  The Kochen-Specker theorem proves the inability to assign, simultaneously,
noncontextual definite values to all (of a finite set of) quantum mechanical
observables in a consistent manner. If one assumes that any definite values
behave noncontextually, one can nonetheless only conclude that some observables
(in this set) are value indefinite.
  In this paper we prove a variant of the Kochen-Specker theorem showing that,
under the same assumption of noncontextuality, if a single one-dimensional
projection observable is assigned the definite value 1, then no one-dimensional
projection observable that is incompatible (i.e., non-commuting) with this one
can be assigned consistently a definite value. Unlike standard proofs of the
Kochen-Specker theorem, in order to localise and show the extent of value
indefiniteness this result requires a constructive method of reduction between
Kochen-Specker sets.
  If a system is prepared in a pure state $|\psi\rangle$, then it is reasonable
to assume that any value assignment (i.e., hidden variable model) for this
system assigns the value 1 to the observable projecting onto the
one-dimensional linear subspace spanned by $|\psi\rangle$, and the value 0 to
those projecting onto linear subspaces orthogonal to it. Our result can be
interpreted, under this assumption, as showing that the outcome of a
measurement of any other incompatible one-dimensional projection observable
cannot be determined in advance, thus formalising a notion of quantum
randomness.
