= Solution
Suppose $D\equiv A'+E'$ as in (c), and set $P=D-E'$. Its numerical class is ample, so $P$ is an <ample real divisor>. Fix an ample Cartier divisor $A$. The <ample cone> is open by <Kleiman's criterion>, so choose $\varepsilon>0$ with $P-\varepsilon A$ still ample. Express this ample real divisor as a positive real combination $\sum b_iB_i$ of ample Cartier divisors. Some positive multiple of each $B_i$ has an effective representative; dividing by that multiple gives $B_i\sim_{\mathbb Q}G_i$ with $G_i$ effective. Hence
$$
D\sim_{\mathbb R}\varepsilon A+\left(E'+\sum_i b_iG_i\right),
$$
which is (b). The equivalence of numerical ampleness with a positive combination of ample Cartier representatives follows by rational approximation inside the open ample cone; the <rational approximation of an ample-plus-effective real divisor> proof also accounts for principal-divisor directions.
Finally, if $D'\equiv D$, the same expression $D'\equiv A'+E'$ satisfies (c). By the equivalences just proved, $D'$ satisfies all three conditions exactly when $D$ does. Thus \b[bigness is invariant under numerical equivalence], for real as well as Cartier divisors.
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