# Examples of interesting false proofs

According to Wikipedia False proof

For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.

The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics.

What are some examples of interesting false proofs?

• Is this a duplicate? – Bruce Westbury Apr 21 '12 at 17:19
• the answers to this will turn out to replicate many of the responses to Gowers' famous question on "false beliefs", so I am not so sure if this question should remain open. – Suvrit Apr 22 '12 at 5:46
• A false proof is not the same as a false belief. One can read a false proof, know for certain that the conclusion is false (so there is no false belief), and still have trouble pinpointing the error. – Steven Landsburg Apr 22 '12 at 15:36
• I'm surprised no one has mentioned Stallings's false proof of the Poincare Conjecture, in his paper "How Not to Prove the Poincare Conjecture". – Steve D Apr 30 '12 at 22:13
• There are no false proofs, by definition. – Fernando Muro Mar 19 '13 at 17:32

My favorite example is the following proof of the Cayley-Hamilton theorem, which caused me some disconcertion when I was a student. Let $A$ be a square matrix, and call $p(t) = \det(tI - A)$ its characteristic polynomial. Then $p(A) = \det(AI-A) = 0$.

• Somehow this can be made into a correct proof with the Zariski topology. – Martin Brandenburg Apr 21 '12 at 15:46
• It can be made into a correct proof in several ways; unfortunately, they all spoil the pristine elegance of the false proof. – Angelo Apr 21 '12 at 15:48
• This false proof is so good I've got used to proposing to my students $q(t)=tr(tI-A)$ as an antidote. – domenico fiorenza Apr 21 '12 at 17:15
• @domenico: that's as close to a funny mathematical joke as we are going to get :D – Mariano Suárez-Álvarez Apr 23 '12 at 2:25
• Another antidote is the following: if $\det(B-A)=0$, it does not imply that $p(B)=0$. So why should it imply for $B=A$? – Fedor Petrov Apr 23 '12 at 6:49

$$e^i = (e^i)^{(2\pi/2\pi)} = (e^{2\pi i})^{1/2\pi} = 1^{1/2\pi} = 1.$$

I first saw this one many years ago, written on the wall of a bathroom stall in the Princeton University math department.

• Math departments have the best bathroom graffiti. – Noam D. Elkies Jan 15 '13 at 4:51
• Oh, this is really good. – Newb Nov 27 '13 at 16:20
• This is not conceptually different from $-1=(-1)^{2/2}=((-1)^2)^{1/2}=1^{1/2}=1$ – Qfwfq Apr 5 '19 at 18:51

I like this one, invented by T.Clausen in 1827: since $e^{2\pi i n}=1$ for all integers $n$, we have $e^{2\pi i n+1}=e$, which implies $e^{(2\pi i n+1)^2}=(e^{2\pi i n+1})^{2\pi i n+1}=e^{2\pi i n+1}=e$. Now expanding the square at the exponent gives $$e^{1-4\pi^2n^2+4\pi n i}=e$$ and after simplifying $$e^{-4\pi^2n^2}=1$$ for all $n$.

• How easily we forget that everything must be defined! – some guy on the street Jun 16 '12 at 16:22

In the definition of an equivalence relation $\sim$, the reflexivity of $\sim$ is redundant: Indeed, for any $x$, by the symmetric property we have $x \sim y$ implies $y \sim x$. By transitivity we have $x \sim y$ and $y \sim x$ imply $x \sim x$. Therefore, using only symmetry and transitivity, we obtain reflexivity.

• But this proves the result if there is at least one equivalence? – David Corwin Mar 20 '13 at 18:02
• As Davidac says you only need that for any $x$ there exists at least one $y$ such that $x \sim y$. I set this as a homework question for my undergraduate groups course every year and the answers systematically ignore the necessary assumption – Paul Levy Mar 20 '13 at 20:33
• A very similar fallacy: a subset $H$ of a group $G$ is a subgroup if it contains the unit, is closed under multiplication, and is closed under inverses. CLAIM: the second and third condition imply the first. Indeed, take any $x\in H$. Then $x^{-1}$ is also in $H$, so $xx^{-1}=e$ is in $H$. – Dan Petersen Jan 10 '19 at 17:10
• @DanPetersen Interesting. How is this a fallacy? – trisct Dec 27 '19 at 5:14
• @trisct It is a fallacy since $H=\varnothing$ is closed under multiplication and inverses, but is not a subgroup; in this case the first step "take any $x \in H$" fails. – Dan Petersen Mar 31 at 19:45

Ethan Akin's "proof" that all vector bundles are stably trivial, and hence the $$K$$-theory of any space must vanish:

Let $$V$$ be a vector bundle over the base space $$B$$. Let $$T$$ be a trivial bundle of the same rank as $$V$$. To show that $$V$$ is stably trivial, it suffices to prove that $$V\oplus V=V\oplus T$$.

Let $$P$$ be the principal bundle associated with $$V$$. Pull $$P$$ back over itself to get a bundle $$Q$$:

Then $$Q$$ (together with the map to $$B$$) is the principal bundle associated to $$V\oplus V$$. But the bundle $$Q\rightarrow P$$ clearly has a section, namely the diagonal map (viewing $$Q$$ as a subspace of $$P\times P$$). Thus $$Q=P\times GL_n$$, which (together with the same map to $$B$$) is the principal bundle associated to $$V\oplus T$$.

(Reference: Ethan Akin, K-theory doesn’t exist, JPAA 12 (1978) pp.177–179.)

• Was the paper peer reviewed? – joro Apr 22 '12 at 14:49
• I like this. It shows how easy it is to fool yourself and others by drawing a diagram and saying ''the natural map'' and ''canonically isomorphic'' a few times! Apparently, the paper was peer-reviewed, but it states clearly that the purpose was to discuss a fallacious proof. – Johannes Ebert Apr 30 '12 at 19:55
• Every this question bubbles back up to the front page again, this answer is the one that stops me in my tracks for 5 minutes trying to find the error. – Peter LeFanu Lumsdaine Jan 10 '19 at 8:42

True Theorem The symmetric groups (consisting of all permutations) on infinite sets of different cardinalities are not isomorphic.

False proof: The two groups have different cardinalities, since there are $2^\kappa$ many permutations of an infinite set of size $\kappa$, and $\kappa\lt\lambda$ implies $2^\kappa\lt 2^\lambda$. QED

See the question: Can the symmetric groups on sets of differing infinite cardinalities be isomorphic? for further information and a correct proof.

I find the false proof illuminating, since it shows the limitation of a naive treatment of the continuum function $\kappa\mapsto 2^\kappa$. It simply isn't necessarily the case that the two groups have different cardinalities, even though it is necessarily the case that they are not isomorphic.

Theorem. $\int_0^\infty \sin x \phantom. dx/x = \pi/2$.

Poof. For $x>0$ write $1/x = \int_0^\infty e^{-xt} \phantom. dt$, and deduce that $\int_0^\infty \sin x \phantom. dx/x$ is $$\int_0^\infty \sin x \int_0^\infty e^{-xt} \phantom. dt \phantom. dx = \int_0^\infty \left( \int_0^\infty e^{-tx} \sin x \phantom. dx \right) \phantom. dt = \int_0^\infty \frac{dt}{t^2+1},$$ which is the arctangent integral for $\pi/2$, QED.

The theorem is correct, and usually obtained as an application of contour integration, or of Fourier inversion ($\sin x / x$ is a multiple of the Fourier transform of the characteristic function of an interval). The poof, which is the first one I saw (given in a footnote in an introductory textbook on quantum physics), is not correct, because the integral does not converge absolutely. One can rescue it by writing $\int_0^M \sin x \phantom. dx/x$ as a double integral in the same way, obtaining $$\int_0^M \sin x \frac{dx}{x} = \int_0^\infty \frac{dt}{t^2+1} - \int_0^\infty e^{-Mt} (\cos M + t \cdot \sin M) \frac{dt}{t^2+1}$$ and showing that the second integral approaches $0$ as $M \rightarrow \infty$; but this detour makes for a much less appealing alternative to the usual proof by complex or Fourier analysis.

Still the double-integral trick can be used legitimately to evaluate $\int_0^\infty \sin^m x \phantom. dx/x^n$ for integers $m,n$ such that the integral converges absolutely (that is, with $2 \leq n \leq m$; NB unlike the contour or Fourier approach this technique applies also when $m \not\equiv n \bmod 2$). Write $(n-1)!/x^n = \int_0^\infty t^{n-1} e^{-xt} \phantom. dt$ to obtain $$\int_0^\infty \sin^m x \frac{dx}{x^n} = \frac1{(n-1)!} \int_0^\infty t^{n-1} \left( \int_0^\infty e^{-tx} \sin^m x \phantom. dx \right) \phantom. dt,$$ in which the inner integral is a rational function of $t$, and then the integral with respect to $t$ is elementary. For example, when $m=n=2$ we find $$\int_0^\infty \sin^2 x \frac{dx}{x^2} = \int_0^\infty t \frac2{t^3+4t} dt = 2 \int_0^\infty \frac{dt}{t^2+4} = \frac\pi2.$$ As a bonus, we recover a correct proof of our starting theorem by integration by parts:

$$\frac\pi2 = \int_0^\infty \sin^2 x \frac{dx}{x^2} = \int_0^\infty \sin^2 x \phantom. d(-1/x) = \int_0^\infty \frac1x d(\sin^2 x) = \int_0^\infty 2 \sin x \cos x \frac{dx}{x};$$ since $2 \sin x \cos x = \sin 2x$, the desired $\int_0^\infty \sin x \phantom. dx/x = \pi/2$ follows by a linear change of variable.

Exercise Use this technique to prove that $\int_0^\infty \sin^3 x \phantom. dx/x^2 = \frac34 \log 3$, and more generally $$\int_0^\infty \sin^3 x \frac{dx}{x^\nu} = \frac{3-3^{\nu-1}}{4} \cos \frac{\nu\pi}{2} \Gamma(1-\nu)$$ when the integral converges. [Both are in Gradshteyn and Ryzhik, page 449, formula 3.827; the $\nu=2$ case is 3.827#3, credited to D. Bierens de Haan, Nouvelles tables d'intégrales définies, Amsterdam 1867; the general case is 3.827#1, from Gröbner and Hofreiter's Integraltafel II, Springer: Vienna and Innsbruck 1958.]

• +1: "Poof" is a great new term for an "incorrect proof," whether you intended it or not ;) – David Corwin Mar 20 '13 at 18:37
• Thanks. Yes, it was intentional (I repeated it in the text); it's not new, though apparently not well-known, and I don't remember where I got it from. – Noam D. Elkies Mar 20 '13 at 21:32

Theorem: All people have the same eye color.

Proof by induction: we prove the statement "All members of any set of people have the same eye color". This is clearly true for any singleton set.

Now, assume we have a set $S$ of people, and the inductive hypothesis is true for all smaller sets. Choose an ordering on the set, and let $S_1$ be the set formed by removing the first person, and $S_2$ be the set formed by removing the last person.

All members of $S_1$ have the same eye color, and also for $S_2$. However, $S_1 \cap S_2$ has members from both sets, so all members of $S$ have the same eyecolor. $\square$

Theorem: Every bounded differentiable function $$f\colon \mathbb{R}\to \mathbb{R}$$ is constant.

Proof. By assumption there exist real numbers $$M,N$$ such that
$$N\leq f(x) \leq M.$$ Taking derivatives we get $$0\leq f'(x)\leq 0.$$ Hence $$f'(x)=0$$ so $$f$$ is constant. QED

• For $\mathbb C$ it’s true but the proof is a little different. – Zach Teitler Oct 1 '19 at 21:50

Here's a nice false proof of the continuum hypothesis.

Consider the rational numbers $\mathbb{Q}$ as a totally ordered field. We can add an indeterminate $T_0$ and make it positive but infinitely small (i.e., smaller than positive any element of $\mathbb{Q}$), that is, order $\mathbb{Q}(T_0)$ by lexicographic order of the Laurent series expansion at $0$. Then we can add another indeterminate $T_1$ and make it positive but infinitely small (i.e., smaller than any positive element of $\mathbb{Q}(T_0)$). This process can be iterated transfinitely and we can add $\aleph_1$ indeterminates $T_\iota$ for $\iota<\omega_1$, each infinitely smaller than all the previous ones. The resulting field $K = \mathbb{Q}(T_\iota)$ has cardinality $\aleph_1$ as is easy to show. Now any positive sequence converging to $0$ in $K$ must be eventually constant because it has to cross uncountably many $T_\iota$. So any Cauchy sequence in $K$ is eventually constant. So any Cauchy sequence in $K$ is convergent. So $K$ is complete. But since $K$ contains $\mathbb{Q}$, it contains $\mathbb{R}$. So we have a set of cardinality $\aleph_1$ containing $\mathbb{R}$, which proves the continuum hypothesis.

(The error, of course, is simply that the notion of "completeness" is wrong and its use is nonsense. But if you tell it quickly enough, many people will fall for it.)

• I wouldn't say its "nonsense" -- there's a perfectly sensible notion of "completeness" and "completion" for ordered fields. It just isn't detectable via sequences in general; I'd say that's the real error here. – Harry Altman Jan 15 '13 at 1:50

Not so much of a proof but rather a computation.

$$\frac{64}{16} = \frac{\not{6}4}{1\not{6}}= \frac{4}{1} = 4$$

by canceling the $6$s.

• This reminds me of a student in one of my classes who simplified $\frac{\sin x}{n} = six$. I almost gave him credit for that. – Donu Arapura Apr 21 '12 at 15:06
• Likewise $19/95 = 1/5$, $26/65 = 2/5$, and (a bit less satisfactory because not in lowest terms) $49/98 = 4/8$. – Noam D. Elkies Apr 21 '12 at 19:59
• For more examples and analysis of these "weird fractions", see A Pumping Lemma for Invalid Reductions of Fractions, Michael N. Fried and Mayer Goldberg, The College Mathematics Journal, Vol. 41, No. 5 (November 2010), pp. 357-364. – Doug Chatham Apr 22 '12 at 21:24
• My algebra students know better than to fall for this, but they will try to reduce $\frac{x+3}{x+4}$ to $\frac{3}{4}$. So then I invoke this, asking them if $\frac{13}{14}$ reduces to $\frac{3}{4}$, and (when they say No) asking them what happens when $x := 10$. – Toby Bartels Jun 16 '12 at 15:25
• Those who were as mystified by cheval/oiseau may find enlightenment at algorythmes.blogspot.com/2009/09/cheval-oiseau-pi.html When that link disappears, just type cheval/oiseau = pi into (whatever search engine has replaced) Google. – Gerry Myerson Jan 10 '19 at 15:20

One night I proved that every module is flat. Let $M$ be an $R$-module and let $\mathfrak{a}$ be any ideal of the ring $R$. Tensoring the natural inclusion $i:\mathfrak{a} \to R$ we obtain $i_\ast : M \otimes \mathfrak{a} \to M \otimes R$ such that $i_\ast(x\otimes y)=x\otimes i(y)=x\otimes y$, for every $x\in M$ and $y \in \mathfrak{a}$. So $i_\ast$ is injective and we conclude that $M$ is flat...

I always liked this proof, from the theory of Umbral Calculus developed by Rota (See "Combinatorics: The Rota Way", by Joseph Kung, Gian Carlo Rota and Catherin Yan, chapter 4.2).

Proposition: Let $(a_n)_{n\geq 0}$ and $(b_n)_{n \geq 0}$ be sequences. Then $$b_n=\sum_{k=0}^n\binom{n}{k} a_k \ \text{ for all } n \Longleftrightarrow a_n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b_k \ \text{ for all } n.$$

The heuristic proof use the notion of "raising and lowering subscripts and superscript". Raising subscripts at the left side we obtain $$b^n=\sum_{k=0}^n\binom{n}{k}a^k=(a+1)^n.$$ Hence, for all $n$, $$a^n=(b-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b^k.$$ Lowering exponents, we obtain the inverse relation.

• This doesn't look like a false proof. Rather, it's a proof that looks absurd at first glance but that can be made rigorous if you set up the right theoretical framework. Sort of like certain kinds of manipulations with divergent series, or arguments using infinitesimals, or the Dirac delta function. – Timothy Chow Apr 23 '12 at 14:32

One usual "proof" of Leopoldt Conjecture is that $\mathbb{Z}_p$ is $\mathbb{Z}$-flat, hence the rank of the $p$-adic completion of the units of a number field has the same rank of the units themselves (which is Leopoldt Conjecture) because you can obtain the completion simply as $\mathcal{O}^\times\otimes\mathbb{Z}_p$.

I came across this one in a book of false proofs, the name of which I can't remember. It stuck out because it's not the usual hidden division by $$0$$ or unestablished base case in an induction.

Theorem: Every implication or its converse must be true.

Proof:

Check the truth table for $$(P\to Q)\vee (Q\to P)$$ and note that it is a tautology.

$$\Box$$

However we know that there are many cases where neither an implication nor its converse is true. For example take $$P$$ to be "$$n$$ is odd" and $$Q$$ to be "$$n$$ is prime."

• Actually, I like this one. Even if universal quantification is implicit, it is better not to forget that it is there. – Alex Gavrilov Jan 10 '19 at 11:14

My favourites are "close" to formal false proofs in Coq.

1) In reply to a challenge by coq developer

Who can address this challenge: find a "simple" statement $T$ (simple in the sense that anyone with a minimal background in logics can understand) such that you can prove both $T$ and $\neg T$ in Coq.

Daniel Schepler solved it here. Daniel's proof was valid and passed coqchk, though it was not enough to prove False in Coq - Coq gave an "Universe inconsistency". AFAICT the proof encoded a paradox.

2) Damien Pous announced and gave link to code

There is a bug with vm_compute and values obtained from functors applications: using the attached code, I can produce an assumption-free proof of False, or Bus errors.

False proofs in Coq are difficult because Coq produces a "certificate" that can be checked for validity (if one doesn't check the certificate and is happy with the compiler as most people do, it is much easier).

Ma & Pa Kettle Math Lesson

Given any $x$, we have (by using the substitution $u=x^2/y$) $$\int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy = \biggl[x e^{-x^2/y}\biggr]_0^1 = x e^{-x^2}.$$ Therefore, for all $x$, \eqalign{e^{-x^2}(1-2x^2) &= {d\over dx}(xe^{-x^2})\cr &= {d\over dx} \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy\cr &= \int_0^1 {\partial \over \partial x} \biggl({x^3\over y^2} e^{-x^2/y}\biggr)\,dy\cr &= \int_0^1 e^{-x^2/y} \biggl({3x^2\over y^2} - {2x^4\over y^3}\biggr)\,dy.\cr} Now set $x=0$; the left-hand side is $e^0(1-0) = 1$, but the right-hand side is $\int_0^1 0\,dy = 0$.

The main idea for this proof comes from an entry in Gelbaum and Olmstead's book Counterexamples in Analysis.

I can't remember where I first saw this: does anybody recognise it?

Let $I$ be the operator, from $C^0(\mathbb{R})$ to itself, which takes $f(x)$ to $\int_0^xf(z)dz$.

Since the exponential function $e(x)$ is its own derivative, we integrate both sides to get $e(x) = I(e(x)) + 1$. Regarding $1$ as the identity operator, we can rearrange to get $$(1-I)e(x) = 1,$$ and hence $$e(x) = \frac{1}{1-I}1 = (1 + I + I^2 + \cdots)1 = 1 + x + \frac{x^2}{2} + \cdots.$$

• As long as you verify that I is a contraction operator on continuous functions on an interval of length less than 1, this works just fine: the series converges in the max norm, i.e. uniformly. Then you can check that this particular series happens to converge everywhere. Although omitting this check is an error, it seems to me that it just exposes an error in the strategy of using a purely algebraic argument to prove an analytic statement. – Ryan Reich May 19 '12 at 22:51
• Indeed, $I$ has zero spectral radius, so the series for $(1-z I)^{-1}$ even converges for all $z$. Notoriously, the exponential series is a particular case of a geometric series. – Pietro Majer Aug 16 '12 at 9:07

Another subtle variant of the induction fallacy suggested by Fedor Petrov.

Theorem: every graph without isolated nodes is connected.

Proof Induction on the number of nodes. Clearly the result is true for graphs with 1 (void statement) and 2 nodes. Now, assume we have proved the statement for graphs with up to $$n$$ nodes. Take a graph with $$n$$ nodes; by induction hypothesis it must be connected. Let's add a non-isolated node to it. As this node is not isolated, it is connected to one of the other $$n$$ nodes. But then it's easy to conclude that the whole graph of $$n+1$$ nodes is connected!

• Wow! Great example! – Behnam Esmayli Jun 29 at 2:24

$\pi$ is irrational : if $\pi=a/b$ is irreducible, and $a$ is divisible by an odd prime $p$, the series for $\sin \pi =\pi-\pi^3/6+\pi^5/120-\dots$ converges in the $p$-adics, and the limit is obviously not zero, absurd (if $a=2^n$, $n>1$ and the convergence is assured in the 2-adics, with the same contradiction).

• True story that I witnessed in a US precalculus class: the teacher told the class that $\pi$ was a rational number, since $\pi = C/d$, where $C$ is the circumference of a circle and $d$ is the diameter. Since $\pi$ can be written as a fraction, it is rational. This still makes me cringe to this day. – John Engbers May 19 '12 at 18:28
• Reminds me of mathoverflow.net/a/81360/88133. – Zach Teitler Oct 1 '19 at 22:16

I have known the following for 45 years: in the Euclidian plane, every triangle is isosceles.

The false proof needs a handmade picture; take your pen, it's easy. Start from a triangle $$ABC$$. Draw the perpendicular bisector of $$BC$$, and the angle bisector from $$A$$. Let $$I$$ be their intersection (if it is not unique, you are done). Let $$J$$ be the projection of $$I$$ over $$AB$$, $$K$$ that over $$AC$$. Considering the right triangles $$AIJ$$ and $$AIK$$, we see that (lengths) $$AJ=AK$$, and that $$IJ=IK$$. Then looking at right triangles $$BIJ$$ and $$CIK$$, we obtain that $$BJ=CK$$. We conclude that $$AB=AJ+JB=AK+KC=AC.$$

The falsity is that one of $$J$$ or $$K$$ is in the triangle, and the other one is out. Therefore one of the sums above (and only one) should be a difference.

Theorem: Every totally disconnected set has the discrete topology.

Proof: Let $X$ be a totally disconnected set. If $X$ has only one element, the conclusion clearly follows. Otherwise, for distinct points $a, b \in X$, we have that {$a, b$} $\subset X$ is not connected. Therefore, {$a, b$} admits a separation; but the only way to write this as a disjoint union of nonempty sets is {$a$} $\cup$ {$b$}. Since this gives a separation, each of {$a$} and {$b$} is open. In particular, {$a$} is open for any $a \in X$; so $X$ has the discrete topology. Q.E.D.

• Well, the proof would prove more and be much simpler if, instead of looking at the subspace $\{a,b\}$, you just look at the subspace $\{a\}$. Now $\{a\}$ is obviously open, so every topological space whatsoever is discrete. – Toby Bartels Jun 16 '12 at 15:36
• UGH this reminds me of when I once wrote about rational points $\mathbb{Q}^d$ being discrete in euclidean space and was interrogated as to why... I must have thought connected components are clopen in their containing space (I had separations in the back of the mind, which I now have it sharply burned that components do not form). – Vandermonde Apr 27 at 3:47

A common mistake in using induction for statements concerning finite sets is the bad logic "prove it for 1-set, and if we have proved this for $$n$$-set, add an element and prove it for $$(n+1)$$-set". I like the following illustrative example proposed by Sergey Berlov:

Theorem. A simple undirected graph with $$n$$ vertices and $$n$$ edges contains a triangle.

Poof. For $$n<3$$ there are simply no such graphs. For $$n=3$$ a triangle exists. Now add a vertex and an edge. The triangle does not diappear, right?

Let me recycle this.

$\phantom{*******}$

I'm fond of the following false proof of the Strong Law of Large Numbers. Let $$X$$ be a random variable with expected value $$\mu$$ and variance $$\sigma^2$$, and let $$X_1, X_2, \dots$$ be i.i.d. copies of $$X$$. Then $$\operatorname{Var} \left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{1}{n^2} \cdot n \sigma^2 = \frac{\sigma^2}{n} \rightarrow 0 \textrm{ as } n\rightarrow\infty$$ and since a random variable with variance 0 takes on a single value with probability 1, we must have $$\lim_{n\rightarrow\infty} \frac{1}{n} \sum_{i=1}^n X_i = \mu \textrm{ almost surely.}$$ (It's a memorable heuristic reason to tell undergraduate probability students, even if not a true argument.)

• It does constitute a proof of the weak law of large numbers, and it shows that if the limit $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n X_i$ exists almost surely, it must equal $\mu$. – Nate Eldredge Mar 20 '13 at 22:54
• For the SLLN, isn't there an issue in exchanging the limit with the integral implicit in the variance? Even for the WLLN, it seems to me that the technical details needed (e.g. Chebyshev's Inequality) rather "ruin the pristine elegance", to quote a previous comment. – Russ Woodroofe Mar 21 '13 at 5:03

This one is interesting also in the sense of having an interesting history: In his writings, Cantor used a principle "every set can be well-ordered." As far as I understand, he claimed that it was obvious.

In 1905, König proposed and published an alleged proof that $$\mathbb{R}$$ cannot be well-ordered:

Suppose that $$\mathbb{R}$$ is well-ordered with an ordering relation $$\preceq$$. Since there are uncountably many reals, there is an undefinable one (say, undefinable even in the language with the symbol $$\preceq$$). Since $$\preceq$$ is a well-ordering, there exists the least one $$x_0$$ which is not definable. But we have just defined it. A contradiction.

In the same year, a paper by Zermelo ''A proof of the principle that every set can be well-ordered'' was published. The author reduced the controversial principle to several reasonably looking statements which became a basis of what is known as Zermelo--Fränkel set theory.

On the other hand, what became known as König paradox had to wait a bit to be resolved until better understanding of truth predicates was obtained.

Here is an interesting false proof as to how to multiply $$2 \cdot 2$$. Taken from this link.

$$\Large\textbf{Another example}$$:

A cavalry sergeant has 24 horses available which he needs to put on 6 carriages. So he needs to divide 24 by 6. He figures that 6 will go into 24 at least once, so he puts down a 1. Subtracting 6 from 24, he gets 18, and he remembers that 18/6=3. So he comes up with the answer 13.

After considerable difficulty with implementing his solution he consults his lieutenant. The lieutenant checks the calculation by evaluating 13*6:

3*6=18 1*6=6

Implementation of the result still remains elusive so they consult the colonel, who uses a different method to check. Write down 13 six times and add.

13

13

13

13

13

13

In adding this up, the colonel arrives at the following sequence of intermediate results: 3,6,9,12,15,18,19,20,21,22,23,24.

• I think something must have gotten lost in the typography--this doesn't make a lot of sense as it appears on my screen. – Dylan Thurston Apr 22 '12 at 11:29
• This is incomprehensible as posted but begins to make sense after you've followed the link in Gerald Edgar's answer. – Steven Landsburg Apr 22 '12 at 16:51
• I could not figure out how to properly line up columns. I know how to do it in TeX, just not in "Math Overflow TeX." I hope the verbal description I substituted is clearer. – Michael Renardy Apr 22 '12 at 22:59
• Essentially the same proof is shown here: youtube.com/watch?v=Lo4NCXOX0p8 (an old Abbot & Costello sketch). – Ketil Tveiten Apr 25 '12 at 11:11

The Graham Pollak theorem is discussed at this link Combinatorial results without known combinatorial proofs . I came up with a nice short and incomplete proof of it. The tricky part for me was to realize it was incomplete. Follow the commentary if you want to see my "D'oh" moment. The induction started by taking an a,b complete bipartite subgraph of an (a+b) complete graph.

Gerhard "The Induction Looked So Pretty" Paseman, 2012.04.21