On 4 Dez., 09:20, Virgil <Vir...@home.esc> wrote: > In article > <949e3b4b-04c5-4d6e-ba64-7f95d37db...@r24g2000yqd.googlegroups.com>, > > > Actual mathematics would say that while every > node and every edge is included in the sequence, there is no member of > the sequence having all of them at once.
Everey member covers the first node. The n-the member covers all nodes including 1 and n. So logic forces us (i.e. those who can) to conclude that if all nodes are covered then there is a path covering all nodes.
But as there is no such path. Hence there is no "all nodes".
Some matheologians claim that there are all nodes and that all nodes are covere by all paths but that there is no path covering all nodes. This would... But it is really useless to try to teach such ...