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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?

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    $\begingroup$ Is this a duplicate? $\endgroup$ Apr 21 '12 at 17:19
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    $\begingroup$ 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 '12 at 5:46
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    $\begingroup$ 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 '12 at 15:36
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    $\begingroup$ 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 '12 at 22:13
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    $\begingroup$ There are no false proofs, by definition. $\endgroup$ Mar 19 '13 at 17:32

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Many years ago, I came up with this stupid proof that all groups are abelian: $$ab^{-1}=a\cdot{1\over b}={a\over b}={1\over b}\cdot a=b^{-1}a$$ I called it The Passing Through Theorem.

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    $\begingroup$ This somewhat reminds me of the Eckermann-Hilton argument. $\endgroup$ Mar 25 at 3:50
  • $\begingroup$ @AntoineLabelle Your remark is a very good mathematical insider joke. $\endgroup$
    – rimu
    May 16 at 10:17
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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.

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    $\begingroup$ 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 '12 at 11:29
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    $\begingroup$ This is incomprehensible as posted but begins to make sense after you've followed the link in Gerald Edgar's answer. $\endgroup$ Apr 22 '12 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 '12 at 22:59
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    $\begingroup$ Essentially the same proof is shown here: youtube.com/watch?v=Lo4NCXOX0p8 (an old Abbot & Costello sketch). $\endgroup$ Apr 25 '12 at 11:11
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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

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  • $\begingroup$ 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 '12 at 16:49
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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.

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    $\begingroup$ 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 '14 at 8:37
  • $\begingroup$ I think so, too. $\endgroup$ Jun 14 '14 at 10:37
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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

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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.

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    $\begingroup$ 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 '12 at 6:16
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    $\begingroup$ 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 '12 at 12:51
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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

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  • $\begingroup$ Interesting. But wrong proof is different from false proof IMHO. $\endgroup$
    – joro
    May 1 '12 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 '12 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 '12 at 11:05
  • $\begingroup$ OK, I didn't know this. $\endgroup$
    – joro
    May 1 '12 at 15:23
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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 $1=0$.

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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 $\xi \to x_0$.

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  • $\begingroup$ 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 at 21:13
  • $\begingroup$ Yep, exactly: you just have to assume that the derivative has (finite) one sided limits. $\endgroup$
    – GaC
    Mar 27 at 8:56
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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.

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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 $0=1$.

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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 $L_1=L_2$.

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    $\begingroup$ 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 at 8:00
  • $\begingroup$ You have a point, of course, and "interesting" is subjective. It seems so to me because I see students using notation this way all the time. In fact even some texbooks providing a correct proof of the above theorem start writing down something like that, so it seems a nice way to point out the incorrect use of the notation. $\endgroup$ Apr 16 at 8:13
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    $\begingroup$ 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 at 8:16
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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: $$(\sqrt{(-1)^2})^\frac{1}{6}=(\sqrt{-1})^\frac{1}{3}\Rightarrow1=-1.$$

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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 $x\in\mathbb{R}$.

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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).

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    $\begingroup$ This is 11 days too early. $\endgroup$ Mar 21 '13 at 15:24
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    $\begingroup$ @Noam: Actually, 11 days minus 4 years too early. $\endgroup$
    – Lee Mosher
    Mar 21 '13 at 15:29
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