[ n=\underbrace{1+\cdots+1}_{n\text{ times}}\in F \] for every positive integer \(n\), and closure under additive inverses gives every negative integer as well. Hence \(\ZZ\subseteq F\).
2. Every rational number belongs to \(F\).
Let \(m/n\in\QQ\) with \(n\ne0\). By 1, \(m,n\in F\). Since \(F\subseteq\CC\), the element \(n\) is nonzero in \(F\), so \(n^{-1}\in F\). Therefore \[ \frac mn=m n^{-1}\in F. \] Thus \(\QQ\subseteq F\).