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# And vice-versa. QED.
(I helped save b3ta! The Alchemist king of the needlessly complicated, Mon 1 Feb 2010, 13:06, archived)
# What? No! THAT'S NOT HOW QED WORKS!
(I helped save b3ta! Mu Dinofiddler, Mon 1 Feb 2010, 13:08, archived)
# Oh. Ok.
So instead I will prove that no simple graph exists whose degree partition is discrete:

Firstly, assume that it does exist. Now, sort the vertices according to their degree. Note that the nth vertex can have a maximum of (n-1) bonds to the others. However, this only leaves (n - 2) other possible degrees for the remaining (n - 1) vertices, so two of the remaining vertices must share the same degree.

Q. E. Fucking D.
(I helped save b3ta! The Alchemist king of the needlessly complicated, Mon 1 Feb 2010, 13:14, archived)
# Fuck.
(I helped save b3ta! Ttssattsr, Mon 1 Feb 2010, 13:16, archived)
# No idea what you just said. So you didn't prove anything to me.
(I helped save b3ta! Mu Dinofiddler, Mon 1 Feb 2010, 13:18, archived)
# DAMMIT, MU!
(I helped save b3ta! The Alchemist king of the needlessly complicated, Mon 1 Feb 2010, 13:20, archived)