Timeline for Is the empty graph a tree?
Current License: CC BY-SA 3.0
6 events
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| Feb 27, 2013 at 2:31 | comment | added | Andreas Blass | @Greg: Yes, but that decomposition doesn't serve as a counterexample to the assertion that the empty graph is connected. That decomposition contains isomorphs of the empty graph itself, as the definition of "connected" would require. What does serve as a counterexample is the decomposition as the disjoint union of the empty family of graphs; that family contains no isomorph of the empty graph (because it contains no graphs at all). | |
| Feb 26, 2013 at 23:09 | comment | added | Greg Martin | Isn't the empty graph the disjoint union of any family (of any cardinality) of empty graphs? | |
| Feb 5, 2013 at 21:28 | comment | added | ACL | Understood. It was kind of a joke! I'm sorry. | |
| Feb 5, 2013 at 14:03 | comment | added | Andreas Blass | I was hoping that it was clear that the definition was for positive integers, even though I admittedly said "positive integer" only once, modifying $p$. Indeed, if you allow factors to be things other than positive integers, then you have not only the problem you exhibited but also $2=(1+i)(1-i)=(4\pi)(\pi/2)$, etc. So please understand "expressed as a product" to mean "expressed as a product of positive integers. | |
| Feb 5, 2013 at 0:57 | comment | added | ACL | Then $2$, being equal to $(-1)\times (-2)$, is not prime? | |
| Feb 2, 2013 at 14:32 | history | answered | Andreas Blass | CC BY-SA 3.0 |