# 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

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.

• 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? – Steven Landsburg Apr 23 '12 at 6:16
• Sure, but if you make it too easy you make it too obvious. Better to obscure the fallacy in lots of irrelevant verbiage. – Gerry Myerson Apr 23 '12 at 12:51

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

• It can be used in the Millenium Prize Problems too ;-) – joro Apr 25 '12 at 11:10
• 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"). – Daniele Apr 25 '12 at 14:10

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

• Interesting. But wrong proof is different from false proof IMHO. – joro May 1 '12 at 5:53
• 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. – Marc Palm May 1 '12 at 10:12
• See e.g. this answer of S.Carnahan: mathoverflow.net/questions/49811/measure-of-adeles-minus-ideles – Marc Palm May 1 '12 at 11:05
• OK, I didn't know this. – joro May 1 '12 at 15:23

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

• 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! – some guy on the street Jun 16 '12 at 16:49

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 $K$."

• What is the right answer? Is it that "why is the collection of all algebraic extensions of a given field K a set?". – knsam Jun 11 '14 at 8:37
• I think so, too. – Torsten Schoeneberg Jun 14 '14 at 10:37

Josh Nicols-Barrer wrote a delightful proof of Fermat's Last Theorem (and much more) here:

In a nutshell: if $$x^n+y^n=z^n$$ then by differentiating and dividing by $$n$$, we get $$x^{n-1}+y^{n-1}=z^{n-1}$$. There are no integer solutions to $$x^0+y^0=z^0$$, so by induction Fermat's Last Theorem holds. As corollaries, there are no Pythagorean triples, and also addition is a lie. (But this is just a summary of Josh's amusing post.)

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

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).

• This is 11 days too early. – Noam D. Elkies Mar 21 '13 at 15:24
• @Noam: Actually, 11 days minus 4 years too early. – Lee Mosher Mar 21 '13 at 15:29