47

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.

Community
  • 1
Jsevillamol
  • 4,759
  • 1
    Are you thinking about proofs from the book? – Michael Burr Dec 06 '16 at 18:01
  • 4
    To be honest most theorem with "trivial proof" has trivial proofs because most of the work needed for that proof was already done before the theorem is proved, and right approach was used. Theorems with "trivial proof" can be nearly impossible to prove without correct approach – user160738 Dec 06 '16 at 18:04
  • @MichaelBurr what book? – Jsevillamol Dec 06 '16 at 18:05
  • 22
    I asked this question several years ago and it got lots of answers: http://mathoverflow.net/questions/28788/nontrivial-theorems-with-trivial-proofs $\qquad$ – Michael Hardy Dec 06 '16 at 18:05
  • @user160738 I think that the bext examples here are not theorem for which the complexity of the proof is hidden in lemmas, but rather that provide a useful construction or way of thinking in their statement. – Jsevillamol Dec 06 '16 at 18:06
  • 4
    "Proofs from THE BOOK" is a particular book in honor of Erdos which focuses on such proofs. – Michael Burr Dec 06 '16 at 18:13
  • 3
    The yoneda lemma springs to mind. The proof is pretty mechanical for the sort of math that it is, yet it's used everywhere and says something pretty deep and meaningful about naturality. – daniel gratzer Dec 06 '16 at 20:32
  • I've always been impressed and awed by the fixed point theorem. – fleablood Dec 06 '16 at 21:05
  • @fleablood Which one among all the fixed point theorems? – Jsevillamol Dec 06 '16 at 21:32
  • 2
    Any of them. But I'd say that Brouwer is the must simple in stating and proving and simple to grasp but surprising in result. – fleablood Dec 06 '16 at 23:17

11 Answers11

35

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.

Nathaniel Bubis
  • 33,425
  • 4
    And it gives a simple proof, if not the simplest possible, of the existence of transcendental numbers. – Akiva Weinberger Dec 06 '16 at 21:51
  • 1
    When I was first exposed to these ideas, what seemed unintuitive to me was that $\mathbb Q$ was countable. It seemed so much larger than $\mathbb Z$. (Of course, it turns out there are easy ways to see that $\mathbb Q$ is countable, but to me it was very surprising.) – littleO Dec 06 '16 at 23:11
  • 5
    @nbubis I take issue with the assertion that the real numbers are commonplace or familiar. For example, almost every real number is normal (i.e., the sequence of its digits are uniformly distributed in every base $b$), but I'm willing to bet that you can't write down a single normal real number. Worse still, there are only countably many computable real numbers! – D Wiggles Dec 07 '16 at 02:30
  • 1
    "the completely unintuitive result that uncountable sets exist" That's not unintuitive at all in my opinion but granted that's tangential. (Instead I find that the rationals are countable is somewhat unintuitive at first.) – quid Dec 07 '16 at 08:00
24

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: $$\int_M\partial\omega=\int_{\partial M}\omega$$

D Wiggles
  • 2,918
  • Nice! never thought about it that way. – Nathaniel Bubis Dec 06 '16 at 22:42
  • Honestly, I've taken diff geometry, but I've never really understood what all those symbols mean or how to use them. How would you use this, for example, to integrate $x^2$ from $0$ to $1$? – goblin GONE Dec 06 '16 at 23:26
  • @goblin -- how would you go about integrating $x^2$ from 0 to 1? Find an antiderivative of $x^2$, then plug in the endpoints? That's the Fundamental Theorem, which is really the 1-dimensional (in the domain) special case of the general Stokes Theorem -- the boundary $\partial M$ is the endpoints of the interval in that case. Green's Theorem is similarly a special case. Part of the power and beauty of the general Stokes Theorem is its ability to cover so many cases so simply. – user128390 Dec 07 '16 at 00:45
  • 1
    But can you actually show the reasoning though, using only Stokes theorem and differential forms, and not the fundamental theorem of calculus? I mean, are we supposed to write $$\int_{[0,1]} x^2 dx = \int_{[0,1]} d(x^3/3) = \int_{\partial [0,1]} \frac{x^3}{3}$$ or something? It's frustrating that everyone claims that Stoke's generalizes the Fundamental Theorem, while never really explaining themselves. – goblin GONE Dec 07 '16 at 01:04
  • @goblin Isn't that it though? ${\partial [0, 1]}$ is defined (earlier in the book) as the set {0, 1} , integration on a set of points is defined... well, in a more thorough way than I'll go into here, but it is defined, and so the end result will be exactly what's expected: $\frac{1}{3}$. (I don't have the book to hand, so I'm going off memory. Maybe I've missed the mark.) – detly Dec 07 '16 at 03:39
  • @detly, I don't think so. For starters, $\partial$ doesn't mean "topological boundary" it means "boundary of a manifold." There's orientation issues, too. Honestly, this stuff is really never explained. – goblin GONE Dec 07 '16 at 04:12
  • @goblin oh that's right, I got my boundaries confused. But doesn't an earlier chapter define boundary orientation and base cases for integrals? – detly Dec 07 '16 at 04:20
  • @goblin I don't know why I'm arguing about a book I don't have access to. I'll take your word for it :P – detly Dec 07 '16 at 04:25
  • @detly, I haven't read Spivak so maybe a good explanation is given. All I know is that my differential geometry class didn't really explain it, and lots of Googling never really turns up anything, either :( – goblin GONE Dec 07 '16 at 04:27
  • 1
    @goblin Ah right. Yes, I had the same experience with a university calculus course, read Spivak and it all became clear! I highly recommend it. The book has something of a cultural context too — it was something of a reaction to mathematicians and physicists of the time playing fast and loose with d's and $\int$'s and mashing symbols together to get a result they liked. So it does approach things with a really good mixture of rigour and clarity. – detly Dec 07 '16 at 04:34
  • @goblin In fact, I got a bit shirty about it all, and sent an email to my lecturer tersely asking for any other resources, that used this "modern" notation he'd just dived into after telling us to throw out That Stewart Book. He referred me to Spivak and a bunch of other incomprehensible books, and Spivak became one of my favourite books (fiction and non-fiction alike). – detly Dec 07 '16 at 04:44
  • @detly, nice :) – goblin GONE Dec 07 '16 at 04:45
  • @goblin $\partial$ means "topological boundary with orientation." The manifold comes with an orientation, and each component of the boundary inherits an orientation from the orientation on the manifold. One way to define an orientation on $M^n$ is a choice of a nowhere vanishing volume form (so it eats up $n$ vectors in the tangent space at a point and spits out a number). Then a "positive basis" at a point is a basis for the tangent space which the form evaluates positively. To get an orientation on a boundary component, you need a nowhere vanishing form on it. – D Wiggles Dec 07 '16 at 04:49
  • @goblin ...continued...If I start with n−1 tangent vectors at a point in ∂M, I need to get a number. So pick the "outward unit normal," and stick all n vectors into my volume form on M and I'm done. So basically it boils down to the fact that at the boundary, there is a well defined notion of pointing outwards. – D Wiggles Dec 07 '16 at 04:59
  • @goblin You may find the description in http://geocalc.clas.asu.edu/html/NFMP.html to be more concrete. If you are familiar with geometric algebra, you can just look at the second chapter, otherwise the first chapter is a reasonable introduction. If you want a more formal presentation see From Clifford Algebra to Geometric Calculus. The GC result generalizes Stokes' Theorem, and in the GC form the boundary operator is the normal boundary of manifolds. Orientation is handled by a pseudoscalar field. – Derek Elkins left SE Dec 07 '16 at 06:25
  • @DerekElkins, thanks. – goblin GONE Dec 07 '16 at 07:35
20

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.

Damian Reding
  • 8,894
  • 2
  • 24
  • 40
  • 7
    Lazlo Babi of the University of Chicago has what he terms the "miracles of linear algebra." This is one of them, as is rank-nullity. I'm blanking on what the third is, but it struck me that between the three of them you can answer almost any introductory level problem in linear algebra. – Stella Biderman Dec 06 '16 at 22:14
  • 2
    @StellaBiderman If you remember the third miracle I'd like to hear it! – littleO Dec 07 '16 at 00:48
  • 4
    @littleO (First Miracle of Linear Algebra). If v1, . . . , vk are linearly independent and they all depend on w1, . . . , wm then k ≤ m. – curious_cat Dec 07 '16 at 03:28
  • 5
    (Second Miracle of Linear Algebra). The row rank of a matrix is equal to its column rank. – curious_cat Dec 07 '16 at 03:29
  • 5
    (Third Miracle of Linear Algebra). Let A ∈ Mn(R). Then the columns of A are orthonormal if and only if the rows of A are orthonormal – curious_cat Dec 07 '16 at 03:31
20

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.

John Gowers
  • 25,678
18

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.

Stella Biderman
  • 31,475
13

The fundamental theorem of arithmetic.

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

Levent
  • 5,240
  • 5
    Glen Stevens is fond of saying that the "real" fundamental theorem of arithmetic is that $p|ab\Rightarrow p|a$ or $p|b$. This key insight (which in algebraic number theory is in fact the definition of primarily) is the source of the simplicity of the inductive proof. In fact, UPF (over a general ring) is equivalent to the assertion that the numbers that satisfy the irreducibility definition and the divisibility definition are the same set. – Stella Biderman Dec 06 '16 at 22:09
  • 1
    (cont.) In number theory we call the former irreducibles and the latter primes, and the proofs that irreducible factorisations always exist and that prime factorization are always unique and both extremely trivial. Thus it's just a question of if they are the same set. All primes are irreducibles, but in $\mathbb{Z}[\sqrt(5)]$ neither $2$ nor $3$ are prime. They are irreducible, but don't divide $(1+\sqrt(-5))$ or $(1-\sqrt(-5))$, despite dividing their product, $6$. – Stella Biderman Dec 06 '16 at 22:11
  • 1
    @StellaBiderman Did you mean the Kummer ring $\mathbb Z[\sqrt{-5}]$ instead of $\mathbb Z[\sqrt{5}]$? – Christian Ivicevic Dec 07 '16 at 11:07
  • @StellaBiderman, how do we know that irreducible factorizations exist in $\mathbb{Z}$ w/o a lot of machinery? – goblin GONE Dec 07 '16 at 12:43
  • @goblin Def: $x$ is irreducible iff $\exists a,b$ such that $|a|,|b|>1$ and $ab=k$. Proof: Let $k$ be a positive integer. If $k=1$ or $k$ is irreducible, we are done. If not, by definition of irreducible, there exists $a,b$ such that $ab=k$ and $|a|\neq1, |b|\neq1$. Now you just need to know that for positive $k$, $a|k \Rightarrow a \leq k$ which isn't hard and apply infinite decent. – Stella Biderman Dec 07 '16 at 17:41
  • @goblin To show prime factorizations are unique, consider two different prime factorizations of $k$, the smallest integer that has two. Let $p$ be a prime in the first factorization. Then it divides the product of the second (since it divides k) and therefore since $p|ab\Rightarrow p|a$ or $p|b$, $p$ divides something in the second list. – Stella Biderman Dec 07 '16 at 17:42
  • (cont.) By minimality, $p$ isn't on the second list. Therefore $p$ divides some prime dividing $k$ that isn't $p$. However, this needs to be true of every prime on both lists since we haven't used any properties of $p$. Let $p$ be the largest prime on either list. But $p|a\Rightarrow |p| \leq |a|$, contradiction. Both proofs (with some work) can be adjusted to work in general rings of integers. It's perhaps most precise to say that the spirit of these proofs hold in general. – Stella Biderman Dec 07 '16 at 17:44
  • 1
    @StellaBiderman: Use \sqrt{-5} to get "$\sqrt{-5}$". – user21820 Dec 19 '16 at 11:32
6

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.

Simply Beautiful Art
  • 76,603
6

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.

DanielWainfleet
  • 59,529
4

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.

g------
  • 196
  • 10
2

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).

fleablood
  • 130,341
  • 4
    I like to write it as $e^{\pi i}+1=0$ to include what are arguably the $5$ most important constants in one equation. – Simply Beautiful Art Dec 06 '16 at 21:14
  • 1
    You aren't alone in that. I kind of like the blunt unexpectedness of the declaration $e^{\pi i} =-1$ myself but I think many many mathematicians prefer your expression. (Then there are a few who prefer $e^{\tau i} -1 =0$ and I have to wonder if they are missing the point ... or maybe they are just yanking by chain) – fleablood Dec 06 '16 at 21:24
  • 1
    Haha, please none of that $\tau$, $\pi$ is just more natural to me. – Simply Beautiful Art Dec 06 '16 at 21:25
  • I would consider the pigeon hole principle a trivial theorem – BlueRaja - Danny Pflughoeft Dec 06 '16 at 21:50
  • @BlueRaja-DannyPflughoeft Yep, you are a prime example of the snobbishness of youth. Of course, the pigeon hole principle is trivial. That's why it works. But it's astonishing how obfiscated and convoluted things can get when we overlook it. And we've all done that at some time in our lives. – fleablood Dec 06 '16 at 22:28
  • 2
    Yes, the $\tau$-ists really irk me. The the entire surprise of euler's formula is that a positive base has a negative value when raised to a transcendental imaginary power. Had it been $e^{2\pi i} = 1$ it would have been "that's odd... oh, well" but not the "wow! really" that $e^{\pi i} =-1$ is. And yet $\tau$-ist do use that as an arguement that it is cleaner and clearer. to me the whole point of $\pi$ is that for odd multiples of $\pi$ you do get toggled negative results of doing a half-circle. It's not a bug it's feature. In fact, it's the whole frikkin' point!!! – fleablood Dec 06 '16 at 22:35
  • @fleablood Perhaps you could give some examples of "how obfuscated things get when we overlook it" and times it's been overlooked in the past, rather than just stating that as fact? – BlueRaja - Danny Pflughoeft Dec 06 '16 at 22:44
  • Well, the classic example is proving two people in new York City have the exact same number of hairs. If we figure we need to figure out which two people and what number we are doomed. – fleablood Dec 06 '16 at 23:09
  • Well, I'm a $τ$-ist. The real whole point is that $e$ to an imaginary power is a rotation in the complex plane. No need to get weird about it. – Daniel R. Collins Dec 06 '16 at 23:13
  • I'm not being weird about it. We're talking aesthetics. Simple Art thinks $e^{\pi i}+1 = 0$ is an aesthetic statement (5 surprising constants). I see his point but I find $e^{\pi i}=-1$ to be my prefered expression (the dramatic off-synchopation nuance). However I find the expression $e^{2\pi i}=1$ to be dull and to be strangely purposely misleading in its flatness. None of these are better statements than any other. They are just personal aesthetics. – fleablood Dec 06 '16 at 23:26
  • @Daniel If I think $2$ is a more important number than $4$, does that make me a $2$-ist? – Matt Samuel Dec 07 '16 at 05:56
2

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 $\mathrm{arg}\,i:=\pi/2$ )

Mehness
  • 607