In order that, for eachx=(xj)(1≤j≤n)and eachε>0, there existmintegersqi(1≤i≤m)andnintegerspj (1≤j≤n) such that
∣q1a1j+q2a2j+⋯+qmamj−pj−xj∣≤ε(1≤j≤n)
it is necessary and sufficient that there exist no finite sequence (rj) of n integers, not all zero, such that each of the m numbers ∑j=1naijrj is an integer.
For if G is dense in Rn, that is if G=Rn, then G∗={0}, and conversely.
In particular (m=1):
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.