# Deepest theorems with simplest proofs [closed]

Which are the deepest theorems with the most elementary proofs?
I give two examples:
i) Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function
ii) Proof that the halting problem is undecidable using diagonalization

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## closed as not a real question by Amitesh Datta, Asaf Karagila, Pete L. Clark, Raskolnikov, Jonas TeuwenOct 23 '11 at 18:37

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This should be community wiki, I think. –  Grumpy Parsnip Oct 22 '11 at 2:33
Isn't depth, by definition, inversely proportional to the elementariness of proofs? –  Henning Makholm Oct 22 '11 at 2:43
How about "reopen, please"?!? Also "PLZ reopen" in the title is obnoxious. –  The Chaz 2.0 Oct 26 '11 at 14:23
@GM2001: Please stop editing titles to contain messages like "AWESOME" or " REOPEN !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...please!!!!!!!!!!‌​!!!!!!!!!!!!!!!!!!!!!". This is not the first time you're doing it. –  ShreevatsaR Oct 26 '11 at 14:36
@GM2001: If you edit this question any more, I will lock it, which will prevent further edits, comments, and answers. –  Zev Chonoles Oct 26 '11 at 14:37

## 2 Answers

I think one should not confuse "important with "deep". The facts that $\sqrt{2}$ is irrational, that there is no surjective map $X\to2^X$, or that there are an infinity of primes, are certainly important or even "fundamental", but their proofs are so simple that one cannot call them "deep". A theorem is "deep" when its proof is really hard and, above all, requires a theory that transcends the realm the problem is formulated in. Consider, e.g., Gauss' theorem about which regular $n$-gons can be constructed with ruler and compass.

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I think the theorem that $\sqrt{2}$ is irrational would have met your criterion for depth at the time it was discovered. Imagine that the Greeks saw $\sqrt{2}$ as the length of the diagonal of a unit square (rather than as the positive solution to $x^2=2$). Then to show the irrationality of $\sqrt{2}$, we need to transcend geometry to go to number theory, where we have available the fundamental theorem of arithmetic. –  Srivatsan Oct 23 '11 at 15:49
To add on Srivatsan's comment, the proof of $|P(X)|>|X|$ while seemingly trivial nowadays required the development of an entire new field in mathematics. I'd say this qualifies as pretty deep. –  Asaf Karagila Oct 23 '11 at 19:35

These perhaps aren't particularly deep, but they are the first that come to mind.

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+1 Because I hadn't been aware of #3 before. –  anon Oct 22 '11 at 2:38
I think the linked proof in (1) is misnamed, and is not a proof by infinite descent (unlike this one)‌​. –  r.e.s. Oct 22 '11 at 3:22