Awe-Inspiring Math: The Most Beautiful Theorem Proofs

  • Thread starter Thread starter jem05
  • Start date Start date
  • Tags Tags
    Proofs Theorem
AI Thread Summary
The discussion highlights the beauty of various mathematical proofs, with participants sharing their favorites. Euclid's proof of the infinitude of primes is frequently mentioned for its simplicity and elegance. The book "Proofs from THE BOOK" by Aigner and Ziegler is recommended as a collection of beautiful proofs. Other notable mentions include Maxwell's proof of the Gaussian distribution and the proof of the change of variables formula, both appreciated for their elegance. The conversation emphasizes the aesthetic appeal of mathematics, particularly in proofs that connect different areas like topology and algebra.
jem05
Messages
54
Reaction score
0
What's the most Beautiful proof of a mathematical theorem you've seen?
 
Mathematics news on Phys.org
The original proof by Euclid for the infinitude of primes is one of the most beautiful proofs that there are. It's beauty because it's simple.

You should have a look at the book "Proofs from THE BOOK" by Aigner and Ziegler. It contains a bunch of very, very, very beautiful proofs.

Proofs that I've personally liked were the proof of Banach fixed point theorem, and the proof of the central limit theorem (which is very ugly, but it's rewarding, so I like it). But I doubt that they're good enough to be included in THE BOOK.
 
I don't know if other people would consider it "beautiful" but a proof that I found fascinating was of the theorem

"Let c be a positive real number. If there exist a function, f(x), such that f and all of its iterated anti-derivatives can be taken to be integer valued at 0 and c, then c is irrational".

The proof involves assuming the concusion is true and deriving "statement A" which does not appear to have anything to do with the hypotheses, then turning around and proving "statement B" which also doesn't appear to have anything to do with the hypostheses but contradicts statement A!

I saw it in a Mathematics Association of America periodical published, I think, in the middle 1980s but cannot cite it now.
 
You should ask Erdos for that. he's dead though
as micromass said, you can find many "beautiful" proofs in the book he mentioned. I personally like Maxwell's proof of the Gaussian distribution for errors and some of Archemedes' proofs. especially the one that he derives a correct formula for the area under a hyperbola using mechanics.
 
Let P(z) be a non-constant polynomial of degree n with no complex roots. Then for all R > 0, \int_{|z|=R} \frac{1}{zP(z)} dz = \frac{2\pi i}{P(0)} \neq 0 by the Residue theorem. By the ML-estimate, \int_{|z|=R} \frac{1}{zP(z)} dz = O\left(\frac{1}{R^n}\right) \to 0. This contradiction amounts to proving the Fundamental Theorem of Algebra.
 
personally, the most elegant proof I've seen is the proof of the change of variables formula. It's crazyyy...
 
I agree with the poster who mentioned the proof of infinite primes. Although I'd imagine most of number theory is very elegant.
 
I love it when there are interactions between topology and algebra (or in fact, any mysterious connections I always feel make for the best proofs) and always consider them very beautiful and interesting, so I like a theorem relating the cohomology of compact lie groups with the rank of their maximal tori.That's more the theorem though and not the proof itself, so I'd have to go for the proof that you cannot effectively classify manifolds of dimension greater than or equal to 4.

What you do basically is take any group G described by certain relations and construct yourself a manifold which has the same fundamental group by invoking some reasonably simple surgeries. But because you can do this for any group presentation, you cannot effectively classify your manifolds, because the word problem for finitely generated groups is not solvable.

I love that proof, very beautiful and elegant.
 
It's very elementary, but I like the proof that for an isosceles triangle ABC with AB = AC and base BC, the base angles are equal. The one where there are no helping lines; you just look at the triangle two different ways:

AB = AC given
AC = AB given
Angle A = Angle A identity
Therefore angle B = angle C by s.a.s = s.a.s.
 
  • #10
my favorite is Gauss finding the equivalent to straight edge and compass construction and finding the 17-gon is constructable with straight edge and compass.
 
  • #11
My answer's going to be lame, but I when I proved that an odd times an odd is an odd, I was amazed.
 

Similar threads

Replies
2
Views
2K
3
Replies
105
Views
6K
Replies
11
Views
2K
Replies
8
Views
2K
Replies
38
Views
4K
Replies
72
Views
7K
Back
Top