Let $K$ be a skew field and $S$ its subset. For $S$ to be a subfield of $K$ it's necessary and sufficient that the following three conditions are fulfilled:

1. $S$ contains a non-zero element of $K$
2. $a\!-\!b \in S$ always when $a,\,b \in S$
3. $ab^{-1} \in S$ always when $a,\,b \in S$ , and $b \neq 0$

Proof. Because the conditions are fulfilled in every skew field, they are necessary. For proving the sufficience, suppose now that the subset $S$ satisfies these conditions. The condition 1 guarantees that $S$ is not empty and the condition 2 that $(S,\, +)$ , is an subgroup of $(K,\, +)$ thus all the required properties of addition for a skew field hold in $S$ If $b$ is a non-zero element of $S$ then, according to the condition 3, we have $0 \neq 1 = bb^{-1} \in S$ Moreover, $a\!\cdot\!1 = 1\!\cdot a = a \in S$ , for all $a \in S \subseteq K$ The laws of multiplication (associativity and left and right distributivity over addition) hold in $S$ since they hold in whole $K$ So $S$ fulfils all the postulates for a skew field.