Deep theorem with trivial proof

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It is the snobbishness of the young to suppose that a theorem is trivial because the proof is trivial.

-- Henry Whitehead

I have been awestruck by the beauty of this quote.

What is in your opinion a good contender to exemplefy the meaning intended by Whitehead?

Off the top of my head I am thinking of Langrange's theorem in Group Theory, which is rather simple to prove but provides a very useful insight.

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This is going to depend on the definitions of deep and trivial, but the following just might qualify:

Every map defined on some basis of a vector space admits a unique linear extension to the whole vector space. What's more, the map is injective/surjective/bijective if and only if it maps the basis to a linearly independent system/ spanning system/basis.

Indeed the proof writes itself straight from the basic definitions, yet the statemente is, in a sense, the essence of linear algebra.

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The fundamental theorem of arithmetic.

The proof is generally shown to freshmen as an application of induction.

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Cantor's diagonal argument shows not only the completely unintuitive result that uncountable sets exist, but that something as everyday as the real numbers belongs to that category.

The proof may be explained even to those with no mathematical background by a simple drawing.

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The fixed point theorem always come to mind. As does the Pigeon Hole Principal.

And $b^{n}b^{m}=b^{n+m}$ is exceedingly profound yet trivially mundane and impossible not to be true as well. As is $\frac {de^x}{dx} = e^x$ and $e^{\pi i} = -1$ (which are really just the same thing result as $b^nb^m = b^{n+m}$ when you get down to it).

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Turing's proof of the undecidability of the halting problem is easy enough to be explained to most students of mathematics who are familiar with the way computer programs work, but it is of huge importance in its field.


It can be used to prove Gödel's incompleteness theorem (if we assume $\omega$-consistency of PA): given a program $P$ in some Turing machine, the statement that $P$ halts can be encoded as a sentence $s_P$ of Peano Arithmetic. If the completeness theorem were false, then there would be a proof of either $s_P$ or $\neg s_P$.

But this now gives us an algorithm for the halting problem, contradicting Turing's proof: given a program $P$, iterate through all possible proofs in Peano Arithmetic until we reach either a proof of $s_P$ or a proof of $\neg s_P$. By the above, this algorithm is guaranteed to terminate.


In a way, though, this is the same answer as the one about the diagonal argument.

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Stokes' Theorem is a "deep theorem" with a trivial proof. Here is a quote from Spivak's Calculus on Manifolds:

Stokes' theorem shares three important attributes with many fully evolved major theorems:

  1. It is trivial.
  2. It is trivial because the terms appearing in it have been properly defined.
  3. It has significant consequences.

Since this entire chapter was little more than a series of definitions which made the statement and proof of Stokes' theorem possible, the reader should be willing to grant the first two of these attributes to Stokes' theorem. The rest of the book is devoted to justifying the third.

Stokes' Theorem includes as a special case:
1. The fundamental theorem of calculus.
2. Green's Theorem.
3. The divergence theorem (from multivariable calculus).

On top of all that, the statement is as elegant as it gets: $$\int_M\partial\omega=\int_{\partial M}\omega$$

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I say this with trepidation as these things can be very polarizing, but publish and be damned I will:

how about that $$(\mathrm{cos}\,\theta + i\, \mathrm{sin} \,\theta)^n = (\mathrm{cos}\,n\theta + i\, \mathrm{sin} \,n\theta)=e^{in\theta}$$ and while we're at it, that $i^i = e^{-\pi/2}$ (after posting this I see that one of the comments has already mentioned $e^{i\pi}+1=0$). In the spirit of one of the comments to OP, rather trivial thanks to Argand etc, but still absolutely captivating. When you learn these things as a kid the simplicity of the proof given the perspective you have no doubt been given belies their import.

If too hackneyed an example please accept my apologies!

(NB: as correctly pointed out, we should define what we mean by $z^\alpha$ in general, namely $\mathrm{exp}(\alpha \log z)$ where $\mathrm{log}\, z =\mathrm{log}\,|z| + i \, \mathrm{arg} \,z$, hence we admit a multiplicity of values unless we restrict $\mathrm{arg} \,z$, which I'll do here to $0\leqslant\mathrm{arg} \,z\lt 2\pi$, so that $\mathrm{arg}\,i:=\pi/2$ )

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Cauchy's Integral Formula is one of the foundations of complex analysis, and from it, the residue theorem also falls in line quite easily. Its proof isn't terribly difficult, at least in comparison to its application.

Also, L'Hospital's rule is even more trivial when you look at the proof, though its impact is quite significant for pesky limit problems.

Recently learning some multivariable calculus, I find the proof of the multivariable chain rule somewhat trivial, though it leads to the fundamental theorem of line integrals concerning path independence.

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Russell's Paradox, that universal set comprehension is inconsistent with the rest of set theory, can be stated in one elegant line, has no hidden lemmas, and was the cause of arguably the single most profound investigation in the history of mathematics: the quest for formal axioms of set theory.

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Pythagorean Theorem. It was proved in the absence of even a hint of algebra, mind you. But its implications immediately shattered the understanding of mathematics at the time.

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1.The Hahn-Banach theorem for real Banach spaces. The proof is a few lines of calculation, which are then baked by Zorn's Lemma.

  1. There is no largest prime (or an infinite set of primes, depending on your choice of axioms). The first chapter of the book Prime Number Records gives about 23 different proofs. One due to Prof. Leo Morse:"It suffices to give a strictly increasing unbounded sequence of pair-wise co-prime natural numbers. For example, the Fermat numbers."

  2. With some modern machinery (Liouville's theorem that a bounded entire function is a constant), the Fundamental Theorem of Algebra is proved as follows: If p is a non-zero polynomial and p(z) is never 0 then 1/p is a bounded entire function.