78
$\begingroup$

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?

$\endgroup$
15
  • 3
    $\begingroup$ Is this a duplicate? $\endgroup$ Apr 21, 2012 at 17:19
  • 3
    $\begingroup$ 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. $\endgroup$
    – Suvrit
    Apr 22, 2012 at 5:46
  • 31
    $\begingroup$ 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. $\endgroup$ Apr 22, 2012 at 15:36
  • 2
    $\begingroup$ 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". $\endgroup$
    – Steve D
    Apr 30, 2012 at 22:13
  • 5
    $\begingroup$ There are no false proofs, by definition. $\endgroup$ Mar 19, 2013 at 17:32

49 Answers 49

1
2
6
$\begingroup$

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

alt text


$\Large\textbf{Another example}$:

alt text

$\endgroup$
5
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ 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 $\mu$. $\endgroup$ Mar 20, 2013 at 22:54
  • $\begingroup$ 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. $\endgroup$ Mar 21, 2013 at 5:03
4
$\begingroup$

An excelent example is the iscosceles triangle fallacy. Here is a link to it in wikipedia http://en.wikipedia.org/wiki/Mathematical_fallacy#Fallacy_of_the_isosceles_triangle

$\endgroup$
1
  • $\begingroup$ That silly article appeals to "accurate instruments", when all that's needed is the circumscribed circle and the central/peripheral angle identity. Check: The location of D depends only on the angle at A! $\endgroup$ Jun 16, 2012 at 16:49
4
$\begingroup$

In S. Bosch's Algebra, exercise 3.4.2 is to find an error in the following existence proof of an algebraic closure of a field $K$ (my translation):
"Consider all algebraic extensions of $K$. Since for a totally ordered (w.r.t. inclusion) family $(K_i)_{i \in I}$ of algebraic extensions of $K$, the union $\bigcup_{i \in I} K_i$ is an algebraic extension of $K$, Zorn's lemma shows the existence of a maximal algebraic extension, i.e. of an algebraic closure of $K$."

Added: Cf. https://math.stackexchange.com/q/621944/96384 for various discussions around, and actually working variants of, this flawed proof.

$\endgroup$
2
  • 4
    $\begingroup$ What is the right answer? Is it that "why is the collection of all algebraic extensions of a given field K a set?". $\endgroup$
    – knsam
    Jun 11, 2014 at 8:37
  • $\begingroup$ I think so, too. $\endgroup$ Jun 14, 2014 at 10:37
4
$\begingroup$

I like to amuse calculus students with this trick: let us calculate by integrating by parts: $$ \int \frac{dx}{x}=\int (x')\frac{1}{x}\,dx=x\cdot\frac{1}{x}-\int x\cdot \left(\frac{1}{x}\right)'\,dx=1+\int\frac{dx}{x}, $$ and we simplify to $0=1$.

$\endgroup$
4
$\begingroup$

I just found the following false proof of the (correct!) Skolem-Mahler-Lech theorem, which I think is interesting.

Statement of (correct) theorem: Suppose $f(z) := \sum_{n=0}^{+\infty} a_n z^n \in \mathbb{C}[[z]]$ is rational. Let $b_n$ equal $1$ when $a_n\neq 0$ and $0$ when $a_n = 0$. Then $g(z) := \sum_{n=0}^{+\infty} b_n z^n$ is also rational.

False proof: Since $f$ is defined by a linear recurrence relation, correcting for the uninteresting constant term, we can interpret it as the series recognized by a weighted finite automaton on the unary language (i.e., consisting of words over the single letter $z$; so the automaton is just a digraph with complex “multiplicities” associated to edges, and $a_n$ is the number of paths of length $n$, each counted with a multiplicity given by the product of the multiplicities of the edges, from an initial node to a final node: see, e.g., Bousquet-Mélou, “Rational and algebraic series in combinatorial enumeration”, §2). Now make this automaton deterministic (or at least unambiguous) while forgetting multiplicities: in the new automaton, the number of paths of length $n$ from an initial node to a final node is simply $b_n$, i.e., $1$ or $0$ according as there is or isn't such a path in the original automaton. But for the same reason (backwards), $g$ is now given by a linear recurrence relation, so it is rational.

Comment: The error is simply that when forgetting multiplicities we also forget possible cancellations between them: two paths could have multiplicities summing to zero. But the proof does work, and generalize to more variables, when $f$ is $\mathbb{N}$-rational, because no cancellation is possible: see the correct statements in Berstel & Reutenauer, Noncommutative Rational Series with Applications, esp. chapter 3 lemma 1.4. So the idea of the proof isn't stupid and gives related theorems, and the conclusion as stated is correct, yet the proof probably can't be fixed to yield that exact conclusion (because it would then work over any field, which isn't true), so I think this qualifies as an interesting proof.

$\endgroup$
3
$\begingroup$

Timothy Chow's answer has a nice application. Let $n,x,y,z$ be natural numbers such that $x^n+y^n-z^n=0$. It follows that $e^{x^n+y^n-z^n}=1=e^i$ and the absurd $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$

$\endgroup$
2
  • $\begingroup$ It can be used in the Millenium Prize Problems too ;-) $\endgroup$
    – joro
    Apr 25, 2012 at 11:10
  • $\begingroup$ We must take seriously that $e^i=1$ was written on a wall of Princeton University math department! Of course, I'm enjoying of the friendly tone of your question (the tag is "recreational"). $\endgroup$
    – Daniele
    Apr 25, 2012 at 14:10
3
$\begingroup$

I have always found interesting, as a student as well as teacher, the "proof" that every derivative is continuous:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Fix any $x_0 \in \mathbb{R}$ and $h > 0$, by the mean value theorem we find $\xi \in (x_0,x_0+h)$ such that:

$$ f'(\xi) = \frac{f(x_0+h) - f(x_0)}{h} \implies \lim_{h \to 0} f'(\xi) = \lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} =f'(x_0),$$

where in the last equality we used that $f$ is differentiable. The conclusion follows since $h \to 0$ entails $\xi \to x_0$.

$\endgroup$
2
  • $\begingroup$ I guess this is basically the valid proof that $f'$ doesn't have removable or jump discontinuities (if you modify a little bit to use one-sided limits). $\endgroup$ Mar 26, 2021 at 21:13
  • $\begingroup$ Yep, exactly: you just have to assume that the derivative has (finite) one sided limits. $\endgroup$
    – GaC
    Mar 27, 2021 at 8:56
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

Add them: 24.

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.

$\endgroup$
4
  • 3
    $\begingroup$ I think something must have gotten lost in the typography--this doesn't make a lot of sense as it appears on my screen. $\endgroup$ Apr 22, 2012 at 11:29
  • 5
    $\begingroup$ This is incomprehensible as posted but begins to make sense after you've followed the link in Gerald Edgar's answer. $\endgroup$ Apr 22, 2012 at 16:51
  • $\begingroup$ 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. $\endgroup$ Apr 22, 2012 at 22:59
  • 2
    $\begingroup$ Essentially the same proof is shown here: youtube.com/watch?v=Lo4NCXOX0p8 (an old Abbot & Costello sketch). $\endgroup$ Apr 25, 2012 at 11:11
2
$\begingroup$

Some years ago, I came up with this false proof of the irrationality of $\pi$.

It suffices to prove that $x=\pi-3$ is irrational.

For real $y$ with $0\le y\lt1$, and positive integer $j$, define $d_j(y)$ to be the $j$th digit in the decimal expansion of $y$.

Let $r_1,r_2,\dots$ be an enumeration of the rationals in $[0,1)$. The $\it value$ of this enumeration is $n$ if $d_n(r_n)=d_n(x)$ and $d_j(r_j)\ne d_j(x)$ for $j\lt n$. If there is no such $n$, then the value of the enumeration is infinite. Note that if there is an enumeration of infinite value, then $x$ is irrational; it cannot equal any of the enumerated rationals, as it differs from the first rational in (at least) the first decimal place, from the second in the second, etc.

Note also that there are enumerations of arbitrarily large value. For, given any $n$, you can find $n$ rationals such that the first differs from $x$ in the first decimal, the second differs from $x$ in the second decimal, and so on, and then any enumeration that starts off with these $n$ rationals will have value greater than $n$.

Now, the set of all enumerations of the rationals can be partially ordered by value; if $E_1$ and $E_2$ are enumerations, then $E_1>E_2$ if the value of $E_1$ exceeds the value of $E_2$. By Zorn's Lemma, there is an enumeration maximal with respect to this order. This maximal enumeration cannot have a finite value --- as we have seen, there are enumerations of arbitrarily great finite value. So, it must have infinite value. So, $x$ is irrational.

An alternative use for this argument is to apply it to prove that $1/3$ is irrational, the contradiction with the known rationality of $1/3$ thereby establishing that Zorn's Lemma is false.

$\endgroup$
2
  • 5
    $\begingroup$ Wouldn't it be easier (and pretty much equivalent) to prove that Zorn's Lemma is false by noting that it implies the existence of a largest natural number? $\endgroup$ Apr 23, 2012 at 6:16
  • 9
    $\begingroup$ Sure, but if you make it too easy you make it too obvious. Better to obscure the fallacy in lots of irrelevant verbiage. $\endgroup$ Apr 23, 2012 at 12:51
2
$\begingroup$

I think that the history of this wrong proof of the Riemann hypothesis is pretty interesting:

http://www.math.columbia.edu/~woit/wordpress/?p=707

In the end, it motivated a paper by Bombieri and Lagarias

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.3791

$\endgroup$
4
  • $\begingroup$ Interesting. But wrong proof is different from false proof IMHO. $\endgroup$
    – joro
    May 1, 2012 at 5:53
  • $\begingroup$ However, I do not think that the wrong proof is to far away from what you call false proof. In the end, the nitpick was an issue of well-definedness of a function, which was nonzero on a measure zero set. This is pretty close to "dividing by zero" for my taste. $\endgroup$
    – Marc Palm
    May 1, 2012 at 10:12
  • $\begingroup$ See e.g. this answer of S.Carnahan: mathoverflow.net/questions/49811/measure-of-adeles-minus-ideles $\endgroup$
    – Marc Palm
    May 1, 2012 at 11:05
  • $\begingroup$ OK, I didn't know this. $\endgroup$
    – joro
    May 1, 2012 at 15:23
2
$\begingroup$

We have $$\int \text{sec}^2(x)\tan(x)dx=\int \text{sec}^2(x)\tan(x)dx$$ $$\int \tan(x)d(\tan(x))=\int \text{sec}(x)d(\text{sec}(x))$$ $$\frac{\tan^2(x)}{2}+C=\frac{\text{sec}^2(x)}{2}+C$$ $$\frac{\tan^2(x)}{2}=\frac{\text{sec}^2(x)}{2}$$ $$\tan^2(x)=\text{sec}^2(x)$$ for all $x\in\mathbb{R}$.

$\endgroup$
1
  • $\begingroup$ Properly, of course, $\tan^2(x)+C_1=\sec^2(x)+C_2$ for some appropriately chosen constants of integration $C_1,C_2$. Forgetting the constants in indefinite integrals is a bane of many a mathematician. $\endgroup$ Dec 1, 2023 at 20:40
2
$\begingroup$

The celebrated hook length formula (https://en.wikipedia.org/wiki/Hook_length_formula) says that the number of standard Young tableaux of shape $\lambda \vdash n$ is $n! \cdot \prod_{u\in \lambda}h_u^{-1}$ where $h_u$ is the hook length of the box $u$ of $\lambda$.

A heuristic argument (I hesitate to call it a "proof") put forward by Knuth for this formula goes as follows. In a random injective filling of the boxes of $\lambda$ with the numbers $1,2,\ldots,n$, the probability that a given box $u$ has the smallest number in its hook is $h_u^{-1}$. Moreover, such a filling is an SYT if and only if each box is filled with the smallest number in its hook. "Thus", the probability that a random filling is an SYT is $\prod_{u\in \lambda}h_u^{-1}$, and so the number of SYTs is $n! \cdot \prod_{u\in \lambda}h_u^{-1}$. The error with this false proof is of course that the events of boxes being filled with the smallest numbers in their hooks are not independent. But it is quite interesting that we arrive at the correct probability treating these events as independent.

$\endgroup$
0
1
$\begingroup$

Mostly based on mlk's comments here

Lemma 1 $\lim_{x\to 0^+} x^0=1$, so $0^0=1$.

Lemma 2 $\lim_{x\to 0^+} 0^x=0$, so $0^0=0$.

Therefore $1=0$.

$\endgroup$
0
1
$\begingroup$

Theorem 1: All integers solutions to $a^2+b^2=c^2$ are given by $a,b,c=(2 x y , x^2 - y^2 , x^2 + y^2)$

Proof: We use sagemath to parametrize the conic:

sage: K.<a,b,c>=QQ[]
sage: co=Conic(a^2+b^2-c^2);pa=co.parametrization();pa
(Scheme morphism:
   From: Projective Space of dimension 1 over Rational Field
   To:   Projective Conic Curve over Rational Field defined by a^2 + b^2 - c^2
   Defn: Defined on coordinates by sending (x : y) to
     (2*x*y : x^2 - y^2 : x^2 + y^2),
 Scheme morphism:
   From: Projective Conic Curve over Rational Field defined by a^2 + b^2 - c^2
   To:   Projective Space of dimension 1 over Rational Field
   Defn: Defined on coordinates by sending (a : b : c) to
     (1/2*a : -1/2*b + 1/2*c))

Counterexample to Theorem 1:

$a,b,c=(9,12,15)$.

Proof: $15$ is not the sum of two integer squares.

$\endgroup$
2
  • $\begingroup$ Does this proof hinge on not understanding the meaning of projective? $\endgroup$ Apr 12, 2023 at 11:28
  • $\begingroup$ @PeterTaylor I think projective might be omitted in the proof and we can work only with expressions over the rationals. $\endgroup$
    – joro
    Apr 12, 2023 at 11:50
0
$\begingroup$

I think nobody point to these interesting false proof:

Let $i=\sqrt{-1}$ be the complex number.

$1)$ $1=\sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=i\times i=-1$.

$2)$ We know that $x^\frac{2}{6}=x^\frac{1}{3}\Rightarrow (\sqrt{x^2})^\frac{1}{6}=(\sqrt{x})^\frac{1}{3}$. Now, let $x=-1$ and so we have: $$(\sqrt{(-1)^2})^\frac{1}{6}=(\sqrt{-1})^\frac{1}{3}\Rightarrow1=-1.$$

$\endgroup$
-1
$\begingroup$

The limit of a function, if it exists, is unique. Indeed, from $\lim_{x\to x_0} f(x)=L_1$ and $\lim_{x\to x_0} f(x)=L_2$, exploiting symmetry and transitivity of the equality you readily deduce $L_1=L_2$.

$\endgroup$
7
  • 2
    $\begingroup$ Rather than an "interesting" false proof, this seems to me a notational ambiguity. Writing "$\lim_{x\to x_0} f(x)=L_1$" it seems that you are already assuming uniqueness, whereas (in general) limits can form a set with more than one element. $\endgroup$ Apr 16, 2021 at 8:00
  • 1
    $\begingroup$ Yes, in fact my personal idea of "interesting false proof" is more something like "proof which is flawed for some subtle conceptual reason". But I understand what you mean. $\endgroup$ Apr 16, 2021 at 8:16
  • 3
    $\begingroup$ Years ago a student came to me asking for clarifications on some exercises on limits I had left. “I did this limit and I got 1” I checked and said “Correct!”. “And then I did exercise 2 and I got 0”. I checked again and said “Correct!”. He said: “But doesn’t this contradict the uniqueness of the limit?” $\endgroup$ Sep 28, 2021 at 21:16
  • 2
    $\begingroup$ @AlessandroDellaCorte The more powerful technique, which I've seen too many students use, is to remember a slogan, like "uniqueness of limits" or some formula highlighted in the textbook, while forgetting the surrounding words, like the hypotheses that underlie the slogan. $\endgroup$ Sep 28, 2021 at 23:54
  • 4
    $\begingroup$ All limits are unique, but some limits are more unique than others. $\endgroup$ Nov 2, 2021 at 23:40
-2
$\begingroup$

Doron Zeilberger proved that P is equal to NP

Abstract: Using 3000 hours of CPU time on a CRAY machine, we settle the notorious P vs. NP problem in the affirmative, by presenting a “polynomial” time algorithm for the NP-complete subset sum problem. Alas the complexity of our algorithm is $O(n^{10^{10000}})$ (with the implied constant being larger than the Skewes number).

$\endgroup$
2
  • 1
    $\begingroup$ This is 11 days too early. $\endgroup$ Mar 21, 2013 at 15:24
  • 2
    $\begingroup$ @Noam: Actually, 11 days minus 4 years too early. $\endgroup$
    – Lee Mosher
    Mar 21, 2013 at 15:29
1
2

Not the answer you're looking for? Browse other questions tagged or ask your own question.