Yes, that’s one of the best examples I know. I plan to include a link to the in my essay. Jim On Mon, Apr 20, 2020 at 11:17 AM Tom Karzes <karzes@sonic.net> wrote:
Here's a Mathologer video of the "windmill proof":
https://www.youtube.com/watch?v=DjI1NICfjOk
Tom
Cris Moore via math-fun writes:
That “windmill” proof is beautiful!
This is too easy, but one could also consider the fact that any
polynomial of odd degree (and real coefficients) has at least one real root...
- Cris
On Jan 29, 2020, at 11:03 AM, Neil Bickford <techie314@gmail.com>
wrote:
This might have been one of the things that prompted the question,
but I
think Don Zagier’s one-sentence proof of Fermat’s theorem on sums of two squares (
https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares#Zagier's_ "one-sentence_proof")
uses this technique! Given a prime p = 4k+1, it defines a finite set S and two involutions. One involution’s fixed points (if they exist) are representations of p as a sum of two squares, while the other has exactly one fixed point. But replacing one involution with another either adds or removes fixed points in pairs, so the first involution has an odd and thus nonzero number of fixed points. (It may be too early for me to do math properly, but it seems like the second involution also shows that S has an odd number of elements, and so the first involution must have an odd number of fixed points?)
(On a related note, in case anyone hasn’t seen it, https://mathoverflow.net/a/299696 has a nice geometric method of showing where the involution comes from!)
--Neil Bickford
On Wed, Jan 29, 2020 at 9:46 AM James Propp <jamespropp@gmail.com> wrote:
Sounds interesting. Do you have a reference?
Jim
On Wed, Jan 29, 2020 at 11:21 AM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
Maybe this is not quite what you want - but Another Hamiltoniain
Path says
that if there is a Hamiltonian path in G, then there is another one either in G or its complement \bar{G}.
The idea is to define an (exponentially large) graph in which Hamiltonian paths correspond to the number of odd-degree vertices. There are an even number of these.
So… this is sorta an example of odd parity, but it’s really showing that the set plus one has even parity :-)
Cris
On Jan 29, 2020, at 7:44 AM, James Propp <jamespropp@gmail.com> wrote:
What are people’s favorite examples of existence proofs that show that a set is not empty by showing that its cardinality is odd?
Jim Propp
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