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Can you make an example of a great proof by induction or construction by recursion?

Given that you already have your own idea of what "great" means, here it can also be taken to mean that the chosen technique :

  • is vital to the argument;
  • sheds new light on the result itself;
  • yields an elegant way to fulfill the task;
  • conveys a powerful and simple view of an intricate matter;
  • is just the only natural way to deal with the problem.

Here induction and recursion are meant in the broadest sense of the words, they can span from induction on natural numbers to well-founded recursion to transfinite induction, and so on...

Elementary examples are especially appreciated, but non-elementary ones are welcome too!

Lorenzo Lami
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    Structure Theorem for finitely generated abelian groups. – Sungjin Kim Mar 30 '12 at 17:02
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    My favorite induction proof is the joke proof that all natural numbers are interesting. Clearly 0 is interesting. Suppose m is interesting for all $0\leq m\leq n$. If n+1 were not interesting, then it would be the smallest non-interesting number, which is pretty interesting, a contradiction. So n+1 is interesting. – Benjamin Steinberg Apr 02 '12 at 20:32
  • @Benjamin: Your proof has earned my first laughter of the day! Nice proof! – Lorenzo Lami Apr 03 '12 at 10:04
  • See also https://math.stackexchange.com/questions/145189/examples-of-mathematical-induction – Gerry Myerson Aug 11 '19 at 13:09

14 Answers14

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A Classic: Fix a positive integer $n$. Show that it is possible to tile any $2^n \times 2^n$ grid with exactly one square removed using 'L'-shaped tiles of three squares.

It serves as a wonderful introductory example to proof by induction. Indeed, the proof can almost be represented with two appropriate figures. Yet, for those just learning induction, it is a significant problem where the application of the inductive hypothesis is far from obvious.

Asher M. Kach
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    It's a great nontrivial exercise for those learning induction; it's also a nice, entertaining fact for those who already know induction well. Great answer! – Lorenzo Lami Apr 08 '12 at 22:16
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    There's no reason to insist $n > 0$; works for $n = 0$ as well! – I. J. Kennedy Oct 28 '12 at 16:54
  • My number theory professor gave us this problem the first semester of college, and it is honestly one of my favorite proofs I've written in my undergraduate career. – Juan Sebastian Lozano Dec 08 '16 at 12:12
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    This is one of those results that is easier to prove when you replace the statement with something stronger. – Jeff Strom Aug 11 '19 at 18:00
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Tsirelson's space (1974) is a good example from Banach space theory. His space is the completion of a $c_{00}$ (all finitely supported scalar sequences) under an inductively defined norm. The base norm is the sup-norm $\|\cdot \|_0$.

For $n \in \mathbb{N}$ the norm $\|x\|_{n+1}$ norm is defined by

$ \|x\|_{n+1}= \sup\{\frac{1}{2} \sum^k_i \|E_ix\|_{n} \} $

where the supremum is taken over all sets $(E_i)_{i=1}^k$, where $E_i$ is a finite interval in $\mathbb{N}$, $\max E_i < \min E_{i+1}$ and $k \leq E_1$ (here $Ex$ denotes the restriction of $x$ to the coordinates of $E$). The Tsirelson norm is $\|x\|_T = \sup_n \|x\|_n$ and satisfies the following implicit equation

$ \|x\|_T= \max ( \|x\|_0 , \sup \frac{1}{2} \sum^k_i \|E_ix\|_T ).$

The space $T$ does not contain a copy of any $\ell_p$ or $c_0$. This solved a major open problem at the time (I should point out that Tsirelson actually defined the dual of $T$ which also has the property).

The, inductive, method he devised for producing this space eventually lead to the solutions of numerous problems in Banach space theory (way to numerous to mention). Moreover, the `necessity' of the inductive construction to produce spaces not containing any $\ell_p$ of $c_0$ is a problem that has been considered by Gowers as a polymath project (unfortunately not much progress here): http://gowers.wordpress.com/2009/02/17/must-an-explicitly-defined-banach-space-contain-c_0-or-ell_p/

Check out Boris Tsirelson's website for more info on his space: http://www.math.tau.ac.il/~tsirel/Research/myspace/main.html

Kevin Beanland
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The $n$-level Tower of Hanoi can be solved in $2^n - 1$ moves.

Not only does induction prove this, it actually shows you the solution!

Frank Thorne
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1)Proof of Euler's formula, V-E+F=2, with induction on F (number of faces).

2)Backward induction proof of generalized AM-GM inequality.

3)Proof of Heine-Borel theorem using Transfinite Topological induction.

Uday
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  • I am surprised at the idea of proving Heine-Borel by transfinite induction. What proof do you have in mind? (Or perhaps I should ask, what version of Heine-Borel do you have in mind?) – John Stillwell Mar 30 '12 at 23:02
  • To show [a,b]⊂R is compact. Consider a open cover. Then, base case {a} can be covered with one set from the cover. Let us assume that the interval [a,x] forever x∈[a,c),c≤b has a finite subcover. Then, we can show that the result holds at c also. Here the transfinite induction is on c. – Uday Mar 31 '12 at 05:26
  • OK, this is not transfinite induction in the usual sense, because it is not based on an underlying well-ordering. However, I agree that it is something like induction. Thurston and others have called it "topological induction." – John Stillwell Mar 31 '12 at 08:05
  • @John Thank you. Any references for 'topological induction'? – Uday Mar 31 '12 at 09:34
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    I first saw "topological induction" in one of the early drafts of Thurston's notes, around 1980. A more recent study of the idea is Iraj Kalantari's "Induction over the Continuum", which can be partly viewed at http://www.springerlink.com/content/uj314q217n7ln2n3/ – John Stillwell Mar 31 '12 at 11:09
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    Sorry, I might be dense, but as far as I understand, an induction that is "not based on an underlying well ordering" is just a bogus mathematical argument. – André Henriques Apr 01 '12 at 14:36
  • @André Can you substantiate? – Uday Apr 01 '12 at 15:11
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    @John I learnt this method of "topological induction" in Dieudonne's "Calcul infinitesimal", where he uses it to prove for example the uniform continuity of continuous functions on closed intervals. I actually used this in class, and students seemed to like it, visualizing the proof as a zipper that you can zip from a to b without being blocked in the middle... – Sylvain Bonnot Apr 01 '12 at 16:28
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    @Andre: you are not correct, induction can be performed on any well-founded relation, it need not be linear. For example, proof theorists and computer scientists often use induction on well-founded trees. – Andrej Bauer Apr 02 '12 at 06:30
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    Following up on "topological induction" or "induction over the continuum," there is a recent paper on it by Pete L. Clark at

    http://math.uga.edu/~pete/induction_completeness_brief.pdf

    He traces versions of the idea back as far as papers by Khinchin and Perron in the 1920s.

     – John Stillwell Apr 02 '12 at 07:09
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    @Andrej: I don't see how to recast this topological induction as well-founded induction. What would be the well-founded relation here? – Emil Jeřábek Apr 02 '12 at 11:32
  • @EmilJeřábek This is exceedingly late to the party but: there's a way to turn the argument into a classical transfinite induction argument (or perhaps more appropriately, "transfinite infinite descent" :P). Suppose $\mathcal{C}$ is a cover of $[0,1]$ by open intervals. We can recursively construct an ordinal-indexed sequence $S=(C_\eta){\eta<\alpha}$ of elements of $\mathcal{C}$ with the following properties: (1) $0\in C_0$, (2) $\sup(\bigcup{\delta<\eta}C_\delta)\in C_\eta\setminus \bigcup_{\delta<\eta}C_\delta$ for every $\eta<\alpha$, and (3) $1\in\bigcup_{\eta<\alpha}C_\eta$. (continued) – Noah Schweber Aug 12 '19 at 10:56
  • Note that we must have $\alpha$ be a successor ordinal (set $\alpha=\beta+1$) and moreover for each $\eta<\alpha$ the set $\bigcup_{\delta<\eta}C_\delta$ is closed downwards. Now walk down this sequence. Supposing for a contradiction that $\mathcal{C}$ has no finite subcover, we can now define an infinite descending sequence $(\theta_i){i\in\omega}$ of ordinals as follows: $\theta_0=\beta$, and if $0\not\in C{\theta_i}$ then $\theta_{i+1}$ is the least $\gamma<\theta_i$ such that $\inf(C_{\theta_i})\in C_\gamma$. I think this morally amounts to the usual argument, but "well-foundifies" it. – Noah Schweber Aug 12 '19 at 10:56
  • @AndréHenriques I think the argument here is that there is a principle applicable to a larger class of relations than just the well-founded ones which yields classical induction as a direct corollary and which has a very similar form. Calling this stronger principle something involving the word "induction" is at least somewhat justified, in my opinion. – Noah Schweber Aug 12 '19 at 10:58
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The following famous puzzle is a great example:

http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/

Steven Gubkin
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Goodstein's theorem hasn't yet been mentioned. A straightforward-looking arithmetic theorem with a surprise proof using transfinite induction. Also (the main interesting characteristic of the theorem), there is NO proof from ordinary first-order Peano arithmetic. It's actually equivalent to the formalized $\Sigma^0_1$-soundness (aka 1-consistency) of PA.

Emil Jeřábek
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none
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  • Nice pick, I should have thought about that (and about other statements not provable in PA)! – Lorenzo Lami Apr 03 '12 at 10:09
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    Goodstein's theorem, which is a $\Pi^0_2$ statement whose existential quantifier is witnessed by a function growing faster than any provably total recursive function of PA, is not equivalent to the consistency of PA, which is merely a $\Pi^0_1$ statement. I'll fix that. – Emil Jeřábek Apr 03 '12 at 10:35
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A problem I enjoyed in my undergraduate algorithms course is as follows:

Suppose you have a computing machine with the following architecture. There are $k$ stacks (for some $k$), input can be pushed onto the first stack, output is popped off of the last, and intermediate operations pop from one stack and push to the next in a line. The top of the stack may also be inspected and compared. Given a permutation of $\{1,\ldots,n\}$ in order as input, how many stacks $k$ do you require to sort the permutation? Describe an algorithm that achieves this bound.

One can prove the bound ($\log_2 n$) by induction, and then just state that this gives a natural recursive algorithm. The same technique was useful for a couple of other problems in a similar vein.

I think this certainly fits the bill of an elegant way to fulfill the task (prove a bound and give an achieving algorithm) in a nice class of cases.

The problem is originally from Knuth Vol. 1, and stack sorting is further elaborated on in this survey.

Edgar A. Bering IV
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Simultaneous induction as used in combinatorial group theory, for example in the proof of the Adyan-Novikov theorem providing the counterexample to the General Burnside Problem:

Some nice references about the nuts and bolts of this were supplied by Mark Sapir in an answer to one of my questions about this proof:

A synopsis of Adyan’s solution to the general Burnside problem?

Certainly the simultaneous induction technique is an important idea in constructing such monster groups.

Community
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Jon Bannon
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The "Ercules and the Hydra" problem, as found in "L. Kirby and J. Paris. Accessible independence results for peano arithmetic. London Mathematical Society, 4:285 293, 1982.".

Using transfinite induction, it is possible to show that Hercules will always kill the hydra (with a finite number of blows) regardless of the strategy chosen to chop off hydra's heads. Moreover, this fact is not provable within Peano Arithmetic.

Lorenzo Lami
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Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that $\sqrt 2$ is irrational. All descent arguments are basically induction.

Dan Piponi
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There are infinifely many positive integers $n$ such that $2^n+1$ is divisible by $n$. One can show this by proving that if $n$ divides $2^n+1$, then $2^n+1$ divides $2^{2^n+1}+1$.

vgmath
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The original proof of Van der Waerden's theorem on monochromatic arithmetic progressions comes to mind. Well, the more recent ones too by the way.

Johan Wästlund
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In pro-algebraic geometry you get to see some nice arguments by induction. For example, M. Kim proves that the continuous cohomology $$H^1(G_{\mathbf{Q}_p},\pi_{1,et}^{uni}(X))$$ is representable by induction on the terms in the lower central series of the $\mathbf{Q}_p$ pro-unipotent algebraic group associated to the etale fundamental group of a curve $X$. Not very surprising, but still crucial for the argument.

For a reference, see page 639 in http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf

Harry
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I came across this one browsing old questions under the logic tag.

Here's proof of the Principle of Induction:

https://openmathematics.org/#induction

The GitHub repository URL is available in the 'More information' section. There is also a link to a presentation on the subject.

Interestingly, you need recursion only in the definition of the context.

I'm not claiming this is a great proof, obviously. But I think it's cute, because you derive the Principle of Induction from nothing but a construction of the natural numbers and the only trick you need, if you want to call it that, is, well, recursion.

James Smith
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