= Solution
Irrationality of $\sqrt2$ makes the expression $q+r\sqrt2$ unique for rational $q,r$: equality of two such expressions with different $r$ would express $\sqrt2$ as a <rational number>. Therefore the rule selecting the coefficient $r$ defines a <function> on $A$. Its domain is a <vector subspace> over $\mathbb Q$, and direct coefficient <addition> and rational scaling show it is $\mathbb Q$-linear, hence weakly linear. In particular
$$
f(1)=0,\qquad f(\sqrt2)=1.
$$
Use <Zorn's lemma> on the nonempty family of weakly <linear extensions> of this particular <function>. <Chain> <unions> remain extensions by (a); hence there is a maximal extension $h$. Any further extension in $\mathcal F$ would also contain the original <function>, so this $h$ is maximal in $\mathcal F$. Part (b) therefore gives $\operatorname{dom}h=\mathbb R$.
A <real-linear map> $\mathbb R\to\mathbb R$ is <multiplication> by its value at $1$. Were $h$ real-linear, it would satisfy
$$
h(\sqrt2)=\sqrt2\,h(1)=0,
$$
contradicting the inherited value $h(\sqrt2)=1$. Thus \b[a <weakly linear function> on all of $\mathbb R$ need not be real-linear]. Its <rational homogeneity of an additive function> is insufficient for real homogeneity. This particular extension is also discontinuous: <rational numbers> approaching $\sqrt2$ have value <zero>, while the value at their <limit> is one.
Back to article page
