ON THE THEORY OF NUMBERS. 769 
00’ produced indefinitely both ways, and let PP’ be drawn parallel and equal 
to 00"; then P’ is anode, and OO'P’P is a parallelogram of which the vertices 
are nodes, and which has no other node either on its contour or in its interior ; 
such a parallelogram we shall call an elementary parallelogram. It is then 
evident from the characteristic property of the system, that every elementary 
parallelogram supplies us with a parallelism of the system; also we can 
obtain an infinite number of dissimilar elementary parallelograms ; for if Ox 
and Oy are the two lines of the given parallelism which intersect in O, and 
if m and m are any two integers relatively prime, the intersection of the mth 
parallel from Ow with the nth parallel from Oy will give a node O' such 
that no node can lie on OO' between O and O’; and, again, instead of P in the 
preceding constructiou, we may take any node lying on either of the two lines 
of the system which are the nearest to OO’. The areas, however, of all 
elementary parallelograms are equal. To prove this, we observe that if 
AOB is an elementary triangle (7. ¢. a triangle of which the vertices are 
nodes, but which has no other node cither on its contour or inside it), the 
parallelogram OAO'B obtained by drawing parallels to any two of its sides 
OA and OB through the opposite vertices B and A is an elementary parallelo- 
gram. For if AO and BO are produced to A’ and B’, so that O bisects AA’ 
and BB’, A’ and B’ are nodes, and the triangle A'OB! is elementary ; because 
if there were a node w' (other than its vertices) in A’OB’, we could imme- 
diately construct a node w (other than its vertices) in AOB. But A/OB! can be 
made to coincide with BO'A by a displacement without rotation ; therefore 
BO’A is elementary as well as AOB ; or the parallelogram AOBO’' is elemen- 
tary. Hence, if two elementary triangles have a common base, they are cer- 
tainly equal. For if through the vertex of either triangle we draw a parallel 
to the base, an elementary parallelogram will be contained between that 
_ parallel and the base; that is, the altitude of either triangle will be the dis- 
tance of the base from the parallel nearest to the base; or the triangles will 
be equal. Again, let AOB, «0d, be any two elementary triangles, which we 
may suppose to have a common vertex ; if BOw is an elementary triangle, they 
are each of them equal to it and to one another; if not, let a be that node 
contained in BOa which lies the nearest to OB, then BOw is elementary, and 
has the side BO in common with AOB ; by proceeding in this manner we shall 
form a series of elementary triangles, of which the first is AOB, and the last 
aOB, each triangle having a side in common with that preceding it, whence 
AOB=a0b; i.e. any two elementary parallelograms are equal. 
We shall next show that it is always possible to find a reduced paral- 
lelogram, 7. e. an elementary parallelogram, the sides of which are not 
greater than its diagonals. Let O be any node; A a node as near to O as 
any other; B a node on one of the parallels nearest to OA, and as near to O 
as any node on either of those parallels; complete the elementary parallelo- 
gram OAO'B; it will have the property required. Produce O'B to 0", making 
BO"=0'B; then AB=OO"; but by hypothesis OA < OB, and OB < 00’, 
OB < 00"; 2. ¢, the sides of OAO'B are not greater than its diagonals. 
Again, if OAO’B is a reduced parallelogram in which OA < OB, it can be 
proved that no node lies nearer to O than A, and that no node, out of the 
line OA, lies nearer to O than B; for, first, no node on the line OBO! lies 
nearer to O than B, because by hypothesis OB < 00', OB < 00", and because 
the extremity of the perpendicular drawn from O to OO’ falls between the 
points of bisection of the segments O0’’B and BO’, or on one of those points : 
secondly, no node on any parallel beyond OBO’ can lie as near to O as B, 
1863. 3D 
