Examples of interesting false proofs

Question

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?

Answer 1

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

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$\Large\textbf{Another example}$:

Answer 2

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.)

Answer 3

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

Answer 4

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.

Answer 5

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 $0=1$.

Answer 6

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.

Answer 7

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 $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$

Answer 8

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 $\xi \to x_0$.

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

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 $x\in\mathbb{R}$.

Answer 14

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.

Answer 15

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 $1=0$.

Answer 16

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.

Answer 17

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: $$(\sqrt{(-1)^2})^\frac{1}{6}=(\sqrt{-1})^\frac{1}{3}\Rightarrow1=-1.$$

Answer 18

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 $L_1=L_2$.

Answer 19

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).